Anton - Elementary Linear Algebra With Applications 10e Student Solutions Manual

Chapter 1 Systems of Linear Equations and Matrices Section 1.1 Exercise Set 1.1 1. (a), (c), and (f) are linear equations in x1 , x2 , and x3 . (b) is not linear because of the term x1 x3 . (d) is not linear because of the term x −2 . (e) is not linear because of the term x13/ 5 . 3. (a) and (d) are linear systems. (b) is not a linear system because the first and second equations are not linear. (c) is not a linear system because the second equation is not linear. 5. By inspection, (a) and (d) are both consistent; x1 = 3, x2 = 2, x3 = −2, x4 = 1 is a solution of (a) and

x1 = 1, x2 = 3, x3 = 2, x4 = 2 is a solution of (d). Note that both systems have infinitely many solutions. 7. (a), (d), and (e) are solutions. (b) and (c) do not satisfy any of the equations. 9. (a) 7 x − 5 y = 3 5 3 x= y+ 7 7 Let y = t. The solution is 5 3 x = y+ 7 7 y=t (b) −8 x1 + 2 x2 − 5 x3 + 6 x4 = 1 1 5 3 1 x1 = x2 − x3 + x4 − 4 8 4 8 Let x2 = r , x3 = s, and x4 = t . The solution is

1 5 3 1 r− s+ t− 4 8 4 8 =r =s =t

x1 =

x2 x3 x4

⎡2 0 0⎤ 11. (a) ⎢ 3 −4 0 ⎥ corresponds to ⎢ ⎥ 1 1⎥⎦ ⎢⎣ 0 2 x1 =0 3x1 − 4 x2 = 0. x2 = 1

1

Chapter 1: Systems of Linear Equations and Matrices

SSM: Elementary Linear Algebra

⎡ 3 0 −2 5⎤ (b) ⎢7 1 4 −3⎥⎥ corresponds to ⎢ 1 7 ⎥⎦ ⎢⎣ 0 −2 3x1 − 2 x3 = 5 7 x1 + x2 + 4 x3 = −3. −2 x2 + x3 = 7

15. If (a, b, c) is a solution of the system, then

ax12 + bx1 + c = y1 , ax22 + bx2 + c = y2 , and ax32 + bx3 + c = y3 which simply means that the points are on the curve. 17. The solutions of x1 + kx2 = c are x1 = c − kt ,

x2 = t where t is any real number. If these satisfy x1 + lx2 = d , then c − kt + lt = d or

⎡ ⎤ (c) ⎢7 2 1 −3 5⎥ corresponds to ⎣ 1 2 4 0 1⎦ 7 x1 + 2 x2 + x3 − 3x4 = 5 . x1 + 2 x2 + 4 x3 =1 ⎡1 ⎢ (d) ⎢0 ⎢0 ⎢⎣0 x1

0 1 0 0

0 0 1 0

c − d = (k − l) t for all real numbers t. In particular, if t = 0, then c = d, and if t = 1, then k = l. True/False 1.1

0 7⎤ 0 −2 ⎥ corresponds to ⎥ 0 3⎥ 1 4 ⎥⎦

x2 x3 x4

(a) True; x1 = x2 =

(b) False; only multiplication by nonzero constants is acceptable.

= 7 = −2 . 3 = = 4

(c) True; if k = 6 the system has infinitely many solutions, while if k ≠ 6, the system has no solution. (d) True; the equation can be solved for one variable in terms of the other(s), yielding parametric equations that give infinitely many solutions.

13. (a) The augmented matrix for −2 x1 = 6 3x1 = 8 9 x1 = −3

(e) False; the system 3x − 5 y = −7 has the 2 x + 9 y = 20 6 x − 10 y = −14 solution x = 1, y = 2.

⎡ −2 6 ⎤ is ⎢ 3 8⎥ . ⎢ ⎥ ⎢⎣ 9 −3⎥⎦ (b) The augmented matrix for ⎡ ⎤ 6 x1 − x2 + 3x3 = 4 is ⎢6 −1 3 4 ⎥ . ⎣0 5 −1 1⎦ 5x − x = 1 2

(f) False; multiplying an equation by a nonzero constant c does not change the solutions of the system.

3

(g) True; subtracting one equation from another is the same as multiplying an equation by −1 and adding it to another.

(c) The augmented matrix for 2 x2 − 3x4 + x5 = 0 −3x1 − x2 + x3 = −1 6 x1 + 2 x2 − x3 + 2 x4 − 3x5 = 6

(h) False; the second row corresponds to the equation 0 x1 + 0 x2 = −1 or 0 = −1 which is false.

⎡ 0 2 0 −3 1 0⎤ is ⎢ −3 −1 1 0 0 −1⎥ . ⎢ ⎥ ⎢⎣ 6 2 −1 2 −3 6 ⎥⎦

Section 1.2 Exercise Set 1.2

(d) The augmented matrix for x1 − x5 = 7 is

[1

0

= xn = 0 will be a solution.

0

0

−1

1. (a) The matrix is in both row echelon and reduced row echelon form.

7].

(b) The matrix is in both row echelon and reduced row echelon form.

2


Section 1.2

(c) The matrix is in both row echelon and reduced row echelon form.

(d) The last line of the matrix corresponds to the equation 0 x1 + 0 x2 + 0 x3 = 1, which is not

(d) The matrix is in both row echelon and reduced row echelon form.

satisfied by any values of x1 , x2 , and x3 , so the system is inconsistent. ⎡ 1 1 2 8⎤ 5. The augmented matrix is ⎢⎢ −1 −2 3 1⎥⎥ . ⎢⎣ 3 −7 4 10 ⎥⎦ Add the first row to the second row and add −3 times the first row to the third row. ⎡1 8⎤ 1 2 ⎢0 9⎥ −1 5 ⎢ ⎥ ⎢⎣0 −10 −2 −14⎥⎦

(e) The matrix is in both row echelon and reduced row echelon form. (f) The matrix is in both row echelon and reduced row echelon form. (g) The matrix is in row echelon form. 3. (a) The matrix corresponds to the system x1 − 3x2 + 4 x3 = 7 x1 = 7 + 3x2 − 4 x3 x2 + 2 x3 = 2 or x2 = 2 − 2 x3 . x3 = 5 x3 = 5

Multiply the second row by −1. ⎡1 8⎤ 1 2 ⎢0 1 −5 −9⎥⎥ ⎢ ⎢⎣0 −10 −2 −14⎥⎦

Thus x3 = 5, x2 = 2 − 2(5) = −8, and x1 = 7 + 3(−8) − 4(5) = −37. The solution is x1 = −37, x2 = −8, x3 = 5.

Add 10 times the second row to the third row. ⎡1 1 8⎤ 2 ⎢ 0 1 −5 −9 ⎥ ⎢ ⎥ ⎣⎢0 0 −52 −104 ⎦⎥

(b) The matrix corresponds to the system x1 + 8 x3 − 5 x4 = 6 x2 + 4 x3 − 9 x4 = 3 or x3 + x4 = 2

Multiply the third row by −

x1 = 6 − 8 x3 + 5 x4 x2 = 3 − 4 x3 + 9 x4 . x3 = 2 − x4

1 . 52

⎡1 1 2 8⎤ ⎢0 1 −5 −9 ⎥ ⎢ ⎥ 1 2 ⎥⎦ ⎢⎣0 0 Add 5 times the third row to the second row and −2 times the third row to the first row. ⎡ 1 1 0 4⎤ ⎢0 1 0 1⎥ ⎢ ⎥ ⎢⎣0 0 1 2⎥⎦

Let x4 = t , then x3 = 2 − t , x2 = 3 − 4(2 − t ) + 9t = 13t − 5, and x1 = 6 − 8(2 − t ) + 5t = 13t − 10. The solution is x1 = 13t − 10, x2 = 13t − 5, x3 = −t + 2, x4 = t .

Add −1 times the second row to the first row. ⎡ 1 0 0 3⎤ ⎢0 1 0 1⎥ ⎢ ⎥ ⎢⎣0 0 1 2⎥⎦

(c) The matrix corresponds to the system x1 + 7 x2 − 2 x3 − 8 x5 = −3 x3 + x4 + 6 x5 = 5 or x4 + 3x5 = 9

x1 = −3 − 7 x2 + 2 x3 + 8 x5 x3 = 5 − x4 − 6 x5 x4 = 9 − 3 x5

The solution is x1 = 3, x2 = 1, x3 = 2. 7. The augmented matrix is ⎡ 1 −1 2 −1 −1⎤ ⎢ 2 1 −2 −2 −2 ⎥ ⎢ ⎥. 1 1⎥ ⎢ −1 2 −4 ⎣⎢ 3 0 0 −3 −3⎦⎥

Let x2 = s and x5 = t , then x4 = 9 − 3t , x3 = 5 − (9 − 3t ) − 6t = −3t − 4, and x1 = −3 − 7 s + 2(−3t − 4) + 8t = −7 s + 2t − 11. The solution is x1 = −7 s + 2t − 11, x2 = s,

Add −2 times the first row to the second row, 1 times the first row to the third row, and −3 times the first row to the fourth row.

x3 = −3t − 4, x4 = −3t + 9, x5 = t .

3



⎡ 1 −1 2 −1 −1⎤ ⎢0 3 −6 0 0 ⎥ ⎢ ⎥ ⎢0 1 −2 0 0 ⎥ ⎢⎣0 3 −6 0 0 ⎥⎦

13. Since the system has more unknowns (4) than equations (3), it has nontrivial solutions. 15. Since the system has more unknowns (3) than equations (2), it has nontrivial solutions.

1 and then add −1 3 times the new second row to the third row and add −3 times the new second row to the fourth row. ⎡ 1 −1 2 −1 −1⎤ ⎢0 1 −2 0 0 ⎥ ⎢ ⎥ ⎢0 0 0 0 0 ⎥ ⎣⎢0 0 0 0 0 ⎦⎥ Add the second row to the first row. ⎡ 1 0 0 −1 −1⎤ ⎢ 0 1 −2 0 0 ⎥ ⎢ ⎥ ⎢0 0 0 0 0⎥ ⎢⎣0 0 0 0 0⎥⎦ The corresponding system of equations is x = w −1 − w = −1 x or . =0 y = 2z y − 2z Let z = s and w = t. The solution is x = t − 1, y = 2s, z = s, w = t. Multiply the second row by

⎡2 1 3 0⎤ 17. The augmented matrix is ⎢⎢ 1 2 0 0 ⎥⎥ . ⎢⎣ 0 1 1 0 ⎥⎦ Interchange the first and second rows, then add −2 times the new first row to the second row. ⎡ 1 2 0 0⎤ ⎢0 −3 3 0 ⎥ ⎢ ⎥ 1 1 0 ⎥⎦ ⎢⎣0 1 Multiply the second row by − , then add −1 3 times the new second row to the third row. ⎡ 1 2 0 0⎤ ⎢0 1 −1 0 ⎥ ⎢ ⎥ ⎣⎢0 0 2 0 ⎦⎥ 1 , then add the new 2 third row to the second row. ⎡ 1 2 0 0⎤ ⎢0 1 0 0⎥ ⎢ ⎥ ⎢⎣0 0 1 0⎥⎦ Multiply the third row by

9. In Exercise 5, the following row echelon matrix occurred. ⎡1 1 2 8⎤ ⎢0 1 −5 −9 ⎥ ⎢ ⎥ 1 2 ⎥⎦ ⎢⎣0 0

Add −2 times the second row to the first row. ⎡ 1 0 0 0⎤ ⎢0 1 0 0⎥ ⎢ ⎥ ⎣⎢0 0 1 0⎦⎥

The corresponding system of equations is x1 + x2 + 2 x3 = 8 x1 = − x2 − 2 x3 + 8 x2 − 5 x3 = −9 or x2 = 5 x3 − 9 . x3 = 2 x3 = 2

The solution, which can be read from the matrix, is x1 = 0, x2 = 0, x3 = 0.

Since x3 = 2, x2 = 5(2) − 9 = 1, and x1 = −1 − 2(2) + 8 = 3. The solution is x1 = 3,

⎡ ⎤ 19. The augmented matrix is ⎢3 1 1 1 0⎥ . ⎣ 5 −1 1 −1 0 ⎦ 1 Multiply the first row by , then add −5 times 3 the new first row to the second row. 1 1 1 0⎤ ⎡1 3 3 3 ⎢ ⎥ ⎢⎣0 − 83 − 23 − 83 0 ⎥⎦ 3 Multiply the second row by − . 8

x2 = 1, x3 = 2. 11. From Exercise 7, one row echelon form of the ⎡ 1 −1 2 −1 −1⎤ ⎢ ⎥ augmented matrix is ⎢0 1 −2 0 0 ⎥ . ⎢0 0 0 0 0 ⎥ ⎢⎣0 0 0 0 0 ⎥⎦ The corresponding system is x = y − 2z + w −1 x − y + 2 z − w = −1 or . y = 2z y − 2z =0 Let z = s and w = t. Then y = 2s and x = 2s − 2s + t − 1 = t − 1. The solution is x = t − 1, y = 2s, z = s, w = t.

⎡1 1 3 ⎢ ⎢⎣0 1 4

1 3 1 4

0⎤ ⎥ 1 0 ⎥⎦

1 3


Add −

Section 1.2 w= y =0 = 0 or x = − y . z=0 z=0 Let y = t. The solution is w = t, x = −t, y = t, z = 0. w

1 times the second row to the first row. 3

⎡ 1 0 1 0 0⎤ 4 ⎢ ⎥ ⎢⎣0 1 14 1 0 ⎥⎦ This corresponds to the system 1 1 + x3 =0 x1 x1 = − x3 4 4 or . 1 1 x2 + x3 + x4 = 0 x2 = − x3 − x4 4 4 Let x3 = 4s and x4 = t . Then x1 = − s and

−y x+ y

⎡ 2 −1 3 4 9 ⎤ ⎢ 1 0 −2 7 11⎥ 23. The augmented matrix is ⎢ ⎥. 1 5 8⎥ ⎢ 3 −3 1 4 4 10⎦⎥ ⎣⎢ 2 Interchange the first and second rows. ⎡ 1 0 −2 7 11⎤ ⎢ 2 −1 3 4 9 ⎥ ⎢ ⎥ 1 5 8⎥ ⎢ 3 −3 1 4 4 10⎥⎦ ⎢⎣ 2 Add −2 times the first row to the second and fourth rows, and add −3 times the first row to the third row. 7 11⎤ ⎡ 1 0 −2 ⎢0 −1 7 −10 −13⎥ ⎢ ⎥ ⎢0 −3 7 −16 −25⎥ 1 8 −10 −12⎦⎥ ⎣⎢0

x2 = − s − t . The solution is x1 = − s, x2 = − s − t , x3 = 4s, x4 = t .

⎡ 0 2 2 4 0⎤ ⎢ 1 0 −1 −3 0 ⎥ 21. The augmented matrix is ⎢ ⎥. ⎢ 2 3 1 1 0⎥ ⎢⎣ −2 1 3 −2 0 ⎥⎦ Interchange the first and second rows. ⎡ 1 0 −1 −3 0 ⎤ ⎢ 0 2 2 4 0⎥ ⎢ ⎥ ⎢ 2 3 1 1 0⎥ ⎣⎢ −2 1 3 −2 0 ⎦⎥

Multiply the second row by −1, then add 3 times the new second row to the third row and −1 times the new second row to the fourth row. 7 11⎤ ⎡ 1 0 −2 ⎢0 1 −7 10 13⎥ ⎢ ⎥ ⎢0 0 −14 14 14 ⎥ 15 −20 −25⎦⎥ ⎣⎢0 0

Add −2 times the first row to the third row and 2 times the first row to the fourth row. ⎡ 1 0 −1 −3 0 ⎤ ⎢0 2 2 4 0 ⎥ ⎢ ⎥ ⎢0 3 3 7 0 ⎥ ⎢⎣0 1 1 −8 0 ⎥⎦ 1 Multiply the second row by , then add −3 2 times the new second row to the third row and −1 times the new second row to the fourth row. ⎡ 1 0 −1 −3 0 ⎤ ⎢0 1 1 2 0⎥ ⎢ ⎥ 1 0⎥ ⎢0 0 0 ⎣⎢0 0 0 −10 0 ⎦⎥

1 , then add −15 14 times the new third row to the fourth row. 11⎤ ⎡ 1 0 −2 7 ⎢0 1 −7 10 13⎥ ⎢ ⎥ 1 −1 −1⎥ ⎢0 0 ⎣⎢0 0 0 −5 −10 ⎦⎥ Multiply the third row by −

1 Multiply the fourth row by − , then add the 5 new fourth row to the third row, add −10 times the new fourth row to the second row, and add −7 times the new fourth row to the first row. ⎡ 1 0 −2 0 −3⎤ ⎢ 0 1 −7 0 −7 ⎥ ⎢ ⎥ 1 0 1⎥ ⎢0 0 0 1 2 ⎦⎥ ⎣⎢0 0 Add 7 times the third row to the second row and 2 times the third row to the first row.

Add 10 times the third row to the fourth row, −2 times the third row to the second row, and 3 times the third row to the first row. ⎡ 1 0 −1 0 0 ⎤ ⎢0 1 1 0 0 ⎥ ⎢ ⎥ ⎢0 0 0 1 0 ⎥ ⎢⎣0 0 0 0 0 ⎥⎦ The corresponding system is

5



⎡2 1 a ⎤ 29. The augmented matrix is ⎢ . ⎣ 3 6 b ⎥⎦ 1 Multiply the first row by , then add −3 times 2 the new first row to the second row. a ⎡1 1 ⎤ 2 2 ⎢ ⎥ ⎢⎣0 92 − 32a + b ⎥⎦ 2 1 Multiply the second row by , then add − 2 9 times the new second row to the first row. ⎡1 0 2 a − b ⎤ 3 9 ⎥ ⎢ ⎢0 1 − a3 + 29b ⎥ ⎣ ⎦ 2a b a 2b − , y=− + . The solution is x = 3 9 3 9

⎡ 1 0 0 0 −1⎤ ⎢0 1 0 0 0 ⎥ ⎢ ⎥ ⎢0 0 1 0 1⎥ ⎢⎣0 0 0 1 2 ⎥⎦ The solution, which can be read from the matrix, is I1 = −1, I 2 = 0, I 3 = 1, I 4 = 2. 25. The augmented matrix is ⎡1 2 −3 4 ⎤ ⎢ 3 −1 5 2 ⎥⎥ ⎢ ⎢⎣ 4 1 a 2 − 14 a + 2⎥⎦

Add −3 times the first row to the second row and −4 times the first row to the third row. ⎡1 2 −3 4 ⎤ ⎢0 −7 14 −10 ⎥⎥ ⎢ 2 ⎣⎢0 −7 a − 2 a − 14 ⎦⎥ Add −1 times the second row to the third row. ⎡1 2 −3 4 ⎤ ⎢0 −7 14 −10 ⎥⎥ ⎢ 2 ⎢⎣0 0 a − 16 a − 4 ⎥⎦

31. Add −2 times the first row to the second row. ⎡ 1 3⎤ ⎢⎣0 1⎥⎦ is in row echelon form. Add −3 times the second row to the first row. ⎡ 1 0⎤ ⎢⎣0 1⎥⎦ is another row echelon form for the matrix.

The third row corresponds to the equation

(a 2 − 16) z = a − 4 or (a + 4)(a − 4)z = a − 4. If a = −4, this equation is 0z = −8, which has no solution. If a = 4, this equation is 0z = 0, and z is a free variable. For any other value of a, the solution of 1 . Thus, if a = 4, the this equation is z = a+4 system has infinitely many solutions; if a = −4, the system has no solution; if a ≠ ±4, the system has exactly one solution.

33. Let x = sin α, y = cos β, and z = tan γ; then the x + 2 y + 3z = 0 system is 2 x + 5 y + 3 z = 0 . − x − 5 y + 5z = 0

⎡ 1 2 3 0⎤ The augmented matrix is ⎢ 2 5 3 0⎥ . ⎢ ⎥ ⎣ −1 −5 5 0 ⎦ Add −2 times the first row to the second row, and add the first row to the third row. ⎡ 1 2 3 0⎤ ⎢0 1 −3 0 ⎥ ⎢ ⎥ ⎣ 0 −3 8 0 ⎦

2 1 ⎤ ⎡1 27. The augmented matrix is ⎢ ⎥. 2 ⎣ 2 a − 5 a − 1⎦ Add −2 times the first row to the second row. 2 1 ⎤ ⎡1 ⎢ ⎥ 2 ⎣ 0 a − 9 a − 3⎦ The second row corresponds to the equation

Add 3 times the second row to the third row. ⎡ 1 2 3 0⎤ ⎢0 1 −3 0 ⎥ ⎢ ⎥ ⎣0 0 −1 0 ⎦

(a 2 − 9) y = a − 3 or (a + 3)(a − 3)y = a − 3. If a = −3, this equation is 0y = −6, which has no solution. If a = 3, this equation is 0y = 0, and y is a free variable. For any other value of a, the 1 solution of this equation is y = . a+3 Thus, if a = −3, the system has no solution; if a = 3, the system has infinitely many solutions; if a ≠ ±3, the system has exactly one solution.

Multiply the third row by −1 then add 3 times the new third row to the second row and −3 times the new third row to the first row. ⎡ 1 2 0 0⎤ ⎢0 1 0 0⎥ ⎢ ⎥ ⎣0 0 1 0⎦ Add −2 times the second row to the first row. 6


Section 1.2 X = 1 ⇒ x = ±1 Y=3⇒ y=± 3

⎡ 1 0 0 0⎤ ⎢0 1 0 0⎥ ⎢ ⎥ ⎣0 0 1 0⎦ The system has only the trivial solution, which corresponds to sin α = 0, cos β = 0, tan γ = 0. For 0 ≤ α ≤ 2π , sin α = 0 ⇒ α = 0, π , 2π . For 0 ≤ β ≤ 2π , cos β = 0 ⇒ β =

Z=2⇒ z=± 2 The solutions are x = ±1, y = ± 3, z = ± 2. 37. (0, 10): d = 10 (1, 7): a + b + c + d = 7 (3, −11): 27a + 9b + 3c + d = −11 (4, −14): 64a + 16b + 4c + d = −14 The system is a +b +c+d = 7 27a + 9b + 3c + d = −11 and the augmented 64a + 16b + 4c + d = −14 d = 10 1 1 1 1 7⎤ ⎡ ⎢ 27 9 3 1 −11⎥ matrix is ⎢ ⎥. ⎢ 64 16 4 1 −14⎥ ⎣⎢ 0 0 0 1 10⎦⎥

π 3π

, . 2 2 For 0 ≤ γ ≤ 2π , tan γ = 0 ⇒ γ = 0, π , 2π . Thus, the original system has 3 ⋅ 2 ⋅ 3 = 18 solutions. 35. Let X = x 2 , Y = y 2 , and Z = z 2 , then the system is X +Y + Z = 6 X − Y + 2Z = 2 2X + Y − Z = 3 ⎡ 1 1 1 6⎤ The augmented matrix is ⎢ 1 −1 2 2⎥ . ⎢ ⎥ ⎣ 2 1 −1 3⎦ Add −1 times the first row to the second row and −2 times the first row to the third row. ⎡ 1 1 1 6⎤ ⎢ 0 −2 1 −4 ⎥ ⎢ ⎥ ⎣ 0 −1 −3 −9 ⎦

Add −27 times the first row to the second row and −64 times the first row to the third row. 1 1 1 7⎤ ⎡1 ⎢0 −18 −24 −26 −200 ⎥ ⎢ ⎥ ⎢0 −48 −60 −63 −462 ⎥ 0 0 1 10 ⎦⎥ ⎣⎢0 1 , then add 48 18 times the new second row to the third row. 7 ⎤ ⎡1 1 1 1 ⎢0 1 4 13 100 ⎥ ⎢ 3 9 9 ⎥ ⎢0 0 4 19 214 ⎥ 3 3 ⎥ ⎢ ⎣⎢0 0 0 1 10 ⎦⎥

Multiply the second row by −

1 Multiply the second row by − , then add the 2 new second row to the third row. ⎡1 1 1 6⎤ ⎢ ⎥ 2⎥ ⎢0 1 − 12 ⎢ ⎥ 7 ⎢⎣0 0 − 2 −7 ⎥⎦ 2 1 Multiply the third row by − , then add 2 7 times the new third row to the second row and −1 times the new third row to the first row. ⎡ 1 1 0 4⎤ ⎢ 0 1 0 3⎥ ⎢ ⎥ ⎣0 0 1 2⎦

Multiply the third row by ⎡1 ⎢0 ⎢ ⎢0 ⎢ ⎣⎢0

X Y

0 1 0 0

Add −

Add −1 times the second row to the first row. ⎡ 1 0 0 1⎤ ⎢ 0 1 0 3⎥ . ⎢ ⎥ ⎣0 0 1 2⎦ This corresponds to the system

1 1 1 43

1 13 9 19 12

1

1 . 4

7 ⎤

100 ⎥ 9 ⎥ 107 ⎥ 6 ⎥

10 ⎦⎥

19 times the fourth row to the third row, 12

13 times the fourth row to the second row, 9 and −1 times the fourth row to the first row. −

=1 =3. Z =2 7



1 1 0 −3 ⎤ ⎥ 1 43 0 − 10 3⎥ 0 1 0 2 ⎥ 0 0 1 10 ⎥⎦ 4 Add − times the third row to the second row 3 and −1 times the third row to the first row. ⎡ 1 1 0 0 −5⎤ ⎢0 1 0 0 −6 ⎥ ⎢ ⎥ ⎢0 0 1 0 2 ⎥ ⎢⎣0 0 0 1 10 ⎥⎦ Add −1 times the second row to the first row. 1⎤ ⎡1 0 0 0 ⎢0 1 0 0 −6 ⎥ ⎢ ⎥ ⎢0 0 1 0 2 ⎥ ⎣⎢0 0 0 1 10 ⎦⎥ ⎡1 ⎢0 ⎢ ⎢0 ⎢0 ⎣

43. (a) There are eight possibilities. p and q are any real numbers. ⎡ 1 0 0 ⎤ ⎡ 1 0 p ⎤ ⎡ 1 p 0⎤ ⎢0 1 0 ⎥ , ⎢0 1 q ⎥ , ⎢0 0 1⎥ , ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 0 1⎦ ⎣0 0 0 ⎦ ⎣0 0 0⎦

⎡0 1 0 ⎤ ⎢0 0 1⎥ , ⎢ ⎥ ⎣0 0 0 ⎦

⎡ 1 p q ⎤ ⎡0 1 p ⎤ ⎢0 0 0 ⎥ , ⎢0 0 0 ⎥ , ⎢ ⎥ ⎢ ⎥ ⎣0 0 0 ⎦ ⎣0 0 0 ⎦

⎡0 0 0 ⎤ ⎡0 0 1⎤ ⎢0 0 0 ⎥ , and ⎢0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 0 0 ⎦ ⎣0 0 0 ⎦ (b) There are 16 possibilities. p, q, r, and s are any real numbers. ⎡ 1 0 0 0⎤ ⎡ 1 0 0 p ⎤ ⎢0 1 0 0⎥ ⎢ 0 1 0 q ⎥ ⎢ ⎥, ⎢ ⎥, ⎢0 0 1 0⎥ ⎢ 0 0 1 r ⎥ ⎢⎣0 0 0 1⎥⎦ ⎢⎣ 0 0 0 0 ⎥⎦ ⎡ 1 0 p 0 ⎤ ⎡ 1 p 0 0⎤ ⎢0 1 q 0 ⎥ ⎢0 0 1 0⎥ ⎢ ⎥, ⎢ ⎥, ⎢0 0 0 1⎥ ⎢0 0 0 1⎥ ⎢⎣0 0 0 0 ⎥⎦ ⎢⎣0 0 0 0⎥⎦ ⎡0 1 0 0⎤ ⎡ 1 0 p q ⎤ ⎢0 0 1 0⎥ ⎢ 0 1 r s ⎥ ⎢ ⎥, ⎢ ⎥, ⎢0 0 0 1⎥ ⎢ 0 0 0 0 ⎥ ⎣⎢0 0 0 0⎦⎥ ⎣⎢ 0 0 0 0 ⎦⎥

The coefficients are a = 1, b = −6, c = 2, d = 10. 39. Since the homogeneous system has only the trivial solution, using the same steps of GaussJordan elimination will reduce the augmented matrix of the nonhomogeneous system to the ⎡ 1 0 0 d1 ⎤ form ⎢0 1 0 d 2 ⎥ . ⎢ ⎥ ⎣ 0 0 1 d3 ⎦ Thus the system has exactly one solution.

⎡ 1 p 0 q⎤ ⎢0 0 1 r ⎥ ⎢ ⎥, ⎢0 0 0 0 ⎥ ⎢⎣0 0 0 0 ⎥⎦

41. (a) If a ≠ 0, the sequence of matrices is b ⎤ ⎡1 ⎡ a b ⎤ ⎡1 ba ⎤ a ⎥ ⎢ → → ⎢⎣ c d ⎥⎦ ⎢ c d ⎥ −bc ⎥ ad ⎢0 ⎣⎢ ⎦⎥ a ⎦ ⎣ ⎡1 b ⎤ 1 0⎤ a⎥ → ⎡ →⎢ ⎢⎣0 1 ⎥⎦ . ⎢⎣ 0 1 ⎥⎦

⎡1 p ⎢0 0 ⎢ ⎢0 0 ⎢⎣0 0

q 0⎤ 0 1⎥ ⎥, 0 0⎥ 0 0 ⎥⎦

⎡0 1 0 p ⎤ ⎡ 0 1 p 0 ⎤ ⎢0 0 1 q ⎥ ⎢ 0 0 0 1⎥ ⎢ ⎥, ⎢ ⎥, ⎢0 0 0 0 ⎥ ⎢ 0 0 0 0 ⎥ ⎢⎣0 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 0 ⎥⎦

If a = 0, the sequence of matrices is ⎡1 d ⎤ ⎡0 b ⎤ ⎡ c d ⎤ ⎡1 dc ⎤ c ⎢⎣ c d ⎥⎦ → ⎢⎣0 b ⎥⎦ → ⎢ 0 b ⎥ → ⎢ 0 1 ⎥ ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ ⎡1 0 ⎤ . →⎢ ⎣ 0 1 ⎥⎦

⎡0 ⎢0 ⎢ ⎢0 ⎣⎢0

0 1 0⎤ ⎡ 1 p 0 0 1⎥ ⎢ 0 0 ⎥, ⎢ 0 0 0⎥ ⎢ 0 0 0 0 0⎦⎥ ⎢⎣ 0 0

⎡0 1 p ⎢0 0 0 ⎢ ⎢0 0 0 ⎣⎢0 0 0

(b) Since ad − bc ≠ 0, the lines are neither identical nor parallel, so the augmented ⎡a b k ⎤ matrix ⎢ will reduce to ⎣ c d l ⎥⎦ ⎡ 1 0 k1 ⎤ ⎢0 1 l ⎥ and the system has exactly one 1⎦ ⎣ solution.

⎡0 ⎢0 ⎢ ⎢0 ⎢⎣0 8

0 0 0 0

q⎤ 0⎥ ⎥, 0⎥ 0 ⎦⎥

⎡0 ⎢0 ⎢ ⎢0 ⎣⎢0

q 0 0 0

r⎤ 0⎥ ⎥, 0⎥ 0 ⎦⎥

0 1 p⎤ 0 0 0⎥ ⎥, 0 0 0⎥ 0 0 0 ⎦⎥

0 1⎤ ⎡0 0 ⎢0 0 0 0⎥ ⎥ , and ⎢ 0 0⎥ ⎢0 0 0 0 ⎥⎦ ⎢⎣0 0

0 0 0 0

0⎤ 0⎥ ⎥ 0⎥ 0⎥⎦


Section 1.3

5 + 1 2 + 3⎤ ⎡ 1+ 6 3. (a) D + E = ⎢ −1 + (−1) 0 + 1 1 + 2 ⎥ ⎢ ⎥ 2 + 1 4 + 3⎦ ⎣ 3+ 4 ⎡ 7 6 5⎤ = ⎢ −2 1 3⎥ ⎢ ⎥ ⎣ 7 3 7⎦

True/False 1.2 (a) True; reduced row echelon form is a row echelon form. (b) False; for instance, adding a nonzero multiple of the first row to any other row will result in a matrix that is not in row echelon form.

⎡ 1− 6 (b) D − E = ⎢ −1 − (−1) ⎢ ⎣ 3− 4 ⎡ −5 4 = ⎢ 0 −1 ⎢ ⎣ −1 1

(c) False; see Exercise 31. (d) True; the sum of the number of leading 1’s and the number of free variables is n. (e) True; a column cannot contain more than one leading 1.

5 − 1 2 − 3⎤ 0 −1 1 − 2 ⎥ ⎥ 2 − 1 4 − 3⎦ −1⎤ −1⎥ ⎥ 1⎦

⎡ 15 0 ⎤ (c) 5 A = ⎢ −5 10 ⎥ ⎢ ⎥ ⎣ 5 5⎦

(f) False; in row echelon form, there can be nonzero entries above the leading 1 in a column. (g) True; since there are n leading 1’s, there are no free variables, so only the trivial solution is possible.

⎡ −7 −28 −14 ⎤ (d) −7C = ⎢ ⎣ −21 −7 −35⎥⎦

(h) False; if the system has more equations than unknowns, a row of zeros does not indicate infinitely many solutions.

(e) 2B − C is not defined since 2B is a 2 × 2 matrix and C is a 2 × 3 matrix.

(i) False; the system could be inconsistent.

⎡ 24 (f) 4 E − 2 D = ⎢ −4 ⎢ ⎣ 16 ⎡ 22 = ⎢ −2 ⎢ ⎣ 10

Section 1.3 Exercise Set 1.3 1. (a) BA is undefined. (b) AC + D is 4 × 2.

4 12 ⎤ ⎡ 2 10 4 ⎤ 4 8 ⎥ − ⎢ −2 0 2 ⎥ ⎥ ⎢ ⎥ 4 12 ⎦ ⎣ 6 4 8⎦ −6 8 ⎤ 4 6⎥ ⎥ 0 4⎦

(g) −3( D + 2 E ) ⎛ ⎡ 1 5 2⎤ ⎡ 12 2 6 ⎤ ⎞ = −3 ⎜ ⎢ −1 0 1⎥ + ⎢ −2 2 4 ⎥ ⎟ ⎜⎜ ⎢ ⎥ ⎢ ⎥ ⎟⎟ ⎝ ⎣ 3 2 4⎦ ⎣ 8 2 6 ⎦ ⎠ ⎡ 13 7 8⎤ = −3 ⎢ −3 2 5 ⎥ ⎢ ⎥ ⎣ 11 4 10 ⎦ ⎡ −39 −21 −24 ⎤ = ⎢ 9 −6 −15⎥ ⎢ ⎥ ⎣ −33 −12 −30 ⎦

(c) AE + B is undefined, since AE is 4 × 4 not 4 × 5. (d) AB + B is undefined since AB is undefined. (e) E(A + B) is 5 × 5. (f) E(AC) is 5 × 2. (g) E T A is undefined since ET is 4 × 5. (h) ( AT + E ) D is 5 × 2.

⎡0 0⎤ (h) A − A = ⎢ 0 0 ⎥ ⎢ ⎥ ⎣0 0⎦ (i) tr(D) = 1 + 0 + 4 = 5

9


(j)


⎛ ⎡ 1 5 2 ⎤ ⎡ 18 3 9 ⎤ ⎞ tr( D − 3E ) = tr ⎜ ⎢ −1 0 1⎥ − ⎢ −3 3 6 ⎥ ⎟ ⎜⎜ ⎢ ⎥ ⎢ ⎥ ⎟⎟ ⎝ ⎣ 3 2 4 ⎦ ⎣ 12 3 9 ⎦ ⎠ ⎛ ⎡ −17 2 −7 ⎤ ⎞ = tr ⎜ ⎢ 2 −3 −5⎥ ⎟ ⎜⎜ ⎢ ⎥ ⎟⎟ ⎝ ⎣ −9 −1 −5⎦ ⎠ = −17 − 3 − 5 = −25

⎛ ⎡ 28 −7 ⎤ ⎞ (k) 4 tr(7 B) = 4 tr ⎜ ⎢ ⎥⎟ ⎝ ⎣ 0 14 ⎦ ⎠ = 4(28 + 14) = 4(42) = 168 (l) tr(A) is not defined because A is not a square matrix.

⎡ 3 0⎤ ⎡ 4 −1⎤ 5. (a) AB = ⎢ −1 2 ⎥ ⎢ ⎢ ⎥ ⎣ 0 2 ⎥⎦ ⎣ 1 1⎦ ⎡ (3 ⋅ 4) + (0 ⋅ 0) −(3 ⋅1) + (0 ⋅ 2) ⎤ = ⎢ −(1 ⋅ 4) + (2 ⋅ 0) (1⋅1) + (2 ⋅ 2) ⎥ ⎢ ⎥ ⎣ (1 ⋅ 4) + (1 ⋅ 0) −(1 ⋅1) + (1 ⋅ 2) ⎦ ⎡ 12 −3⎤ = ⎢ −4 5 ⎥ ⎢ ⎥ 1⎦ ⎣ 4 (b) BA is not defined since B is a 2 × 2 matrix and A is 3 × 2.

⎡ 18 3 9 ⎤ ⎡ 1 5 2 ⎤ (c) (3E ) D = ⎢ −3 3 6 ⎥ ⎢ −1 0 1⎥ ⎢ ⎥⎢ ⎥ ⎣ 12 3 9 ⎦ ⎣ 3 2 4 ⎦ ⎡ (18 ⋅1) − (3 ⋅1) + (9 ⋅ 3) (18 ⋅ 5) + (3 ⋅ 0) + (9 ⋅ 2) (18 ⋅ 2) + (3 ⋅1) + (9 ⋅ 4) ⎤ = ⎢ −(3 ⋅1) − (3 ⋅1) + (6 ⋅ 3) −(3 ⋅ 5) + (3 ⋅ 0) + (6 ⋅ 2) −(3 ⋅ 2) + (3 ⋅1) + (6 ⋅ 4) ⎥ ⎢ ⎥ ⎣ (12 ⋅1) − (3 ⋅1) + (9 ⋅ 3) (12 ⋅ 5) + (3 ⋅ 0) + (9 ⋅ 2) (12 ⋅ 2) + (3 ⋅1) + (9 ⋅ 4) ⎦ ⎡ 42 108 75⎤ = ⎢ 12 −3 21⎥ ⎢ ⎥ ⎣ 36 78 63⎦ ⎡ 12 −3⎤ ⎡1 4 (d) ( AB)C = ⎢ −4 5⎥ ⎢ ⎢ ⎥ ⎣3 1 1⎦ ⎣ 4 ⎡ (12 ⋅1) − (3 ⋅ 3) = ⎢ −(4 ⋅1) + (5 ⋅ 3) ⎢ ⎣ (4 ⋅1) + (1 ⋅ 3) ⎡ 3 45 9 ⎤ = ⎢11 −11 17 ⎥ ⎢ ⎥ ⎣ 7 17 13⎦

2⎤ 5⎥⎦ (12 ⋅ 4) − (3 ⋅1) (12 ⋅ 2) − (3 ⋅ 5) ⎤ −(4 ⋅ 4) + (5 ⋅1) −(4 ⋅ 2) + (5 ⋅ 5) ⎥ ⎥ (4 ⋅ 4) + (1 ⋅1) (4 ⋅ 2) + (1 ⋅ 5) ⎦

10


Section 1.3

⎡ 3 0⎤ ⎛ ⎡ 4 −1⎤ ⎡ 1 4 2 ⎤ ⎞ (e) A( BC ) = ⎢ −1 2 ⎥ ⎜ ⎢ ⎢ ⎥ ⎝ ⎣ 0 2 ⎥⎦ ⎢⎣3 1 5⎥⎦ ⎟⎠ ⎣ 1 1⎦ ⎡ 3 0⎤ ⎡ (4 ⋅1) − (1 ⋅ 3) (4 ⋅ 4) − (1 ⋅1) (4 ⋅ 2) − (1 ⋅ 5) ⎤ = ⎢ −1 2 ⎥ ⎢ ⎢ ⎥ ⎣(0 ⋅1) + (2 ⋅ 3) (0 ⋅ 4) + (2 ⋅1) (0 ⋅ 2) + (2 ⋅ 5) ⎥⎦ ⎣ 1 1⎦ ⎡ 3 0⎤ ⎡ 1 15 3⎤ = ⎢ −1 2 ⎥ ⎢ ⎢ ⎥ ⎣6 2 10⎥⎦ ⎣ 1 1⎦ ⎡ (3 ⋅1) + (0 ⋅ 6) (3 ⋅15) + (0 ⋅ 2) (3 ⋅ 3) + (0 ⋅10) ⎤ = ⎢ −(1 ⋅1) + (2 ⋅ 6) −(1⋅15) + (2 ⋅ 2) −(1 ⋅ 3) + (2 ⋅10) ⎥ ⎢ ⎥ (1⋅15) + (1 ⋅ 2) (1 ⋅ 3) + (1 ⋅10) ⎦ ⎣ (1 ⋅1) + (1 ⋅ 6) ⎡ 3 45 9⎤ = ⎢11 −11 17 ⎥ ⎢ ⎥ ⎣ 7 17 13⎦ ⎡ 1 3⎤ ⎡ 1 4 2⎤ ⎢ 4 1⎥ (f) CCT = ⎢ ⎣3 1 5⎥⎦ ⎢ 2 5⎥ ⎣ ⎦ ⎡(1 ⋅1) + (4 ⋅ 4) + (2 ⋅ 2) (1 ⋅ 3) + (4 ⋅1) + (2 ⋅ 5) ⎤ =⎢ ⎣ (3 ⋅1) + (1 ⋅ 4) + (5 ⋅ 2) (3 ⋅ 3) + (1⋅1) + (5 ⋅ 5) ⎥⎦ ⎡ 21 17 ⎤ =⎢ ⎣17 35⎥⎦ T

⎛ ⎡ 1 5 2⎤ ⎡ 3 0⎤ ⎞ (g) ( DA) = ⎜ ⎢ −1 0 1⎥ ⎢ −1 2 ⎥ ⎟ ⎜⎜ ⎢ ⎥⎢ ⎥ ⎟⎟ ⎝ ⎣ 3 2 4⎦ ⎣ 1 1⎦ ⎠ T

T

⎛ ⎡ (1⋅ 3) − (5 ⋅1) + (2 ⋅1) (1 ⋅ 0) + (5 ⋅ 2) + (2 ⋅1) ⎤ ⎞ = ⎜ ⎢ −(1 ⋅ 3) − (0 ⋅1) + (1 ⋅1) −(1 ⋅ 0) + (0 ⋅ 2) + (1 ⋅1) ⎥ ⎟ ⎜⎜ ⎢ ⎥ ⎟⎟ ⎝ ⎣ (3 ⋅ 3) − (2 ⋅1) + (4 ⋅1) (3 ⋅ 0) + (2 ⋅ 2) + (4 ⋅1) ⎦ ⎠ T

⎛ ⎡ 0 12⎤ ⎞ = ⎜ ⎢ −2 1⎥ ⎟ ⎜⎜ ⎢ ⎥ ⎟⎟ ⎝ ⎣ 11 8⎦ ⎠ ⎡ 0 −2 11⎤ =⎢ 1 8⎥⎦ ⎣12

11



⎛ ⎡ 1 3⎤ ⎞ ⎡ 4 −1⎤ ⎟ ⎡ 3 −1 1⎤ (h) (C T B) AT = ⎜ ⎢ 4 1⎥ ⎢ ⎜⎜ ⎢ ⎥ ⎣ 0 2⎥⎦ ⎟⎟ ⎢⎣ 0 2 1⎥⎦ ⎝ ⎣ 2 5⎦ ⎠ ⎡ (1 ⋅ 4) + (3 ⋅ 0) −(1 ⋅1) + (3 ⋅ 2) ⎤ ⎡ 3 −1 1⎤ = ⎢ (4 ⋅ 4) + (1 ⋅ 0) −(4 ⋅1) + (1 ⋅ 2) ⎥ ⎢ ⎢ ⎥ ⎣ 0 2 1⎥⎦ ⎣(2 ⋅ 4) + (5 ⋅ 0) −(2 ⋅1) + (5 ⋅ 2) ⎦ 5⎤ ⎡ 4 ⎡ 3 −1 1⎤ = ⎢16 −2⎥ ⎢ ⎢ ⎥ ⎣ 0 2 1⎥⎦ ⎣ 8 8⎦ ⎡ (4 ⋅ 3) + (5 ⋅ 0) −(4 ⋅1) + (5 ⋅ 2) (4 ⋅1) + (5 ⋅1) ⎤ = ⎢(16 ⋅ 3) − (2 ⋅ 0) −(16 ⋅1) − (2 ⋅ 2) (16 ⋅1) − (2 ⋅1) ⎥ ⎢ ⎥ ⎣ (8 ⋅ 3) + (8 ⋅ 0) −(8 ⋅1) + (8 ⋅ 2) (8 ⋅1) + (8 ⋅1) ⎦ 6 9⎤ ⎡ 12 = ⎢ 48 −20 14 ⎥ ⎢ ⎥ 8 16 ⎦ ⎣ 24

(i)

⎛ ⎡ 1 5 2 ⎤ ⎡ 1 − 1 3⎤ ⎞ tr( DDT ) = tr ⎜ ⎢ −1 0 1⎥ ⎢ 5 0 2 ⎥ ⎟ ⎜⎜ ⎢ ⎥⎢ ⎥ ⎟⎟ ⎝ ⎣ 3 2 4⎦ ⎣2 1 4⎦ ⎠ ⎛ ⎡ (1 ⋅1) + (5 ⋅ 5) + (2 ⋅ 2) −(1 ⋅1) + (5 ⋅ 0) + (2 ⋅1) (1⋅ 3) + (5 ⋅ 2) + (2 ⋅ 4) ⎤ ⎞ = tr ⎜ ⎢ −(1 ⋅1) + (0 ⋅ 5) + (1 ⋅ 2) (1⋅1) + (0 ⋅ 0) + (1 ⋅1) −(1 ⋅ 3) + (0 ⋅ 2) + (1⋅ 4) ⎥ ⎟ ⎜⎜ ⎢ ⎥ ⎟⎟ ⎝ ⎣ (3 ⋅1) + (2 ⋅ 5) + (4 ⋅ 2) −(3 ⋅1) + (2 ⋅ 0) + (4 ⋅1) (3 ⋅ 3) + (2 ⋅ 2) + (4 ⋅ 4) ⎦ ⎠ ⎛ ⎡30 1 21⎤ ⎞ 1⎥ ⎟ = tr ⎜ ⎢ 1 2 ⎜⎜ ⎢ ⎥ ⎟⎟ ⎝ ⎣ 21 1 29 ⎦ ⎠ = 30 + 2 + 29 = 61

(j)

⎛ ⎡ 24 −4 16 ⎤ ⎡ 1 5 2 ⎤ ⎞ tr(4 ET − D) = tr ⎜ ⎢ 4 4 4 ⎥ − ⎢ −1 0 1⎥ ⎟ ⎜⎜ ⎢ ⎥ ⎢ ⎥⎟ 8 12 ⎦ ⎣ 3 2 4 ⎦ ⎟⎠ ⎝ ⎣ 12 ⎛ ⎡ 23 −9 14 ⎤ ⎞ = tr ⎜ ⎢ 5 4 3⎥ ⎟ ⎜⎜ ⎢ ⎥ ⎟⎟ ⎝ ⎣ 9 6 8⎦ ⎠ = 23 + 4 + 8 = 35

12


Section 1.3

⎛ ⎡ 1 3⎤ ⎡ 3 −1 (k) tr(C T AT + 2 ET ) = tr ⎜ ⎢ 4 1⎥ ⎢ ⎜⎜ ⎢ ⎥ ⎣0 2 ⎝ ⎣ 2 5⎦ ⎛ ⎡ (1⋅ 3) + (3 ⋅ 0) = tr ⎜ ⎢ (4 ⋅ 3) + (1 ⋅ 0) ⎜⎜ ⎢(2 ⋅ 3) + (5 ⋅ 0) ⎝⎣ ⎛⎡ 3 5 = tr ⎜ ⎢12 −2 ⎜⎜ ⎢ 8 ⎝⎣ 6 ⎛ ⎡15 3 = tr ⎜ ⎢14 0 ⎜⎜ ⎢ ⎝ ⎣12 12 = 15 + 0 + 13 = 28

(l)

⎡12 −2 1⎤ ⎢ + 2 2 1⎥⎦ ⎢ 6 4 ⎣ −(1⋅1) + (3 ⋅ 2) −(4 ⋅1) + (1 ⋅ 2) −(2 ⋅1) + (5 ⋅ 2)

8⎤ ⎞ 2⎥ ⎟ ⎥⎟ 6 ⎦ ⎟⎠ (1 ⋅1) + (3 ⋅1) ⎤ ⎡12 −2 8⎤ ⎞ (4 ⋅1) + (1 ⋅1) ⎥ + ⎢ 2 2 2 ⎥ ⎟ ⎥ ⎢ ⎥⎟ (2 ⋅1) + (5 ⋅1) ⎦ ⎣ 6 4 6 ⎦ ⎟⎠

4 ⎤ ⎡12 −2 8⎤ ⎞ 5⎥ + ⎢ 2 2 2 ⎥ ⎟ ⎥ ⎢ ⎥⎟ 7 ⎦ ⎣ 6 4 6 ⎦ ⎟⎠ 12⎤ ⎞ 7⎥ ⎟ ⎥⎟ 13⎦ ⎟⎠

⎛ ⎛ ⎡ 6 1 3⎤ ⎡ 1 3⎤ ⎞T ⎡ 3 0 ⎤ ⎞ ⎜ ⎟ tr(( ECT )T A) = tr ⎜ ⎜ ⎢ −1 1 2 ⎥ ⎢ 4 1⎥ ⎟ ⎢ −1 2 ⎥ ⎟ ⎜⎜ ⎢ ⎟ ⎜ ⎝ ⎣ 4 1 3⎥⎦ ⎢⎣ 2 5⎥⎦ ⎟⎠ ⎢⎣ 1 1⎥⎦ ⎟ ⎝ ⎠ ⎛ ⎛ ⎡ (6 ⋅1) + (1 ⋅ 4) + (3 ⋅ 2) (6 ⋅ 3) + (1 ⋅1) + (3 ⋅ 5) ⎤ ⎞T ⎜ = tr ⎜ ⎜ ⎢ −(1⋅1) + (1 ⋅ 4) + (2 ⋅ 2) −(1 ⋅ 3) + (1⋅1) + (2 ⋅ 5) ⎥ ⎟ ⎜ ⎟ ⎜ ⎝⎜ ⎢⎣ (4 ⋅1) + (1⋅ 4) + (3 ⋅ 2) (4 ⋅ 3) + (1 ⋅1) + (3 ⋅ 5) ⎥⎦ ⎟⎠ ⎝ ⎛ ⎛ ⎡16 34 ⎤T ⎞ ⎡ 3 0 ⎤ ⎞ ⎜⎜ ⎟ ⎟ = tr ⎜ ⎜ ⎢ 7 8⎥ ⎟ ⎢ −1 2 ⎥ ⎟ ⎢ ⎥ ⎢ ⎥ ⎜ ⎜ ⎣14 28⎦ ⎟ ⎣ 1 1⎦ ⎟ ⎠ ⎝⎝ ⎠ ⎛ ⎡ 3 0⎤ ⎞ ⎡16 7 14 ⎤ ⎢ = tr ⎜ ⎢ −1 2 ⎥ ⎟ ⎜⎜ ⎣34 8 28⎥⎦ ⎢ ⎥ ⎟⎟ ⎣ 1 1⎦ ⎠ ⎝

⎡ 3 0 ⎤ ⎞⎟ ⎢ −1 2 ⎥ ⎟ ⎢ ⎥ ⎣ 1 1⎦ ⎟⎠

⎛ ⎡ (16 ⋅ 3) − (7 ⋅1) + (14 ⋅1) (16 ⋅ 0) + (7 ⋅ 2) + (14 ⋅1)⎤ ⎞ = tr ⎜ ⎢ ⎥⎟ ⎝ ⎣(34 ⋅ 3) − (8 ⋅1) + (28 ⋅1) (34 ⋅ 0) + (8 ⋅ 2) + (28 ⋅1) ⎦ ⎠ ⎛ ⎡ 55 28⎤ ⎞ = tr ⎜ ⎢ ⎥⎟ ⎝ ⎣122 44 ⎦ ⎠ = 55 + 44 = 99 7. (a) The first row of AB is the first row of A times B. ⎡ 6 −2 4 ⎤ 1 3⎥ [a1 B] = [3 −2 7] ⎢ 0 ⎢ ⎥ ⎣ 7 7 5⎦ = [18 + 0 + 49 −6 − 2 + 49 12 − 6 + 35] = [67 41 41]

13



(f) The third column of AA is A times the third column of A. ⎡ 3 −2 7 ⎤ ⎡7 ⎤ [ A a3 ] = ⎢ 6 5 4 ⎥ ⎢ 4 ⎥ ⎢ ⎥⎢ ⎥ ⎣ 0 4 9 ⎦ ⎣9 ⎦ ⎡ 21 − 8 + 63 ⎤ = ⎢ 42 + 20 + 36 ⎥ ⎢ ⎥ ⎣ 0 + 16 + 81 ⎦ ⎡ 76 ⎤ = ⎢ 98 ⎥ ⎢ ⎥ ⎣97 ⎦

(b) The third row of AB is the third row of A times B. [ a3 B ] ⎡ 6 −2 4 ⎤ 1 3⎥ = [0 4 9] ⎢ 0 ⎢ ⎥ ⎣ 7 7 5⎦ = [0 + 0 + 63 0 + 4 + 63 0 + 12 + 45] = [63 67 57]

(c) The second column of AB is A times the second column of B. ⎡ 3 −2 7 ⎤ ⎡ −2 ⎤ [ A b 2 ] = ⎢6 5 4 ⎥ ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎣0 4 9 ⎦ ⎣ 7 ⎦ ⎡ −6 − 2 + 49 ⎤ = ⎢ −12 + 5 + 28⎥ ⎢ ⎥ ⎣ 0 + 4 + 63 ⎦ ⎡ 41⎤ = ⎢ 21⎥ ⎢ ⎥ ⎣67 ⎦

⎡ −3 12 76 ⎤ 9. (a) AA = ⎢ 48 29 98⎥ ⎢ ⎥ ⎣ 24 56 97 ⎦ ⎡ −3 ⎤ ⎡ 3⎤ ⎡ −2 ⎤ ⎢ 48⎥ = 3 ⎢6 ⎥ + 6 ⎢ 5⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 24 ⎦ ⎣0 ⎦ ⎣ 4⎦ ⎡ 12 ⎤ ⎡ 3⎤ ⎡ −2 ⎤ ⎡7 ⎤ ⎢ 29 ⎥ = −2 ⎢ 6⎥ + 5 ⎢ 5⎥ + 4 ⎢ 4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 56 ⎦ ⎣ 0⎦ ⎣ 4⎦ ⎣9⎦

(d) The first column of BA is B times the first column of A. ⎡ 6 −2 4 ⎤ ⎡ 3 ⎤ [ B a1 ] = ⎢ 0 1 3⎥ ⎢ 6 ⎥ ⎢ ⎥⎢ ⎥ 7 7 5⎦ ⎣ 0 ⎦ ⎣ ⎡18 − 12 + 0 ⎤ = ⎢ 0+6+0 ⎥ ⎢ ⎥ ⎣ 21 + 42 + 0⎦ ⎡ 6⎤ = ⎢ 6⎥ ⎢ ⎥ ⎣63⎦

⎡76 ⎤ ⎡ 3⎤ ⎡ −2 ⎤ ⎡7 ⎤ ⎢ 98⎥ = 7 ⎢6 ⎥ + 4 ⎢ 5⎥ + 9 ⎢ 4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣97 ⎦ ⎣0 ⎦ ⎣ 4⎦ ⎣9⎦ ⎡ 64 14 38⎤ (b) BB = ⎢ 21 22 18⎥ ⎢ ⎥ ⎣77 28 74⎦ ⎡ 64 ⎤ ⎡6⎤ ⎡ 4⎤ ⎢ 21⎥ = 6 ⎢ 0 ⎥ + 7 ⎢ 3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣77 ⎦ ⎣7 ⎦ ⎣ 5⎦

(e) The third row of AA is the third row of A times A. [a3 A] ⎡ 3 −2 7 ⎤ = [ 0 4 9] ⎢ 6 5 4 ⎥ ⎢ ⎥ ⎣0 4 9⎦ = [0 + 24 + 0 0 + 20 + 36 0 + 16 + 81] = [ 24 56 97]

⎡ 14 ⎤ ⎡ 6 ⎤ ⎡ −2 ⎤ ⎡4⎤ ⎢ 22 ⎥ = −2 ⎢ 0 ⎥ + 1 ⎢ 1⎥ + 7 ⎢ 3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 28⎦ ⎣7 ⎦ ⎣ 7⎦ ⎣ 5⎦ ⎡ 38⎤ ⎡6⎤ ⎡ −2 ⎤ ⎡ 4⎤ ⎢ 18⎥ = 4 ⎢ 0 ⎥ + 3 ⎢ 1⎥ + 5 ⎢ 3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣74 ⎦ ⎣7 ⎦ ⎣ 7⎦ ⎣ 5⎦ ⎡ x1 ⎤ ⎡ 2 −3 5⎤ ⎡ 7⎤ 11. (a) A = ⎢ 9 −1 1⎥ , x = ⎢ x2 ⎥ , b = ⎢ −1⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0⎦ ⎣ 1 5 4⎦ ⎣ x3 ⎦ ⎡ 2 −3 5⎤ ⎡ x1 ⎤ ⎡ 7 ⎤ The equation is ⎢ 9 −1 1⎥ ⎢ x2 ⎥ = ⎢ −1⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 1 5 4⎦ ⎣ x3 ⎦ ⎣ 0 ⎦

14


Section 1.3 21. The diagonal entries of A + B are aii + bii , so tr( A + B) = (a11 + b11 ) + (a22 + b22 ) + + (ann + bnn ) = (a11 + a22 + + ann ) + (b11 + b22 + + bnn ) = tr(A) + tr( B)

⎡ x1 ⎤ 1⎤ ⎡ 1⎤ ⎡ 4 0 −3 ⎢x ⎥ ⎢5 ⎢ 3⎥ 1 0 −8 ⎥ (b) A = ⎢ ⎥, x = ⎢ 2⎥, b = ⎢ ⎥ 2 − 5 9 − 1 x ⎢ 3⎥ ⎢ ⎥ ⎢0⎥ ⎣⎢ 0 3 −1 7 ⎦⎥ ⎣⎢ 2 ⎦⎥ ⎣⎢ x4 ⎦⎥ The equation is 1⎤ ⎡ x1 ⎤ ⎡ 1⎤ ⎡ 4 0 −3 ⎢5 1 0 −8⎥ ⎢ x2 ⎥ ⎢ 3⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥. ⎢ 2 −5 9 −1⎥ ⎢ x3 ⎥ ⎢ 0 ⎥ 3 −1 7 ⎦⎥ ⎣⎢ x4 ⎦⎥ ⎣⎢ 2 ⎦⎥ ⎣⎢ 0

⎡ a11 0 ⎢ 0 a 22 ⎢ 0 0 23. (a) ⎢ ⎢ 0 0 ⎢ 0 0 ⎢ 0 ⎣⎢ 0

0 0 a33 0 0 0

0 0 0 a44 0 0

0 0 0 0 a55 0

0 ⎤ 0 ⎥ ⎥ 0 ⎥ 0 ⎥ 0 ⎥ ⎥ a66 ⎦⎥

(b) The system is x1 + x2 + x3 = 2 =2 2 x1 + 3x2 5 x1 − 3x2 − 6 x3 = −9

⎡ a11 a12 ⎢ 0 a 22 ⎢ 0 0 (b) ⎢ ⎢ 0 0 ⎢ 0 0 ⎢ 0 ⎢⎣ 0

a13 a23 a33 0 0 0

a14 a24 a34 a44 0 0

a15 a25 a35 a45 a55 0

a16 ⎤ a26 ⎥ ⎥ a36 ⎥ a46 ⎥ a56 ⎥⎥ a66 ⎥⎦

⎡ 1 1 0⎤ ⎡k ⎤ 15. [ k 1 1] ⎢ 1 0 2 ⎥ ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎣ 0 2 −3⎦ ⎣ 1⎦ ⎡k ⎤ = [ k + 1 k + 2 −1] ⎢ 1⎥ ⎢ ⎥ ⎣ 1⎦

⎡ a11 0 ⎢a a ⎢ 21 22 a a (c) ⎢ 31 32 ⎢ a41 a42 ⎢a a ⎢ 51 52 ⎢⎣ a61 a62

0 0 a33 a43 a53 a63

0 0 0 a44 a54 a64

0 0 0 0 a55 a65

0 ⎤ 0 ⎥ ⎥ 0 ⎥ 0 ⎥ 0 ⎥ ⎥ a66 ⎥⎦

⎡ a11 a12 ⎢a a ⎢ 21 22 0 a32 (d) ⎢ ⎢ 0 0 ⎢ 0 0 ⎢ 0 0 ⎢⎣

0 a23 a33 a43 0 0

0 0 a34 a44 a54 0

0 0 0 a45 a55 a65

0 ⎤ 0 ⎥ ⎥ 0 ⎥ 0 ⎥ a56 ⎥ ⎥ a66 ⎦⎥

13. (a) The system is 5 x1 + 6 x2 − 7 x3 = 2 − x1 − 2 x2 + 3 x3 = 0 4 x2 − x3 = 3

= [ k 2 + k + k + 2 − 1] = [ k 2 + 2k + 1] The only solution of k 2 + 2k + 1 = 0 is k = −1. 3 ⎤ ⎡ 4 d − 2c ⎤ ⎡a 17. ⎢ = gives the −2 ⎥⎦ ⎣ −1 a + b ⎥⎦ ⎢⎣ d + 2c equations a=4 d − 2c = 3 d + 2c = −1 a + b = −2 a = 4 ⇒ b = −6 d − 2c = 3, d + 2c = −1 ⇒ 2d = 2, d = 1 d = 1 ⇒ 2c = −2, c = −1 The solution is a = 4, b = −6, c = −1, d = 1.

25.

⎛ x ⎞ ⎡1 1⎤ ⎡ x1 ⎤ ⎛ x1 + x2 ⎞ f ⎜ 1⎟=⎢ ⎟ ⎥⎢ ⎥ =⎜ ⎝ x2 ⎠ ⎣ 0 1⎦ ⎣ x2 ⎦ ⎝ x2 ⎠ ⎛1⎞ ⎛1 + 1⎞ ⎛ 2 ⎞ (a) f ⎜ ⎟ = ⎜ ⎟=⎜ ⎟ ⎝1⎠ ⎝ 1 ⎠ ⎝ 1 ⎠ y 1

f(x)

x

19. The ij-entry of kA is kaij . If kaij = 0 for all i, j

x

then either k = 0 or aij = 0 for all i, j, in which

1

case A = 0.

15

2


⎛ 2⎞ ⎛ 2 + 0⎞ ⎛ 2⎞ (b) f ⎜ ⎟ = ⎜ ⎟=⎜ ⎟ ⎝ 0⎠ ⎝ 0 ⎠ ⎝ 0⎠

⎡5 0 ⎤ (b) The four square roots of ⎢ are ⎣0 9 ⎥⎦ ⎡ 5 0⎤ ⎡− 5 0⎤ ⎡ 5 0 ⎤ , , and ⎢ 0 3⎥ , ⎢ 0 3⎥⎦ ⎢⎣ 0 −3⎥⎦ ⎣ ⎦ ⎣ ⎡− 5 0 ⎤ . ⎢ 0 −3⎥⎦ ⎣

y 1 f(x) = x 1

(c)

x

2

⎛ 4 ⎞ ⎛ 4 + 3⎞ ⎛ 7 ⎞ f ⎜ ⎟=⎜ ⎟=⎜ ⎟ ⎝ 3⎠ ⎝ 3 ⎠ ⎝ 3⎠ 5


True/False 1.3

y

(a) True; only square matrices have main diagonals. f(x)

x

(b) False; an m × n matrix has m row vectors and n column vectors.

x 8

4

(c) False; matrix multiplication is not commutative.

⎛ 2⎞ ⎛ 2 − 2⎞ ⎛ 0 ⎞ (d) f ⎜ ⎟ = ⎜ ⎟=⎜ ⎟ ⎝ −2 ⎠ ⎝ −2 ⎠ ⎝ −2 ⎠

(d) False; the ith row vector of AB is found by multiplying the ith row vector of A by B.

y

(e) True x 1

⎡ 1 3⎤ (f) False; for example, if A = ⎢ and ⎣ 4 −1⎥⎦ ⎡ 1 6⎤ ⎡1 0 ⎤ and B=⎢ , then AB = ⎢ ⎥ 0 2 ⎣ ⎦ ⎣ 4 −2⎥⎦ tr(AB) = −1, while tr(A) = 0 and tr(B) = 3.

2

–1 –2

f(x) x

27. Let A = [ aij ].

a13 ⎤ ⎡ x ⎤ ⎡ a11 x + a12 y + a13 z ⎤ a23 ⎥ ⎢ y ⎥ = ⎢ a21 x + a22 y + a23 z ⎥ ⎥⎢ ⎥ ⎢ ⎥ a33 ⎦⎥ ⎣ z ⎦ ⎣⎢ a31 x + a32 y + a33 z ⎦⎥ ⎡x + y⎤ = ⎢x − y⎥ ⎢ ⎥ ⎣ 0 ⎦ The matrix equation yields the equations a11 x + a12 y + a13 z = x + y a21 x + a22 y + a23 z = x − y . a31 x + a32 y + a33 z = 0 For these equations to be true for all values of x, y, and z it must be that a11 = 1, a12 = 1, a21 = 1, ⎡ a11 a12 ⎢a a ⎢ 21 22 a ⎣⎢ 31 a32

⎡ 1 3⎤ (g) False; for example, if A = ⎢ and ⎣ 4 −1⎥⎦ ⎡ 1 4⎤ ⎡ 1 0⎤ , while B=⎢ , then ( AB)T = ⎢ ⎥ 0 2 ⎣6 −2 ⎥⎦ ⎣ ⎦ ⎡ 1 8⎤ AT BT = ⎢ . ⎣3 −2 ⎥⎦ (h) True; for a square matrix A the main diagonals

of A and AT are the same. (i) True; if A is a 6 × 4 matrix and B is an m × n

matrix, then AT is a 4 × 6 matrix and BT is an

a22 = −1, and aij = 0 for all other i, j. There is

n × m matrix. So BT AT is only defined if m = 4, and it can only be a 2 × 6 matrix if n = 2.

⎡ 1 1 0⎤ one such matrix, A = ⎢ 1 −1 0 ⎥ . ⎢ ⎥ ⎣0 0 0⎦

(j) True, tr(cA) = ca11 + ca22 + = c(a11 + a22 + = c tr( A)

⎡1 1⎤ ⎡ −1 −1⎤ 29. (a) Both ⎢ and ⎢ are square roots ⎥ 1 1 ⎣ ⎦ ⎣ −1 −1⎥⎦ ⎡2 2⎤ of ⎢ . ⎣ 2 2 ⎥⎦

+ cann + ann )

(k) True; if A − C = B − C, then aij − cij = bij − cij , so it follows that aij = bij .

16


Section 1.4

⎡ 1 2⎤ ⎡ 1 5⎤ (l) False; for example, if A = ⎢ , B=⎢ , ⎣3 4 ⎥⎦ ⎣3 7 ⎥⎦ ⎡ 1 0⎤ ⎡ 1 0⎤ and C = ⎢ , then AC = BC = ⎢ . ⎥ 0 0 ⎣ ⎦ ⎣3 0 ⎥⎦

⎡ 2 4⎤ 11. BT = ⎢ ; ⎣ −3 4⎥⎦ ( BT ) − 1 =

(m) True; if A is an m × n matrix, then for both AB and BA to be defined B must be an n × m matrix. Then AB will be an m × m matrix and BA will be an n × n matrix, so m = n in order for the sum to be defined.

1 1 ⎡ 4 −4 ⎤ ⎡ 5 ⎢ = 8 + 12 ⎢⎣ 3 2 ⎥⎦ ⎢ 3 ⎣ 20

⎡ 70 45⎤ 13. ABC = ⎢ ⎣122 79 ⎥⎦ 1 ⎡ 79 −45⎤ 70 ⋅ 79 − 122 ⋅ 45 ⎣⎢ −122 70 ⎦⎥ 1 ⎡ 79 −45⎤ = 40 ⎢⎣ −122 70⎥⎦ 1 ⎡ −1 −4 ⎤ 1 ⎡ −1 −4 ⎤ C −1 = = 6(−1) − 4(−2) ⎢⎣ 2 6 ⎥⎦ 2 ⎢⎣ 2 6 ⎥⎦ 1 ⎡ 4 3⎤ 1 ⎡ 4 3 ⎤ B −1 = = 2 ⋅ 4 − (−3)4 ⎢⎣ −4 2 ⎥⎦ 20 ⎢⎣ −4 2 ⎥⎦ ( ABC )−1 =

(n) True; since the jth column vector of AB is A[jth column vector of B], a column of zeros in B will result in a column of zeros in AB.

⎡ 1 0 ⎤ ⎡ 1 5⎤ ⎡ 1 5 ⎤ (o) False; ⎢ = ⎣0 0⎥⎦ ⎢⎣3 7 ⎥⎦ ⎢⎣ 0 0 ⎥⎦

1 ⎡ 2 −1⎤ ⎡ 2 −1⎤ = 3 ⋅ 2 − 1 ⋅ 5 ⎢⎣ −5 3⎥⎦ ⎢⎣ −5 3⎥⎦ 1 ⎛ ⎡ −1 −4 ⎤ ⎡ 4 3⎤ ⎡ 2 −1⎤ ⎞ C −1 B −1 A−1 = ⎜ ⎟ 40 ⎝ ⎢⎣ 2 6 ⎥⎦ ⎢⎣ −4 2 ⎥⎦ ⎢⎣ −5 3⎥⎦ ⎠ 1 ⎡ 79 −45⎤ = 40 ⎢⎣ −122 70⎥⎦

Section 1.4

A−1 =

Exercise Set 1.4

1 ⎡ 4 3⎤ 2 ⋅ 4 − (−3)4 ⎢⎣ −4 2 ⎥⎦ 1 ⎡ 4 3⎤ = 20 ⎢⎣ −4 2 ⎥⎦ 3 ⎤ = ⎡ 15 20 ⎢ ⎥ 1⎥ ⎢⎣ − 15 10 ⎦

5. B −1 =

7. D −1 =

15. 7 A = ((7 A)−1 )−1 1 ⎡ −2 −7 ⎤ = (−3)(−2) − 1⋅ 7 ⎢⎣ −1 −3⎥⎦ ⎡ −2 −7 ⎤ = −1 ⎢ ⎣ −1 −3⎥⎦ ⎡2 7⎤ =⎢ ⎣ 1 3⎥⎦ ⎡ 2 1⎤ ⎥. Thus A = ⎢ 7 ⎢⎣ 17 73 ⎥⎦

1 1 ⎡ 3 0 ⎤ 1 ⎡ 3 0 ⎤ ⎡ 2 0⎤ ⎢ ⎥ = = 2 ⋅ 3 − 0 ⎢⎣0 2 ⎥⎦ 6 ⎢⎣0 2 ⎥⎦ ⎢ 0 1 ⎥ 3⎦ ⎣

1 9. Here, a = d = (e x + e − x ) and 2 1 x −x b = c = (e − e ), so 2 1 1 ad − bc = (e x + e − x )2 − (e x − e− x )2 4 4 1 2x 1 −2 x = (e + 2 + e ) − (e2 x − 2 + e −2 x ) 4 4 = 1. ⎡ 1 (e x + e − x ) Thus, ⎢ 2 x − x ⎢ 1 (e − e ) ⎣2

17. I + 2 A = (( I + 2 A)−1 )−1 1 ⎡ 5 −2 ⎤ = −1⋅ 5 − 2 ⋅ 4 ⎢⎣ −4 −1⎥⎦ 1 ⎡ 5 −2 ⎤ =− ⎢ 13 ⎣ −4 −1⎥⎦ ⎡− 5 2 ⎤ = ⎢ 13 13 ⎥ 4 1⎥ ⎢⎣ 13 13 ⎦

−1

− e− x ) ⎤ ⎥ + e− x ) ⎥⎦ ⎡ 1 (e x + e − x ) − 1 (e x − e − x ) ⎤ 2 ⎥. = ⎢ 2 x −x 1 (e x + e − x ) ⎥ ⎢ − 1 (e − e ) ⎣ 2 ⎦ 2 1 (e x 2 1 (e x 2

− 15 ⎤ ⎥ = ( B −1 )T 1⎥ 10 ⎦

17


1⎤ 2⎤ ⎡− 9 ⎡ − 5 2 ⎤ ⎡ 1 0 ⎤ ⎡ − 18 13 ⎥ . 13 13 ⎥ and A = ⎢ 13 ⎢ Thus 2 A = ⎢ 13 13 ⎥ − ⎢ = 4 1 ⎥ ⎣ 0 1⎦⎥ ⎢ 4 − 12 ⎥ 2 − 6⎥ ⎢ 13 ⎢⎣ 13 13 ⎦ 13 ⎦ 13 ⎦ ⎣ 13 ⎣

⎛ ⎡3 19. (a) A3 = ⎜ ⎢ ⎝ ⎣2 ⎡11 =⎢ ⎣8 ⎡ 41 =⎢ ⎣30

1⎤ ⎡ 3 1⎥⎦ ⎢⎣ 2 4⎤ ⎡ 3 3⎥⎦ ⎢⎣ 2 15⎤ 11⎥⎦

1⎤ ⎞ ⎡ 3 1⎤ 1⎥⎦ ⎟⎠ ⎢⎣ 2 1⎥⎦ 1⎤ 1⎥⎦

(b) A−3 = ( A3 ) −1 1 ⎡ 11 −15⎤ = 41⎥⎦ 41⋅11 − 15 ⋅ 30 ⎢⎣ −30 ⎡ 11 −15⎤ =⎢ 41⎥⎦ ⎣ −30

⎡11 (c) A2 − 2 A + I = ⎢ ⎣8 ⎡6 =⎢ ⎣4

4 ⎤ ⎡ 6 2 ⎤ ⎡ 1 0⎤ − + 3⎥⎦ ⎢⎣ 4 2 ⎥⎦ ⎢⎣0 1⎥⎦ 2⎤ 2 ⎥⎦

⎡ 3 1⎤ ⎡ 2 0 ⎤ ⎡ 1 1⎤ (d) p( A) = A − 2 I = ⎢ − = ⎣ 2 1⎥⎦ ⎢⎣ 0 2 ⎥⎦ ⎢⎣ 2 −1⎥⎦ (e)

p( A) = 2 A2 − A + I ⎡ 22 8⎤ ⎡ 3 1⎤ ⎡ 1 0 ⎤ =⎢ − + ⎣ 16 6 ⎥⎦ ⎢⎣ 2 1⎥⎦ ⎢⎣0 1⎥⎦ ⎡ 20 7 ⎤ =⎢ ⎣ 14 6⎥⎦

(f)

p( A) = A3 − 2 A + 4 I ⎡ 41 15⎤ ⎡ 6 2 ⎤ ⎡ 4 0⎤ =⎢ − + ⎣30 11⎥⎦ ⎢⎣ 4 2 ⎥⎦ ⎢⎣ 0 4⎥⎦ ⎡ 39 13⎤ =⎢ ⎣ 26 13⎥⎦

⎛ ⎡ 3 0 0⎤ ⎡ 3 21. (a) A3 = ⎜ ⎢0 −1 3⎥ ⎢ 0 ⎜⎜ ⎢ ⎥⎢ ⎝ ⎣0 −3 −1⎦ ⎣ 0 ⎡9 0 0⎤ ⎡ 3 = ⎢ 0 −8 −6 ⎥ ⎢ 0 ⎢ ⎥⎢ ⎣ 0 6 −8 ⎦ ⎣ 0 0⎤ ⎡ 27 0 = ⎢ 0 26 −18⎥ ⎢ ⎥ ⎣ 0 18 26⎦

0 0⎤ ⎞ ⎡ 3 0 0⎤ −1 3⎥ ⎟ ⎢0 −1 3⎥ ⎥⎟⎢ ⎥ −3 −1⎦ ⎟⎠ ⎣0 −3 −1⎦ 0 0⎤ − 1 3⎥ ⎥ −3 −1⎦

18



Section 1.4

1 1 ⎡ −1 −3⎤ ⎡ − 10 ⎡ −1 3⎤ ⎢ is (b) Note that the inverse of ⎢ = 10 ⎢⎣ 3 −1⎥⎦ ⎢ 3 ⎣ −3 −1⎥⎦ ⎣ 10

A−3

⎛ ⎡1 ⎜ ⎢3 = ( A− 1 ) 3 = ⎜ ⎢ 0 ⎜⎢ ⎜ ⎢0 ⎝⎣ ⎡1 ⎢9 = ⎢0 ⎢ ⎢⎣ 0 ⎡1 ⎢ 27 =⎢0 ⎢0 ⎣

⎡1 0 0⎤ ⎢3 ⎥ 1 − 3 ⎥. Thus A−1 = ⎢ 0 − 10 10 ⎢ ⎥ 3 − 1 ⎢⎣ 0 10 10 ⎥⎦

0⎤ ⎡ 13 0 0 ⎤ ⎞ ⎡ 13 0 0⎤ ⎥⎢ ⎥⎟⎢ ⎥ 1 − 3 ⎥ ⎢0 − 1 − 3 ⎥ ⎟ ⎢0 − 1 − 3 ⎥ − 10 10 10 10 ⎟ 10 10 ⎥⎢ ⎥ ⎢ ⎥ 3 − 1 3 − 1 ⎟ 0 3 − 1 0 10 10 ⎥⎦ ⎢⎣ 10 10 ⎥⎦ ⎠ ⎢⎣ 10 10 ⎥⎦ 0 0 ⎤ ⎡ 13 0 0⎤ ⎥⎢ ⎥ 3 ⎥ ⎢0 − 1 − 3 ⎥ 2 − 25 50 10 10 ⎥⎢ ⎥ 3 3 − 1 2 − 50 − 25 ⎥⎦ ⎢⎣ 0 10 10 ⎥⎦ 0 0 ⎤ ⎥ 0.026 0.018 ⎥ −0.018 0.026 ⎥ ⎦ 0

⎡9 0 0⎤ ⎡6 0 0 ⎤ ⎡ 1 0 0⎤ (c) A2 − 2 A + I = ⎢0 −8 −6⎥ − ⎢0 −2 6 ⎥ + ⎢ 0 1 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 6 −8⎦ ⎣0 −6 −2 ⎦ ⎣ 0 0 1⎦ 0⎤ ⎡4 0 = ⎢ 0 −5 −12⎥ ⎢ ⎥ ⎣ 0 12 −5⎦ ⎡3 (d) p( A) = A − 2 I = ⎢0 ⎢ ⎣0 ⎡1 = ⎢0 ⎢ ⎣0

3⎤ − 10 ⎥. 1⎥ − 10 ⎦

0 0⎤ ⎡ 2 0 0 ⎤ −1 3⎥ − ⎢ 0 2 0 ⎥ ⎥ ⎢ ⎥ −3 −1⎦ ⎣ 0 0 2 ⎦ 0 0⎤ −3 3⎥ ⎥ −3 −3⎦

(e)

p( A) = 2 A2 − A + I 0 0⎤ ⎡ 3 0 0 ⎤ ⎡ 1 0 0⎤ ⎡18 = ⎢ 0 −16 −12⎥ − ⎢0 −1 3⎥ + ⎢ 0 1 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 12 −16⎦ ⎣0 −3 −1⎦ ⎣ 0 0 1⎦ 0 0⎤ ⎡16 = ⎢ 0 −14 −15⎥ ⎢ ⎥ 15 −14 ⎦ ⎣ 0

(f)

p( A) = A3 − 2 A + 4 I 0 ⎤ ⎡6 0 0⎤ ⎡ 4 0 0 ⎤ ⎡ 27 0 ⎢ = 0 26 −18⎥ − ⎢0 −2 6⎥ + ⎢ 0 4 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 18 26 ⎦ ⎣0 −6 −2⎦ ⎣ 0 0 4 ⎦ 0⎤ ⎡ 25 0 ⎢ = 0 32 −24 ⎥ ⎢ ⎥ ⎣ 0 24 32 ⎦ 19

Chapter 1: Systems of Linear Equations and Matrices 27. Since a11a22 ann ≠ 0, ⎡1 0 ⎢ a11 1 ⎢0 a22 Let B = ⎢ ⎢ ⎢ 0 ⎢⎣ 0


none of the diagonal entries are zero. 0 ⎤ ⎥ 0 ⎥ ⎥. ⎥ 1 ⎥ ann ⎥⎦

AB = BA = I so A is invertible and B = A−1 . 29. (a) If A has a row of zeros, then by formula (9) of Section 1.3, AB will have a row of zeros, so A cannot be invertible. (b) If B has a column of zeros, then by formula (8) of Section 1.3, AB will have a column of zeros, and B cannot be invertible.

CT B −1 A2 BAC −1DA−2 BT C −2 = C T

31.

CA−1 B −1 A−2 B(CT )−1 ⋅ CT B −1 A2 BAC −1DA−2 BT C −2 = CA−1B −1 A−2 B(C T ) −1 C T DA−2 BT C −2 = CA−1B −1 A−2 B

DA−2 BT C −2 ⋅ C 2 ( BT )−1 A2 = CA−1B −1 A−2 B ⋅ C 2 ( BT )−1 A2 D = CA−1B −1 A−2 BC 2 ( BT )−1 A2

33. ( AB)−1 ( AC −1 )( D −1C −1 )−1 D −1 = ( B −1 A−1 )( AC −1 )(CD) D −1 = B −1 ( A−1 A)(C −1C )( DD −1 ) = B −1

⎡ b11 b12 35. Let X = ⎢b21 b22 ⎢ ⎣⎢b31 b32

b13 ⎤ b23 ⎥ . ⎥ b33 ⎦⎥

AX = I ⎡ b11 + b31 b12 + b32 b13 + b33 ⎤ ⎡1 0 0⎤ ⎢b + b b +b b + b ⎥ = ⎢ 0 1 0⎥ ⎢ 11 21 12 22 13 23 ⎥ ⎢ ⎥ ⎣⎢b21 + b31 b22 + b32 b23 + b33 ⎦⎥ ⎣ 0 0 1 ⎦ Consider the equations from the first columns of the matrices. b11 + b31 = 1 b11 + b21 = 0 b21 + b31 = 0 1 1 1 Thus b11 = b31 = −b21 , so b11 = , b21 = − , and b31 = . 2 2 2

⎡ 1 ⎢ 2 Working similarly with the other columns of both matrices yields A−1 = ⎢ − 12 ⎢ 1 ⎢⎣ 2

20

1 2 1 2 − 12

− 12 ⎤ ⎥ 1⎥. 2 1⎥ 2 ⎥⎦


⎡ b11 b12 37. Let X = ⎢b21 b22 ⎢ ⎣⎢b31 b32

Section 1.4

b13 ⎤ b23 ⎥ . ⎥ b33 ⎦⎥

AX = I b b b ⎡ ⎤ ⎡1 0 0 ⎤ 31 32 33 ⎢ b +b ⎥ = ⎢0 1 0 ⎥ b12 + b22 b13 + b23 11 21 ⎢ ⎥ ⎢ ⎥ ⎢⎣ −b11 + b21 + b31 −b12 + b22 + b32 −b13 + b23 + b33 ⎥⎦ ⎣0 0 1 ⎦ Consider the equations from the first columns of the matrices. b31 = 1 b11 + b21 = 0 −b11 + b21 + b31 = 0 1 1 Thus, b21 = −b11 and −2b11 + 1 = 0, so b11 = , b21 = − , and b31 = 1. 2 2 Working similarly with the other columns of both matrices yields ⎡ 1 1 − 1⎤ 2⎥ ⎢ 2 2 −1 1 1 1 A = ⎢− 2 2 ⎥. 2 ⎢ ⎥ 1 0 0 ⎢⎣ ⎥⎦

⎡ 3 −2 ⎤ ⎡ x1 ⎤ ⎡ −1⎤ , so the solution is 39. The matrix form of the system is ⎢ = 5⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣ 3⎥⎦ ⎣4 ⎡ 3 −2 ⎤ ⎢⎣ 4 5⎥⎦

−1

⎡ x1 ⎤ ⎡ 3 −2 ⎤ ⎢ x ⎥ = ⎢4 5⎥⎦ ⎣ 2⎦ ⎣

−1

⎡ −1⎤ ⎢⎣ 3⎥⎦ .

−1

⎡ 0⎤ ⎢⎣ −2 ⎥⎦ .

1 ⎡ 5 2⎤ 3 ⋅ 5 − (−2)4 ⎢⎣ −4 3⎥⎦ 1 ⎡ 5 2⎤ = 23 ⎢⎣ −4 2 ⎥⎦ =

⎡ x1 ⎤ 1 ⎡ 5 2 ⎤ ⎡ −1⎤ 1 ⎡ −5 + 6⎤ 1 ⎡ 1 ⎤ ⎢ x ⎥ = 23 ⎢ −4 3⎥ ⎢ 3⎥ = 23 ⎢ 4 + 9⎥ = 23 ⎢13⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 2⎦ 1 13 The solution is x1 = , x2 = . 23 23 1⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ ⎡6 , so the solution is 41. The matrix form of the system is ⎢ ⎢ ⎥= 4 3 − ⎣ ⎦⎥ ⎣ x2 ⎦ ⎣⎢ −2 ⎦⎥ 1⎤ ⎡6 ⎢⎣ 4 −3⎥⎦

−1

1 ⎡ −3 −1⎤ 6(−3) − 1 ⋅ 4 ⎢⎣ −4 6 ⎥⎦ 1 ⎡ −3 −1⎤ =− ⎢ 22 ⎣ −4 6 ⎥⎦ =

1 ⎡ −3 −1⎤ ⎡ 0⎤ ⎡ x1 ⎤ ⎢ x ⎥ = − 22 ⎢ −4 6⎥ ⎢ −2⎥ ⎣ ⎦⎣ ⎦ ⎣ 2⎦ 1 ⎡ 0+2 ⎤ =− ⎢ 22 ⎣0 − 12 ⎥⎦ 1 ⎡ 2⎤ =− ⎢ 22 ⎣ −12⎥⎦ 1 6 The solution is x1 = − , x2 = . 11 11 21

1⎤ ⎡ x1 ⎤ ⎡ 6 ⎢ x ⎥ = ⎢ 4 −3⎥ ⎦ ⎣ 2⎦ ⎣



51. (a) If A is invertible, then A−1 exists. AB = AC A−1 AB = A−1 AC IB = IC B=C

(b) The matrix A in Example 3 is not invertible. 53. (a) A( A−1 + B −1 ) B( A + B)−1 = ( AA−1 + AB −1 ) B( A + B)−1 = ( I + AB −1 ) B( A + B)−1

= ( IB + AB −1 B)( A + B) −1 = ( B + A)( A + B)−1 = ( A + B)( A + B)−1 =I

(b) A−1 + B −1 ≠ ( A + B)−1 55. ( I − A)( I + A + A2 +

+ Ak −1 ) = ( I − A) I + ( I − A) A + ( I − A) A2 + = I − A + A − A2 + A2 − A3 + =I−A = I −0 =I

+ ( I − A) Ak −1

+ Ak −1 − Ak

k

True/False 1.4 (a) False; A and B are inverses if and only if AB = BA = I. (b) False; ( A + B)2 = ( A + B)( A + B) = A2 + AB + BA + B 2 Since AB ≠ BA the two terms cannot be combined.

(c) False; ( A − B)( A + B) = A2 + AB − BA − B 2 . Since AB ≠ BA, the two terms cannot be combined. (d) False; AB is invertible, but ( AB)−1 = B −1 A−1 ≠ A−1 B −1 . (e) False; if A is an m × n matrix and B is and n × p matrix, with m ≠ p, then AB and ( AB)T are defined but AT BT is not defined. (f) True; by Theorem 1.4.5. (g) True; (kA + B)T = (kA)T + BT = kAT + BT . (h) True; by Theorem 1.4.9. (i) False; p( I ) = (a0 + a1 + a2 +

+ am ) I which is a matrix, not a scalar.

22


Section 1.5

(j) True, if the square matrix A has a row of zeros, the product of A and any matrix B will also have a row or column of zeros (depending on whether the product is AB or BA), so AB = I is impossible. A similar statement can be made if A has a column of zeros.

5. (a) E interchanges the first and second rows of A. 5 −1⎤ ⎡ 0 1 ⎤ ⎡ −1 −2 EA = ⎢ ⎥ ⎢ − − −6 ⎥⎦ 1 0 3 6 6 ⎣ ⎦⎣ ⎡ 3 −6 −6 −6 ⎤ =⎢ 5 −1⎥⎦ ⎣ −1 −2

⎡ 1 0⎤ ⎡ −1 0 ⎤ (k) False; for example, ⎢ and ⎢ are ⎥ 0 1 ⎣ ⎦ ⎣ 0 −1⎥⎦ ⎡0 0 ⎤ both invertible, but their sum ⎢ is not ⎣0 0 ⎥⎦ invertible.

(b) E adds −3 times the second row to the third row. ⎡ 1 0 0⎤ ⎡ 2 −1 0 −4 −4 ⎤ EA = ⎢0 1 0⎥ ⎢ 1 −3 −1 5 3⎥ ⎢ ⎥⎢ ⎥ ⎣0 −3 1⎦ ⎣ 2 0 1 3 −1⎦ ⎡ 2 −1 0 −4 −4 ⎤ 5 3⎥ = ⎢ 1 −3 −1 ⎢ ⎥ ⎣ −1 9 4 −12 −10 ⎦

Section 1.5 Exercise Set 1.5 1. (a) The matrix results from adding −5 times the first row of I 2 to the second row; it is an elementary matrix.

(c) E adds 4 times the third row to the first row. ⎡ 1 0 4 ⎤ ⎡ 1 4 ⎤ ⎡13 28⎤ EA = ⎢0 1 0 ⎥ ⎢ 2 5⎥ = ⎢ 2 5⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 1⎦ ⎣ 3 6 ⎦ ⎣ 3 6⎦

(b) The matrix is not elementary; more than one elementary row operation on I 2 is required.

7. (a) To obtain B from A, the first and third rows must be interchanged. ⎡0 0 1⎤ E = ⎢0 1 0 ⎥ ⎢ ⎥ ⎣ 1 0 0⎦

(c) The matrix is not elementary; more than one elementary row operation on I3 is required. (d) The matrix is not elementary; more than one elementary row operation on I 4 is required.

(b) To obtain A from B, the first and third rows must be interchanged. ⎡0 0 1⎤ E = ⎢0 1 0 ⎥ ⎢ ⎥ ⎣ 1 0 0⎦

3. (a) The operation is to add 3 times the second ⎡ 1 3⎤ row to the first row. The matrix is ⎢ . ⎣0 1⎥⎦ (b) The operation is to multiply the first row by ⎡− 1 0 0⎤ ⎢ 7 ⎥ 1 − . The matrix is ⎢ 0 1 0 ⎥ . 7 ⎢ 0 0 1⎥ ⎣ ⎦

(c) To obtain C from A, add −2 times the first row to the third row. ⎡ 1 0 0⎤ E = ⎢ 0 1 0⎥ ⎢ ⎥ ⎣ −2 0 1⎦

(c) The operation is to add 5 times the first row ⎡ 1 0 0⎤ to the third row. The matrix is ⎢0 1 0 ⎥ . ⎢ ⎥ ⎣ 5 0 1⎦

(d) To obtain A from C, add 2 times the first row to the third row. ⎡ 1 0 0⎤ E = ⎢0 1 0⎥ ⎢ ⎥ ⎣ 2 0 1⎦

(d) The operation is to interchange the first and ⎡0 0 1 0⎤ ⎢0 1 0 0⎥ third rows. The matrix is ⎢ ⎥. ⎢1 0 0 0 ⎥ ⎢⎣0 0 0 1 ⎥⎦

⎡ 1 4 1 0⎤ 9. ⎢ ⎣ 2 7 0 1⎥⎦ Add −2 times the first row to the second.

23


1 0⎤ ⎡1 4 ⎢⎣0 −1 −2 1⎥⎦ Multiply the second row by −1. ⎡ 1 4 1 0⎤ ⎢⎣0 1 2 −1⎥⎦ Add −4 times the second row to the first. ⎡ 1 0 −7 4 ⎤ ⎢⎣0 1 2 −1⎥⎦ ⎡1 4 ⎤ ⎢⎣ 2 7 ⎥⎦

−1


3 0 1 0⎤ ⎡1 0 ⎢0 5 −10 0 −2 1⎥ ⎢ ⎥ ⎣0 4 −10 1 −3 0 ⎦ Multiply the second row by

1 . 5

3 0 1 0⎤ ⎡1 0 ⎢ 0 1 −2 0 − 2 1 ⎥ 5 5⎥ ⎢ ⎢⎣0 4 −10 1 −3 0⎥⎦ Add −4 times the second row to the third. ⎡1 0 3 0 1 0⎤ ⎢ ⎥ 2 1⎥ ⎢ 0 1 −2 0 − 5 5 ⎢ ⎥ 7 4 ⎢⎣0 0 −2 1 − 5 − 5 ⎥⎦ 1 Multiply the third row by − . 2 ⎡1 0 3 0 1 0⎤ ⎢ ⎥ 2 0 − 5 15 ⎥ ⎢ 0 1 −2 ⎢ ⎥ 7 2 1 − 12 10 ⎢⎣0 0 5 ⎥⎦ Add −3 times the third row to the first and 2 times the third row to the second. 3 − 11 − 6 ⎤ ⎡1 0 0 2 10 5⎥ ⎢ 1 1⎥ ⎢0 1 0 −1 7 2⎥ ⎢0 0 1 − 1 2 10 5⎦ ⎣

⎡ −7 4 ⎤ =⎢ ⎣ 2 −1⎥⎦

⎡ −1 3 1 0 ⎤ 11. ⎢ ⎣ 3 −2 0 1⎥⎦ Multiply the first row by −1. ⎡ 1 −3 −1 0 ⎤ ⎢⎣3 −2 0 1⎥⎦ Add −3 times the first row to the second. ⎡ 1 −3 −1 0 ⎤ ⎢⎣0 7 3 1⎥⎦ 1 Multiply the second row by . 7 ⎡ 1 −3 −1 0 ⎤ ⎢0 1 73 17 ⎥⎥ ⎣⎢ ⎦ Add 3 times the second row to the first. ⎡1 0 2 3⎤ 7 7⎥ ⎢ ⎢0 1 73 17 ⎥ ⎣ ⎦ −1 ⎡ 2 3 ⎤ ⎡ −1 3 ⎤ 7 7 ⎢⎣ 3 −2 ⎥⎦ = ⎢⎢ 3 1 ⎥⎥ ⎣7 7⎦

⎡ 3 4 −1⎤ ⎢1 0 3⎥ ⎢ ⎥ 2 5 4 − ⎣ ⎦

−1

⎡ 3 − 11 − 6 ⎤ 10 5⎥ ⎢ 2 1 1⎥ = ⎢ −1 7 2⎥ ⎢− 1 10 5⎦ ⎣ 2

⎡ −1 3 −4 1 0 0 ⎤ 1 0 1 0⎥ 15. ⎢ 2 4 ⎢ ⎥ ⎣ −4 2 −9 0 0 1⎦ Multiply the first row by −1. ⎡ 1 −3 4 −1 0 0 ⎤ ⎢ 2 4 1 0 1 0⎥ ⎢ ⎥ 4 2 9 0 0 1⎦ − − ⎣

⎡ 3 4 −1 1 0 0 ⎤ 3 0 1 0⎥ 13. ⎢ 1 0 ⎢ ⎥ ⎣ 2 5 −4 0 0 1⎦ Interchange the first and second rows. 3 0 1 0⎤ ⎡1 0 ⎢ 3 4 −1 1 0 0 ⎥ ⎢ ⎥ ⎣ 2 5 −4 0 0 1⎦

Add −2 times the first row to the second and 4 times the first row to the third. ⎡ 1 −3 4 −1 0 0 ⎤ ⎢0 10 −7 2 1 0⎥ ⎢ ⎥ ⎣0 −10 7 −4 0 1⎦

Add −3 times the first row to the second and −2 times the first row to the third. 3 0 1 0⎤ ⎡1 0 ⎢0 4 −10 1 −3 0 ⎥ ⎢ ⎥ ⎣0 5 −10 0 −2 1⎦

Multiply the second row by

Interchange the second and third rows.

24

1 . 10


Section 1.5

⎡ 1 3 3 1 0 0⎤ 2 ⎢ ⎥ 2 7 6 0 1 0⎥ ⎢ ⎢ 2 7 7 0 0 1⎥ ⎣ ⎦ Add −2 times the first row to both the second and third rows. ⎡ 1 3 3 1 0 0⎤ 2 ⎢ ⎥ ⎢ 0 1 0 −1 1 0 ⎥ ⎢0 1 1 −1 0 1⎥ ⎣ ⎦ Add −1 times the second row to the third. ⎡1 3 3 1 0 0⎤ 2 ⎢ ⎥ ⎢0 1 0 −1 1 0 ⎥ ⎢0 0 1 0 −1 1⎥ ⎣ ⎦ Add −3 times the third row to the first. ⎡1 3 0 1 3 −3⎤ 2 ⎢ ⎥ ⎢0 1 0 −1 1 0 ⎥ ⎢0 0 1 0 −1 1⎥ ⎣ ⎦ Add −3 times the second row to the first. ⎡ 1 0 0 7 0 −3⎤ 2 ⎢ ⎥ ⎢0 1 0 −1 1 0 ⎥ ⎢0 0 1 0 −1 1⎥ ⎣ ⎦ −1 ⎡ 7 0 −3⎤ ⎡2 6 6⎤ 2 ⎢ ⎢ 2 7 6 ⎥ = ⎢ −1 1 0 ⎥⎥ ⎢ ⎥ ⎢ 0 −1 1⎥ ⎣2 7 7 ⎦ ⎣ ⎦

4 −1 0 0 ⎤ ⎡ 1 −3 ⎢0 7 1 1 0⎥ 1 − 10 ⎢ ⎥ 5 10 ⎢⎣0 −10 7 −4 0 1⎥⎦ Add 10 times the second row to the third. 4 −1 0 0 ⎤ ⎡ 1 −3 ⎢0 7 1 1 0⎥ 1 − 10 ⎢ ⎥ 5 10 0 −2 1 1⎦⎥ ⎣⎢0 0 Since there is a row of zeros on the left side, ⎡ −1 3 −4 ⎤ ⎢ 2 4 1⎥ is not invertible. ⎢ ⎥ ⎣ −4 2 −9 ⎦

⎡ 1 0 1 1 0 0⎤ 17. ⎢0 1 1 0 1 0 ⎥ ⎢ ⎥ ⎣ 1 1 0 0 0 1⎦ Add −1 times the first row to the third. ⎡ 1 0 1 1 0 0⎤ ⎢0 1 1 0 1 0 ⎥ ⎢ ⎥ ⎣0 1 −1 −1 0 1⎦ Add −1 times the second row to the third. 1 1 0 0⎤ ⎡1 0 ⎢0 1 1 0 1 0⎥ ⎢ ⎥ ⎣0 0 −2 −1 −1 1⎦ 1 Multiply the third row by − . 2 ⎡1 0 1 1 0 ⎤ 0 ⎢ ⎥ 0 1 1 0 1 0 ⎢ ⎥ ⎢0 0 1 1 1 − 1 ⎥ 2 2 2⎦ ⎣

⎡ 2 −4 0 0 1 0 0 0 ⎤ ⎢ 1 2 12 0 0 1 0 0 ⎥ 21. ⎢ ⎥ ⎢0 0 2 0 0 0 1 0⎥ ⎢⎣ 0 −1 −4 −5 0 0 0 1⎥⎦ Interchange the first and second rows. ⎡ 1 2 12 0 0 1 0 0 ⎤ ⎢ 2 −4 0 0 1 0 0 0 ⎥ ⎢ ⎥ ⎢0 0 2 0 0 0 1 0⎥ ⎣⎢ 0 −1 −4 −5 0 0 0 1⎦⎥

Add −1 times the third row to both the first and second rows. 1 −1 1⎤ ⎡1 0 0 2 2 2⎥ ⎢ 1 1⎥ ⎢0 1 0 − 12 2 2 ⎢ 1 1 − 1⎥ ⎢⎣0 0 1 2 2 2 ⎥⎦ 1⎤ −1 ⎡ 1 − 1 2 2⎥ ⎡ 1 0 1⎤ ⎢ 2 1 1⎥ ⎢0 1 1⎥ = ⎢ − 1 2 2 2 ⎢ ⎥ ⎢ 1 1 − 1⎥ ⎣ 1 1 0⎦ ⎢⎣ 2 2 2 ⎥⎦

Add −2 times the first row to the second. 1 0 0⎤ ⎡ 1 2 12 0 0 ⎢0 −8 −24 0 1 −2 0 0 ⎥ ⎢ ⎥ 2 0 0 0 1 0⎥ ⎢0 0 ⎣⎢0 −1 −4 −5 0 0 0 1⎦⎥ Interchange the second and fourth rows. 1 0 0⎤ ⎡ 1 2 12 0 0 ⎢0 −1 −4 −5 0 0 0 1⎥ ⎢ ⎥ 2 0 0 0 1 0⎥ ⎢0 0 ⎢⎣0 −8 −24 0 1 −2 0 0 ⎥⎦

⎡2 6 6 1 0 0⎤ 19. ⎢ 2 7 6 0 1 0 ⎥ ⎢ ⎥ ⎣ 2 7 7 0 0 1⎦ Multiply the first row by

1 . 2

25

Chapter 1: Systems of Linear Equations and Matrices Multiply the second row by −1. 1 0 0⎤ ⎡ 1 2 12 0 0 ⎢0 1 4 5 0 0 0 −1⎥ ⎢ ⎥ 2 0 0 0 1 0⎥ ⎢0 0 ⎢⎣0 −8 −24 0 1 −2 0 0 ⎥⎦ Add 8 times the second row to the fourth. 1 0 0⎤ ⎡ 1 2 12 0 0 ⎢0 1 4 5 0 0 0 −1⎥ ⎢ ⎥ ⎢0 0 2 0 0 0 1 0 ⎥ ⎣⎢0 0 8 40 1 −2 0 −8⎦⎥


⎡1 ⎢ ⎢0 ⎢ ⎢0 ⎢ 0 ⎣⎢

0 0 0 1 0 0 0

1 0

0 0

1

1 4 − 18

1 2 1 4

0

0

1 40

1 − 20

⎡ 2 −4 0 0 ⎤ ⎢ 1 2 12 0 ⎥ ⎢ ⎥ ⎢0 0 2 0⎥ ⎣⎢ 0 −1 −4 −5⎦⎥

1 . 2 1 0 0⎤ ⎡ 1 2 12 0 0 ⎢0 1 4 5 0 0 0 −1⎥ ⎢ ⎥ 1 0⎥ ⎢0 0 1 0 0 0 2 ⎢0 0 8 40 1 −2 0 −8⎥ ⎣ ⎦ Add −8 times the third row to the fourth. 1 0 0⎤ ⎡ 1 2 12 0 0 ⎢0 1 4 5 0 0 0 −1⎥ ⎢ ⎥ 1 0⎥ ⎢0 0 1 0 0 0 2 ⎢0 0 0 40 1 −2 −4 −8⎥ ⎣ ⎦ 1 . Multiply the fourth row by 40 1 0 0⎤ ⎡ 1 2 12 0 0 ⎢0 1 4 5 0 0 0 −1⎥ ⎢ ⎥ 1 0 0⎥ ⎢0 0 1 0 0 2 ⎢ 1 1 1 1⎥ ⎢⎣0 0 0 1 40 − 20 − 10 − 5 ⎥⎦ Add −5 times the fourth row to the second. 0 1 0 0⎤ ⎡ 1 2 12 0 ⎢0 1 4 0 − 1 1 1 0⎥ 8 4 2 ⎢ ⎥ 1 ⎢0 0 1 0 ⎥ 0 0 0 2 ⎢ ⎥ ⎢0 0 0 1 1 − 1 − 1 − 1 ⎥ 40 20 10 5⎦ ⎣ Add −12 times the third row to the first and −4 times the third row to the second. 0 1 −6 0⎤ ⎡1 2 0 0 ⎢0 1 0 0 − 1 3 1 0 ⎥⎥ −2 ⎢ 8 4 1 ⎢0 0 1 0 0 0 0⎥ 2 ⎢ ⎥ ⎢0 0 0 1 1 − 1 − 1 − 1 ⎥ 40 20 10 5⎦ ⎣ Multiply the third row by

−1

−3 − 32 1 2 1 − 10

0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎥ − 15 ⎥ ⎦

1 ⎡ 1 0⎤ −3 2 ⎢ 4 ⎥ 3 1 − 0 ⎢ − 18 ⎥ 4 2 =⎢ ⎥ 1 0 0 0⎥ ⎢ 2 ⎢ 1 1 1 1⎥ ⎢⎣ 40 − 20 − 10 − 5 ⎥⎦

1 0 1 0 0 0⎤ ⎡ −1 0 ⎢ 2 3 −2 6 0 1 0 0 ⎥ 23. ⎢ ⎥ ⎢ 0 −1 2 0 0 0 1 0 ⎥ 1 5 0 0 0 1⎥⎦ ⎢⎣ 0 0 Multiply the first row by −1. ⎡ 1 0 −1 0 −1 0 0 0 ⎤ ⎢ 2 3 −2 6 0 1 0 0 ⎥ ⎢ ⎥ ⎢ 0 −1 2 0 0 0 1 0 ⎥ 1 5 0 0 0 1⎥⎦ ⎢⎣ 0 0 Add −2 times the first row to the second. ⎡ 1 0 −1 0 −1 0 0 0⎤ ⎢0 3 0 6 2 1 0 0⎥ ⎢ ⎥ ⎢0 −1 2 0 0 0 1 0⎥ ⎣⎢0 0 1 5 0 0 0 1⎦⎥ Interchange the second and third rows. ⎡ 1 0 −1 0 −1 0 0 0⎤ ⎢0 −1 2 0 0 0 1 0⎥ ⎢ ⎥ ⎢0 3 0 6 2 1 0 0⎥ ⎢⎣0 0 1 5 0 0 0 1⎥⎦ Multiply the second row by −1. ⎡ 1 0 −1 0 −1 0 0 0 ⎤ ⎢ 0 1 −2 0 0 0 −1 0 ⎥ ⎢ ⎥ ⎢0 3 0 6 2 1 0 0 ⎥ 1 5 0 0 0 1⎥⎦ ⎢⎣0 0 Add −3 times the second row to the third. ⎡ 1 0 −1 0 −1 0 0 0 ⎤ ⎢ 0 1 −2 0 0 0 −1 0 ⎥ ⎢ ⎥ ⎢0 0 6 6 2 1 3 0 ⎥ 1 5 0 0 0 1⎦⎥ ⎣⎢0 0 Interchange the third and fourth rows. ⎡ 1 0 −1 0 −1 0 0 0 ⎤ ⎢ 0 1 −2 0 0 0 −1 0 ⎥ ⎢ ⎥ 1 5 0 0 0 1⎥ ⎢0 0 ⎢⎣0 0 6 6 2 1 3 0 ⎥⎦ Add −6 times the third row to the fourth.

Add −2 times the second row to the first.

26


⎡1 ⎢0 ⎢ ⎢0 ⎢⎣0

Section 1.5

⎡1 ⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢⎣

0 −1 0 −1 1 −2 0 0 0 1 5 0 0 0 −24 2

0 0 0⎤ 0 −1 0 ⎥ ⎥ 0 0 1⎥ 1 3 −6 ⎥⎦ 1 Multiply the fourth row by − . 24 −1 0 0 0⎤ ⎡ 1 0 −1 0 ⎢ 0 1 −2 0 0 0 −1 0 ⎥ ⎢ ⎥ 1 5 0 0 0 1⎥ ⎢0 0 ⎢0 0 0 1 − 1 − 1 − 1 1 ⎥ 12 24 8 4⎦ ⎣ Add −5 times the fourth row to the third. −1 0 0 0⎤ ⎡ 1 0 −1 0 ⎢ 0 1 −2 0 0 0 −1 0⎥ ⎢ 5 5 5 − 1⎥ 1 0 ⎢0 0 12 24 8 4⎥ ⎢ ⎥ 1 1 1 1 0 0 0 1 − 12 − 24 − 8 4 ⎦⎥ ⎣⎢ Add the third row to the first and 2 times the third row to the second. 5 5 − 1⎤ ⎡1 0 0 0 − 7 12 24 8 4⎥ ⎢ 5 5 1 − 1⎥ ⎢0 1 0 0 6 12 4 2 ⎢ ⎥ 5 5 5 1 − 4⎥ ⎢0 0 1 0 12 24 8 ⎢ 1 − 1 1⎥ 0 0 0 1 − 12 − 81 24 4 ⎦⎥ ⎣⎢ 1 0⎤ ⎡ −1 0 ⎢ 2 3 −2 6 ⎥ ⎢ ⎥ ⎢ 0 −1 2 0 ⎥ 1 5⎥⎦ ⎢⎣ 0 0

−1

⎡− 7 ⎢ 12 ⎢ 56 =⎢ 5 ⎢ 12 ⎢ 1 − ⎣⎢ 12

5 24 5 12 5 24 1 − 24

⎡ k1 0 0 0 1 0 0 ⎢0 k 0 0 0 1 0 2 25. (a) ⎢ k 0 0 0 0 0 1 3 ⎢ ⎢⎣ 0 0 0 k4 0 0 0 1 Multiply the first row by , k1

5 8 1 4 5 8 − 81

0 0 0

1 k1

0

0

1 0 0

0

1 k2

0

0

0

0

1 k3

0

0

0

1 0

0 0

1

⎡ k1 0 0 0⎤ ⎢0 k 0 0⎥ 2 ⎢ ⎥ 0⎥ ⎢ 0 0 k3 ⎢⎣ 0 0 0 k4 ⎥⎦

−1

⎡1 ⎢ k1 ⎢0 =⎢ ⎢0 ⎢ ⎢0 ⎢⎣

⎡k 1 0 0 1 0 0 ⎢0 1 0 0 0 1 0 (b) ⎢ ⎢0 0 k 1 0 0 1 ⎢⎣ 0 0 0 1 0 0 0

0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎥ 1⎥ k4 ⎥⎦ 0

0

1 k2

0

0

1 k3

0

0

0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦

Multiply the first and third rows by ⎡1 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣0

0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎥ 1⎥ k4 ⎥⎦

1 . k

0 0⎤ ⎥ 1 0 0 0 1 0 0⎥ 0 1 1k 0 0 k1 0 ⎥ ⎥ 0 0 1 0 0 0 1⎥⎦ 1 Add − times the fourth row to the third k 1 and − times the second row to the first. k ⎡1 0 0 0 1 − 1 0 0⎤ k k ⎢ ⎥ 1 0 0⎥ ⎢0 1 0 0 0 ⎢0 0 1 0 0 0 k1 − k1 ⎥ ⎢ ⎥ 0 0 1⎦⎥ ⎣⎢0 0 0 1 0

− 14 ⎤ ⎥ − 12 ⎥ ⎥ − 14 ⎥ 1⎥ 4 ⎦⎥

0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦

1 k

0

0

⎡k 1 0 0⎤ ⎢ 0 1 0 0⎥ ⎢ ⎥ ⎢ 0 0 k 1⎥ ⎣⎢ 0 0 0 1⎥⎦

the second

1 1 , the third row by , and the k2 k3 1 fourth row by . k4

row by

1 k

−1

0

⎡1 − 1 k ⎢k 0 1 ⎢ = ⎢0 0 ⎢ 0 ⎣⎢ 0

0⎤ ⎥ 0 0⎥ 1 − 1⎥ k k⎥ 0 1⎦⎥ 0

⎡c c c 1 0 0 ⎤ 27. ⎢ 1 c c 0 1 0 ⎥ ⎢ ⎥ ⎣ 1 1 c 0 0 1⎦ Multiply the first row by requirement).

27

1 (so c ≠ 0 is a c


⎡1 1 1 1 0 0 ⎤ c ⎢ ⎥ c c 1 0 1 0⎥ ⎢ ⎢1 1 c 0 0 0 ⎥ ⎣ ⎦ Add −1 times and the first row to the second and third rows. 1 0 0⎤ ⎡1 1 1 c ⎢ ⎥ ⎢0 c − 1 c − 1 − 1c 1 0⎥ ⎢ ⎥ 0 c − 1 − 1c 0 1⎥ ⎢⎣0 ⎦ 1 Multiply the second and third rows by (so c −1 c ≠ 1 is a requirement). ⎡1 1 1 1 0 0 ⎤ c ⎢ ⎥ 1 ⎢0 1 1 − 1 ⎥ 0 c (c −1) c −1 ⎢ ⎥ 1 ⎥ ⎢0 0 1 − 1 0 c (c −1) c −1 ⎦ ⎣ Add −1 times the third row to the first and second rows. 1 ⎡1 1 0 0 − c1−1 ⎤ c −1 ⎢ ⎥ 1 1 ⎥ ⎢0 1 0 0 − c −1 c −1 ⎥ ⎢ 1 ⎥ ⎢0 0 1 − c(c1−1) 0 c −1 ⎦ ⎣ The matrix is invertible for c ≠ 0, 1.

⎡ −3 1⎤ ⎢⎣ 2 2 ⎥⎦

−1

−1

⎡ 1 0 ⎤ ⎡ 1 −1⎤ −1 ⎡ − 1 0⎤ ⎡ 1 0 ⎤ −1 4 =⎢ ⎥ ⎢ 1⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 2 ⎦ ⎣ 0 1⎦ ⎣ 0 1⎦ ⎣ −1 1⎦ ⎡ 1 0⎤ ⎡ 1 1⎤ ⎡ −4 0 ⎤ ⎡1 0 ⎤ =⎢ ⎣ 0 2⎥⎦ ⎢⎣ 0 1⎥⎦ ⎢⎣ 0 1⎥⎦ ⎢⎣1 1⎥⎦ Note that other answers are possible. ⎡ 1 0 −2 ⎤ 31. Use elementary matrices to reduce ⎢0 4 3⎥ ⎢ ⎥ 1⎦ ⎣0 0 to I3 . ⎡ 1 0 2 ⎤ ⎡ 1 0 −2 ⎤ ⎡ 1 0 0 ⎤ ⎢0 1 0 ⎥ ⎢ 0 4 3 ⎥ = ⎢ 0 4 3⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 1 0 0 1⎦ ⎣0 0 1⎦ ⎣ ⎦⎣ ⎡ 1 0 0⎤ ⎡ 1 0 0⎤ ⎡ 1 0 0⎤ ⎢ 0 1 − 3 ⎥ ⎢ 0 4 3⎥ = ⎢ 0 4 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 1⎦ ⎣0 0 1⎦ ⎣0 0 1⎦ ⎣0 0 ⎡ 1 0 0⎤ ⎡ 1 ⎢0 1 0 ⎥ ⎢0 4 ⎢ ⎥⎢ ⎢⎣0 0 1⎥⎦ ⎣0 Thus ⎡ 1 0 0⎤ ⎡ 1 ⎢0 1 0 ⎥ ⎢0 4 ⎢ ⎥⎢ ⎢⎣0 0 1⎥⎦ ⎣0 ⎡ 1 0 0⎤ = ⎢ 0 1 0⎥ ⎢ ⎥ ⎣ 0 0 1⎦ so ⎡ 1 0 −2 ⎤ ⎢0 4 3⎥ ⎢ ⎥ 0 0 1⎦ ⎣

⎡ −3 1⎤ 29. Use elementary matrices to reduce ⎢ to ⎣ 2 2 ⎥⎦ I2 . ⎡ 1 0⎤ ⎡ −3 1⎤ ⎡ −3 ⎢0 1 ⎥ ⎢ 2 2 ⎥ = ⎢ 1 ⎦ ⎣ ⎣ 2⎦ ⎣ ⎡ 1 −1⎤ ⎡ −3 1⎤ ⎡ −4 ⎢⎣0 1⎥⎦ ⎢⎣ 1 1⎥⎦ = ⎢⎣ 1 ⎡ − 1 0 ⎤ ⎡ −4 0 ⎤ ⎡1 ⎢ 4 ⎥⎢ ⎥=⎢ ⎣ 0 1⎦ ⎣ 1 1⎦ ⎣1


1⎤ 1⎥⎦ 0⎤ 1⎥⎦ 0⎤ 1⎥⎦

0 0⎤ ⎡ 1 0 0 ⎤ 4 0⎥ = ⎢ 0 1 0 ⎥ ⎥ ⎢ ⎥ 0 1⎦ ⎣ 0 0 1⎦

0 0 ⎤ ⎡ 1 0 2 ⎤ ⎡ 1 0 −2 ⎤ 1 −3 ⎥ ⎢ 0 1 0 ⎥ ⎢ 0 4 3⎥ ⎥⎢ ⎥⎢ ⎥ 0 1⎦ ⎣0 0 1⎦ ⎣ 0 0 1⎦

−1 −1 ⎡ 1 0 2 ⎤ ⎡ 1 0 0 ⎤ ⎡ 1 0 0⎤ = ⎢ 0 1 0 ⎥ ⎢0 1 −3⎥ ⎢ 0 14 0⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1⎦ ⎢⎣ 0 0 1⎥⎦ ⎣ 0 0 1⎦ ⎣0 0 ⎡ 1 0 −2 ⎤ ⎡ 1 0 0 ⎤ ⎡ 1 0 0⎤ = ⎢ 0 1 0 ⎥ ⎢ 0 1 3⎥ ⎢ 0 4 0 ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ 1⎦ ⎣ 0 0 1⎦ ⎣0 0 1⎦ ⎣0 0 Note that other answers are possible.

⎡ 1 0 ⎤ ⎡1 0 ⎤ ⎡ 1 0 ⎤ ⎢⎣ −1 1⎥⎦ ⎢⎣1 1⎥⎦ = ⎢⎣0 1⎥⎦ Thus, ⎡ 1 0 ⎤ ⎡ − 14 0 ⎤ ⎡ 1 −1⎤ ⎡ 1 0⎤ ⎡ −3 1⎤ ⎢⎣ −1 1⎥⎦ ⎢ 0 1⎥ ⎢⎣0 1⎥⎦ ⎢0 1 ⎥ ⎢⎣ 2 2⎥⎦ ⎣ ⎦ ⎣ 2⎦ ⎡ 1 0⎤ =⎢ ⎣ 0 1⎥⎦ so

28

−1


Section 1.6

33. From Exercise 29,

⎡ −3 1⎤ ⎢⎣ 2 2 ⎥⎦

−1

(c) True; since A and B are row equivalent, there exist elementary matrices E1 , E2 , … , Ek such

0⎤ ⎡ − 14 ⎢

0 ⎤ ⎡ 1 −1⎤ ⎡ 1 0 ⎤ ⎡ 1 =⎢ ⎥ ⎢ 1⎥ ⎥ ⎣ −1 1⎦ ⎣ 0 1⎦ ⎢⎣0 1⎥⎦ ⎣0 2 ⎦ ⎡− 1 1 ⎤ = ⎢ 4 83 ⎥ ⎢⎣ 14 8 ⎥⎦

that B = Ek

E2 E1 A. Similarly there exist elementary matrices E1′, E2′ , …, El′ such that C = El′ E2′ E1′B = El′ E2′ E1′Ek and C are row equivalent.

35. From Exercise 31,

⎡1 0 ⎢0 4 ⎢ ⎣0 0 ⎡1 = ⎢0 ⎢ ⎢⎣ 0 ⎡1 = ⎢0 ⎢ ⎢⎣ 0

E2 E1 A so A

(d) True; a homogeneous system has either exactly one solution or infinitely many solutions. Since A is not invertible, Ax = 0 cannot have exactly one solution.

−1

−2 ⎤ 3⎥ ⎥ 1⎦ 0 0⎤ ⎡ 1 0 0 ⎤ ⎡ 1 0 2 ⎤ 1 0 ⎥ ⎢ 0 1 −3⎥ ⎢ 0 1 0 ⎥ 4 ⎥⎢ ⎥⎢ ⎥ 1⎦ ⎣ 0 0 1⎦ 0 1⎥⎦ ⎣ 0 0 0 2⎤ 1 − 3⎥ 4 4⎥ 0 1⎥⎦

(e) True; interchanging two rows is an elementary row operation, hence it does not affect whether a matrix is invertible. (f) True; adding a multiple of the first row to the second row is an elementary row operation, hence it does not affect whether a matrix is invertible. (g) False; since the sequence of row operations that convert an invertible matrix A into I n is not unique, the expression of A as a product of elementary matrices is not unique. For instance, ⎡ −3 1⎤ ⎡ 1 0 ⎤ ⎡ 1 1⎤ ⎡ −4 0 ⎤ ⎡1 0⎤ ⎢⎣ 2 2 ⎥⎦ = ⎢⎣0 2 ⎥⎦ ⎢⎣0 1⎥⎦ ⎢⎣ 0 1⎥⎦ ⎢⎣1 1⎥⎦ ⎡ 1 −2 ⎤ ⎡ 1 0 ⎤ ⎡ 1 0 ⎤ ⎡ 1 5⎤ =⎢ 1⎥⎦ ⎢⎣ 2 1⎥⎦ ⎢⎣0 −8⎥⎦ ⎢⎣0 1⎥⎦ ⎣0 (This is the matrix from Exercise 29.)

⎡ 1 2 3⎤ 37. A = ⎢ 1 4 1⎥ ⎢ ⎥ ⎣2 1 9⎦ Add −1 times the first row to the second row. 3⎤ ⎡1 2 ⎢ 0 2 −2 ⎥ ⎢ ⎥ ⎣2 1 9⎦ Add −1 times the first row to the third row. 3⎤ ⎡1 2 ⎢0 2 −2 ⎥ ⎢ ⎥ ⎣ 1 −1 6 ⎦


Add −1 times the second row to the first row. 5⎤ ⎡1 0 ⎢0 2 −2 ⎥ ⎢ ⎥ ⎣ 1 −1 6 ⎦

1. The matrix form of the system is ⎡ 1 1⎤ ⎡ x1 ⎤ ⎡ 2 ⎤ ⎢⎣5 6 ⎥⎦ ⎢ x ⎥ = ⎢⎣ 9 ⎥⎦ . ⎣ 2⎦ −1

Add the second row to the third row. 5⎤ ⎡1 0 ⎢0 2 −2 ⎥ = B ⎢ ⎥ ⎣ 1 1 4⎦

1 ⎡ 6 −1⎤ ⎡ 6 −1⎤ ⎡ 1 1⎤ ⎢⎣5 6 ⎥⎦ = 6 − 5 ⎢⎣ −5 1⎥⎦ = ⎢⎣ −5 1⎥⎦ ⎡ 6 −1⎤ ⎡ 2 ⎤ ⎡ 12 − 9 ⎤ ⎡ 3⎤ ⎢⎣ −5 1⎥⎦ ⎢⎣ 9 ⎥⎦ = ⎢⎣ −10 + 9 ⎥⎦ = ⎢⎣ −1⎥⎦ The solution is x1 = 3, x2 = −1.

True/False 1.5 (a) False; the product of two elementary matrices is not necessarily elementary.

3. The matrix form of the system is ⎡ 1 3 1⎤ ⎡ x1 ⎤ ⎡ 4 ⎤ ⎢ 2 2 1⎥ ⎢ x ⎥ = ⎢ −1⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎣ 2 3 1⎦ ⎣ x3 ⎦ ⎣ 3⎦

(b) True; by Theorem 1.5.2.

Find the inverse of the coefficient matrix.

29



⎡ 1 3 1 1 0 0⎤ ⎢2 2 1 0 1 0⎥ ⎢ ⎥ ⎣ 2 3 1 0 0 1⎦

⎡ 1 1 1 1 0 0⎤ ⎢0 5 5 4 0 1⎥ ⎢ ⎥ ⎣ 0 0 −5 −1 1 0 ⎦

⎡ 1 3 1 1 0 0⎤ ⎢0 −4 −1 −2 1 0⎥ ⎢ ⎥ ⎣0 −3 −1 −2 0 1⎦

⎡ 1 1 1 1 0 0⎤ ⎢0 1 1 4 0 1 ⎥ 5 5⎥ ⎢ ⎣⎢0 0 −5 −1 1 0 ⎥⎦

0 0⎤ ⎡1 3 1 1 ⎢0 1 1 1 4 − 14 0 ⎥ 2 ⎢ ⎥ ⎢⎣0 −3 −1 −2 0 1⎥⎦

⎡1 1 1 1 0 ⎢ 4 0 ⎢0 1 1 5 ⎢ 1 0 0 1 5 − 15 ⎣⎢

0⎤ ⎥ 1⎥ 5 ⎥ 0⎥ ⎦

⎡1 1 0 ⎢ ⎢0 1 0 ⎢ 0 0 1 ⎣⎢

0⎤ ⎥ 1⎥ 5 ⎥ 0⎥ ⎦

⎡1 3 1 1 0 0⎤ ⎢ ⎥ 1 1 1 − 0 ⎢0 1 ⎥ 4 2 4 ⎢ ⎥ 3 1 1 ⎢⎣0 0 − 4 − 2 − 4 1⎥⎦ 0 0⎤ ⎡1 3 1 1 ⎢0 1 1 1 − 1 0⎥ 4 2 4 ⎢ ⎥ ⎢⎣0 0 1 2 3 −4 ⎥⎦

4 5 3 5 1 5

⎡1 0 0 1 5 ⎢ ⎢0 1 0 53 ⎢ 1 ⎢⎣0 0 1 5

⎡ 1 3 0 −1 −3 4 ⎤ ⎢0 1 0 0 −1 1⎥ ⎢ ⎥ ⎣ 0 0 1 2 3 −4 ⎦

1 5 1 5 − 15

0 − 15 ⎤ ⎥ 1 1⎥ 5 5 ⎥ 0⎥ − 15 ⎦

⎡1 0 − 15 ⎤ 5 ⎢5 ⎥ ⎡ ⎤ ⎡ 1 ⎤ ⎡ 1⎤ 1 1 ⎥ ⎢10 ⎥ = ⎢3 + 2 ⎥ = ⎢ 5 ⎥ ⎢ 53 5 5 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢1 ⎥ 0 1 ⎣ ⎦ ⎣1 − 2 ⎦ ⎣ −1⎦ 0 − ⎢⎣ 5 ⎥ 5 ⎦ The solution is x = 1, y = 5, z = −1.

1⎤ ⎡ 1 0 0 −1 0 ⎢0 1 0 0 −1 1⎥ ⎢ ⎥ ⎣ 0 0 1 2 3 −4 ⎦ 1⎤ ⎡ 4 ⎤ ⎡ −4 + 3 ⎤ ⎡ −1⎤ ⎡ −1 0 ⎢ 0 −1 1⎥ ⎢ −1⎥ = ⎢ 1 + 3 ⎥ = ⎢ 4 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 2 3 −4⎦ ⎣ 3⎦ ⎣8 − 3 − 12⎦ ⎣ −7 ⎦

7. The matrix form of the system is ⎡3 5⎤ ⎡ x1 ⎤ ⎡ b1 ⎤ ⎢⎣ 1 2 ⎥⎦ ⎢ x ⎥ = ⎢b ⎥ . ⎣ 2⎦ ⎣ 2⎦

The solution is x1 = −1, x2 = 4, x3 = −7.

−1

1 ⎡ 2 −5⎤ ⎡ 2 −5⎤ = 6 − 5 ⎢⎣ −1 3⎥⎦ ⎢⎣ −1 3⎥⎦ ⎡ 2 −5⎤ ⎡ b1 ⎤ ⎡ 2b1 − 5b2 ⎤ ⎢⎣ −1 3⎥⎦ ⎢b ⎥ = ⎢ −b + 3b ⎥ ⎣ 2⎦ ⎣ 1 2⎦ The solution is x1 = 2b1 − 5b2 , x2 = −b1 + 3b2 . ⎡3 5 ⎤ ⎢⎣ 1 2 ⎥⎦

5. The matrix form of the equation is ⎡ 1 1 1⎤ ⎡ x ⎤ ⎡ 5 ⎤ ⎢ 1 1 −4 ⎥ ⎢ y ⎥ = ⎢10 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −4 1 1⎦ ⎣ z ⎦ ⎣ 0 ⎦

Find the inverse of the coefficient matrix. ⎡ 1 1 1 1 0 0⎤ ⎢ 1 1 −4 0 1 0 ⎥ ⎢ ⎥ ⎣ −4 1 1 0 0 1⎦

=

9. Write and reduce the augmented matrix for both systems. ⎡ 1 −5 1 −2 ⎤ ⎢⎣3 2 4 5⎥⎦

⎡ 1 1 1 1 0 0⎤ ⎢ 0 0 −5 −1 1 0 ⎥ ⎢ ⎥ ⎣0 5 5 4 0 1⎦

⎡ 1 −5 1 −2 ⎤ ⎢⎣0 17 1 11⎥⎦ 30


Section 1.6

⎡ 1 −5 1 −2 ⎤ ⎢0 1 11 ⎥ 1 17 17 ⎦ ⎣ ⎡ 1 0 22 17 ⎢ 1 ⎢⎣0 1 17

⎡ 1 3 b1 ⎤ 13. ⎢ ⎥ ⎣ −2 1 b2 ⎦ b1 ⎤ ⎡1 3 ⎢0 7 2b + b ⎥ ⎣ 1 2⎦

21 ⎤ 17 ⎥ 11 ⎥ 17 ⎦

b1 ⎡1 3 ⎤ ⎢ ⎥ 2 1 ⎢⎣0 1 7 b1 + 7 b2 ⎥⎦

22 1 (i) The solution is x1 = , x2 = . 17 17 (ii) The solution is x1 =

⎡ 1 0 1 b1 − 3 b2 ⎤ 7 7 ⎥ ⎢ ⎢⎣0 1 72 b1 + 17 b2 ⎥⎦ There are no restrictions on b1 and b2 .

21 11 , x2 = . 17 17

11. Write and reduce the augmented matrix for the four systems. ⎡ 4 −7 0 −4 −1 −5⎤ ⎢⎣ 1 2 1 6 3 1⎥⎦

⎡ 1 −2 5 b1 ⎤ 15. ⎢ 4 −5 8 b2 ⎥ ⎢ ⎥ ⎣ −3 3 −3 b3 ⎦

1⎤ ⎡1 2 1 6 3 ⎢⎣ 4 −7 0 −4 −1 −5⎥⎦

5 b1 ⎤ ⎡ 1 −2 ⎢0 3 −12 −4b1 + b2 ⎥ ⎢ ⎥ 3b1 + b3 ⎦ ⎣0 −3 12

2 1 6 3 1⎤ ⎡1 ⎢⎣0 −15 −4 −28 −13 −9 ⎥⎦

5 b1 ⎡ 1 −2 ⎤ ⎢0 3 −12 −4b1 + b2 ⎥ ⎢ ⎥ 0 −b1 + b2 + b3 ⎦ ⎣0 0

⎡ 1 2 1 6 3 1⎤ ⎢0 1 4 28 13 3 ⎥ ⎢⎣ 15 15 15 5 ⎥⎦ ⎡1 0 7 15 ⎢ 4 ⎢0 1 15 ⎣

34 15 28 15

19 15 13 15

5 b1 ⎡ 1 −2 ⎤ ⎢ ⎥ 4 1 1 −4 − 3 b1 + 3 b2 ⎥ ⎢0 ⎢⎣0 0 0 −b1 + b2 + b3 ⎥⎦

− 15 ⎤ ⎥ 3⎥ 5⎦

(i) The solution is x1 =

7 4 , x2 = . 15 15

(ii) The solution is x1 =

34 28 , x2 = . 15 15

(iii) The solution is x1 =

19 13 , x2 = . 15 15

⎡ 1 0 −3 − 5 b1 + 2 b2 ⎤ 3 3 ⎢ ⎥ 4 1b ⎥ 0 1 4 b − − + ⎢ 1 2 3 3 ⎢ ⎥ ⎢⎣0 0 0 −b1 + b2 + b3 ⎥⎦ The only restriction is from the third row: −b1 + b2 + b3 = 0 or b3 = b1 − b2 .

1 3 (iv) The solution is x1 = − , x2 = . 5 5

31


⎡ 1 −1 ⎢ −2 1 17. ⎢ − 3 2 ⎢ ⎣⎢ 4 −3

b1 ⎤ b2 ⎥ ⎥ b3 ⎥ b4 ⎦⎥

3 2 5 1 2 −1 1 3

⎡ 1 −1 1 1 0 0⎤ ⎢ ⎥ 1 0⎥ ⎢0 1 − 52 − 52 5 ⎢ ⎥ 1 4 −2 1 0 0 − ⎥⎦ 5 5 5 ⎣⎢ 1 1 ⎡ 1 −1 ⎢0 1 − 2 − 2 ⎢ 5 5 ⎢⎣0 0 1 −4

b1 ⎤ 3 2 ⎡ 1 −1 ⎢0 −1 11 5 2b + b ⎥ 1 2⎥ ⎢ 3b1 + b3 ⎥ ⎢0 −1 11 5 ⎣⎢0 1 −11 −5 −4b1 + b4 ⎦⎥

−b1 − b2 ⎤ 0 −8 −3 1 −11 −5 −2b1 − b2 ⎥ ⎥ 0 0 0 b1 − b2 + b3 ⎥ 0 0 0 −2b1 + b2 + b4 ⎥⎦

invertible. Thus, Ak is also invertible and Ak x = 0 has only the trivial solution.

−2b1 + b2 + b4 = 0. Thus b3 = −b1 + b2 and b4 = 2b1 − b2 . Expressing b1 and b2 in terms of b3 and b4 gives b1 = b3 + b4 and b2 = 2b3 + b4 .

Note that ( Ak )−1 = ( A−1 ) k . 23. Let x1 be a fixed solution of Ax = b, and let x be any other solution. Then A(x − x1 ) = Ax − Ax1 = b − b = 0. Thus

⎡ 2 −1 5 7 8⎤ ⎢ 4 0 −3 0 1⎥ ⎢ ⎥ ⎣ 3 5 −7 2 1⎦

x0 = x − x1 is a solution to Ax = 0, so x can be

expressed as x = x1 + x0 . Also A(x1 + x0 ) = Ax1 + Ax0 = b + 0 = b, so every matrix of the form x1 + x0 is a solution of Ax = b.

⎡ 1 −1 1 1 0 0 ⎤ ⎢ 2 3 0 0 1 0⎥ ⎢ ⎥ ⎣ 0 2 −1 0 0 1⎦

True/False 1.6

⎡ 1 −1 1 1 0 0⎤ ⎢0 5 −2 −2 1 0⎥ ⎢ ⎥ ⎣0 2 −1 0 0 1⎦

1 1 ⎡ 1 −1 ⎢0 1 − 2 − 2 ⎢ 5 5 ⎢⎣0 2 −1 0

5 7 8⎤ −3 0 1⎥ ⎥ −7 2 1⎦ 26 ⎤ −17 ⎥ ⎥ −35⎦

21. Since Ax = 0 has only the trivial solution, A is

From the bottom two rows, b1 − b2 + b3 = 0 and

−1

1 5

3 −1 3 ⎤ ⎡1 0 0 ⎢0 1 0 −2 1 −2⎥ ⎢ ⎥ ⎣0 0 1 −4 2 −5⎦ ⎡ 3 −1 3⎤ ⎡ 2 −1 X = ⎢ −2 1 −2 ⎥ ⎢ 4 0 ⎢ ⎥⎢ ⎣ −4 2 −5 ⎦ ⎣ 3 5 ⎡ 11 12 −3 27 = ⎢ −6 −8 1 −18 ⎢ ⎣ −15 −21 9 −38

b1 3 2 ⎡ 1 −1 ⎤ ⎢0 1 −11 −5 −2b1 − b2 ⎥ ⎢ ⎥ b1 − b2 + b3 ⎥ 0 0 ⎢0 0 0 0 −2b1 + b2 + b4 ⎦⎥ ⎣⎢0 0

⎡ 1 −1 1⎤ 19. X = ⎢ 2 3 0 ⎥ ⎢ ⎥ ⎣ 0 2 −1⎦

0⎤ 0 ⎥⎥ 2 −5⎥⎦ 0

5 −2 5⎤ ⎡ 1 −1 0 ⎢ 0 1 0 −2 1 −2 ⎥ ⎢ ⎥ ⎣ 0 0 1 −4 2 −5 ⎦

b1 ⎡ 1 −1 3 2 ⎤ ⎢0 −1 11 5 2b1 + b2 ⎥ ⎢ ⎥ b1 − b2 + b3 ⎥ ⎢0 0 0 0 ⎣⎢0 0 0 0 −2b1 + b2 + b4 ⎦⎥

⎡1 ⎢0 ⎢ ⎢0 ⎢⎣0


(a) True; if a system of linear equations has more than one solution, it has infinitely many solutions.

0 0⎤ 1 0⎥ ⎥ 5 0 1⎥⎦

(b) True; if Ax = b has a unique solution, then A is invertible and Ax = c has the unique solution

A−1c.

32


Section 1.7

(c) True; if AB = I n , then B = A−1 and BA = I n also.

⎡12 9. A2 = ⎢ ⎢⎣ 0 ⎡1−2 A−2 = ⎢ ⎢⎣ 0

(d) True; elementary row operations do not change the solution set of a linear system, and row equivalent matrices can be obtained from one another by elementary row operations.

⎡1− k A− k = ⎢ ⎢⎣ 0

(e) True; since ( S −1 AS )x = b, then

S ( S −1 AS )x = Sb or A(Sx) = Sb, Sx is a solution of Ay = Sb.

⎡ ⎢ ⎢ 11. A2 = ⎢ ⎢ ⎢ ⎢⎣

⎤

⎡1 0 ⎤ =⎢ 1⎥ (−2) ⎥⎦ ⎣0 4 ⎦ 0 ⎤ 0 ⎤ ⎡1 = ⎢ 1 ⎥ ⎥ (−2)− k ⎥⎦ ⎢⎣0 ( −2)k ⎥⎦ 0

−2 ⎥

⎤ 0 ⎥ ⎡1 0 ⎥ ⎢4 2 1 0 0 ⎥ = ⎢ 0 91 3 ⎥ ⎢ 2 1 ⎥ ⎢⎣ 0 0 0 0 4 ⎥ ⎦ ⎡ 1 −2 ⎤ 0 0 ⎥ ⎢ 2 ⎢ ⎥ ⎡4 −2 1 A−2 = ⎢ 0 0 ⎥ = ⎢0 3 ⎢ ⎥ ⎢0 −2 1 ⎢ 0 ⎥ ⎣ 0 4 ⎢⎣ ⎦⎥ − k ⎡ 1 ⎤ 0 0 ⎥ ⎢ 2 ⎢ ⎥ −k 1 0 A− k = ⎢ 0 ⎥ 3 ⎢ ⎥ −k 1 ⎢ 0 ⎥ 0 4 ⎣⎢ ⎦⎥ ⎡ 2k 0 ⎤ 0 ⎢ ⎥ k =⎢0 3 0⎥ ⎢0 0 4k ⎥⎥ ⎣⎢ ⎦

(f) True; the system Ax = 4x is equivalent to the system (A − 4I)x = 0. If the system Ax = 4x has a unique solution, then so will (A − 4I)x = 0, hence A − 4I is invertible. If A − 4I is invertible, then the system (A − 4I)x = 0 has a unique solution and so does the equivalent system Ax = 4x.

( 12 )

2

()

(g) True; if AB were invertible, then both A and B would have to be invertible.

()

Section 1.7 Exercise Set 1.7 1. The matrix is a diagonal matrix with nonzero entries on the diagonal, so it is invertible. ⎡1 0⎤ ⎥. The inverse is ⎢ 2 ⎢⎣ 0 − 15 ⎥⎦ 3. The matrix is a diagonal matrix with nonzero entries on the diagonal, so it is invertible. ⎡ −1 0 0 ⎤ The inverse is ⎢ 0 12 0 ⎥ . ⎢ ⎥ ⎢⎣ 0 0 3⎥⎦

0

()

()

()

()

()

0⎤ ⎥ 0⎥ ⎥ 1 16 ⎥⎦

0 0⎤ 9 0⎥ ⎥ 0 16 ⎦

()

13. The matrix is not symmetric since a12 = −8 ≠ 0 = a21 . 15. The matrix is symmetric. 17. The matrix is not symmetric, since a23 = −6 ≠ 6 = a32 .

⎡ 3 0 0 ⎤ ⎡ 2 1⎤ ⎡ 6 3⎤ 5. ⎢0 −1 0 ⎥ ⎢ −4 1⎥ = ⎢ 4 −1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 2 ⎦ ⎣ 2 5⎦ ⎣ 4 10 ⎦ ⎡ 5 0 0⎤ ⎡ −3 2 0 7. ⎢0 2 0⎥ ⎢ 1 −5 3 ⎢ ⎥⎢ ⎣0 0 −3⎦ ⎣ −6 2 2 ⎡ −15 10 0 20 = ⎢ 2 −10 6 0 ⎢ ⎣ 18 −6 −6 −6

0 ⎤ ⎡1 0 ⎤ ⎥=⎢ ⎥ (−2) 2 ⎥⎦ ⎣0 4 ⎦

19. The matrix is not symmetric, since a13 = 1 ≠ 3 = a31 .

4 −4 ⎤ 0 3⎥ ⎥ 2 2⎦ −20 ⎤ 6⎥ ⎥ −6 ⎦

21. The matrix is not invertible because it is upper triangular and has a zero on the main diagonal. 23. For A to be symmetric, a12 = a21 or −3 = a + 5, so a = −8. 25. For A to be invertible, the entries on the main diagonal must be nonzero. Thus x ≠ 1, −2, 4. 33


⎡ 1 0 0⎤ 27. By inspection, A = ⎢0 −1 0 ⎥ . ⎢ ⎥ ⎣0 0 −1⎦

1 1 ( A − AT )T = ( AT − ( AT )T ) 2 2 1 T = ( A − A) 2 1 = − ( A − AT ) 2 1 Thus ( A − AT ) is skew-symmetric. 2

33. If AT A = A, then AT = ( AT A)T = AT ( AT )T = AT A = A, so A is

symmetric. Since A = AT , then A2 = AT A = A.

35. (a) A is symmetric since aij = a ji for all i, j.

39. From AT = − A, the entries on the main diagonal must be zero and the reflections of entries across the main diagonal must be opposites. ⎡ 0 0 −8 ⎤ A = ⎢ 0 0 −4 ⎥ ⎢ ⎥ ⎣ 8 4 0⎦

(b) i 2 − j 2 = j 2 − i 2 ⇒ 2i 2 = 2 j 2 or i 2 = j 2 ⇒ i = j, since i, j > 0 Thus, A is not symmetric unless n = 1.

(c) A is symmetric since aij = a ji for all i, j.

41. No; if A and B are commuting skew-symmetric matrices, then

(d) Consider a12 and a21. 2


( AB)T = ( BA)T = AT BT = (− A)(− B) = AB so the product of commuting skew-symmetric matrices is symmetric rather than skewsymmetric.

3

a12 = 2(1) + 2(2) = 18 while a21 = 2(2)2 + 2(1)3 = 10. A is not symmetric unless n = 1.

⎡x y⎤ 43. Let A = ⎢ . Then ⎣ 0 z ⎥⎦ ⎡ x3 y( x 2 + xz + z 2 ) ⎤ 3 A3 = ⎢ ⎥ , so x = 1 and 3 z ⎣⎢ 0 ⎦⎥

37. (a) If A is invertible and skew-symmetric then ( A−1 )T = ( AT )−1 = (− A)−1 = − A−1 so A−1 is also skew-symmetric.

(b) Let A and B be skew-symmetric matrices.

z 3 = −8 or x = 1, z = −2. Then 3y = 30 and y = 10. ⎡ 1 10⎤ A=⎢ ⎣0 −2⎥⎦

( AT )T = (− A)T = − AT ( A + B)T = AT + BT = − A − B = −( A + B) ( A − B)T = AT − BT = − A + B = −( A − B) (kA)T = k ( AT ) = k (− A) = −kA

True/False 1.7

(c) From the hint, it’s sufficient to prove that 1 1 ( A + AT ) is symmetric and ( A − AT ) is 2 2 skew-symmetric. 1 1 1 ( A + AT )T = ( AT + ( AT )T ) = ( A + AT ) 2 2 2 1 T Thus ( A + A ) is symmetric. 2

(a) True; since a diagonal matrix must be square and have zeros off the main diagonal, its transpose is also diagonal. (b) False; the transpose of an upper triangular matrix is lower triangular. ⎡ 1 2 ⎤ ⎡ 3 0⎤ ⎡ 4 2 ⎤ (c) False; for example, ⎢ . + = ⎣0 1⎥⎦ ⎢⎣ 2 1⎥⎦ ⎢⎣ 2 2 ⎥⎦

(d) True; the entries above the main diagonal determine the entries below the main diagonal in a symmetric matrix. (e) True; in an upper triangular matrix, the entries below the main diagonal are all zero. 34


Section 1.8

x1 + x2 x1 + x3 x3 x2

(f) False; the inverse of an invertible lower triangular matrix is lower triangular. (g) False; the diagonal entries may be negative, as long as they are nonzero.

By inspection, the solution is x1 = 40, x2 = 10,

(h) True; adding a diagonal matrix to a lower triangular matrix will not create nonzero entries above the main diagonal.

x3 = −10 and the completed network is shown. 50

(i) True; since the entries below the main diagonal must be zero, so also must be the entries above the main diagonal.

40

3. (a) Consider the nodes from left to right on the top, then the bottom. 300 + x2 = 400 + x3 750 + x3 = 250 + x4 100 + x1 = 400 + x2 200 + x4 = 300 + x1 The system is x2 − x3 = 100 x3 − x4 = −500 x1 − x2 = 300 − x1 + x4 = 100

2

(m) True; if (kA)T = kA, then 0 = (kA)T − kA = kAT − kA = k ( AT − A) since k ≠ 0, then A = AT .

Section 1.8

(b) Reduce the augmented matrix. ⎡ 0 1 −1 0 100 ⎤ ⎢ 0 0 1 −1 −500 ⎥ ⎢ ⎥ ⎢ 1 −1 0 0 300 ⎥ ⎣⎢ −1 0 0 1 100 ⎦⎥

Exercise Set 1.8 1. Label the network with nodes, flow rates, and flow directions as shown. 50

B x3

C

⎡ 1 −1 0 0 300 ⎤ ⎢ 0 0 1 −1 −500 ⎥ ⎢ ⎥ ⎢ 0 1 −1 0 100 ⎥ ⎢⎣ −1 0 0 1 100 ⎥⎦

x2 D

50

40

⎡1 0 ⎤ ⎡ 1 0⎤ . (l) False; for example ⎢ =⎢ ⎣1 −1⎥⎦ ⎣ 0 1⎥⎦

30

60 10

⎡ 1 2 ⎤ ⎡ 1 3⎤ ⎡ 2 5 ⎤ (k) False; for example, ⎢ + = ⎣5 1⎥⎦ ⎢⎣ −5 1⎥⎦ ⎢⎣ 0 2 ⎥⎦ which is upper triangular.

A

10

30

⎡ 1 2 ⎤ ⎡ 5 0⎤ ⎡ 6 2 ⎤ (j) False; for example, ⎢ + = ⎣0 1⎥⎦ ⎢⎣ 2 6⎥⎦ ⎢⎣ 2 7 ⎥⎦ which is symmetric.

x1

= 50 = 30 = −10 = 10

60

50

⎡ 1 −1 0 0 300 ⎤ ⎢0 0 1 −1 −500 ⎥ ⎢ ⎥ ⎢0 1 −1 0 100 ⎥ ⎣⎢0 −1 0 1 400 ⎦⎥

40

Node A: x1 + x2 = 50 Node B: x1 + x3 = 30 Node C: x3 + 50 = 40 Node D: x2 + 50 = 60 The linear system is

⎡ 1 −1 0 0 300 ⎤ ⎢0 1 −1 0 100 ⎥ ⎢ ⎥ ⎢0 0 1 −1 −500 ⎥ ⎣⎢0 −1 0 1 400 ⎥⎦ 35



⎡ 1 1 −1 0 ⎤ ⎢0 1 2 4⎥ ⎢ ⎥ ⎣ 0 −2 1 3⎦

⎡ 1 −1 0 0 300 ⎤ ⎢0 1 −1 0 100 ⎥ ⎢ ⎥ ⎢0 0 1 −1 −500 ⎥ ⎢⎣0 0 −1 1 500 ⎥⎦

⎡ 1 1 −1 0 ⎤ ⎢0 1 2 4 ⎥ ⎢ ⎥ ⎣0 0 5 11⎦ From the last row we conclude that 11 I3 = = 2.2 A, and from back substitution it 5 then follows that 22 2 I 2 = 4 − 2 I3 = 4 − = − = −0.4 A, and 5 5 11 2 13 I1 = I3 − I 2 = + = = 2.6 A. Since I 2 is 5 5 5 negative, its direction is opposite to that indicated in the figure.

⎡ 1 −1 0 0 300 ⎤ ⎢0 1 −1 0 100 ⎥ ⎢ ⎥ ⎢0 0 1 −1 −500 ⎥ 0 ⎦⎥ ⎣⎢0 0 0 0 ⎡ 1 0 0 −1 −100 ⎤ ⎢0 1 0 −1 −400 ⎥ ⎢ ⎥ ⎢0 0 1 −1 −500 ⎥ 0 ⎥⎦ ⎢⎣0 0 0 0 Since x4 is a free variable, let x4 = t . The solution to the system is x1 = −100 + t , x2 = −400 + t , x3 = −500 + t , x4 = t .

7. From application of the current law at each of the four nodes (clockwise from upper left) we have I1 = I 2 + I 4 , I 4 = I3 + I5 , I 6 = I3 + I5 , and I1 = I 2 + I 6 . These four equations can be reduced to the following: I1 = I 2 + I 4 , I 4 = I3 + I5 , I 4 = I 6 From application of the voltage law to the three inner loops (left to right) we have 10 = 20 I1 + 20 I 2 , 20 I 2 = 20 I3 , and 10 + 20 I3 = 20 I 5 . These equations can be simplified to: 2 I1 + 2 I 2 = 1, I3 = I 2 , 2 I5 − 2 I3 = 1 This gives us six equations in the unknown currents I1 , … , I 6 . But, since I3 = I 2 and I 6 = I 4 , this can be simplified to a system of only four equations in the variables I1, I 2 , I 4 , and I5 :

(c) The flow from A to B is x4 . To keep the traffic flowing on all roads, each flow rate must be nonnegative. Thus, from x3 = −500 + t , the minimum flow from A to B is 500 vehicles per hour. Thus x1 = 400, x2 = 100, x3 = 0, x4 = 500. 5. From Kirchoff’s current law, applied at either of the nodes, we have I1 + I 2 = I3 . From Kirchoff’s voltage law, applied to the left and right loops, we have 2 I1 = 2 I 2 + 6 and 2 I 2 + 4 I3 = 8. Thus the currents I1, I 2 , and I3 satisfy the following system of equations: I1 + I 2 − I 3 = 0 =6 2 I1 − 2 I 2 2 I 2 + 4I3 = 8 Reduce the augmented matrix. ⎡ 1 1 −1 0 ⎤ ⎢ 2 −2 0 6 ⎥ ⎢ ⎥ ⎣ 0 2 4 8⎦

=0 I1 − I 2 − I 4 I 2 − I 4 + I5 = 0 =1 2 I1 + 2 I 2 −2 I 2 + 2 I5 = 1 Reduce the augmented matrix. ⎡ 1 −1 −1 0 0⎤ ⎢0 1 −1 1 0 ⎥ ⎢ ⎥ ⎢ 2 2 0 0 1⎥ ⎣⎢ 0 −2 0 2 1⎦⎥

⎡ 1 1 −1 0 ⎤ ⎢ 1 −1 0 3⎥ ⎢ ⎥ ⎣0 1 2 4 ⎦ ⎡ 1 1 −1 0 ⎤ ⎢ 0 −2 1 3⎥ ⎢ ⎥ 1 2 4⎦ ⎣0 36


⎡ 1 −1 −1 0 ⎢0 1 −1 1 ⎢ 0 4 2 0 ⎢ ⎢⎣0 −2 0 2

Section 1.8

homogeneous linear system. 3x1 − x3 =0 8 x1 − 2 x4 = 0 2 x2 − 2 x3 − x4 = 0 Reduce the augmented matrix. ⎡ 3 0 −1 0 0⎤ ⎢ 8 0 0 −2 0⎥ ⎢ ⎥ ⎣ 0 2 −2 −1 0 ⎦

0⎤ 0⎥ ⎥ 1⎥ 1⎥⎦

⎡ 1 −1 −1 0 ⎢ 0 1 −1 1 ⎢ ⎢0 0 6 −4 ⎣⎢0 0 −2 4

0⎤ 0⎥ ⎥ 1⎥ 1⎦⎥

⎡ 1 −1 −1 0 ⎢ 0 1 −1 1 ⎢ ⎢0 0 −2 4 ⎢⎣0 0 6 −4

0⎤ 0⎥ ⎥ 1⎥ 1⎥⎦

⎡1 0 − 1 0 0⎤ 3 ⎢ ⎥ 0 −2 0 ⎥ ⎢8 0 ⎢0 2 −2 −1 0 ⎥ ⎣ ⎦ ⎡1 0 − 1 0 0⎤ 3 ⎢ ⎥ 8 −2 0 ⎥ ⎢0 0 3 ⎢ ⎥ ⎢⎣0 2 −2 −1 0 ⎥⎦

0⎤ ⎡ 1 −1 −1 0 ⎢0 1 −1 1 0⎥ ⎢ 1⎥ ⎢0 0 1 −2 − 2 ⎥ ⎢0 0 6 −4 1⎥⎦ ⎣

⎡1 0 − 1 0 0⎤ 3 ⎢ ⎥ ⎢0 2 −2 −1 0 ⎥ 8 −2 0 ⎥ ⎢0 0 3 ⎣ ⎦

0⎤ ⎡ 1 −1 −1 0 ⎢0 1 −1 1 0⎥ ⎢ 1⎥ ⎢0 0 1 −2 − 2 ⎥ ⎢0 0 0 8 4 ⎥⎦ ⎣

⎡1 0 − 1 0 0⎤ 3 ⎢ ⎥ ⎢0 1 −1 − 12 0⎥ ⎢ ⎥ 8 −2 0 ⎥ ⎢⎣0 0 3 ⎦

0⎤ ⎡ 1 −1 −1 0 ⎢0 1 −1 1 0⎥ ⎢ 1⎥ ⎢0 0 1 −2 − 2 ⎥ ⎢ 1⎥ 1 ⎢⎣0 0 0 2 ⎥⎦

⎡1 0 − 1 0 0⎤ 3 ⎢ ⎥ ⎢0 1 −1 − 12 0⎥ ⎢ ⎥ 1 − 34 0⎥⎦ ⎢⎣0 0 From this we conclude that x4 is a free variable, and that the general solution of the system is 3 3 given by x4 = t , x3 = x4 = t , 4 4 1 3 1 5 x2 = x3 + x4 = t + t = t , and 2 4 2 4 1 1 x1 = x3 = t . The smallest positive integer 3 4 solutions are obtained by taking t = 4, in which case we obtain x1 = 1, x2 = 5, x3 = 3, and x4 = 4. Thus the balanced equation is C3 H8 + 5O 2 → 3CO 2 + 4H 2 O.

1 From this we conclude that I5 = , 2 1 1 1 I 6 = I 4 = − + 2 I5 = − + 1 = , 2 2 2 1 1 I3 = I 2 = I 4 − I5 = − = 0, and 2 2 1 1 I1 = I 2 + I 4 = 0 + = . 2 2

9. We seek positive integers x1 , x2 , x3 , and x4 that will balance the chemical equation x1 (C3 H8 ) + x2 (O2 ) → x3 (CO 2 ) + x4 (H 2 O) For each of the atoms in the equation (C, H, and O), the number of atoms on the left must be equal to the number of atoms on the right. This leads to the equations 3x1 = x3 , 8 x1 = 2 x4 , and 2 x2 = 2 x3 + x4 . These equations form the 37


a0 − a1 + a2 − a3 = −1 a0 =1 a0 + a1 + a2 + a3 = 3 a0 + 4a1 + 16a2 + 64a3 = −1 The augmented matrix of this system is ⎡1 −1 1 −1 −1⎤ ⎢1 0 0 0 1⎥ ⎢ ⎥ ⎢1 1 1 1 3⎥ ⎣⎢1 4 16 64 −1⎦⎥

11. We must find positive integers x1 , x2 , x3 , and x4 that balance the chemical equation x1 (CH3COF) + x2 (H 2 O) → x3 (CH3COOH) + x4 (HF) For each of the elements in the equation (C, H, O, and F), the number of atoms on the left must equal the number of atoms on the right. This leads to the equations x1 = x3 ,

3x1 + 2 x2 = 4 x3 + x4 , x1 + x2 = 2 x3 , and x1 = x4 . This is a very simple system of equations having (by inspection) the general solution x1 = x2 = x3 = x4 = t . Thus, taking t = 1, the balanced equation is CH3COF + H 2 O → CH3COOH + HF.

⎡1 0 0 0 1⎤ ⎢1 −1 1 −1 −1⎥ ⎢ ⎥ ⎢1 1 1 1 3⎥ ⎣⎢1 4 16 64 −1⎦⎥ 1⎤ ⎡1 0 0 0 ⎢0 −1 1 −1 −2 ⎥ ⎢ ⎥ ⎢0 1 1 1 2 ⎥ ⎣⎢0 4 16 64 −2 ⎦⎥

13. The graph of the polynomial p( x) = a0 + a1 x + a2 x 2 passes through the points (1, 1), (2, 2), and (3, 5) if and only if the coefficients a0 , a1 , and a2 satisfy the equations a0 + a1 + a2 = 1 a0 + 2a1 + 4a2 = 2 a0 + 3a1 + 9a2 = 5 Reduce the augmented matrix. ⎡1 1 1 1⎤ ⎢1 2 4 2 ⎥ ⎢ ⎥ ⎣1 3 9 5⎦

Add −1 times row 1 to rows 2 and 3. ⎡ 1 1 1 1⎤ ⎢0 1 3 1⎥ ⎢ ⎥ ⎣0 2 8 4 ⎦ Add −2 times row 2 to row 3. ⎡ 1 1 1 1⎤ ⎢0 1 3 1⎥ ⎢ ⎥ ⎣0 0 2 2 ⎦

⎡1 ⎢0 ⎢ ⎢0 ⎣⎢0

0 0 0 1⎤ 1 −1 1 2 ⎥ ⎥ 1 1 1 2⎥ 4 16 64 −2 ⎦⎥

⎡1 ⎢0 ⎢ ⎢0 ⎢⎣0

0 0 0 1⎤ 1 −1 1 2⎥ ⎥ 0 2 0 0⎥ 0 20 60 −10⎥⎦

⎡1 ⎢0 ⎢ ⎢0 ⎣⎢0

0 0 0 1⎤ 1 −1 1 2⎥ ⎥ 0 1 0 0⎥ 0 20 60 −10⎦⎥

1⎤ ⎡1 0 0 0 ⎢ 0 1 −1 1 2⎥ ⎢ ⎥ 0⎥ ⎢0 0 1 0 ⎢⎣0 0 0 60 −10 ⎥⎦ From the first row we see that a0 = 1, and from the last two rows we conclude that a2 = 0 and

We conclude that a2 = 1, and from back substitution it follows that a1 = 1 − 3a2 = −2, and a0 = 1 − a1 − a2 = 2. Thus the quadratic polynomial is p( x) = x 2 − 2 x + 2.

1 a3 = − . Finally, from back substitution, it 6 13 follows that a1 = 2 + a2 − a3 = . Thus the 6 1 3 13 polynomial is p( x) = − x + x + 1. 6 6

15. The graph of the polynomial 2


3

p( x) = a0 + a1 x + a2 x + a3 x passes through the points (−1, −1), (0, 1), (1, 3), and (4, −1), if and only if the coefficients a0 , a1 , a2 , and a3 satisfy the equations

38


Section 1.9

−1

17. (a) The quadratic polynomial

⎡ 0.50 −0.25 ⎤ (b) We have ( I − C ) −1 = ⎢ ⎣ −0.25 0.90 ⎥⎦ 1 ⎡ 0.90 0.25⎤ , = 0.3875 ⎢⎣ 0.25 0.50⎥⎦ and so the production needed to provide customers $7000 worth of mechanical work and $14,000 worth of body work is given by x = ( I − C )−1 d 1 ⎡0.9 0.25⎤ ⎡ 7000 ⎤ = 0.3875 ⎢⎣0.25 0.5 ⎥⎦ ⎢⎣14, 000 ⎥⎦ ⎡$25, 290 ⎤ ≈⎢ ⎣ $22, 581⎥⎦

2

p( x) = a0 + a1 x + a2 x passes through the points (0, 1) and (1, 2) if and only if the coefficients a0 , a1 , and a2 satisfy the equations a0 =1 a0 + a1 + a2 = 2 The general solution of this system, using a1 as a parameter, is a0 = 1, a1 = k , a2 = 1 − k . Thus this family of polynomials is represented by the equation p( x) = 1 + kx + (1 − k ) x 2 where −∞ < k < ∞.

3. (a) A consumption matrix for this economy is ⎡ 0 .1 0 .6 0 .4 ⎤ C = ⎢ 0 .3 0 .2 0 .3 ⎥ . ⎢ ⎥ ⎣ 0 .4 0 .1 0 .2 ⎦

(b) Below are graphs of four curves in the 2

family: y = 1 + x (k = 0), y = 1 + x (k = 1), y = 1 + 2 x − x 2 (k = 2), and y = 1 + 3x − 2 x 2 (k = 3). 4

y

k=0

(b) The Leontief equation (I − C)x = d is represented by the augmented matrix ⎡ 0.9 −0.6 −0.4 1930 ⎤ ⎢ −0.3 0.8 −0.3 3860 ⎥ . ⎢ ⎥ ⎣ −0.4 −0.1 0.8 5790 ⎦

k=1 3 2

–2

–1

1 k=3

2

The reduced row echelon form of this matrix is ⎡1 0 0 31, 500 ⎤ ⎢0 1 0 26, 500 ⎥ . ⎢ ⎥ ⎣0 0 1 26, 300 ⎦

x 3 k=2

True/False 1.8

Thus the required production vector is ⎡ $31, 500 ⎤ x = ⎢$26, 500⎥ . ⎢ ⎥ ⎣$26, 300⎦

(a) True; only such networks were considered. (b) False; when a current passes through a resistor, there is a drop in electrical potential in the circuit.

5. The Leontief equation (I − C)x = d for this ⎡ 0.9 −0.3 ⎤ ⎡ x1 ⎤ ⎡50 ⎤ ; thus the economy is ⎢ ⎢ ⎥= ⎣ −0.5 0.6 ⎥⎦ ⎣ x2 ⎦ ⎢⎣60 ⎥⎦ required production vector is

(c) True (d) False; the number of atoms of each element must be the same on each side of the equation.

x = ( I − C )−1 d 1 ⎡0.6 0.3⎤ ⎡50 ⎤ = 0.39 ⎢⎣0.5 0.9⎥⎦ ⎢⎣ 60 ⎥⎦ 1 ⎡ 48⎤ = 0.39 ⎢⎣79 ⎥⎦ ⎡123.08 ⎤ ≈⎢ . ⎣ 202.56 ⎥⎦

(e) False; this is only true if the n points have distinct x-coordinates. Section 1.9 Exercise Set 1.9 1. (a) A consumption matrix for this economy is ⎡0.50 0.25 ⎤ C=⎢ . ⎣0.25 0.10 ⎥⎦

39



Chapter 1 Supplementary Exercises

7. (a) The Leontief equation for this economy is ⎡ 1 0 ⎤ ⎡ x1 ⎤ ⎡ d1 ⎤ ⎢2 ⎥ ⎢ ⎥ = ⎢ ⎥ . If d1 = 2 and d 2 = 0, ⎣ 0 0 ⎦ ⎣ x2 ⎦ ⎣ d 2 ⎦ the system is consistent with general solution x1 = 4, x2 = t (0 ≤ t < ∞). If

1. The corresponding system is 3 x1 − x2 + 4 x4 = 1 2 x1 + 3x3 + 3x4 = −1 ⎡ 3 −1 0 4 1⎤ ⎢⎣ 2 0 3 3 −1⎥⎦

d1 = 2 and d 2 = 1 the system is inconsistent. ⎡ 1 0⎤ (b) The consumption matrix C = ⎢ 2 ⎥ ⎣ 0 1⎦ indicates that the entire output of the second sector is consumed in producing that output; thus there is nothing left to satisfy any outside demand. Mathematically, the ⎡ 1 0⎤ Leontief matrix I − C = ⎢ 2 ⎥ is not ⎣ 0 0⎦ invertible.

⎡1 − 1 0 3 ⎢ 0 3 ⎣2

4 3

1⎤ 3⎥

⎡1 − 1 0 3 ⎢ 2 3 ⎢⎣0 3

4 3 1 3

1⎤ 3⎥ − 35 ⎥⎦

3 −1⎦

1⎤ ⎡1 − 1 0 4 3 3 3⎥ ⎢ 1 92 12 − 25 ⎥⎦ ⎢⎣0 Thus, x3 and x4 are free variables.

9. Since c21c12 < 1 − c11, it follows that (1 − C11 ) − C21C12 > 0. From this we conclude

Let x3 = s and x4 = t .

that the matrix I − C is invertible, and c12 ⎤ 1 ⎡ 1 ( I − C ) −1 = has ⎢ 1 − c c11 ⎥⎦ (1 − C11 ) − C21C12 ⎣ 21 nonnegative entries. This shows that the economy is productive. Thus the Leontief equation (I − C)x = d has a unique solution for every demand vector d.

9 1 5 9 1 5 x2 = − x3 − x4 − = − s − t − 2 2 2 2 2 2 1 4 1 x1 = x2 − x4 + 3 3 3 1⎛ 9 1 5⎞ 4 1 = ⎜− s − t − ⎟ − t + 3⎝ 2 2 2⎠ 3 3 3 3 1 =− s− t− 2 2 2 3 3 1 The solution is x1 = − s − t − , 2 2 2 9 1 5 x2 = − s − t − , x3 = s, x4 = t . 2 2 2

11. If C has row sums less than 1, then CT has column sums less than 1. Thus, from Theorem 1.9.1, it follows that I − C T is invertible and that ( I − C T )−1 has nonnegative entries. Since I − C = ( I T − C T )T = ( I − C T )T , we can conclude that I − C is invertible and that

3. The corresponding system is 2 x1 − 4 x2 + x3 = 6 −4 x1 + 3x3 = −1 x2 − x3 = 3

( I − C ) −1 = (( I − C T )T )−1 = (( I − CT )−1 )T has nonnegative entries.

⎡ 2 −4 1 6 ⎤ ⎢ −4 0 3 −1⎥ ⎢ ⎥ 1 −1 3⎦ ⎣ 0

True/False 1.9 (a) False; open sectors are those that do not produce output.

⎡ 1 −2 1 3⎤ 2 ⎢ ⎥ ⎢ −4 0 3 −1⎥ ⎢ 0 1 −1 3 ⎥ ⎣ ⎦

(b) True (c) True (d) True; by Theorem 1.9.1. (e) True 40



⎡ 1 −2 1 3⎤ 2 ⎢ ⎥ 0 8 5 11⎥ − ⎢ ⎢0 1 −1 3⎥ ⎣ ⎦

⎡1 0 3 x + 4 y ⎤ 5 5 ⎥ ⎢ ⎢0 1 − 45 x + 53 y ⎥ ⎣ ⎦ The solution is x ′ =

⎡ 1 −2 1 3⎤ 2 ⎢ ⎥ 1 −1 3⎥ ⎢0 ⎢0 −8 5 11⎥ ⎣ ⎦

7. Reduce the augmented matrix. ⎡1 1 1 9 ⎤ ⎢⎣1 5 10 44 ⎥⎦ ⎡ 1 1 1 9⎤ ⎢⎣0 4 9 35⎥⎦

⎡ 1 −2 1 3⎤ 2 ⎢ ⎥ 1 −1 3⎥ ⎢0 ⎢0 0 −3 35⎥ ⎣ ⎦

⎡1 1 ⎢0 1 ⎣

⎡ 1 −2 1 3⎤ 2 ⎢ ⎥ 1 −1 3⎥ ⎢0 ⎢0 0 1 − 35 ⎥ 3⎦ ⎣ 35 26 Thus x3 = − , x2 = x3 + 3 = − , and 3 3 1 52 35 17 x1 = 2 x2 − x3 + 3 = − + + 3 = − . 2 3 6 2 17 26 The solution is x1 = − , x2 = − , 2 3 35 x3 = − . 3

5 3

⎡1 − 4 3 ⎢ ⎢0 53 ⎣

5x 3 − 43 x +

⎡1 − 4 3 ⎢ ⎢0 1 ⎣

⎤ ⎥ − 54 x + 35 y ⎥ ⎦

9⎤

35 ⎥ 4⎦

⎡a 0 b 2⎤ 9. ⎢ a a 4 4 ⎥ ⎢ ⎥ ⎣0 a 2 b⎦ Add −1 times the first row to the second. b 2⎤ ⎡a 0 ⎢0 a 4 − b 2⎥ ⎢ ⎥ b⎦ 2 ⎣0 a

x⎤ ⎥ y⎥ ⎦

5 3

1 9 4

⎡1 0 − 5 1 ⎤ 4 4 ⎥ ⎢ 9 35 ⎢⎣0 1 ⎥ 4 4⎦ Thus z is a free variable. Let z = s. Then 9 35 5 1 y =− s+ and x = s + . For x, y, and z to 4 4 4 4 be positive integers, s must be a positive integer such that 5s + 1 and −9s + 35 are positive and divisible by 4. Since −9s + 35 > 0, s < 4, so s = 1, 2, or 3. The only possibility is s = 3. Thus, x = 4, y = 2, z = 3.

5. Reduce the augmented matrix. ⎡3 − 4 x⎤ 5 ⎢5 ⎥ ⎢ 45 35 y ⎥ ⎣ ⎦ ⎡1 − 4 3 ⎢ ⎢ 45 35 ⎣

3 4 4 3 x + y, y ′ = − x + y. 5 5 5 5

Add −1 times the second row to the third. b 2 ⎤ ⎡a 0 ⎢0 a 4 − b 2 ⎥ ⎢ ⎥ ⎣0 0 b − 2 b − 2⎦

⎤ ⎥ y⎥ ⎦

(a) The system has a unique solution if the corresponding matrix can be put in reduced row echelon form without division by zero. Thus, a ≠ 0, b ≠ 2 is required.

x

(b) If a ≠ 0 and b = 2, then x3 is a free variable and the system has a one-parameter solution. (c) If a = 0 and b = 2, the system has a twoparameter solution. 41


SSM: Elementary Linear Algebra 13. (a) Here X must be a 2 × 3 matrix. Let ⎡ a b c⎤ X =⎢ . ⎣ d e f ⎥⎦ Then ⎡ −1 0 1⎤ ⎡ a b c⎤ ⎢ ⎥ ⎡ 1 2 0⎤ ⎢⎣ d e f ⎥⎦ ⎢ 1 1 0⎥ = ⎢⎣ −3 1 5⎥⎦ or ⎣ 3 1 −1⎦

(d) If b ≠ 2 but a = 0, the system has no solution. 9. Note that K must be a 2 × 2 matrix. Let ⎡a b ⎤ K =⎢ . ⎣ c d ⎥⎦ Then ⎡ 1 4⎤ ⎡ 8 6 −6 ⎤ ⎡a b ⎤ ⎡ 2 0 0⎤ ⎢ ⎢ −2 3⎥ ⎢ = 6 −1 1⎥ ⎢ ⎥ ⎣ c d ⎦⎥ ⎣⎢ 0 1 −1⎦⎥ ⎢ ⎥ ⎣ 1 −2 ⎦ ⎣ −4 0 0 ⎦

⎡ −a + b + 3c b + c a − c ⎤ ⎢⎣ − d + e + 3 f e + f d − f ⎥⎦ ⎡ 1 2 0⎤ . =⎢ ⎣ −3 1 5⎥⎦ Equating entries in the first row leads to the system − a + b + 3c = 1 b +c = 2 −c = 0 a Equating entries in the second row yields the system − d + e + 3 f = −3 e + f =1 d − f =5 These systems can be solved simultaneously by reducing the following augmented matrix. ⎡ −1 1 3 1 −3⎤ ⎢ 0 1 1 2 1⎥ ⎢ ⎥ ⎣ 1 0 −1 0 5⎦

⎡ 1 4⎤ ⎡ 8 6 −6 ⎤ ⎡ 2 a b −b ⎤ ⎢ or ⎢ −2 3⎥ ⎢ = 6 −1 1⎥ or ⎢ ⎥ ⎣ 2c d −d ⎥⎦ ⎢ ⎥ 1 − 2 − ⎣ ⎦ ⎣ 4 0 0⎦ −b − 4 d ⎤ b + 4d ⎡ 2a + 8c ⎢ −4a + 6c −2b + 3d 2b − 3d ⎥ ⎢ ⎥ −b + 2d ⎦ b − 2d ⎣ 2 a − 4c ⎡ 8 6 −6 ⎤ = ⎢ 6 −1 1⎥ . ⎢ ⎥ ⎣ −4 0 0 ⎦ Thus 2a + 8c =8 b + 4d = 6 −4 a + 6c =6 −2b + 3d = −1 2a − 4c = −4 − 2d = 0 b Note that we have omitted the 3 equations obtained by equating elements of the last columns of these matrices because the information so obtained would be just a repeat of that gained by equating elements of the second columns. The augmented matrix of the above system is 8 0 8⎤ ⎡ 2 0 ⎢ 0 1 0 4 6⎥ ⎢ ⎥ ⎢ −4 0 6 0 6 ⎥ . 3 −1⎥ ⎢ 0 −2 0 ⎢ 2 0 −4 0 −4 ⎥ ⎢ 0 1 0 −2 0 ⎥⎦ ⎣ The reduced row-echelon form of this matrix is ⎡ 1 0 0 0 0⎤ ⎢0 1 0 0 2 ⎥ ⎢ ⎥ ⎢0 0 1 0 1⎥ . ⎢0 0 0 1 1⎥ ⎢0 0 0 0 0 ⎥ ⎢0 0 0 0 0 ⎥ ⎣ ⎦ Thus a = 0, b = 2, c = 1, and d = 1. ⎡0 2 ⎤ The matrix is K = ⎢ . ⎣1 1 ⎥⎦

⎡ 1 0 −1 0 5⎤ ⎢ 0 1 1 2 1⎥ ⎢ ⎥ − − 1 1 3 1 3 ⎣ ⎦ ⎡ 1 0 −1 0 5 ⎤ ⎢0 1 1 2 1⎥ ⎢ ⎥ ⎣0 1 2 1 2 ⎦ ⎡ 1 0 −1 0 5 ⎤ ⎢0 1 1 2 1⎥ ⎢ ⎥ ⎣0 0 1 −1 1⎦ ⎡ 1 0 0 −1 6 ⎤ ⎢0 1 0 3 0 ⎥ ⎢ ⎥ ⎣0 0 1 −1 1⎦ Thus, a = −1, b = 3, c = −1, d = 6, e = 0, ⎡ −1 3 −1⎤ . f = 1 and X = ⎢ ⎣ 6 0 1⎥⎦

42



(b) X must be a 2 × 2 matrix. ⎡x y⎤ Let X = ⎢ . Then ⎣ z w⎥⎦ ⎡ 1 −1 2 ⎤ ⎡ x + 3 y − x 2 x + y ⎤ = X⎢ . ⎣3 0 1⎥⎦ ⎢⎣ z + 3w − z 2 z + w⎥⎦ If we equate matrix entries, this gives us the equations z + 3w = 6 x + 3 y = −5 − z = −3 − x = −1 2z + w = 7 2x + y = 0 Thus x = 1 and z = 3, so that the top two equations give y = −2 and w = 1. Since these values are consistent with the bottom two equations, we have that ⎡1 − 2 ⎤ X =⎢ . ⎣3 1 ⎥⎦

15. Since the coordinates of the given points must satisfy the polynomial, we have a +b+c = 2 p(1) = 2 ⇒ a −b +c = 6 p(−1) = 6 ⇒ p(2) = 3 ⇒ 4a + 2b + c = 3 This system has augmented matrix ⎡ 1 1 1 2⎤ ⎢ 1 −1 1 6 ⎥ which reduces to ⎢ ⎥ ⎣ 4 2 1 3⎦

1⎤ ⎡1 0 0 ⎢0 1 0 −2 ⎥ . ⎢ ⎥ ⎣ 0 0 1 3⎦ Thus, a = 1, b = −2, and c = 3. 19. First suppose that AB −1 = B −1 A. Note that all matrices must be square and of the same size.

Therefore ( AB −1 ) B = ( B −1 A) B or A = B −1 AB

(c) Again, X must be a 2 × 2 matrix. Let ⎡x y⎤ X =⎢ , so that the matrix equation ⎣ z w⎥⎦ becomes 3y + w ⎤ ⎡ x + 2 y 4x⎤ ⎡ 3x + z ⎢⎣ − x + 2 z − y + 2w⎥⎦ − ⎢⎣ z + 2 w 4 z ⎥⎦ ⎡ 2 −2 ⎤ . =⎢ ⎣ 5 4 ⎥⎦ This yields the system of equations 2x − 2 y + z =2 −4 x + 3 y + w = −2 −x + z − 2w = 5 − y − 4 z + 2w = 4

so that BA = B( B −1 AB) = ( BB −1 )( AB) = AB. A similar argument shows that if AB = BA then AB −1 = B −1 A. 21. Consider the ith row of AB. By the definition of matrix multiplication, this is ai1b11 + ai 2b21 + + ain bn1 a + a + + ain = i1 i 2 n = ri

1 since all entries bi1 = . n

1 0 2⎤ ⎡ 2 −2 ⎢ −4 3 0 1 −2 ⎥ with matrix ⎢ ⎥ which 1 −2 5 ⎥ ⎢ −1 0 ⎣⎢ 0 −1 −4 2 4 ⎦⎥

⎡1 0 ⎢ ⎢0 1 reduces to ⎢ ⎢0 0 ⎢ ⎢⎣0 0 113 , Hence, x = − 37

⎤ 0 0 − 113 37 ⎥ 0 0 − 160 ⎥ 37 . ⎥ 20 1 0 − 37 ⎥ 46 ⎥ 0 1 − 37 ⎥⎦ 160 20 y=− , z=− , 37 37 ⎡ − 113 − 160 ⎤ 46 37 ⎥ . w = − , and X = ⎢ 37 20 46 ⎥ 37 ⎢ − 37 − 37 ⎣ ⎦

43

Chapter 10 Applications of Linear Algebra 3. Using Equation (10) we obtain

Section 10.1

x2 0 0 4 4 16

Exercise Set 10.1 1. (a) Substituting the coordinates of the points x y 1 into Equation (4) yields 1 −1 1 = 0 2 2 1 which, upon cofactor expansion along the first row, yields −3x + y + 4 = 0; that is, y = 3x − 4.

2x + y − 1 = 0 or y = −2x + 1. 2. (a) Equation (9) yields

x +y x y 1 40 2 6 1 = 0 which, upon first4 2 0 1 34 5 3 1 row cofactor expansion, yields

x 2 + 2 xy + y 2 − 2 x + y = 0, which is the equation of a parabola.

2

dividing by 18, x + y − 4 x − 6 y + 4 = 0. Completing the squares in x and y yields the standard form ( x − 2)2 + ( y − 3)2 = 9. x y 2 −2 3 5 −4 6

1 1 1 = 0 which is the 1 1 1

160 x 2 + 320 xy + 160( y 2 + y) − 320 x = 0; that is,

18( x 2 + y 2 ) − 72 x − 108 y + 72 = 0 or,

x2 + y 2 8 (b) As in (a), 34 52

x y 0 0 0 −1 2 0 2 −5 4 −1

x 2 xy y 2 + y x 4 0 0 2 = 0. 4 −10 20 2 16 −4 0 4 Now expand along the first row and get

2

2

y2 0 1 0 25 1

x 2 xy y 2 x y 0 0 1 0 −1 same as 4 0 0 2 0 = 0 by 4 −10 25 2 −5 16 −4 1 4 −1 expansion along the second row (taking advantage of the zeros there). Add column five to column three and take advantage of another row of all but one zero to get

x y 1 (b) As in (a), 0 1 1 = 0 yields 1 −1 1

2

xy 0 0 0 −10 −4

4. (a) From Equation (11), the equation of the x y z 1 1 1 −3 1 plane is = 0. 1 −1 1 1 0 −1 2 1 Expansion along the first row yields −2x − 4y − 2z = 0; that is, x + 2y + z = 0.

1 1 = 0 yields 1 1

50( x 2 + y 2 ) + 100 x − 200 y − 1000 = 0; that

x y z 2 3 1 (b) As in (a), 2 −1 −1 1 2 1

is, x 2 + y 2 + 2 x − 4 y − 20 = 0. In standard form this is ( x + 1)2 + ( y − 2)2 = 25.

1 1 = 0 yields 1 1

−2x + 2y − 4z + 2 = 0; that is −x + y − 2z + 1 = 0 for the equation of the plane.

211

Chapter 10: Applications of Linear Algebra


5. (a) Equation (11) involves the determinant of the coefficient matrix of the system y z 1⎤ ⎡ c1 ⎤ ⎡ 0⎤ ⎡x ⎢x y1 z1 1⎥ ⎢ c2 ⎥ ⎢ 0⎥ 1 ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥. ⎢ x2 y2 z2 1⎥ ⎢ c3 ⎥ ⎢ 0⎥ ⎣⎢ x3 y3 z3 1⎦⎥ ⎢⎣ c4 ⎥⎦ ⎢⎣ 0⎥⎦ Rows 2 through 4 show that the plane passes through the three points ( xi , yi , zi ), i = 1, 2, 3, while row 1 gives the equation c1 x + c2 y + c3 z + c4 = 0. For the plane passing through the origin parallel to the plane passing through the three points, the constant term in the final equation will be 0, which is accomplished by using x y z 0 x1 y1 z1 1 =0 x2 y2 z2 1 x3 y3 z3 1

x2 + y 2 + z 2 5 (b) As in (a), 11 5 11 yields

x y z 0 1 −2 1 3 1 2 −1 0 3 1 −1

1 1 1 =0 1 1

−24( x 2 + y 2 + z 2 ) + 48 x + 48 y + 72 = 0; that is, x 2 + y 2 + z 2 − 2 x − 2 y − 3 = 0 or in standard form, ( x − 1) 2 + ( y − 1)2 + z 2 = 5. 7. Substituting each of the points ( x1 , y1 ),

( x2 , y2 ), ( x3 , y3 ), ( x4 , y4 ), and ( x5 , y5 ) into the equation c1 x 2 + c2 xy + c3 y 2 + c4 x + c5 y + c6 = 0 yields

c1 x12 + c2 x1 y1 + c3 y12 + c4 x1 + c5 y1 + c6 = 0 # # # c1 x52 + c2 x5 y5 + c3 y52 + c4 x5 + c5 y5 + c6 = 0. These together with the original equation form a homogeneous linear system with a non-trivial solution for c1, c2 , ..., c6 . Thus the determinant of the coefficient matrix is zero, which is exactly Equation (10).

(b) The parallel planes passing through the origin are x + 2y + z = 0 and −x + y − 2z = 0, respectively. 6. (a) Using Equation (12), the equation of the

x2 + y2 + z 2 x y z 1 14 1 2 3 1 sphere is 6 −1 2 1 1 = 0. 2 1 0 1 1 6 1 2 −1 1 Expanding by cofactors along the first row yields

8. As in the previous problem, substitute the coordinates ( xi , yi , zi ) of each of the three

points into the equation c1 x + c2 y + c3 z + c4 = 0 to obtain a homogeneous system with nontrivial solution for c1, ..., c4 . Thus the determinant of the coefficient matrix is zero, which is exactly Equation (11).

16( x 2 + y 2 + z 2 ) − 32 x − 64 y − 32 z + 32 = 0; that is, ( x 2 + y 2 + z 2 ) − 2 x − 4 y − 2 x + 2 = 0. Completing the squares in each variable yields the standard form

9. Substituting the coordinates ( xi , yi , zi ) of the four points into the equation

c1 ( x 2 + y 2 + z 2 ) + c2 x + c3 y + c4 z + c5 = 0 of the sphere yields four equations, which together with the above sphere equation form a homogeneous linear system for c1, ..., c5 with a nontrivial solution. Thus the determinant of this system is zero, which is Equation (12).

( x − 1) 2 + ( y − 2)2 + ( z − 1)2 = 4. Note: When evaluating the cofactors, it is useful to take advantage of the column of ones and elementary row operations; for example, the cofactor of x 2 + y 2 + z 2 above can be evaluated as follows: 1 2 3 1 1 2 3 1 −1 2 1 1 −2 0 −2 0 = = 16 by 1 0 1 1 0 −2 −2 0 1 2 −1 1 0 0 −4 0 cofactor expansion of the latter determinant along the last column.

212


Section 10.2

10. Upon substitution of the coordinates of the three points ( x1 , y1 ), ( x2 , y2 ) and ( x3 , y3 ), we obtain the equations:


c1 y + c2 x 2 + c3 x + c4 = 0 c1 y1 + c2 x12 + c3 x1 + c4 c1 y2 + c2 x22 + c3 x2 + c4 c1 y3 + c2 x32 + c3 x3 + c4

=0 =0

= 0. This is a homogeneous system with a nontrivial solution for c1, c2 , c3 , c4 , so the determinant of the coefficient matrix is zero; that is, y

x2

x

1

y1

x12 x22 x32

x1

1

y2 y3

x2 1

1 ,1 2

3 ,1 2 2,

(0, 0)

(2, 0)

x2 = 1 2 3 2 x1 + 3 x2 = 6 5 x1

In the figure the feasible region is shown and the extreme points are labeled. The values of the objective function are shown in the following table:

= 0.

x3 1

11. Expanding the determinant in Equation (9) by cofactors of the first row makes it apparent that

the coefficient of x 2 + y 2 in the final equation is x1 x2 x3

x 2 2 x1 − x2 = 0 x1 = 2

1.

y1 1 y2 1 . If the points are collinear, then the y3 1

columns are linearly dependent ( yi = mxi + b), so the coefficient of x 2 + y 2 = 0 and the resulting equation is that of the line through the three points. 12. If the three distinct points are collinear then two of the coordinates can be expressed in terms of the third. Without loss of generality, we can say that y and z can be expressed in terms of x, i.e., x is the parameter. If the line is (x, ax + b, cx + d), then the determinant in Equation (11) is x −ax − b −cx − d 1 x 0 0 1 . equivalent to 1 x2 0 0 1 x3 0 0 1 Expanding along the first row, it is clear that the determinant is 0 and Equation (11) becomes 0 = 0.

Extreme point ( x1 , x2 )

Value of z = 3x1 + 2 x2

(0, 0)

0

( 12 , 1)

7 2

( 32 , 1) ( 2, 23 )

22 3

(2, 0)

6

13 2

Thus the maximum,

22 , is attained when 3

2 x1 = 2 and x2 = . 3 2.

x2 x2 = 3

3

2 x1 − x2 = −2 4 x1 − x2 = 0

13. As in Exercise 11, the coefficient of

x 2 + y 2 + z 2 will be 0, so Equation (12) gives the equation of the plane in which the four points lie.

2

x1

The intersection of the five half-planes defined by the constraints is empty. Thus this problem has no feasible solutions.

213



x2

3.

Thus, it has a minimum, which is attained at the 14 25 and x2 = . The value of point where x1 = 9 18 335 . In the problem’s terms, if we use z there is 18 7 25 cup of milk and ounces of corn flakes, a 9 18 minimum cost of 18.6¢ is realized.

5 − x1 + x2 = 1

3x1 − x2 = −5 2 x1 + 4 x2 = 12

x1

6. (a) Both x1 ≥ 0 and x2 ≥ 0 are nonbinding. Of the remaining constraints, only 2 x1 + 3x2 ≤ 24 is binding.

The feasible region for this problem, shown in the figure, is unbounded. The value of z = −3x1 + 2 x2 cannot be minimized in this region since it becomes arbitrarily negative as we travel outward along the line − x1 + x2 = 1; i.e., the value of z is −3x1 + 2 x2 = −3x1 + 2( x1 + 1) = − x1 + 2 and x1 can be arbitrarily large. 4.

x1 + x2 = 10, 000

x2

(b) x1 − x2 ≤ v will be binding if the line

x1 − x2 = v intersects the line x2 = 6 with 0 < x1 < 3, thus if −6 < v < −3. If v < −6, then the feasible region is empty.

x1 = 6000

(c) x2 ≤ v will be nonbinding for v ≥ 8 and the feasible region will be empty for v < 0.

(5000, 5000)

7. Letting x1 be the number of Company A’s containers shipped and x2 the number of Company B’s, the problem is: Maximize z = 2.20 x1 + 3.00 x2 subject to

(6000, 4000) (6000, 2000) x2 = 2000

(2000, 2000)

40 x1 + 50 x2 ≤ 37, 000 2 x1 + 3x2 ≤ 2000 x1 ≥ 0 x2 ≥ 0.

x1

x1 − x2 = 0

The feasible region and vertices for this problem are shown in the figure. The maximum value of the objective function z = .1x1 + .07 x2 is attained

x2

when x1 = 6000 and x2 = 4000, so that z = 880. In other words, she should invest $6000 in bond A and $4000 in bond B for annual yield of $880. 5.

x2 5

14 25 , 9 18

x1 =

0, 666

3 3x1 − 2 x1 = 0 2 3 9 , 2 4

40 20 , 21 21 4 x1 + 2 x2 = 9

40 x1 + 50 x2 = 37, 000

(550, 300) 2 x1 + 3x2 = 2000

(0, 0)

x1 − 2 x2 = 0

3 3 , 2 2

2 3

(925, 0)

x1

The feasible region is shown in the figure. The vertex at which the maximum is attained is x1 = 550 and x2 = 300, where z = 2110. A truck should carry 550 containers from Company A and 300 containers from Company B for maximum shipping charges of $2110.

x1

1 1 1 x8 + x2 = 8 1 10 3

The feasible region and its extreme points are shown in the figure. Though the region is unbounded, x1 and x2 are always positive, so

8. We must now maximize z = 2.50 x1 + 3.00 x2 . The feasible region is as in Exercise 7, but now the maximum occurs for x1 = 925 and x2 = 0, where z = 2312.50.

the objective function z = 7.5 x1 + 5.0 x2 is also. 214


Section 10.3

925 containers from Company A and no containers from Company B should be shipped for maximum shipping charges of $2312.50.

Section 10.3 Exercise Set 10.3 1. The number of oxen is 50 per herd, and there are 7 herds, so there are 350 oxen. Hence the total number of oxen and sheep is 350 + 350 = 700.

9. Let x1 be the number of pounds of ingredient A used and x2 the number of pounds of ingredient B. Then the problem is: Minimize z = 8 x1 + 9 x2 subject to 2 x1 + 5 x2 ≥ 10 2 x1 + 3x2 ≥ 8 6 x1 + 4 x2 ≥ 12 x1 ≥ 0 x2 ≥ 0

2. (a) The equations are B = 2A, C = 3(A + B), D = 4(A + B + C), 300 = A + B + C + D. Solving this linear system gives A = 5 (and B = 10, C = 45, D = 240). (b) The equations are B = 2A, C = 3B, D = 4C, 132 = A + B + C + D. Solving this linear system gives A = 4 (and B = 8, C = 24, D = 96).

x 2 (0, 3) 2 12 , 5 5

3. Note that this is, effectively, Gaussian elimination applied to the augmented matrix ⎡1 1 10 ⎤ ⎢1 1 7 ⎥ . ⎣ 4 ⎦

2 x1 + 3x2 = 8

5 ,1 2 2 x1 + 5 x2 = 10

6 x1 + 4 x2 = 12

4. (a) Let x represent oxen and y represent sheep, then the equations are 5x + 2y = 10 and 2x + 5y = 8. The corresponding array is

(5, 0) 5 x1

Though the feasible region shown in the figure is unbounded, the objective function is always positive there and hence must have a minimum. This minimum occurs at the vertex where 2 12 x1 = and x2 = . The minimum value of z is 5 5 124 or 24.8¢. 5 Each sack of feed should contain 0.4 lb of ingredient A and 2.4 lb of ingredient B for a minimum cost of 24.8¢.

2

5

5

2

8

10

and the elimination step subtracts twice column 2 from five times column 1, giving 5

10. It is sufficient to show that if a linear function has the same value at two points, then it has that value along the line connecting the two points. Let ax1 + bx2 be the function, and ax1′ + bx2′ = ax1′′ + bx2′′ = c. Then

21

2

20

10

and so y =

(tx1′ + (1 − t ) x1′′, tx2′ + (1 − t ) x2′′ ) is an arbitrary point on the line connecting ( x1′ , x2′ ) to ( x1′′, x2′′ ) and a(tx1′ + (1 − t ) x1′′) + b(tx2′ + (1 − t ) x2′′ ) = t (ax1′ + bx2′ ) + (1 − t )(ax1′′ + bx2′′ ) = tc + (1 − t )c = c.

x=

20 unit for a sheep, and 21

34 units for an ox. 21

(b) Let x, y, and z represent the number of bundles of each class. Then the equations are 2x + y =1 3y + z = 1 x 4z = 1 and the corresponding array is 215


2 3

(b) Exercise 7(b) may be solved using this technique. x1 represents gold, x2 represents

1

1

1 1


brass, x3 represents tin, and x4 represents much-wrought iron, so n = 4 and a1 = 60, 2 a2 = (60) = 40, 3 3 a3 = (60) = 45, 4 3 a4 = (60) = 36. 5 (a2 + a3 + a4 ) − a1 x1 = n−2 40 + 45 + 36 − 60 = 4−2 61 = 2 61 19 x2 = a2 − x1 = 40 − = 2 2 61 29 x3 = a3 − x1 = 45 − = 2 2 61 11 x4 = a4 − x1 = 36 − = 2 2 1 The crown was made with 30 minae of 2 1 1 gold, 9 minae of brass, 14 minae of 2 2 1 tin, and 5 minae of iron. 2

4 1

1

Subtract two times the numbers in the third column from the second column to get 1 3

1

1

−8

4

1 −1 Now subtract three times the numbers in the second column from the first column to get 1

1 1 25

−8

4

4

−1

1

This is equivalent to the linear system x + 4z = 1 y − 8 z = −1. 25 z = 4 From this, the solution is that a bundle of the 9 measure, second first class contains 25 7 measure, and third class class contains 25 4 measure. contains 25

6. (a) We can write this as 5x + y + z − K = 0 x + 7y + z − K = 0 x + y + 8 − K = 0 (a 3 × 4 system). Since the coefficient matrix of equations (5) is invertible (its determinant is 262), there is a unique solution x, y, z for every K; hence, K is an arbitrary parameter.

5. (a) From equations 2 through n, x j = a j − x1

(j = 2, ..., n). Using these equations in equation 1 gives x1 + (a2 − x1 ) + (a3 − x1 ) + " + (an − x1 ) = a1

(b) Gaussian elimination gives x =

a + a + " + an − a1 x1 = 2 3 . n−2 First find x1 in terms of the known quantities n and the ai . Then we can use

21K , 131

14 K 12 K , z= , for any choice of K. 131 131 Since 131 is prime, we must choose K to be an integer multiple of 131 to get integer solutions. The obvious choice is K = 131, giving x = 21, y = 14, z = 12, and K = 131. y=

x j = a j − x1 (j = 2, ..., n) to find the other xi .

216


Section 10.4

6( y1 − 2 y2 + y3 )

(c) This solution corresponds to K = 262 = 2 ⋅ 131.

h2 6( y2 − 2 y3 + y4 )

7. (a) The system is x + y = 1000, ⎛1⎞ ⎛1⎞ ⎜ ⎟ x − ⎜ ⎟ y = 10, with solution x = 577 ⎝4⎠ ⎝5⎠ 7 2 and , y = 422 and . The legitimate son 9 9 7 receives 577 staters, the illegitimate son 9 2 receives 422 . 9

h2 6( y3 − 2 y4 + y5 ) h2 6( y4 − 2 y5 + y6 )

= −1.1880 = −2.3295 = −3.3750

= −4.2915 h2 and the linear system (21) for the parabolic runout spline becomes ⎡5 1 0 0 ⎤ ⎡ M 2 ⎤ ⎡ −1.1880 ⎤ ⎢1 4 1 0 ⎥ ⎢ M 3 ⎥ ⎢ −2.3295 ⎥ ⎥=⎢ ⎢ ⎥⎢ ⎥. ⎢0 1 4 1 ⎥ ⎢ M 4 ⎥ ⎢ −3.3750 ⎥ ⎣⎢0 0 1 5 ⎦⎥ ⎢⎣ M 5 ⎥⎦ ⎣⎢ −4.2915 ⎦⎥

⎛2⎞ (b) The system is G + B = ⎜ ⎟ 60, ⎝3⎠ ⎛3⎞ ⎛3⎞ G + T = ⎜ ⎟ 60, G + I = ⎜ ⎟ 60, ⎝4⎠ ⎝5⎠ G + B + T + I = 60, with solution G = 30.5, B = 9.5, T = 14.5, and I = 5.5. The crown 1 was made with 30 minae of gold, 2 1 1 9 minae of brass, 14 minae of tin, and 2 2 1 5 minae of iron. 2

Solving this system yields M 2 = −.15676, M 3 = −.40421, M 4 = −.55592, M 5 = −.74712. From (19) and (20) we have M1 = M 2 = −.15676, M 6 = M 5 = −.74712. The specific interval .4 ≤ x ≤ .6 is the third interval. Using (14) to solve for a3 , b3 , c3 , and d3 gives (M 4 − M 3 ) = −.12643 6h M b3 = 3 = −.20211 2 ( y4 − y3 ) ( M 4 + 2M 3 )h − = .92158 c3 = h 6 d3 = y3 = .38942. The interpolating parabolic runout spline for .4 ≤ x ≤ .6 is thus a3 =

⎛1 ⎞ (c) The system is A = B + ⎜ C , ⎟ ⎝3 ⎠ ⎛1⎞ ⎛1⎞ B = C + ⎜ ⎟ A, C = ⎜ ⎟ B + 10, with 3 ⎝ ⎠ ⎝3⎠ solution A = 45, B = 37.5, and C = 22.5. The first person has 45 minae, the second 1 1 has 37 , and the third has 22 . 2 2

S ( x) = −.12643( x − .4)3 − .20211( x − .4) 2 + .92158( x − .4) + .38942.

Section 10.4

(b) S (.5) = −.12643(.1)3 − .20211(.1) 2 + .92158(.1) + .38942 = .47943. Since sin(.5) = S(.5) = .47943 to five decimal places, the percentage error is zero.

Exercise Set 10.4 2. (a) Set h = .2 and x1 = 0, y1 = .00000 x2 = .2, y2 = .19867 x3 = .4, y3 = .38942 x4 = .6, y4 = .56464 x5 = .8, y5 = .71736 x6 = 1.0, y6 = .84147 Then

3. (a) Given that the points lie on a single cubic curve, the cubic runout spline will agree exactly with the single cubic curve.

217


(b) Set h = 1 and x1 = 0, y1 = 1 x2 = 1, y2 = 7 x3 = 2, y3 = 27 x4 = 3, y4 = 79 x5 = 4, y5 = 181. Then 6( y1 − 2 y2 + y3 )

h2 6( y2 − 2 y3 + y4 ) h2 6( y3 − 2 y4 + y5 )

SSM: Elementary Linear Algebra d 4 = y4 = 79. For 0 ≤ x ≤ 1 we have S ( x) = S1 ( x) = 3x3 − 2 x 2 + 5 x + 1. For 1 ≤ x ≤ 2 we have S ( x) = S2 ( x)

= 3( x − 1)3 + 7( x − 1)2 + 10( x − 1) + 7 = 3x3 − 2 x 2 + 5 x + 1. For 2 ≤ x ≤ 3 we have S ( x) = S3 ( x)

= 84 = 192

= 3( x − 2)3 + 16( x − 2)2 + 33( x − 2) + 27 = 3x3 − 2 x 2 + 5 x + 1. For 3 ≤ x ≤ 4 we have S ( x) = S 4 ( x)

= 300

h2 and the linear system (24) for the cubic runout spline becomes ⎡6 0 0 ⎤ ⎡ M 2 ⎤ ⎡ 84 ⎤ ⎢1 4 1 ⎥ ⎢ M 3 ⎥ = ⎢192 ⎥ . ⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎣0 0 6 ⎦ ⎣ M 4 ⎦ ⎣300 ⎦ Solving this system yields M 2 = 14, M 3 = 32, M 4 = 50. From (22) and (23) we have M 1 = 2 M 2 − M 3 = −4 M 5 = 2 M 4 − M 3 = 68. Using (14) to solve for the ai’s, bi’s, ci’s, and ( M 2 − M1 ) di’s we have a1 = = 3, 6h (M 3 − M 2 ) (M 4 − M 3 ) a2 = = 3, a3 = = 3, 6h 6h (M 5 − M 4 ) a4 = = 3, 6h

= 3( x − 3)3 + 25( x − 3)2 + 74( x − 3) + 79 = 3x3 − 2 x 2 + 5 x + 1. Thus S1 ( x) = S2 ( x) = S3 ( x) = S4 ( x), or S ( x) = 3x3 − 2 x 2 + 5 x + 1 for 0 ≤ x ≤ 4. 4. The linear system (16) for the natural spline ⎡ 4 1 0 ⎤ ⎡ M 2 ⎤ ⎡ −.0001116 ⎤ becomes ⎢1 4 1 ⎥ ⎢ M 3 ⎥ = ⎢ −.0000816 ⎥ . ⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎣ 0 1 4 ⎦ ⎣ M 4 ⎦ ⎣ −.0000636 ⎦

Solving this system yields M 2 = −.0000252, M 3 = −.0000108, M 4 = −.0000132. From (17) and (18) we have M1 = 0, M 5 = 0. Solving for the ai’s, bi’s, ci’s, and di’s from Equations (14) we have ( M 2 − M1 ) a1 = = −.00000042, 6h (M 3 − M 2 ) a2 = = .00000024, 6h (M 4 − M 3 ) a3 = = −.00000004, 6h (M 5 − M 4 ) a4 = = −.00000022, 6h

M1 M = −2, b2 = 2 = 7, 2 2 M3 M4 b3 = = 16, b4 = = 25, 2 2 b1 =

( y2 − y1 ) ( M 2 + 2M1 )h − = 5, h 6 ( y − y2 ) ( M 3 + 2M 2 )h = 3 − = 10, h 6 ( y − y3 ) ( M 4 + 2M 3 )h = 4 − = 33, h 6 ( y − y4 ) ( M 5 + 2 M 4 )h = 5 − = 74, h 6

c1 = c2 c3 c4

M1 M = 0, b2 = 2 = −.0000126, 2 2 M3 M = = −.0000054, b4 = 4 = −.0000066, 2 2 M5 = = 0. 2

b1 = b3 b5

c1 =

d1 = y1 = 1, d 2 = y2 = 7, d3 = y3 = 27, 218

( y2 − y1 ) ( M 2 + 2 M1 )h − = .000214, h 6


Section 10.4

( y3 − y2 ) ( M 3 + 2M 2 )h ( y − y3 ) ( M 4 + 2M 3 )h − = .000088, c3 = 4 − = −.000092, h h 6 6 ( y − y4 ) ( M 5 + 2M 4 )h c4 = 5 − = −.000212, h 6 c2 =

d1 = y1 = .99815, d 2 = y2 = .99987, d3 = y3 = .99973, d 4 = y4 = .99823. The resulting natural spline is ⎧−.00000042( x + 10)3 + .000214( x + 10) + .99815, − 10 ≤ x ≤ 0 ⎪ ⎪.00000024( x)3 − .0000126( x)2 + .000088( x) + .99987, 0 ≤ x ≤ 10 S ( x) = ⎨ 3 2 ⎪−.00000004( x − 10) − .0000054( x − 10) − .000092( x − 10) + .99973, 10 ≤ x ≤ 20 ⎪−.00000022( x − 20)3 − .0000066( x − 20)2 − .000212( x − 20) + .99823, 20 ≤ x ≤ 30. ⎩ Assuming the maximum is attained in the interval [0, 10] we set S ′( x) equal to zero in this interval: S ′( x) = .00000072 x 2 − .0000252 x + .000088 = 0. To three significant digits the root of this quadratic equation in the interval [0, 10] is x = 3.93, and S(3.93) = 1.00004. ⎡6 0 0 ⎤ ⎡ M 2 ⎤ ⎡ −.0001116 ⎤ 5. The linear system (24) for the cubic runout spline becomes ⎢1 4 1 ⎥ ⎢ M 3 ⎥ = ⎢ −.0000816 ⎥ . ⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎣0 0 6 ⎦ ⎣ M 4 ⎦ ⎣ −.0000636 ⎦ Solving this system yields M 2 = −.0000186, M 3 = −.0000131, M 4 = −.0000106. From (22) and (23) we have M1 = 2M 2 − M 3 = −.0000241, M 5 = 2 M 4 − M 3 = −.0000081. ( M 2 − M1 ) = .00000009, 6h (M 3 − M 2 ) (M 4 − M 3 ) (M 5 − M 4 ) a2 = = .00000009, a3 = = .00000004, a4 = = .00000004. 6h 6h 6h

Solving for the ai’s, bi’s, ci’s, and di’s from Equations (14) we have a1 =

b1 =

M M1 M M = −.0000121, b2 = 2 = −.0000093, b3 = 3 = −.0000066, b4 = 4 = −.0000053, 2 2 2 2

( y − y2 ) ( M 3 + 2M 2 )h ( y2 − y1 ) ( M 2 + 2 M1 )h − = .000282, c2 = 3 − = .000070, h 6 h 6 ( y − y3 ) ( M 4 + 2M 3 )h ( y − y4 ) ( M 5 + 2M 4 )h c3 = 4 − = −.000087, c4 = 5 − = −.000207, h 6 h 6

c1 =

d1 = y1 = .99815, d 2 = y2 = .99987, d3 = y3 = .99973, d 4 = y4 = .99823. The resulting cubic runout spline is ⎧.00000009( x + 10)3 − .0000121( x + 10) 2 + .000282( x + 10) + .99815, − 10 ≤ x ≤ 0 ⎪ ⎪.00000009( x)3 − .0000093( x) 2 + .000070( x) + .99987, 0 ≤ x ≤ 10 S ( x) = ⎨ 3 2 ⎪.00000004( x − 10) − .0000066( x − 10) − .000087( x − 10) + .99973, 10 ≤ x ≤ 20 ⎪.00000004( x − 20)3 − .0000053( x − 20)2 − .000207( x − 20) + .99823, 20 ≤ x ≤ 30. ⎩ Assuming the maximum is attained in the interval [0, 10], we set S ′( x) equal to zero in this interval: S ′( x) = .00000027 x 2 − .0000186 x + .000070 = 0. To three significant digits the root of this quadratic equation in the interval [0, 10] is 4.00 and S(4.00) = 1.00001.

219


SSM: Elementary Linear Algebra For 1 ≤ x ≤ 1.5, we have S ( x) = S2 ( x) = −2( x − 1) + 0 = −2 x + 2. The resulting natural spline is ⎧ − 2 x + 2 0 .5 ≤ x ≤ 1 S ( x) = ⎨ . ⎩−2 x + 2 1 ≤ x ≤ 1.5

6. (a) Set h = .5 and x1 = 0, y1 = 0

x2 = .5, y2 = 1 x3 = 1, y3 = 0 For a natural spline with n = 3, M1 = M 3 = 0 and 6( y1 − 2 y2 + y3 ) h2 M 2 = −12.

(c) The three data points are collinear, so the spline is just the line the points lie on.

= −48 = 4M 2 . Thus

7. (b) Equations (15) together with the three equations in part (a) of the exercise statement give 6( yn −1 − 2 y1 + y2 ) 4M1 + M 2 + M n −1 = h2 6( y1 − 2 y2 + y3 ) M1 + 4 M 2 + M 3 = h2 6( y2 − 2 y3 + y4 ) M 2 + 4M 3 + M 4 = h2 # 6( yn −3 − 2 yn − 2 + yn −1 ) M n −3 + 4 M n −2 + M n −1 = h2 6( yn − 2 − 2 yn −1 + y1 ) M1 + M n −2 + 4 M n −1 = . h2 The linear system for M1 , M 2 , ..., M n −1 in matrix form is ⎡ 4 1 0 0 " 0 0 0 1 ⎤ ⎡ M1 ⎤ ⎢1 4 1 0 " 0 0 0 0 ⎥ ⎢ M 2 ⎥ ⎥ ⎢ ⎥⎢ ⎢0 1 4 1 " 0 0 0 0 ⎥ ⎢ M 3 ⎥ # # # # ⎥⎢ # ⎥ ⎢# # # # ⎢0 0 0 0 " 0 1 4 1 ⎥ ⎢ M n −2 ⎥ ⎥ ⎢1 0 0 0 " 0 0 1 4 ⎥ ⎢ ⎣ ⎦ ⎣⎢ M n −1 ⎦⎥ ⎡ yn −1 − 2 y1 + y2 ⎤ ⎢ ⎥ y1 − 2 y2 + y3 ⎢ ⎥ 6 ⎢ y2 − 2 y3 + y4 ⎥ = . ⎥ # h2 ⎢ ⎢y − 2 yn −2 + yn −1 ⎥ ⎢ n −3 ⎥ ⎣⎢ yn − 2 − 2 yn −1 + y1 ⎦⎥

M − M2 M 2 − M1 = −4, a2 = 3 = 4, 6h 6h M M b1 = 1 = 0, b2 = 2 = −6, 2 2 y2 − y1 ( M 2 + 2M1 )h c1 = − = 3, h 6 y − y2 ( M 3 + 2 M 2 )h c2 = 3 − = 0, d1 = 0, h 6 d2 = 1 For 0 ≤ x ≤ .5, we have a1 =

S ( x) = S1 ( x) = −4 x3 + 3x. For .5 ≤ x ≤ 1, we have S ( x) = S2 ( x) = 4( x − .5)3 − 6( x − .5)2 + 1 = 4 x3 − 12 x 2 + 9 x − 1. The resulting natural spline is ⎧⎪−4 x3 + 3 x 0 ≤ x ≤ 0.5 S ( x) = ⎨ . 3 2 ⎪⎩4 x − 12 x + 9 x − 1 0.5 ≤ x ≤ 1 (b) Again h = .5 and x1 = .5, y1 = 1 x2 = 1, y2 = 0 x3 = 1.5, y3 = −1

4M 2 =

6( y1 − 2 y2 + y3 )

=0 h2 Thus M1 = M 2 = M 3 = 0, hence all ai and bi are also 0. y − y2 y −y = −2 c1 = 2 1 = −2, c2 = 3 h h d1 = y1 = 1, d 2 = y2 = 0 For .5 ≤ x ≤ 1, we have S ( x) = S1 ( x) = −2( x − .5) + 1 = −2 x + 2.

220


Section 10.5

8. (b) Equations (15) together with the two equations in part (a) give 6( y2 − y1 − hy1′ ) 2 M1 + M 2 = h2 6( y1 − 2 y2 + y3 ) M1 + 4 M 2 + M 3 = h2 6( y2 − 2 y3 + y4 ) M 2 + 4M 3 + M 4 = h2 # 6( yn − 2 − 2 yn −1 + yn ) M n − 2 + 4 M n −1 + M n = h2 6( yn −1 − yn + hyn′ ) M n −1 + 2M n = . h2 This linear system for M1 , M 2 , ..., M n in matrix form is ⎡2 1 0 0 " 0 0 0 0 ⎤ ⎡ M ⎤ ⎢1 4 1 0 " 0 0 0 0 ⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ M2 ⎥ ⎢0 1 4 1 " 0 0 0 0 ⎥ ⎢ M 3 ⎥ ⎢0 0 1 4 " 0 0 0 0 ⎥ ⎢ # ⎥ ⎢# # # # # # # # ⎥⎢ ⎥ ⎢0 0 0 0 " 1 4 1 0 ⎥ ⎢ M n− 2 ⎥ ⎢ ⎥ ⎢ M n −1 ⎥ ⎢0 0 0 0 " 0 1 4 1 ⎥ ⎢ ⎥ ⎢⎣0 0 0 0 " 0 0 1 2 ⎥⎦ ⎣ M n ⎦ ⎡ −hy1′ − y1 + y2 ⎤ ⎢ y − 2y + y ⎥ 2 3 ⎥ ⎢ 1 6 ⎢ y2 − 2 y3 + y4 ⎥ . = ⎥ # h2 ⎢ ⎢y − 2 yn −1 + yn ⎥ ⎥ ⎢ n−2 ⎣⎢ yn −1 − yn + hyn′ ⎦⎥

.6q1 − .5q2 = 0, or q1 =

5 q2 . Solutions are 6

⎡5⎤ thus of the form q = s ⎢ 6 ⎥ . Set ⎢⎣ 1 ⎥⎦

s=

1 5 6

+1

=

⎡5⎤ 6 to obtain q = ⎢ 11 ⎥ . 6⎥ 11 ⎢⎣ 11 ⎦

⎡. 7 ⎤ ⎡.23⎤ 2. (a) x(1) = Px(0) = ⎢.2 ⎥ ; likewise x( 2) = ⎢.52 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣.25⎦ ⎣ .1⎦ ⎡.273⎤ and x(3) = ⎢.396⎥ . ⎢ ⎥ ⎣ .331⎦ (b) P is regular because all of its entries are positive. To solve (I − P)q = 0, i.e. ⎡ .8 −.1 −.7 ⎤ ⎡ q1 ⎤ ⎡ 0 ⎤ ⎢ −.6 .6 −.2 ⎥ ⎢ q2 ⎥ = ⎢ 0 ⎥ , reduce the ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −.2 −.5 .9 ⎦ ⎣ q3 ⎦ ⎣ 0 ⎦ coefficient matrix to row-echelon form: 5 −9 ⎤ ⎡ 8 −1 −7 ⎤ ⎡2 ⎢ −6 6 −2⎥ → ⎢ 0 −21 29 ⎥ ⎢ ⎥ ⎢ ⎥ 0 0⎦ ⎣ −2 −5 9⎦ ⎣0 ⎡ 1 0 − 22 ⎤ 21 ⎥ ⎢ → ⎢ 0 1 − 29 ⎥ . 21 ⎢ ⎥ 0⎦ ⎣0 0 This yields solutions (setting q3 = s) of the ⎡ 22 ⎤ ⎢ 21 ⎥ form ⎢ 29 ⎥ s. ⎢ 21 ⎥ ⎣1⎦ To obtain a probability vector, take ⎡ 22 ⎤ ⎢ 72 ⎥ 1 21 s= = yielding q = ⎢ 29 ⎥. 72 22 + 29 + 1 72, ⎢ ⎥ 21 21 21 ⎢⎣ 72 ⎥⎦


⎡. 4 ⎤ ⎡.46 ⎤ 1. (a) x(1) = Px(0) = ⎢ ⎥ , x( 2) = Px(1) = ⎢ ⎥ . ⎣. 6 ⎦ ⎣.54 ⎦ Continuing in this manner yields ⎡.454⎤ . x(3) = ⎢ ⎣.546⎥⎦ ⎡.4546 ⎤ ⎡.45454⎤ . x ( 4) = ⎢ and x(5) = ⎢ ⎥ . 5454 ⎣ ⎦ ⎣.54546⎥⎦

3. (a) Solve (I − P)q = 0, i.e., ⎡ 2 − 3⎤ ⎡q ⎤ 4 ⎥ 1 = ⎡0⎤ . ⎢ 3 3 ⎥ ⎢ q ⎥ ⎢⎣ 0 ⎥⎦ 2 ⎢− 3 ⎣ 2⎦ 4⎦ ⎣ The only independent equation is

(b) P is regular because all of the entries of P are positive. Its steady-state vector q solves ⎡ .6 −.5⎤ ⎡ q1 ⎤ ⎡0 ⎤ . (I − P)q = 0; that is, ⎢ ⎢ ⎥= ⎣ −.6 .5⎥⎦ ⎣ q2 ⎦ ⎢⎣0 ⎥⎦ This yields one independent equation,

221



⎡ 1 n ⎤ ⎢ 2 x1 ⎥ n P x=⎢ ⎥. n x1 + x2 ⎥ ⎢ 1 − 12 ⎣ ⎦ Already true for n = 1, 2. If true for n − 1,

()

⎡9 ⎤ 2 3 q1 = q2 , yielding q = ⎢ 8 ⎥ s. Setting 3 4 ⎢⎣ 1 ⎥⎦

(

⎡9⎤ 8 s= yields q = ⎢ 17 ⎥ . 8⎥ 17 ⎢ 17 ⎣ ⎦

()

(

() ) ( )( ) ( () () () ( ())

⎡ 26 ⎤ ⎡ 26 ⎤ 19 to get q = ⎢ 45 ⎥ . q = ⎢ 19 ⎥ s. Set s = 45 ⎢ 19 ⎥ ⎢⎣ 1 ⎥⎦ ⎣ 45 ⎦ ⎡ 2 −1 0⎤ ⎡ q ⎤ 2 ⎡0 ⎤ ⎢ 3 ⎥ 1 1 − 14 ⎥ ⎢ q2 ⎥ = ⎢0 ⎥ by (c) Again, solve ⎢ − 13 ⎢ ⎥ ⎢ ⎥ ⎢ 1 1 1 ⎥ ⎣ q3 ⎦ ⎣ 0 ⎦ ⎢⎣ − 3 − 2 ⎥ 4⎦ reducing the coefficient matrix to row⎡1 0 − 1⎤ 4⎥ ⎢ echelon form: ⎢0 1 − 1 ⎥ yielding 3 ⎢ ⎥ 0⎦ ⎣0 0 ⎡1⎤ ⎢4⎥ solutions of the form q = ⎢ 1 ⎥ s. ⎢3⎥ ⎣1 ⎦

⎡1⎤ ⎢k ⎥ ⎢1⎥ 5. Let q = ⎢ k ⎥ . Then ⎢# ⎥ ⎢1⎥ ⎣k ⎦ 1 1 k 1 pij = ∑ pij = , k j =1 k j =1 j =1 k since the row sums of P are 1. Thus ( Pq)i = qi for all i. ( Pq ) i =

( n) 4. (a) Prove by induction that p12 = 0: Already

true for n = 1. If true for n − 1, we have P n = P n −1 P, so

( n) ( n −1) ( n −1) p12 = p11 p12 + p12 p22 . But

p12 =

= 0 so

)

(c) The Theorem says that the entries of the steady state vector should be positive; they ⎡0 ⎤ are not for ⎢ ⎥ . ⎣1 ⎦

⎡3⎤ ⎢ 19 ⎥ 12 4 ⎥. to get q = ⎢ 19 Set s = 19 ⎢ 12 ⎥ ⎢⎣ 19 ⎥⎦

( n) p12

)

then P n = P( P n −1x) ⎡ 1 n −1 ⎤ x1 ⎢ 2 ⎥ = P⎢ ⎥ n −1 x1 + x2 ⎥ ⎢ 1 − 12 ⎣ ⎦ ⎡ 1 1 n −1 ⎤ x1 ⎢ 2 2 ⎥ =⎢ ⎥ n −1 n x1 + x2 ⎥ + 12 ⎢ 1 − 12 ⎣ ⎦ n ⎡ 1 ⎤ ⎢ 2 x1 ⎥ =⎢ ⎥. n 1 x x + ⎢ 1− 2 1 2⎥ ⎣ ⎦ ⎡ 0 ⎤ ⎡0 ⎤ if x is a Since lim P n x = ⎢ ⎥= n→∞ ⎣ x1 + x2 ⎦ ⎢⎣1 ⎥⎦ state vector.

⎡ .19 −.26 ⎤ ⎡ q1 ⎤ ⎡0 ⎤ (b) As in (a), solve ⎢ ⎢ ⎥= ⎣ −.19 .26 ⎥⎦ ⎣ q2 ⎦ ⎢⎣0 ⎥⎦ i.e., .19q1 = .26q2 . Solutions have the form

( n −1) p12

()

k

k

∑ pij q j = ∑

6. Since P has zeros entries, consider ⎡1 1 1⎤ ⎢2 4 4⎥ P 2 = ⎢ 14 12 14 ⎥ , so P is regular. Note that all ⎢1 1 1⎥ ⎢⎣ 4 4 2 ⎥⎦ the rows of P sum to 1. Since P is 3 × 3, ⎡1 ⎤ ⎢3⎥ Exercise 5 implies q = ⎢ 13 ⎥ . ⎢1 ⎥ ⎢⎣ 3 ⎥⎦

= 0 + 0 = 0. Thus,

no power of P can have all positive entries, so P is not regular.

⎡ 1 x1 ⎤ ⎡x ⎤ ⎥, (b) If x = ⎢ 1 ⎥ , Px = ⎢ 2 1 ⎢⎣ 2 x1 + x2 ⎥⎦ ⎣ x2 ⎦ ⎡ 1 x1 ⎤ ⎥ etc. We use P2x = ⎢ 4 ⎢⎣ 14 x1 + 12 x1 + x2 ⎥⎦ induction to show 222


7. Let x = [ x1

Section 10.6

Section 10.6

x2 ]T be the state vector, with

x1 = probability that John is happy and

Exercise Set 10.6

x2 = probability that John is sad. The transition

1. Note that the matrix has the same number of rows and columns as the graph has vertices, and that ones in the matrix correspond to arrows in the graph.

⎡4 2⎤ matrix P will be P = ⎢ 5 3 ⎥ since the columns ⎢⎣ 15 13 ⎥⎦ must sum to one. We find the steady state vector ⎡ 1 − 2 ⎤ ⎡ q ⎤ ⎡0⎤ 3⎥ 1 = for P by solving ⎢ 5 , i.e., 2 ⎥ ⎢ q ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ − 15 ⎣ 2⎦ 3⎦ ⎡ 10 ⎤ 1 2 3 q1 = q2 , so q = ⎢ 3 ⎥ s. Let s = and get 5 3 13 ⎣⎢ 1 ⎦⎥ ⎡ 10 ⎤ 10 is the probability that John will q = ⎢ 13 ⎥ , so 3 13 ⎢ 13 ⎥ ⎣ ⎦ be happy on a given day. 8. The state vector x = [ x1 x2 x3 ]T will represent the proportion of the population living in regions 1, 2, and 3, respectively. In the transition matrix, pij will represent the proportion of the people in region j who move to ⎡.90 .15 .10 ⎤ region i, yielding P = ⎢.05 .75 .05⎥ . ⎢ ⎥ ⎣.05 .10 .85⎦

⎡0 ⎢1 (a) ⎢ ⎢1 ⎣⎢0

0 0 1 0

0 1 0 0

1⎤ 1⎥ ⎥ 1⎥ 0⎦⎥

⎡0 ⎢0 ⎢ (b) ⎢1 ⎢0 ⎢⎣0

1 0 0 0 0

1 0 0 1 1

0 0 1 0 0

0⎤ 1⎥ ⎥ 0⎥ 0⎥ 0 ⎥⎦

⎡0 ⎢1 ⎢ 0 (c) ⎢ ⎢0 ⎢0 ⎢0 ⎣

1 0 1 0 0 0

0 0 0 0 0 1

1 0 1 0 0 0

0 0 1 0 0 1

0⎤ 0⎥ ⎥ 1⎥ 1⎥ 1⎥ 0 ⎥⎦

2. See the remark in problem 1; we obtain

⎡ .10 −.15 −.10⎤ ⎡ q1 ⎤ ⎡0 ⎤ ⎢ −.05 .25 −.05⎥ ⎢ q2 ⎥ = ⎢0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −.05 −.10 .15⎦ ⎣ q3 ⎦ ⎣0 ⎦

(a)

P1

P2

⎡1 0 − 13 ⎤ 7⎥ ⎢ First reduce to row echelon form ⎢0 1 − 4 ⎥ , 7 ⎢ ⎥ ⎣0 0 0 ⎦ ⎡ 13 ⎤ ⎢7⎥ 7 and get yielding q = ⎢ 4 ⎥ s. Set s = 24 ⎢7⎥ ⎣1⎦ q=

⎡ 13 ⎤ ⎢ 24 ⎥ 4 ⎥, ⎢ 24 ⎢7⎥ ⎢⎣ 24 ⎥⎦

i.e., in the long run

(b)

P4

P3

P1

13 ⎛ 1 ⎞ ⎜ or 54 % ⎟ 24 ⎝ 6 ⎠ P5

of the people reside in region 1, in region 2, and

P4

4 24

2 ⎞ ⎛ ⎜ or 16 % ⎟ 3 ⎠ ⎝

P3

7 ⎛ 1 ⎞ ⎜ or 29 % ⎟ in region 3. 24 ⎝ 6 ⎠

223

P2


(c)

P1


P2

⎡1 ⎢0 ⎢ T 4. (a) M M = ⎢0 ⎢0 ⎢⎣0

P3

P6

P5

3. (a)

0 1 0 0 0

0 0 1 1 0

0 0 1 2 1

0⎤ 0⎥ ⎥ 0⎥ 1⎥ 2⎥⎦

(b) The kth diagonal entry of M T M is

P4

5

∑ mik 2 ,

P1

i.e., the sum of the squares of the

i =1

entries in column k of M. These entries are 1 if family member i influences member k and 0 otherwise. P4 P2

(c) The ij entry of M T M is the number of family members who influence both member i and member j.

P3

(b) m12 = 1, so there is one 1-step connection

5. (a) Note that to be contained in a clique, a vertex must have “two-way” connections with at least two other vertices. Thus, P4

from P1 to P2 . ⎡1 ⎢0 M =⎢ ⎢1 ⎢⎣1

2 1 1 1

1 1 1 0

1⎤ 1⎥ ⎥ and 0⎥ 1 ⎥⎦

⎡2 ⎢1 M3 = ⎢ ⎢1 ⎣⎢1

3 2 2 2

2 1 1 2

2⎤ 1⎥ ⎥. 2⎥ 1 ⎦⎥

2

could not be in a clique, so {P1 , P2 , P3} is the only possible clique. Inspection shows that this is indeed a clique. (b) Not only must a clique vertex have two-way connections to at least two other vertices, but the vertices to which it is connected must share a two-way connection. This consideration eliminates P1 and P2 , leaving

( 2) (3) = 2 and m12 = 3 meaning there So m12

{P3 , P4 , P5} as the only possible clique. Inspection shows that it is indeed a clique.

are two 2-step and three 3-step connections from P1 to P2 by Theorem 10.6.1. These are: 1-step: P1 → P2

(c) The above considerations eliminate P1 , P3

and P7 from being in a clique. Inspection

2-step: P1 → P4 → P2 and P1 → P3 → P2

shows that each of the sets {P2 , P4 , P6},

3-step: P1 → P2 → P1 → P2 ,

{P4 , P6 , P8}, {P2 , P6 , P8}, {P2 , P4 , P8}

P1 → P3 → P4 → P2 , and

and {P4 , P5 , P6 } satisfy conditions (i) and (ii) in the definition of a clique. But note that P8 can be added to the first set and we

P1 → P4 → P3 → P2 . ( 2) (3) (c) Since m14 = 1, m14 = 1 and m14 = 2, there

still satisfy the conditions. P5 may not be

are one 1-step, one 2-step and two 3-step connections from P1 to P4 . These are:

added, so {P2 , P4 , P6 , P8} is a clique, containing all the other possibilities except {P4 , P5 , P6 }, which is also a clique.

1-step: P1 → P4 2-step: P1 → P3 → P4 3-step: P1 → P2 → P1 → P4 and P1 → P4 → P3 → P4 .

224


Section 10.7

6. (a) With the given M we get ⎡0 1 0 1 0 ⎤ ⎢1 0 1 0 0 ⎥ ⎢ ⎥ S = ⎢0 1 0 0 1 ⎥ , ⎢1 0 0 0 1 ⎥ ⎢⎣0 0 1 1 0 ⎥⎦ ⎡0 3 1 3 1 ⎤ ⎢3 0 3 1 1 ⎥ ⎥ 3 ⎢ S = ⎢1 3 0 1 3 ⎥ . ⎢ 3 1 1 0 3⎥ ⎢⎣1 1 3 3 0 ⎥⎦

8. Associating vertex P1 with team A, P2 with

B, ..., P5 with E, the game results yield the following dominance-directed graph: P1

P2

P3

in the graph represented by M. 1 0 1 0 1 0

0 1 0 1 0 1

1 0 1 0 1 1

0 1 0 1 0 0

0⎤ 0⎥ ⎥ 1⎥ , 1⎥ 0⎥ 0 ⎥⎦

⎡0 6 1 7 0 2⎤ ⎢6 0 7 1 6 3⎥ ⎢ ⎥ 1 7 2 8 1 4⎥ S3 = ⎢ . ⎢7 1 8 2 7 5 ⎥ ⎢0 6 1 7 0 2⎥ ⎢2 3 4 5 2 2⎥ ⎣ ⎦ The elements along the main diagonal tell us that only P3 , P4 , and P6 are members of a clique. Since a clique contains at least three vertices, we must have {P3 , P4 , P6 } as the only clique. ⎡0 ⎢1 7. M = ⎢ ⎢0 ⎣⎢0

0 0 1 1

1 0 0 0

⎡0 2 ⎢0 Then M = ⎢ ⎢1 ⎢⎣1 ⎡0 ⎢1 M +M =⎢ ⎢1 ⎢⎣1 2

1⎤ 0⎥ ⎥ 1⎥ 0⎦⎥ 2 0 1 0 2 0 2 1

P4

which has vertex matrix ⎡0 1 1 1 0 ⎤ ⎢0 0 1 0 1 ⎥ ⎢ ⎥ M = ⎢0 0 0 1 1 ⎥ . ⎢0 1 0 0 0 ⎥ ⎢⎣1 0 0 1 0 ⎥⎦ ⎡0 1 1 1 2⎤ ⎢1 0 0 2 1 ⎥ ⎥ 2 ⎢ Then M = ⎢1 1 0 1 0 ⎥ , ⎢0 0 1 0 1 ⎥ ⎢⎣0 2 1 1 0 ⎥⎦ ⎡0 2 2 2 2 ⎤ ⎢1 0 1 2 2 ⎥ ⎥ 2 ⎢ M + M = ⎢1 1 0 2 1 ⎥ . ⎢0 1 1 0 1 ⎥ ⎢⎣1 2 1 2 0 ⎥⎦ Summing the rows, we get that the power of A is 8, of B is 6, of C is 5, of D is 3, and of E is 6. Thus ranking in decreasing order we get A in first place, B and E tie for second place, C in fourth place, and D last.

Since sii(3) = 0 for all i, there are no cliques ⎡0 ⎢1 ⎢ 0 (b) Here S = ⎢ ⎢1 ⎢0 ⎢0 ⎣

P5


0 1 0 0 1 1 0 0

1. (a) From Equation (2), the expected payoff of the game is ⎡1⎤ ⎢4⎥ 1⎤ ⎢ 1 ⎥ ⎡ −4 6 −4 pAq = ⎡⎣ 12 0 12 ⎤⎦ ⎢ 5 −7 3 8⎥ ⎢ 14 ⎥ ⎢ ⎥ ⎣ −8 0 6 −2 ⎦ ⎢ 4 ⎥ ⎢1⎥ ⎣⎢ 4 ⎦⎥ 5 =− . 8

1⎤ 1⎥ ⎥ and 0⎥ 0 ⎥⎦ 2⎤ 1⎥ ⎥. 1⎥ 0 ⎥⎦

By summing the rows of M + M 2 , we get that the power of P1 is 2 + 1 + 2 = 5, the power of P2 is 3, of P3 is 4, and of P4 is 2. 225


(b) If player R uses strategy [ p1

p2


p3 ]

(c) Here, a32 is a saddle point, so optimal ⎡0⎤ strategies are p* = [0 0 1], q* = ⎢1 ⎥ and ⎢ ⎥ ⎣0⎦

⎡1⎤ ⎢4⎥ ⎢1⎥ against player C’s strategy ⎢ 14 ⎥ his payoff ⎢4⎥ ⎢1⎥ ⎣⎢ 4 ⎦⎥

v = a32 = 2. (d) Here, a21 is a saddle point, so

⎛ 1⎞ ⎛9⎞ will be pAq = ⎜ − ⎟ p1 + ⎜ ⎟ p 2 − p3 . Since ⎝ 4⎠ ⎝4⎠ p1, p2 , and p3 are nonnegative and add up to 1, this is a weighted average of the 1 9 numbers − , , and −1. Clearly this is the 4 4 largest if p1 = p3 = 0 and p2 = 1; that is,

⎡1 ⎤ p* = [0 1 0 0], q* = ⎢ 0 ⎥ and ⎢ ⎥ ⎣0⎦

v = a21 = −2. 4. (a) Calling the matrix A, the formulas of Theorem 10.7.2 yield p* = ⎡ 85 83 ⎤ , ⎣ ⎦ 1 ⎡ ⎤ 27 q* = ⎢ 8 ⎥ , v = (A has no saddle points). 8 ⎢⎣ 78 ⎥⎦

p = [0 1 0]. (c) As in (b), if player C uses [ q1 q2

q3

q4 ]T against ⎡⎣ 12

0

1⎤, 2⎦

40 20 ⎤ = ⎡ 2 1 ⎤ , (b) As in (a), p* = ⎡ 60 60 ⎦ ⎣ 3 3 ⎦ ⎣ 10 ⎡ ⎤ ⎡1⎤ 1400 70 q* = ⎢ 60 ⎥ = ⎢ 6 ⎥ , v = = (Again, A 50 5 60 3 ⎢ ⎥ ⎢ ⎥ ⎣ 60 ⎦ ⎣ 6 ⎦ has no saddle points).

1 we get pAq = −6q1 + 3q2 + q3 − q4 . 2 Clearly this is minimized over all strategies by setting q1 = 1 and q2 = q3 = q4 = 0. That is q = [1 0 0 0]T . 2. As per the hint, we will construct a 3 × 3 matrix with two saddle points, say a11 = a33 = 1. Such a

(c) For this matrix, a11 is a saddle point, so ⎡1 ⎤ p* = [1 0], q* = ⎢ ⎥ , and v = a11 = 3. ⎣0⎦

⎡1 2 1 ⎤ matrix is A = ⎢ 0 7 0 ⎥ . Note that ⎢ ⎥ ⎣1 2 1 ⎦ a13 = a31 = 1 are also saddle points.

(d) This matrix has no saddle points, so, as in (a), p* = ⎡ −−53 −−25 ⎤ = ⎡ 35 25 ⎤ , ⎣ ⎦ ⎣ ⎦ − 3 3 ⎡ ⎤ ⎡ ⎤ −19 19 q* = ⎢ −5 ⎥ = ⎢ 5 ⎥ , and v = = . −5 5 ⎢⎣ −−25 ⎥⎦ ⎢⎣ 25 ⎥⎦

3. (a) Calling the matrix A, we see a22 is a saddle point, so the optimal strategies are pure, ⎡0⎤ namely: p* = [0 1], q* = ⎢ ⎥ , the value of ⎣1 ⎦ the game is v = a22 = 3.

(e) Again, A has no saddle points, so as in (a), ⎡1⎤ 3 10 ⎤ , q* = ⎢ 13 ⎥ , and v = −29 . p* = ⎡ 13 13 ⎦ ⎣ 13 ⎢⎣ 12 ⎥ 13 ⎦

(b) As in (a), a21 is a saddle point, so optimal ⎡1 ⎤ strategies are p* = [0 1 0], q* = ⎢ ⎥ , the ⎣0⎦ value of the game is v = a21 = 2.

5. Let a11 = payoff to R if the black ace and black two are played = 3. a12 = payoff to R if the black ace and red three are played = −4. a21 = payoff to R if the red four and black two are played = −6. a22 = payoff to R if the red four and red three 226


Section 10.8

⎡ 1 0 − 78 ⎤ ⎡ p ⎤ ⎡0⎤ 1 79 ⎥ ⎢ which reduces to ⎢0 1 − 54 ⎥ ⎢ p2 ⎥ = ⎢ 0 ⎥ . ⎥ ⎢ ⎥ 79 ⎢ ⎢ ⎥ 0 0 ⎦ ⎣ p3 ⎦ ⎣ ⎦ ⎣0 0 ⎡ 78 ⎤ ⎢ 79 ⎥ Solutions are of the form p = ⎢ 54 ⎥ . Let ⎢ 79 ⎥ ⎣1⎦ ⎡ 78 ⎤ s = 79 to obtain p = ⎢54 ⎥ . ⎢ ⎥ ⎣79 ⎦

are played = 7. So, the payoff matrix for the game is ⎡ 3 −4 ⎤ A=⎢ . ⎣ −6 7 ⎥⎦ A has no saddle points, so from Theorem 10.7.2, ⎡ 11 ⎤ 7 ⎤ , q* = ⎢ 20 ⎥ ; that is, player R p* = ⎡ 13 9 ⎥ ⎣ 20 20 ⎦ ⎢⎣ 20 ⎦ should play the black ace 65 percent of the time, and player C should play the black two 55 percent of the time. The value of the game is 3 − , that is, player C can expect to collect on 20 the average 15 cents per game.

2. (a) By Corollary 10.8.4, this matrix is productive, since each of its row sums is .9.

Section 10.8

(b) By Corollary 10.8.5, this matrix is productive, since each of its column sums is less than one.

Exercise Set 10.8 1. (a) Calling the given matrix E, we need to solve ⎡ 1 − 1 ⎤ ⎡ p ⎤ ⎡0 ⎤ 3⎥ 1 = ( I − E )p = ⎢ 2 . 1 ⎥ ⎢ p ⎥ ⎣⎢ 0 ⎦⎥ ⎢⎣ − 12 ⎣ 2⎦ 3⎦ ⎡1⎤ 1 1 This yields p1 = p2 , that is, p = s ⎢ 3 ⎥ . 2 3 ⎣2⎦ 2 ⎡ ⎤ Set s = 2 and get p = ⎢ ⎥ . ⎣3⎦

⎡2⎤ ⎡1.9 ⎤ (c) Try x = ⎢ 1 ⎥ . Then Cx = ⎢ .9 ⎥ , i.e., x > Cx, ⎢ ⎥ ⎢ ⎥ ⎣1 ⎦ ⎣ .9 ⎦ so this matrix is productive by Theorem 10.8.3. 3. Theorem 10.8.2 says there will be one linearly independent price vector for the matrix E if some positive power of E is positive. Since E is not

positive, try E 2 . ⎡.2 .34 .1⎤ E 2 = ⎢.2 .54 .6⎥ > 0 ⎢ ⎥ ⎣.6 .12 .3⎦

(b) As in (a), solve ⎡ 1 0 − 12 ⎤ ⎡ p ⎤ ⎡0⎤ ⎢ 2 ⎥ 1 1 − 12 ⎥ ⎢ p2 ⎥ = ⎢ 0 ⎥ . ( I − E )p = ⎢ − 13 ⎢ 1 ⎥⎢ ⎥ ⎢ ⎥ 1⎥ ⎣ p3 ⎦ ⎣ 0 ⎦ ⎢⎣ − 6 −1 ⎦ In row-echelon form, this reduces to ⎡ 1 0 −1⎤ ⎡ p1 ⎤ ⎡ 0 ⎤ ⎢0 1 − 5 ⎥ ⎢ p ⎥ = ⎢ 0 ⎥ . ⎢ 6⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣0 0 0 ⎥⎦ ⎣ p3 ⎦ ⎣ 0 ⎦

4. The exchange matrix for this arrangement (using ⎡1 1 1⎤ ⎢2 3 4⎥ A, B, and C in that order) is ⎢ 13 13 14 ⎥ . ⎢1 1 1⎥ ⎢⎣ 6 3 2 ⎥⎦ For equilibrium, we must solve (I − E)p = 0. ⎡ 1 − 1 − 1⎤ 3 4 ⎥ ⎡ p1 ⎤ ⎡ 0 ⎤ ⎢ 2 1 2 1 That is ⎢ − 3 − 4 ⎥ ⎢ p2 ⎥ = ⎢0 ⎥ . 3 ⎢ ⎥ ⎢ ⎥ ⎢ 1 1 1 ⎥ ⎣ p3 ⎦ ⎣ 0 ⎦ ⎢⎣ − 6 − 3 ⎥ 2⎦ Row reduction yields solutions of the form ⎡ 18 ⎤ ⎢ 16 ⎥ 1600 p = ⎢ 15 ⎥ s. Set s = and obtain 15 ⎢ 16 ⎥ ⎣1⎦

Solutions of this system have the form ⎡1⎤ ⎡6 ⎤ p = s ⎢⎢ 56 ⎥⎥ . Set s = 6 and get p = ⎢5 ⎥ . ⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎣6 ⎦

(c) As in (a), solve ⎡ .65 −.50 −.30 ⎤ ⎡ p1 ⎤ ⎡ 0⎤ ( I − E )p = ⎢ −.25 .80 −.30 ⎥ ⎢ p2 ⎥ = ⎢ 0⎥ , ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −.40 −.30 .60 ⎦ ⎣ p3 ⎦ ⎣ 0⎦

227



⎡120 ⎤ p = ⎢100 ⎥ ; i.e., the price of tomatoes was ⎢ ⎥ ⎣106.67 ⎦ $120, corn was $100, and lettuce was $106.67.

(b) The increase in value is the second column ⎡542 ⎤ 1 ⎢ of ( I − C )−1, 690⎥ . Thus the value of 503 ⎢170 ⎥ ⎣ ⎦ the coal-mining operation must increase by 542 . 503

5. Taking the CE, EE, and ME in that order, we form the consumption matrix C, where cij = the

amount (per consulting dollar) of the i-th engineer’s services purchased by the j-th ⎡ 0 .2 .3 ⎤ engineer. Thus, C = ⎢.1 0 .4 ⎥ . ⎢ ⎥ ⎣.3 .4 0 ⎦

n

7. The i-th column sum of E is

∑ e ji ,

and the

j =1

elements of the i-th column of I − E are the negatives of the elements of E, except for the ii-th, which is 1 − eii . So, the i-th column sum of

We want to solve (I − C)x = d, where d is the demand vector, i.e. ⎡ 1 −.2 −.3⎤ ⎡ x1 ⎤ ⎡500 ⎤ ⎢ −.1 1 −.4 ⎥ ⎢ x2 ⎥ = ⎢700 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −.3 −.4 1 ⎦ ⎣ x3 ⎦ ⎣600 ⎦

n

I − E is 1 − ∑ e ji = 1 − 1 = 0. Now, ( I − E )T has j =1

zero row sums, so the vector x = [1 1 " 1]T

In row-echelon form this reduces to ⎡1 −.2 −.3 ⎤ ⎡ x1 ⎤ ⎡ 500 ⎤ ⎢0 1 −.43877 ⎥ ⎢ x2 ⎥ = ⎢ 765.31⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ 1 ⎣0 0 ⎦ ⎣ x3 ⎦ ⎣1556.19⎦ Back-substitution yields the solution ⎡1256.48 ⎤ x = ⎢1448.12 ⎥ . ⎢ ⎥ ⎣1556.19 ⎦ The CE received $1256, the EE received $1448, and the ME received $1556.

solves ( I − E )T x = 0. This implies det( I − E )T = 0. But det( I − E )T = det( I − E ), so (I − E)p = 0 must have nontrivial (i.e., nonzero) solutions. The CE received $1256, the EE received $1448, and the ME received $1556.

8. Let C be a consumption matrix whose column sums are less than one; then the row sums of C T are less than one. By Corollary 10.8.4, C T is

6. (a) The solution of the system (I − C)x = d is

productive so ( I − C T )−1 ≥ 0. But

−1

x = ( I − C ) d. The effect of increasing the demand di for the ith industry by one unit

( I − C )−1 = ((( I − C )T ))−1 )T

= (( I − C T )−1 )T ≥ 0. Thus, C is productive.

⎡0 ⎤ ⎢# ⎥ ⎢ ⎥ ⎢0 ⎥ is the same as adding ⎢1 ⎥ to d where the 1 ⎢0 ⎥ ⎢# ⎥ ⎢ ⎥ ⎢⎣0 ⎥⎦ is in the ith row. The new solution is ⎛ ⎡0 ⎤ ⎞ ⎡ 0⎤ ⎜ ⎢# ⎥ ⎟ ⎢# ⎥ ⎜ ⎢ ⎥⎟ ⎢ ⎥ 0 ⎜ ⎢0 ⎥ ⎟ −1 −1 ⎢ ⎥ ( I − C ) ⎜ d + ⎢1 ⎥ ⎟ = x + ( I − C ) ⎢ 1 ⎥ ⎜ ⎢0 ⎥ ⎟ ⎢ 0⎥ ⎜ ⎢# ⎥ ⎟ ⎢# ⎥ ⎜ ⎢ ⎥ ⎟⎟ ⎢ ⎥ ⎜ ⎢⎣0 ⎥⎦ ⎠ ⎢⎣ 0⎥⎦ ⎝ which has the effect of adding the ith

Section 10.9 Exercise Set 10.9 1. Using Equation (18), we calculate 30s = 15s Yld 2 = 2 50s 100s . = Yld3 = 7 2+ 3 2

So all the trees in the second class should be harvested for an optimal yield (since s = 1000) of $15,000.

column of ( I − C ) −1 to the original solution. 228


Section 10.10

2. From the solution to Example 1, we see that for the fifth class to be harvested in the optimal case we must have p5 s > 14.7 s, yielding p5 > $222.63. ( .28−1 + .31−1 + .25−1 + .23−1 ) 3. Assume p2 = 1, then Yld 2 = p3 s

(.28

−1

+ .31−1 ) p4 s

s

(.28)−1

= .28s. Thus, for all the yields to be the same we must have

= .28s

(.28−1 + .31−1 + .25−1 ) p5 s

= .28s

(.28−1 + .31−1 + .25−1 + .23−1 ) p6 s

= .28s

= .28s (.28−1 + .31−1 + .25−1 + .23−1 + .37−1 ) Solving these successively yields p3 = 1.90, p4 = 3.02, p5 = 4.24 and p6 = 5.00. Thus the ratio p2 : p3 : p4 : p5 : p6 = 1 : 1.90 : 3.02 : 4.24 : 5.00. n

5. Since y is the harvest vector, N = ∑ yi is the number of trees removed from the forest. Then Equation (7) and i =1

the first of Equations (8) yield N = g1 x1 , and from Equation (17) we obtain N=

g1s

1+

g1 g2

+" +

g1 g k −1

=

s 1 g1

+" +

1 g k −1

.

6. Set g1 = " = g n −1 = g , and p2 = 1. Then from Equation (18), Yld 2 = to be the same, we need to solve Yld k =

pk s

(k − 1) g1

p2 s 1 g1

= gs. Since we want all of the Yld k ’s

= gs for pk for 3 ≤ k ≤ n. Thus pk = k − 1. So the ratio

p2 : p3 : p4 : " : pn = 1 : 2 : 3 : " : (n − 1).

Section 10.10 Exercise Set 10.10 ⎡0 1 1 0⎤ 1. (a) Using the coordinates of the points as the columns of a matrix we obtain ⎢0 0 1 1 ⎥ . ⎢ ⎥ ⎣0 0 0 0⎦

229



⎡ 3 0 0⎤ ⎢2 ⎥ (b) The scaling is accomplished by multiplication of the coordinate matrix on the left by ⎢ 0 1 0 ⎥ , resulting in 2 ⎢ ⎥ ⎣ 0 0 1⎦ ⎡0 3 3 0 ⎤ 2 2 ⎢ ⎥ ⎛3 ⎞ ⎛3 1 ⎞ ⎛ 1 ⎞ the matrix ⎢0 0 1 1 ⎥ , which represents the vertices (0, 0, 0), ⎜ , 0, 0 ⎟ , ⎜ , , 0 ⎟ and ⎜ 0, , 0 ⎟ as 2 2 ⎝2 ⎠ ⎝2 2 ⎠ ⎝ 2 ⎠ ⎢ ⎥ ⎣0 0 0 0 ⎦ shown below.

⎡ −2 −2 −2 −2 ⎤ ⎡ −2 −1 −1 −2 ⎤ (c) Adding the matrix ⎢ −1 −1 −1 −1⎥ to the original matrix yields ⎢ −1 −1 0 0 ⎥ , which represents ⎢ ⎥ ⎢ ⎥ ⎣ 3 3 3 3⎦ ⎣ 3 3 3 3⎦ the vertices (−2, −1, 3), (−1, −1, 3), (−1, 0, 3), and (−2, 0, 3) as shown below.

⎡cos(−30°) − sin(−30°) 0 ⎤ (d) Multiplying by the matrix ⎢ sin(−30°) cos(−30°) 0 ⎥ , we obtain ⎢ ⎥ 0 0 1⎦ ⎣ .866 1.366 .500 ⎤ ⎡0 cos(−30°) cos(−30°) − sin(−30°) − sin(−30°) ⎤ ⎡0 ⎢0 sin(−30°) cos(−30°) + sin(−30°) cos(−30°) ⎥ = ⎢0 −.500 .366 .866 ⎥ . ⎢ ⎥ ⎢ ⎥ 0 0 0 0 0 ⎣0 ⎦ ⎣0 0 ⎦ The vertices are then (0, 0, 0), (.866, −.500, 0), (1.366, .366, 0), and (.500, .866, 0) as shown:

⎡1 1 0 ⎤ ⎡ xi ⎤ ⎡ xi + 1 yi ⎤ 2 ⎥ 2 ⎢ 2. (a) Simply perform the matrix multiplication ⎢0 1 0 ⎥ ⎢ yi ⎥ = ⎢ yi ⎥ . ⎢ ⎥⎢ ⎥ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ zi ⎥⎦ ⎢⎣ zi ⎥⎦

230


Section 10.10

(b) We multiply ⎡1 1 0 ⎤ ⎡ 0 1 1 0 ⎤ ⎡ 0 1 3 1 ⎤ 2 2 2⎥ ⎢ ⎥ ⎢0 0 1 1 ⎥ = ⎢ 0 1 0 0 0 1 1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢⎣0 0 1 ⎥⎦ ⎣0 0 0 0 ⎦ ⎣⎢0 0 0 0 ⎦⎥ yielding the vertices (0, 0, 0), (1, 0, 0), ⎛3 ⎞ ⎛1 ⎞ ⎜ , 1, 0 ⎟ , and ⎜ , 1, 0 ⎟ . ⎝2 ⎠ ⎝2 ⎠

4. (a) The formulas for scaling, translation, and rotation yield the matrices ⎡ 1 0 0⎤ ⎢2 ⎥ M1 = ⎢ 0 2 0 ⎥ , ⎢0 0 1⎥ 3⎦ ⎣ ⎡1 2 M2 = ⎢0 ⎢ ⎣⎢ 0

(c) Obtain the vertices via ⎡ 1 0 0 ⎤ ⎡0 1 1 0 ⎤ ⎡0 1 1 0⎤ ⎢.6 1 0 ⎥ ⎢0 0 1 1 ⎥ = ⎢0 .6 1.6 1 ⎥ , ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 0 1 ⎦ ⎣0 0 0 0 ⎦ ⎣0 0 0 0⎦ yielding (0, 0, 0), (1, .6, 0), (1, 1.6, 0), and (0, 1, 0), as shown:

1 2

0 0

" 12 ⎤ ⎥ 0 " 0⎥ , 0 " 0⎦⎥

1 2

1 0 ⎤ ⎡1 M 3 = ⎢0 cos 20° − sin 20° ⎥ , ⎢ ⎥ ⎣0 sin 20° cos 20° ⎦

⎡ cos(−45°) 0 sin(−45)° ⎤ ⎥ , and M4 = ⎢ 0 1 0 ⎢ ⎥ sin( 45 ) 0 cos( 45 ) − − ° − ° ⎣ ⎦ cos 90 ° − sin 90 ° 0 ⎡ ⎤ M 5 = ⎢ sin 90° cos 90° 0 ⎥ . ⎢ ⎥ 0 1⎦ ⎣ 0

(b) Clearly P ′ = M 5 M 4 M 3 ( M1 P + M 2 ). 3. (a) This transformation looks like scaling by the factors 1, −1, 1, respectively and indeed its ⎡ 1 0 0⎤ matrix is ⎢0 −1 0 ⎥ . ⎢ ⎥ ⎣0 0 1⎦

⎡. 3 0 0 ⎤ 5. (a) As in 4(a), M1 = ⎢ 0 .5 0⎥ , ⎢ ⎥ ⎣ 0 0 1⎦ 0 0 ⎤ ⎡1 M 2 = ⎢0 cos 45° − sin 45° ⎥ , ⎢ ⎥ ⎣0 sin 45° cos 45° ⎦ ⎡1 1 1 " 1 ⎤ M 3 = ⎢0 0 0 " 0⎥ , ⎢ ⎥ ⎣0 0 0 " 0⎦ ⎡ cos 35° 0 sin 35° ⎤ M4 = ⎢ 0 1 0 ⎥, ⎢ ⎥ sin 35 0 cos 35°⎦ − ° ⎣ ⎡cos(−45°) − sin(−45°) 0 ⎤ M 5 = ⎢ sin(−45°) cos(−45°) 0 ⎥ , ⎢ ⎥ 0 0 1⎦ ⎣ ⎡0 0 0 " 0 ⎤ M 6 = ⎢0 0 0 " 0 ⎥ , and ⎢ ⎥ ⎣1 1 1 " 1 ⎦ ⎡2 0 0⎤ M 7 = ⎢0 1 0⎥ . ⎢ ⎥ ⎣0 0 1 ⎦

(b) For this reflection we want to transform ( xi , yi , zi ) to (− xi , yi , zi ) with the matrix ⎡ −1 0 0 ⎤ ⎢ 0 1 0⎥ . ⎢ ⎥ ⎣ 0 0 1⎦ Negating the x-coordinates of the 12 points in view 1 yields the 12 points (−1.000, −.800, .000), (−.500, −.800, −.866), etc., as shown:

(c) Here we want to negate the z-coordinates, ⎡ 1 0 0⎤ with the matrix ⎢0 1 0 ⎥ . ⎢ ⎥ ⎣0 0 −1⎦ This does not change View 1.

(b) As in 4(b), P ′ = M 7 ( M 6 + M 5 M 4 ( M 3 + M 2 M1P)).

231



6. Using the hing given, we have ⎡ cos β 0 sin β ⎤ R1 = ⎢ 0 1 0 ⎥, ⎢ ⎥ ⎣ − sin β 0 cos β ⎦ ⎡cos α − sin α 0 ⎤ R2 = ⎢ sin α cos α 0 ⎥ , ⎢ ⎥ 0 1⎦ ⎣ 0 ⎡ cos θ 0 sin θ ⎤ R3 = ⎢ 0 1 0 ⎥, ⎢ ⎥ sin 0 cos − θ θ⎦ ⎣ ⎡ cos(−α ) − sin(−α ) 0 ⎤ R4 = ⎢ sin(−α ) cos(−α ) 0 ⎥ , and ⎢ ⎥ 0 1⎦ ⎣ 0

Section 10.11 Exercise Set 10.11 1. (a) The discrete mean value property yields the four equations 1 t1 = (t2 + t3 ) 4 1 t2 = (t1 + t4 + 1 + 1) 4 1 t3 = (t1 + t4 ) 4 1 t4 = (t2 + t3 + 1 + 1). 4 Translated into matrix notation, this ⎡0 1 1 0⎤ 0 4 4 ⎡ t1 ⎤ ⎢ ⎥ ⎡ t1 ⎤ ⎡⎢ ⎤⎥ 1 1 ⎢t ⎥ ⎢ 0 0 4 ⎥ ⎢t2 ⎥ 12 becomes ⎢ 2 ⎥ = ⎢ 14 ⎢ ⎥ + ⎢ ⎥. 1⎥ t ⎢0⎥ t 0 0 3 ⎢ 3⎥ ⎢4 4⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣t4 ⎥⎦ ⎢ ⎢ ⎥ t ⎥ 1 1 ⎣ 4 ⎦ ⎢⎣ 12 ⎥⎦ ⎢⎣ 0 4 4 0 ⎥⎦

⎡ cos(− β ) 0 sin(− β ) ⎤ R5 = ⎢ 0 1 0 ⎥. ⎢ ⎥ ⎣ − sin(− β ) 0 cos(− β ) ⎦

7. (a) We rewrite the formula for vi′ as ⎡1 ⋅ xi + x0 ⋅1⎤ ⎢1 ⋅ y + y ⋅1⎥ 0 ⎥ . So vi′ = ⎢ i z z ⋅ + 1 ⎢ i 0 ⋅1 ⎥ ⎢⎣ ⎥⎦ 1 ⋅1 ⎡1 ⎢0 vi′ = ⎢ ⎢0 ⎢⎣0

0 1 0 0

0 0 1 0

(b) To solve the system in part (a), we solve (I − M)t = b for t: ⎡ 1 −1 −1 0⎤ 0 4 4 ⎢ ⎥ ⎡ t1 ⎤ ⎡⎢ ⎤⎥ 1 1 1 ⎢ ⎥ 1 0 − 4 ⎥ t2 ⎢− 4 ⎢2⎥. ⎢ ⎥ = ⎢ 1 ⎥ 0 1 − 14 ⎥ ⎢ t3 ⎥ ⎢ 0 ⎥ ⎢− 4 ⎢ ⎥ ⎢t ⎥ ⎢ 1 ⎥ 1 1 1⎥⎦ ⎣ 4 ⎦ ⎣⎢ 2 ⎦⎥ ⎢⎣ 0 − 4 − 4 In row-echelon form, this is ⎡1 − 1 − 1 0 ⎤ ⎡ t1 ⎤ ⎡ 0 ⎤ 4 4 ⎢ ⎥ ⎢ ⎥ 1 −15 4 ⎥ ⎢t2 ⎥ ⎢ 0 ⎥ ⎢0 ⎢ ⎥= 1 . ⎢0 0 1 − 12 ⎥ ⎢ t3 ⎥ ⎢ − 8 ⎥ ⎢ ⎥ ⎢t ⎥ ⎢ 3 ⎥ 0 0 1⎦⎥ ⎣ 4 ⎦ ⎣⎢ 4 ⎦⎥ ⎣⎢0

x0 ⎤ ⎡ xi ⎤ y0 ⎥ ⎢ yi ⎥ ⎥ ⎢ ⎥. z0 ⎥ ⎢ zi ⎥ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦

(b) We want to translate xi by −5, yi by +9, zi by −3, so x0 = −5, y0 = 9, z0 = −3. ⎡1 ⎢0 The matrix is ⎢ ⎢0 ⎢⎣0

0 1 0 0

0 −5⎤ 0 9⎥ ⎥. 1 −3⎥ 0 1⎥⎦

⎡1⎤ ⎢4⎥ ⎢3⎥ Back substitution yields the result t = ⎢ 14 ⎥ . ⎢4⎥ ⎢3⎥ ⎣⎢ 4 ⎦⎥

8. This can be done most easily performing the multiplication RRT and showing that this is I. For example, for the rotation matrix about the xaxis we obtain 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 RRT = ⎢ 0 cos θ − sin θ ⎥ ⎢0 cos θ sin θ ⎥ ⎢ ⎥⎢ ⎥ ⎣ 0 sin θ cos θ ⎦ ⎣0 − sin θ cos θ ⎦ ⎡1 0 0 ⎤ = ⎢0 1 0⎥ . ⎢ ⎥ ⎣0 0 1 ⎦

232


(c) t (1) = Mt (0) + b ⎡0 1 1 4 4 ⎢ ⎢ 14 0 0 =⎢ 1 ⎢4 0 0 ⎢ 1 1 ⎣⎢ 0 4 4 ⎡ 0⎤ ⎢1⎥ = ⎢2⎥ ⎢ 0⎥ ⎢1⎥ ⎢⎣ 2 ⎥⎦

Section 10.12

−.0371 × 100% = −12.9%, and for .2871 .0371 t2 and t4 was × 100% = 5.2%. .7129 t3 was

0⎤ 0 ⎥ ⎡0 ⎤ ⎡ ⎤ 1⎥ ⎢ ⎥ ⎢1⎥ 4 0 + ⎢2⎥ 1 ⎥ ⎢0 ⎥ ⎢ 0 ⎥ 4⎥ ⎢ ⎥ ⎢1⎥ ⎥ ⎢0 ⎥ 0 ⎦⎥ ⎣ ⎦ ⎣⎢ 2 ⎦⎥

2. The average value of the temperature on the 1 π f (θ )r dθ , where r is the radius circle is 2π r ∫−π of the circle and f(θ) is the temperature at the point of the circumference where the radius to that point makes the angle θ with the horizontal. −π π <θ < and is zero Clearly f(θ) = 1 for 2 2 otherwise. Consequently, the value of the integral above (which equals the temperature at 1 the center of the circle) is . 2

⎡1⎤ ⎢8⎥ ⎢5⎥ t ( 2) = Mt(1) + b = ⎢ 81 ⎥ , ⎢8⎥ ⎢5⎥ ⎢⎣ 8 ⎥⎦ ⎡3⎤ ⎢ 16 ⎥ ⎢ 11 ⎥ t (3) = Mt ( 2) + b = ⎢ 16 ⎥ 3 ⎢ 16 ⎥ ⎢ 11 ⎥ ⎣⎢ 16 ⎦⎥

3. As in 1(c), but using M and b as in the problem statement, we obtain t (1) = Mt (0) + b

= ⎡⎣ 34

5 4

1 2

5 4

1 2

5 4

13 16

7 16

1

3⎤ 4⎦

1

T

t ( 2) = Mt(1) + b

⎡7⎤ ⎢ 32 ⎥ ⎢ 23 ⎥ t ( 4) = Mt(3) + b = ⎢ 32 ⎥ , 7 ⎢ 32 ⎥ ⎢ 23 ⎥ ⎢⎣ 32 ⎥⎦

13 = ⎡ 16 ⎣

9 8

9 16

11 8

21 16

1

15 ⎤ 8⎦

T

.


⎡ 15 ⎤ ⎢ 64 ⎥ ⎢ 47 ⎥ t (5) = Mt( 4) + b = ⎢ 64 ⎥ 15 ⎢ 64 ⎥ ⎢ 47 ⎥ ⎣⎢ 64 ⎦⎥

1. (c) The linear system 1 * * * x31 = [ 28 + x31 − x32 ] 20 1 * * * x32 = [ 24 + 3x31 − 3x32 ] 20 can be rewritten as

⎡ 15 ⎤ ⎡ 1 ⎤ ⎡ − 1 ⎤ ⎢ 64 ⎥ ⎢ 4 ⎥ ⎢ 64 ⎥ ⎢ 47 ⎥ ⎢ 3 ⎥ ⎢ − 1 ⎥ t (5) − t = ⎢ 64 ⎥ − ⎢ 4 ⎥ = ⎢ 64 ⎥ 15 1 1 ⎢ 64 ⎥ ⎢ 4 ⎥ ⎢ − 64 ⎥ ⎢ 47 ⎥ ⎢ 3 ⎥ ⎢ 1 ⎥ − ⎣⎢ 64 ⎦⎥ ⎢⎣ 4 ⎥⎦ ⎣⎢ 64 ⎦⎥

* * 19 x31 + x32 = 28 * * −3x31 + 23x32 = 24, which has the solution 31 * x31 = 22 27 * = . x32 22

(d) Using percentage error computed value − actual value = × 100% actual value we have that the percentage error for t1 and

2. (a) Setting

(

) (

)

(1) (1) = (0) (0) = (0, 0), and x(01) = x01 x31 , x32 , x02

using part (b) of Exercise 1, we have

233



1 [ 28] = 1.40000 20 1 (1) x32 = [ 24] = 1.20000 20 1 ( 2) x31 = [ 28 + 1.4 − 1.2] = 1.41000 20 1 ( 2) x32 = [ 24 + 3(1.4) − 3(1.2)] = 1.23000 20 1 (3) x31 = [ 28 + 1.41 − 1.23] = 1.40900 20 1 (3) x32 = [ 24 + 3(1.41) − 3(1.23)] = 1.22700 20 1 ( 4) x31 = [ 28 + 1.409 − 1.227] = 1.40910 20 1 ( 4) x32 = [ 24 + 3(1.409) − 3(1.227)] 20 = 1.22730 1 (5) x31 = [ 28 + 1.4091 − 1.2273] = 1.40909 20 1 (5) x32 = [ 24 + 3(1.4091) − 3(1.2273)] 20 = 1.22727 1 ( 6) x31 = [ 28 + 1.40909 − 1.22727] 20 = 1.40909 1 ( 6) x32 = [ 24 + 3(1.40909) − 3(1.22727)] 20 = 1.22727. (1) x31 =

(

)

( 0) (0) (b) x(01) = (1, 1) = x31 , x32 1 (1) x31 = [ 28 + 1 − 1] = 1.4 20 1 ( 2) x32 = [ 24 + 3(1) − 3(1)] = 1.2 20

Since x(31) in this part is the same as x(31) in part (a), we will get x(32) as in part (a) and therefore x(33) , ..., x(36) will also be the same as in part (a).

(

( 0) (0) (c) x(01) = (148, − 15) = x31 , x32

)

1 [ 28 + 148 − (−15)] = 9.55000 20 1 (1) x32 = [ 24 + 3(148) − 3(−15)] = 25.65000 20 1 ( 2) x31 = [ 28 + 9.55 − 25.65] = 0.59500 20 (1) x31 =

234


Section 10.12

1 [ 24 + 3(9.55) − 3(25.65)] 20 = −1.21500 1 = [ 28 + 0.595 + 1.215] 20 = 1.49050 1 = [ 24 + 3(0.595) + 3(1.215)] 20 = 1.47150 1 = [ 28 + 1.4905 − 1.4715] = 1.40095 20 1 = [ 24 + 3(1.4905) − 3(1.4715)] 20 = 1.20285 1 = [ 28 + 1.40095 − 1.20285] 20 = 1.40991 1 = [ 24 + 3(1.40095) − 3(1.20285)] 20 = 1.22972 1 = [ 28 + 1.40991 − 1.22972] = 1.40901 20 1 = [ 24 + 3(1.40991) − 3(1.22972)] 20 = 1.22703

( 2) x32 =

(3) x31

(3) x32

( 4) x31 ( 4) x32

(5) x31

(5) x32

( 6) x31 ( 6) x32

4. Referring to the figure below and starting with x(01) = (0, 0) : x(01) is projected to x1(1) on L1 , x1(1) is projected to x(21) on L2 , x(21) is projected to x(31) on L3 , and so on. x2

L1

L3 (2)

(1)

x1

x1

L2

A

x2 = 1 x(1) 3 x1

(1) x0

B (1)

x2 x1 – x2 = 0

x1 – x2 = 2

As seen from the graph the points of the limit cycle are x1∗ = A, x∗2 = B, x∗3 = A. Since x1∗ is the point of intersection of L1 and L3 it follows on solving the system ∗ x12 =1

∗ * x11 − x12 =0 ∗ ∗ ∗ ∗ , x22 ) is on L2 , it follows that x21 − x22 = 2. Now x1∗ x∗2 is perpendicular to L2 , that x1* = (1, 1). Since x∗2 = ( x21

235



⎛ x∗ − 1 ⎞ ∗ ∗ ∗ ∗ therefore ⎜ 22 ⎟ (1) = −1 so we have x22 − 1 = 1 − x21 or x21 + x22 = 2. ⎜ x∗ − 1 ⎟ ⎝ 21 ⎠ Solving the system ∗ ∗ x21 − x22 =2

∗ ∗ x21 + x22 =2 ∗ ∗ = 2 and x22 = 0. Thus the points on the limit cycle are x1∗ = (1, 1), x∗2 = (2, 0), x∗3 = (1, 1). gives x21

7. Let us choose units so that each pixel is one unit wide. Then aij = length of the center line of the i-th beam that lies in the j-th pixel. If the i-th beam crosses the j-th pixel squarely, it follows that aij = 1. From Fig. 10.12.11 in the text, it is then clear that a17 = a18 = a19 = 1 a24 = a25 = a26 = 1 a31 = a32 = a33 = 1 a73 = a76 = a79 = 1 a82 = a85 = a88 = 1 a91 = a94 = a97 = 1 since beams 1, 2, 3, 7, 8, and 9 cross the pixels squarely. Next, the centerlines of beams 5 and 11 lie along the diagonals of pixels 3, 5, 7 and 1, 5, 9, respectively. Since these diagonals have length 2, we have a53 = a55 = a57 = 2 = 1.41421 a11, 1 = a11, 5 = a11, 9 = 2 = 1.41421. In the following diagram, the hypotenuse of triangle A is the portion of the centerline of the 10th beam that lies in the 2nd pixel. The length of this hypotenuse is twice the height of triangle A, which in turn is 2 − 1. Thus,

a10, 2 = 2

(

)

2 − 1 = .82843.

By symmetry we also have a10, 2 = a10, 6 = a12, 4 = a12, 8 = a62 = a64 = a46 = a48 = .82843. Also from the diagram, we see that the hypotenuse of triangle B is the portion of the centerline of the 10th beam that lies in the 3rd pixel. Thus, a10, 3 = 2 − 2 = .58579.

236


Section 10.12

2−1 2 1

A 2( 2 − 1)

B 1

2

2− 2

1 − 2

2 1−

1 − 2

2

1

By symmetry we have a10, 3 = a12, 7 = a61 = a49 = .58579. The remaining aij ’s are all zero, and so the 12 beam equations (4) are

x7 + x8 + x9 = 13.00 x4 + x5 + x6 = 15.00 x1 + x2 + x3 = 8.00 .82843( x6 + x8 ) + .58579 x9 = 14.79 1.41421( x3 + x5 + x7 ) = 14.31 .82843( x2 + x4 ) + .58579 x1 = 3.81 x3 + x6 + x9 x2 + x5 + x8 x1 + x4 + x7 .82843( x2 + x6 ) + .58579 x3

= 18.00 = 12.00 = 6.00 = 10.51

1.41421( x1 + x5 + x9 ) = 16.13 .82843( x4 + x8 ) + .58579 x7 = 7.04

8. Let us choose units so that each pixel is one unit wide. Then aij = area of the i-th beam that lies in the j-th pixel. Since the width of each beam is also one unit it follows that aij = 1 if the i-th beam crosses the j-th pixel squarely. From Fig. 10.12.11 in the text, it is then clear that a17 = a18 = a19 = 1 a24 = a25 = a26 = 1 a31 = a32 = a33 = 1 a73 = a76 = a79 = 1 a82 = a85 = a88 = 1 a91 = a94 = a97 = 1 since beams 1, 2, 3, 7, 8, and 9 cross the pixels squarely. For the remaining aij ’s, first observe from the figure that an isosceles right triangle of height h, as indicated, has area h 2 . From the diagram of the nine pixels, we then have

237



h

Area = h 2

1 − 2

( 2 − 1)

2

B

1

F

D

A

E

C 1 − 2

B

3 − 2

3 −( 2 3 − 2

2 1

⎡1 ⎤ Area of triangle B = ⎢ ( 2 − 1) ⎥ ⎣2 ⎦ 1 = (3 − 2 2 ) 4 = 0.4289

2

2 1 ⎡1⎤ Area of triangle C = ⎢ ⎥ = = .25000 4 ⎣2⎦ 2

⎡3 ⎤ Area of triangle F = ⎢ ( 2 − 1) ⎥ ⎣2 ⎦ 9 = (3 − 2 2 ) 4 = .38604 We also have Area of polygon A = 1 − 2 × (Area of triangle B) 1 = 2− 2 = .91421

238

2 − 1)


Section 10.12

Area of polygon D = 1 − (Area of triangle C ) 1 = 1− 4 3 = 4 = .7500 Area of polygon E = 1 − (Area of triangle F ) 1 = (18 2 − 23) 4 = .61396 Referring back to Fig. 10.12.11, we see that a11, 1 = Area of polygon A = .91421

a10, 1 = Area of triangle B = .04289 a11, 2 = Area of triangle C = .25000 a10, 2 = Area of polygon D = .75000 a10, 3 = Area of polygon E = .61396 By symmetry we then have a11, 1 = a11, 5 = a11, 9 = a53 = a55 = a57 = .91421 a10, 1 = a10, 5 = a10, 9 = a12, 1 = a12, 5 = a12, 9 = a63 = a65 = a67 = a43 = a45 = a47 = .04289

a11, 2 = a11, 4 = a11, 6 = a11, 8 = a52 = a54 = a56 = a58 = .25000 a10, 2 = a10, 6 = a12, 4 = a12, 8 = a62 = a64 = a46 = a48 = .75000 a10, 3 = a12, 7 = a61 = a49 = .61396. The remaining aij ’s are all zero, and so the 12 beam equations (4) are

x7 + x8 + x9 = 13.00 x4 + x5 + x6 = 15.00 x1 + x2 + x3 = 8.00 0.04289( x3 + x5 + x7 ) + 0.75( x6 + x8 ) + 0.61396 x9 = 14.79 0.91421( x3 + x5 + x7 ) + 0.25( x2 + x4 + x6 + x8 ) = 14.31 0.04289( x3 + x5 + x7 ) + 0.75( x2 + x4 ) + 0.61396 x1 = 3.81 x3 + x6 + x9 = 18.00 x2 + x5 + x8 = 12.00 x1 + x4 + x7 = 6.00 0.04289( x1 + x5 + x9 ) + 0.75( x2 + x6 ) + 0.61396 x3 = 10.51 0.91421( x1 + x5 + x9 ) + 0.25( x2 + x4 + x6 + x8 ) = 16.13 0.04289( x1 + x5 + x9 ) + 0.75( x4 + x8 ) + 0.61396 x7 = 7.04

239



Section 10.13 Exercise Set 10.13 12 . Also, the 25 rotation angles for the four subsets are all 0°. The displacement distances can be determined from the figure to find the four similitudes that map the entire set onto the four subsets S1 , S2 , S3 , S4 . These are, respectively,

1. Each of the subsets S1 , S2 , S3 , S4 in the figure is congruent to the entire set scaled by a factor of

y

1

12 25

1 25

S3

S4

S1

S2 x

1 ⎡ 13 ⎤ ⎤ ⎡0⎤ ⎛ ⎡ x ⎤ ⎞ 12 ⎡1 0 ⎤ ⎡ x ⎤ ⎡ ei ⎤ ⎡ ei ⎤ ⎡0 ⎤ ⎡ 13 25 ⎢ 25 ⎥ . , , , and Ti ⎜ ⎢ ⎥ ⎟ = + i = 1, 2, 3, 4, where the four values of , are ⎢f ⎥ ⎢⎣0 ⎥⎦ ⎢ 0 ⎥ ⎢ 13 ⎥ ⎢⎣ 0 1 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢ f ⎥ y 25 ⎣ ⎦ ⎢ 13 ⎥ ⎣ i⎦ ⎣ i⎦ ⎝ ⎠ ⎣⎢ ⎦⎥ ⎣⎢ 25 ⎦⎥ ⎣ 25 ⎦ 12 Because s = and k = 4 in the definition of a self-similar set, the Hausdorff dimension of the set is 25 ln(k ) ln(4) d H (S ) = = = 1.888.... The set is a fractal because its Hausdorff dimension is not an integer. 1 25 ln s ln 12

()

2.

( )

15 ( 16 ) The rough measurements indicated in the figure give an approximate scale factor of s ≈ = .47 to two

decimal places. Since k = 4, the Hausdorff dimension of the set is approximately d H ( S ) =

2 ln(k ) ln

() 1 s

=

ln(4) ln

( .471 )

= 1 .8

to two significant digits. Examination of the figure reveals rotation angles of 180°, 180°, 0°, and −90° for the sets S1 , S2 , S3 , and S4 , respectively.

240


Section 10.13

15 " 16

s3

s4

s1

s2 2"

3. By inspection, reading left to right and top to bottom, the triplets are: (0, 0, 0) none are rotated (1, 0, 0) the upper right iteration is rotated 90° (2, 0, 0) the upper right iteration is rotated 180° (3, 0, 0) the upper right iteration is rotated 270° (0, 0, 1) the lower right iteration is rotated 90° (0, 0, 2) the lower right iteration is rotated 180° (1, 2, 0) the upper right iteration is rotated 90° and the lower left is rotated 180° (2, 1, 3) the upper right iteration is rotated 180°, the lower left is rotated 90°, and the lower right is rotated 270° (2, 0, 1) the upper right iteration is rotated 180° and the lower right is rotated 90° (2, 0, 2) the upper right and lower right iterations are both rotated 180° (2, 2, 0) the upper right and lower left iterations are both rotated 180° (0, 3, 3) the lower left and lower right iterations are both rotated 270° 4. (a) The figure shows the original self-similar set and a decomposition of the set into seven nonoverlapping 1 congruent subsets, each of which is congruent to the original set scaled by a factor s = . By inspection, the 3 rotations angles are 0° for all seven subsets. The Hausdorff dimension of the set is ln(k ) ln(7) d H (S ) = = = 1.771.... Because its Hausdorff dimension is not an integer, the set is a fractal. ln(3) ln 1

(s)

241



(b) The figure shows the original self-similar set and a decomposition of the set into three nonoverlapping 1 congruent subsets, each of which is congruent to the original set scaled by a factor s = . By inspection, the 2 rotation angles are 180° for all three subsets. The Hausdorff dimension of the set is ln(k ) ln(3) d H (S ) = = = 1.584.... Because its Hausdorff dimension is not an integer, the set is a fractal. ln(2) ln 1

(s)

(c) The figure shows the original self-similar set and a decomposition of the set into three nonoverlapping 1 congruent subsets, each of which is congruent to the original set scaled by a factor s = . By inspection, the 2 rotation angles are 180°, 180°, and −90° for S1 , S2 , and S3 , respectively. The Hausdorff dimension of the set is d H ( S ) =

ln(k )

()

ln 1s

=

ln(3) = 1.584.... Because its Hausdorff dimension is not an integer, the set is a fractal. ln(2)

S3

S1

242

S2


Section 10.13

(d) The figure shows the original self-similar set and a decomposition of the set into three nonoverlapping 1 congruent subsets, each of which is congruent to the original set scaled by a factor s = . By inspection, the 2 rotation angles are 180°, 180°, and −90° for S1 , S2 , and S3 , respectively. The Hausdorff dimension of the set is d H ( S ) =

ln(k ) ln

() 1 s

=

ln(3) = 1.584.... Because its Hausdorff dimension is not an integer, the set is a fractal. ln(2)

S3

S2

S1

⎡ .85 .04⎤ 5. The matrix of the affine transformation in question is ⎢ . The matrix portion of a similitude is of the ⎣ −.04 .85⎥⎦ ⎡cos θ − sin θ ⎤ form s ⎢ . Consequently, we must have s cos θ = .85 and cos θ ⎥⎦ ⎣ sin θ

s sin θ = −.04. Solving this pair of equations gives s = (.85)2 + (−.04)2 = .8509... and

−.04 ⎞ ⎟ = −2.69...°. ⎝ .85 ⎠

θ = tan −1 ⎛⎜

⎛ ⎡ x⎤ ⎞ ⎡ x⎤ ⎡x⎤ 6. Letting ⎢ ⎥ be the vector to the tip of the fern and using the hint, we have ⎢ ⎥ = T2 ⎜ ⎢ ⎥ ⎟ or ⎣ y⎦ ⎣ y⎦ ⎝ ⎣ y⎦ ⎠ ⎡ x ⎤ ⎡ .85 .04 ⎤ ⎡ x ⎤ ⎡.075⎤ ⎢⎣ y ⎥⎦ = ⎢⎣ −.04 .85⎥⎦ ⎢⎣ y ⎥⎦ + ⎢⎣.180 ⎥⎦ . Solving this matrix equation gives rounded to three decimal places.

⎡ x ⎤ ⎡.15 −.04 ⎤ ⎢⎣ y ⎥⎦ = ⎢⎣.04 .15⎥⎦

−1

⎡.075⎤ ⎡.766 ⎤ ⎢⎣.180⎥⎦ = ⎢⎣.996 ⎥⎦

7. As the figure indicates, the unit square can be expressed as the union of 16 nonoverlapping congruent squares, 1 each of side length . Consequently, the Hausdorff dimension of the unit square as given by Equation (2) of the 4 ln(k ) ln(16) = = 2. text is d H ( S ) = ln(4) ln 1

(s)

243



1 4

1

8. The similitude T1 maps the unit square (whose vertices are (0, 0), (1, 0), (1, 1), and (0, 1)) onto the square whose ⎛3 ⎞ ⎛3 3⎞ ⎛ 3⎞ vertices are (0, 0), ⎜ , 0 ⎟ , ⎜ , ⎟ , and ⎜ 0, ⎟ . The similitude T2 maps the unit square onto the square whose ⎝4 ⎠ ⎝4 4⎠ ⎝ 4⎠ ⎛1 ⎞ ⎛ 3⎞ ⎛1 3⎞ vertices are ⎜ , 0 ⎟ , (1, 0), ⎜1, ⎟ , and ⎜ , ⎟ . The similitude T3 maps the unit square onto the square whose ⎝4 ⎠ ⎝ 4⎠ ⎝4 4⎠ ⎛ 1⎞ ⎛3 1⎞ ⎛3 ⎞ vertices are ⎜ 0, ⎟ , ⎜ , ⎟ , ⎜ , 1⎟ , and (0, 1). Finally the similitude T4 maps the unit square onto the square ⎝ 4⎠ ⎝4 4⎠ ⎝4 ⎠ 3 ⎛1 1⎞ ⎛ 1⎞ ⎛1 ⎞ whose vertices are ⎜ , ⎟ , ⎜1, ⎟ , (1, 1), and ⎜ , 1⎟ . Each of these four smaller squares has side length of , 4 ⎝4 4⎠ ⎝ 4⎠ ⎝4 ⎠ 3 so that the common scale factor of the similitudes is s = . The right-hand side of Equation (2) of the text gives 4 ln(k ) ln(4) = = 4.818.... This is not the correct Hausdorff dimension of the square (which is 2) because the four ln 1s ln 43

()

()

smaller squares overlap. 9. Because s =

ln(k ) ln(8) 1 = = 3 for the Hausdorff dimension and k = 8, Equation (2) of the text gives d H ( S ) = 2 ln(2) ln 1s

()

of a unit cube. Because the Hausdorff dimension of the cube is the same as its topological dimension, the cube is not a fractal. 10. A careful examination of Figure Ex-10 in the text shows that the Menger sponge can be expressed as the union of 1 1 20 smaller nonoverlapping congruent Menger sponges each of side length . Consequently, k = 20 and s = , 3 3 ln(k ) ln(20) = = 2.726.... Because its and so the Hausdorff dimension of the Menger sponge is d H ( S ) = ln(3) ln 1

(s)

Hausdorff dimension is not an integer, the Menger sponge is a fractal. 11. The figure shows the first four iterates as determined by Algorithm 1 and starting with the unit square as the 1 initial set. Because k = 2 and s = , the Hausdorff dimension of the Cantor set is 3 ln(k ) ln(2) d H (S ) = = = 0.6309.... Notice that the Cantor set is a subset of the unit interval along the x-axis and ln(3) ln 1

(s)

that its topological dimension must be 0 (since the topological dimension of any set is a nonnegative integer less than or equal to its Hausdorff dimension).

244


Section 10.14

Initial set

First Iterate

Second Iterate Third Iterate Fourth Iterate

12. The area of the unit square S0 is, of course, 1. Each of the eight similitudes T1 , T2 , ..., T8 given in Equation (8) of 1 1 the text has scale factor s = , and so each maps the unit square onto a smaller square of area . Because these 3 9 8 eight smaller squares are nonoverlapping, their total area is , which is then the area of the set S1 . By a similar 9 8 argument, the area of the set S2 is -th the area of the set S1 . Continuing the argument further, we find that the 9 areas of S0 , S1, S2 , S3 , S4 , ..., form the geometric sequence 1,

8 ⎛ 8 ⎞ 2 ⎛ 8 ⎞3 ⎛ 8 ⎞ 4 , ⎜ ⎟ , ⎜ ⎟ , ⎜ ⎟ , .... (Notice that this 9 ⎝9⎠ ⎝9⎠ ⎝9⎠ n

⎛8⎞ implies that the area of the Sierpinski carpet is 0, since the limit of ⎜ ⎟ as n tends to infinity is 0.) ⎝9⎠

Section 10.14 Exercise Set 10.14 1. Because 250 = 2 ⋅ 53 it follows from (i) that Π (250) = 3 ⋅ 250 = 750. Because 25 = 52 it follows from (ii) that Π (25) = 2 ⋅ 25 = 50. Because 125 = 53 it follows from (ii) that Π(125) = 2 ⋅125 = 250. Because 30 = 6 ⋅ 5 it follows from (ii) that Π (30) = 2 ⋅ 30 = 60. Because 10 = 2 ⋅ 5 it follows from (i) that Π (10) = 3 ⋅10 = 30.

245

Chapter 10: Elementary Linear Algebra


Because 50 = 2 ⋅ 52 it follows from (i) that Π (50) = 3 ⋅ 50 = 150. Because 3750 = 6 ⋅ 54 it follows from (ii) that Π (3750) = 2 ⋅ 3750 = 7500. Because 6 = 6 ⋅ 50 it follows from (ii) that Π(6) = 2 ⋅ 6 = 12

Because 5 = 51 it follows from (ii) that Π (5) = 2 ⋅ 5 = 10.

⎛m n⎞ 2. The point (0, 0) is obviously a 1-cycle. We now choose another of the 36 points of the form ⎜ , ⎟ , say ⎝ 6 6⎠ 1 ⎛ ⎞ ⎜ 0, ⎟ . Its iterates produce the 12-cycle ⎝ 6⎠ ⎡3⎤ ⎡1⎤ ⎡2⎤ ⎡3⎤ ⎡1⎤ ⎡0⎤ 6⎥ → ⎢6⎥ → ⎢6⎥ → ⎢6⎥ → ⎢6⎥ ⎢ → ⎢1⎥ 5 4 ⎢ 65 ⎥ ⎢⎣ 62 ⎥⎦ ⎢⎣ 16 ⎥⎦ ⎣6⎦ ⎣⎢ 6 ⎦⎥ ⎣⎢ 6 ⎦⎥ ⎣ ⎦ ⎡5⎤ ⎡3⎤ ⎡4⎤ ⎡3⎤ ⎡5⎤ ⎡0⎤ → ⎢5⎥ → ⎢6⎥ → ⎢6⎥ → ⎢6⎥ → ⎢6⎥ → ⎢6⎥. ⎢⎣ 64 ⎥⎦ ⎢⎣ 16 ⎥⎦ ⎢⎣ 65 ⎥⎦ ⎢⎣ 62 ⎥⎦ ⎢⎣ 16 ⎥⎦ ⎣⎢ 6 ⎦⎥ ⎛ 2⎞ So far we have accounted for 13 of the 36 points. Taking one of the remaining points, say ⎜ 0, ⎟ , we arrive at ⎝ 6⎠ ⎡ 2 ⎤ ⎡0⎤ ⎡ 4 ⎤ ⎡0⎤ the 4-cycle ⎢ 2 ⎥ → ⎢ 6 ⎥ → ⎢ 4 ⎥ → ⎢ 6 ⎥ . We continue in this way, each time starting with some point of the form ⎢⎣ 64 ⎥⎦ ⎣ 6 ⎦ ⎢⎣ 62 ⎥⎦ ⎣6⎦ ⎛m n⎞ ⎜ , ⎟ that has not yet appeared in a cycle, until we exhaust all such points. This yields a 3-cycle: ⎝ 6 6⎠ ⎡3⎤ ⎡4⎤ ⎡2⎤ 4 2 0 ⎡3⎤ 6 ⎥ → ⎡ ⎤ ; another 4-cycle: ⎡ 6 ⎤ → ⎢ 6 ⎥ → ⎡ 6 ⎤ → ⎢ 6 ⎥ ; and another 12-cycle: 6 ⎢ → ⎢ ⎥ ⎢3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 63 ⎥ ⎢⎣ 64 ⎥⎦ ⎣ 0 ⎦ ⎢⎣ 62 ⎥⎦ ⎣⎢ 0 ⎦⎥ ⎣⎢ 6 ⎦⎥ ⎣0⎦ ⎣ ⎦ ⎡1⎤ ⎡2⎤ ⎡5⎤ ⎡1⎤ ⎡4⎤ ⎡1⎤ 6 6 6 6 6 ⎢6⎥ → ⎢1⎥ → ⎢3⎥ → ⎢2⎥ → ⎢3⎥ → ⎢1⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎣ ⎦ ⎣6⎦ ⎣6⎦ ⎣6⎦ ⎣6⎦ ⎣ 6 ⎥⎦ ⎡5⎤ ⎡5⎤ ⎡4⎤ ⎡1⎤ ⎡5⎤ → ⎢6⎥ → ⎢6⎥ → ⎢6⎥ → ⎢6⎥ → ⎢6⎥ ⎢ 56 ⎥ ⎢⎣ 63 ⎥⎦ ⎢ 36 ⎥ ⎢⎣ 64 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎣ ⎦ ⎣ ⎦ 2 ⎡ ⎤ → ⎢ 65 ⎥ . ⎢6⎥ ⎣ ⎦ ⎛m n⎞ The possible periods of points for the form ⎜ , ⎟ are thus 1, 3, 4 and 12. The least common multiple of these ⎝ 6 6⎠ four numbers is 12, and so Π (6) = 12.

246


Section 10.14

3. (a) We are given that x0 = 3 and x1 = 7. With p = 15 we have x2 = x1 + x0 mod 15 = 7 + 3 mod 15 = 10 mod 15 = 10, x3 = x2 + x1 mod 15 = 10 + 7 mod 15 = 17 mod 15 = 2, x4 = x3 + x2 mod 15 = 2 + 10 mod 15 = 12 mod 15 = 12 x5 = x4 + x3 mod 15 = 12 + 2 mod 15 = 14 mod 15 = 14, x6 = x5 + x4 mod 15 = 14 + 12 mod 15 = 26 mod 15 = 11, x7 = x6 + x5 mod 15 = 11 + 14 mod 15 = 25 mod 15 = 10 x8 = x7 + x6 mod 15 = 10 + 11 mod 15 = 21 mod 15 = 6, x9 = x8 + x7 mod 15 = 6 + 10 mod 15 = 16 mod 15 = 1, x10 = x9 + x8 mod 15 = 1 + 6 mod 15 = 7 mod 15 = 7, x11 = x10 + x9 mod 15 = 7 + 1 mod 15 = 8 mod 15 = 8 x12 = x11 + x10 mod 15 = 8 + 7 mod 15 = 15 mod 15 = 0, x13 = x12 + x11 mod 15 = 0 + 8 mod 15 = 8 mod 15 = 8, x14 = x13 + x12 mod 15 = 8 + 0 mod 15 = 8 mod 15 = 8, x15 = x14 + x13 mod 15 = 8 + 8 mod 15 = 16 mod 15 = 1, x16 = x15 + x14 mod 15 = 1 + 8 mod 15 = 9 mod 15 = 9,

x17 = x16 + x15 mod 15 = 9 + 1 x18 = x17 + x16 mod 15 = 10 + 9 x19 = x18 + x17 mod 15 = 4 + 10 x20 = x19 + x18 mod 15 = 14 + 4 x21 = x20 + x19 mod 15 = 3 + 14 x22 = x21 + x20 mod 15 = 2 + 3 x23 = x22 + x21 mod 15 = 5 + 2 x24 = x23 + x22 mod 15 = 7 + 5 x25 = x24 + x23 mod 15 = 12 + 7 x26 = x25 + x24 mod 15 = 4 + 12 x27 = x26 + x25 mod 15 = 1 + 4 x28 = x27 + x26 mod 15 = 5 + 1 x29 = x28 + x27 mod 15 = 6 + 5 x30 = x29 + x28 mod 15 = 11 + 6

mod 15 = 10 mod 15 = 10, mod 15 = 19 mod 15 = 4, mod 15 = 14 mod 15 = 14, mod 15 = 18 mod 15 = 3, mod 15 = 17 mod 15 = 2, mod 15 = 5 mod 15 = 5, mod 15 = 7 mod 15 = 7, mod 15 = 12 mod 15 = 12, mod 15 = 19 mod 15 = 4, mod 15 = 16 mod 15 = 1, mod 15 = 5 mod 15 = 5, mod 15 = 6 mod 15 = 6, mod 15 = 11 mod 15 = 11, mod 15 = 17 mod 15 = 2,

x31 = x30 + x29 mod 15 = 2 + 11 mod 15 = 13 mod 15 = 13, x32 = x31 + x30 mod 15 = 13 + 2 mod 15 = 15 mod 15 = 0, x33 = x32 + x31 mod 15 = 0 + 13 mod 15 = 13 mod 15 = 13, x34 = x33 + x32 mod 15 = 13 + 0 mod 15 = 13 mod 15 = 13, x35 = x34 + x33 mod 15 = 13 + 13 mod 15 = 26 mod 15 = 11, x36 = x35 + x34 mod 15 = 11 + 13 mod 15 = 24 mod 15 = 9, x37 = x36 + x35 mod 15 = 9 + 11 mod 15 = 20 mod 15 = 5, x38 = x37 + x36 mod 15 = 5 + 9 mod 15 = 14 mod 15 = 14, x39 = x38 + x37 mod 15 = 14 + 5 mod 15 = 19 mod 15 = 4, x40 = x39 + x38 mod 15 = 4 + 14 mod 15 = 18 mod 15 = 3, x41 = x40 + x39 mod 15 = 3 + 4 mod 15 = 7 mod 15 = 7, and finally: x40 = x0 and x41 = x1 . Thus this sequence is periodic with period 40.

247



⎡ x12 ⎤ ⎡1 1 ⎤ ⎡ 4 ⎤ ⎢ x ⎥ = ⎢1 2⎥ ⎢ 6 ⎥ mod 21 ⎦⎣ ⎦ ⎣ 13 ⎦ ⎣ ⎡10 ⎤ = ⎢ ⎥ mod 21 ⎣16 ⎦ ⎡10 ⎤ =⎢ ⎥ ⎣16 ⎦

(b) Step (ii) of the algorithm is xn +1 = xn + xn −1 mod p. Replacing n in this formula by n + 1 gives xn + 2 = xn +1 + xn mod p = ( xn + xn −1 ) + xn mod p = 2 xn + xn −1 mod p. These equations can be written as xn +1 = xn −1 + xn mod p xn + 2 = xn −1 + 2 xn mod p which in matrix form are ⎡ xn +1 ⎤ ⎡1 1 ⎤ ⎡ xn −1 ⎤ ⎢ x ⎥ = ⎢1 2⎥ ⎢ x ⎥ mod p. ⎦⎣ n ⎦ ⎣ n+2 ⎦ ⎣

⎡ x14 ⎤ ⎡1 1 ⎤ ⎡10 ⎤ ⎢ x ⎥ = ⎢1 2 ⎥ ⎢16 ⎥ mod 21 ⎦⎣ ⎦ ⎣ 15 ⎦ ⎣ ⎡ 26 ⎤ = ⎢ ⎥ mod 21 ⎣ 42 ⎦ ⎡5 ⎤ =⎢ ⎥ ⎣0 ⎦ ⎡ x16 ⎤ ⎡1 1 ⎤ ⎡ 5 ⎤ ⎢ x ⎥ = ⎢1 2⎥ ⎢ 0 ⎥ mod 21 ⎦⎣ ⎦ ⎣ 17 ⎦ ⎣ ⎡5⎤ = ⎢ ⎥ mod 21 ⎣5⎦ ⎡5⎤ =⎢ ⎥ ⎣5⎦

⎡ x ⎤ ⎡5 ⎤ (c) Beginning with ⎢ 0 ⎥ = ⎢ ⎥ , we obtain ⎣ x1 ⎦ ⎣5⎦ ⎡ x2 ⎤ ⎡1 1 ⎤ ⎡5⎤ ⎢ x ⎥ = ⎢1 2⎥ ⎢5⎥ mod 21 ⎦⎣ ⎦ ⎣ 3⎦ ⎣ ⎡10 ⎤ = ⎢ ⎥ mod 21 ⎣15⎦ ⎡10 ⎤ =⎢ ⎥ ⎣15⎦

⎡x ⎤ ⎡x ⎤ and we see that ⎢ 16 ⎥ = ⎢ 0 ⎥ . ⎣ x17 ⎦ ⎣ x1 ⎦

⎡ x4 ⎤ ⎡1 1 ⎤ ⎡10 ⎤ ⎢ x ⎥ = ⎢1 2 ⎥ ⎢15⎥ mod 21 ⎦⎣ ⎦ ⎣ 5⎦ ⎣ ⎡ 25⎤ = ⎢ ⎥ mod 21 ⎣ 40 ⎦ ⎡4⎤ =⎢ ⎥ ⎣19 ⎦ ⎡ x6 ⎤ ⎡1 1 ⎤ ⎡ 4 ⎤ ⎢ x ⎥ = ⎢1 2 ⎥ ⎢19 ⎥ mod 21 ⎦⎣ ⎦ ⎣ 7⎦ ⎣ ⎡ 23⎤ = ⎢ ⎥ mod 21 ⎣ 42 ⎦ ⎡2⎤ =⎢ ⎥ ⎣0⎦ ⎡ x8 ⎤ ⎡1 1 ⎤ ⎡ 2 ⎤ ⎢ x ⎥ = ⎢1 2⎥ ⎢ 0 ⎥ mod 21 ⎦⎣ ⎦ ⎣ 9⎦ ⎣ ⎡2⎤ = ⎢ ⎥ mod 21 ⎣2⎦ ⎡2⎤ =⎢ ⎥ ⎣2⎦

4. (c) We have that

⎡ 1 ⎤ ⎛ ⎡ 1 ⎤ ⎞ ⎡1 1 ⎤ ⎡ 1 ⎤ 101 ⎥ mod 1 = ⎢ 101 ⎥ , C ⎜ ⎢ 101 ⎥ ⎟ = ⎢ ⎢ 1 ⎥ ⎜ 0 ⎟ ⎣1 2 ⎥⎦ 0 ⎢⎣ 101 ⎣ ⎦ ⎝⎣ ⎦⎠ ⎦ 1 2 ⎡ ⎤ ⎡ ⎤ 1 ⎛ ⎡ ⎤ ⎞ ⎡1 1 ⎤ 101 ⎢ 101 ⎥ , ⎢ ⎥ 1 mod C 2 ⎜ ⎢ 101 ⎥ ⎟ = ⎢ = 3 ⎥ ⎜ ⎟ 1 2 ⎥⎦ ⎢ 1 ⎥ ⎢⎣ 101 ⎝⎣ 0 ⎦⎠ ⎣ ⎣ 101 ⎦ ⎦ 2 5 ⎡ ⎤ ⎡ ⎤ 1 ⎛ ⎡ ⎤ ⎞ ⎡1 1 ⎤ 101 ⎢ ⎥ mod 1 = ⎢ 101 ⎥ , C 3 ⎜ ⎢ 101 ⎥ ⎟ = ⎢ ⎜ 0 ⎟ ⎣1 2⎥⎦ ⎢ 3 ⎥ ⎢ 8 ⎥ ⎝⎣ ⎦⎠ ⎣ 101 ⎦ ⎣ 101 ⎦ 5 ⎤ ⎡ 13 ⎤ ⎛ ⎡ 1 ⎤ ⎞ ⎡1 1 ⎤ ⎡ 101 ⎢ ⎥ mod 1 = ⎢ 101 ⎥ , C 4 ⎜ ⎢ 101 ⎥ ⎟ = ⎢ 21 ⎥ ⎜ ⎟ 1 2 ⎥⎦ ⎢ 8 ⎥ ⎢⎣ 101 ⎝⎣ 0 ⎦⎠ ⎣ ⎣ 101 ⎦ ⎦ 34 13 ⎡ ⎤ ⎛ ⎡ 1 ⎤ ⎞ ⎡1 1 ⎤ ⎡ 101 ⎤ ⎢ ⎥ mod 1 = ⎢ 101 ⎥ . C 5 ⎜ ⎢ 101 ⎥ ⎟ = ⎢ ⎥ 55 ⎥ 21 ⎜ ⎟ 1 2⎦ ⎢ ⎥ ⎢ 101 ⎝⎣ 0 ⎦⎠ ⎣ ⎣ 101 ⎦ ⎣ ⎦ Because all five iterates are different, the ⎛ 1 ⎞ period of the periodic point ⎜ , 0 ⎟ must ⎝ 101 ⎠ be greater than 5.

⎡ x10 ⎤ ⎡1 1 ⎤ ⎡ 2 ⎤ ⎢ x ⎥ = ⎢1 2⎥ ⎢ 2 ⎥ mod 21 ⎦⎣ ⎦ ⎣ 11 ⎦ ⎣ ⎡4⎤ = ⎢ ⎥ mod 21 ⎣6⎦ ⎡4⎤ =⎢ ⎥ ⎣6⎦

5. If 0 ≤ x < 1 and 0 ≤ y < 1, then 5 ⎛ ⎞ T ( x, y) = ⎜ x + , y ⎟ mod 1, and so 12 ⎠ ⎝ 10 ⎞ ⎛ T 2 ( x, y) = ⎜ x + , y ⎟ mod 1, ⎝ 12 ⎠ 248


Section 10.14

15 ⎞ ⎛ T 3 ( x, y) = ⎜ x + , y ⎟ mod 1, ..., ⎝ 12 ⎠ 60 ⎞ ⎛ T 12 ( x, y) = ⎜ x + , y ⎟ mod 1 ⎝ 12 ⎠ = ( x + 5, y) mod 1 = ( x, y). Thus every point in S returns to its original position after 12 iterations and so every point in S is a periodic point with period at most 12. Because every point is a periodic point, no point can have a dense set of iterates, and so the mapping cannot be chaotic.

and so by part (ii) of the definition, this is not the matrix of an Anosov automorphism. ⎡ 0 1⎤ (c) The eigenvalues of the matrix ⎢ are ⎣ −1 0⎥⎦ ±i; both of which have magnitude 1. By part (iii) of the definition, this cannot be the matrix of an Anosov automorphism. ⎡ x⎤ Starting with an arbitrary point ⎢ ⎥ in the ⎣ y⎦ interior of S, (that is, with 0 < x < 1 and 0 < y < 1) we obtain ⎡ 0 1⎤ ⎡ x ⎤ ⎡y⎤ ⎡ y ⎤ ⎢⎣ −1 0 ⎥⎦ ⎢⎣ y ⎥⎦ mod 1 = ⎢⎣ − x ⎥⎦ mod 1 = ⎢⎣1 − x ⎥⎦ , ⎡ 0 1⎤ ⎡ y ⎤ ⎡1 − x ⎤ ⎢⎣ −1 0 ⎥⎦ ⎢⎣1 − x ⎥⎦ mod 1 = ⎢⎣ − y ⎥⎦ mod 1 ⎡1 − x ⎤ , =⎢ ⎣1 − y ⎥⎦ ⎡ 0 1⎤ ⎡1 − x ⎤ ⎡ 1− y ⎤ ⎢⎣ −1 0 ⎥⎦ ⎢⎣1 − y ⎥⎦ mod 1 = ⎢⎣ −1 + x ⎥⎦ mod 1 ⎡1 − y ⎤ , =⎢ ⎣ x ⎥⎦ ⎡ 0 1⎤ ⎡1 − y ⎤ ⎡ x ⎤ ⎢⎣ −1 0 ⎥⎦ ⎢⎣ x ⎥⎦ mod 1 = ⎢⎣ −1 + y ⎥⎦ mod 1 ⎡ x⎤ = ⎢ ⎥. ⎣ y⎦ Thus every point in the interior of S is a periodic point with period at most 4. The geometric effect of this transformation, as seen by the iterates, is to rotate each point in the interior of S clockwise by 90° about the ⎡1⎤ center point ⎢ 2 ⎥ of S. Consequently, each ⎢⎣ 12 ⎥⎦ point in the interior of S has period 4 with ⎡1⎤ the exception of the center point ⎢ 2 ⎥ which ⎢⎣ 12 ⎥⎦ is a fixed point. For points not in the interior of S, we first ⎡0 ⎤ observe that the origin ⎢ ⎥ is a fixed point, ⎣0 ⎦ which can easily be verified. Starting with a ⎡ x⎤ point of the form ⎢ ⎥ with 0 < x < 1, we ⎣0 ⎦ obtain a 4-cycle ⎡ x⎤ ⎡ 0 ⎤ ⎡1 − x ⎤ ⎡0⎤ ⎡ x⎤ ⎢⎣ 0 ⎥⎦ → ⎢⎣1 − x ⎥⎦ → ⎢⎣ 0 ⎥⎦ → ⎢⎣ x ⎥⎦ → ⎢⎣ 0 ⎥⎦ if

⎡1 1 ⎤ 6. (a) The matrix of Arnold’s cat map, ⎢ , is ⎣1 2 ⎥⎦ one in which (i) the entries are all integers, (ii) the determinant is 1, and (iii) the 3+ 5 = 2.6180... and eigenvalues, 2 3− 5 = 0.3819..., do not have magnitude 2 1. The three conditions of an Anosov automorphism are thus satisfied. ⎡0 1 ⎤ (b) The eigenvalues of the matrix ⎢ are ⎣1 0 ⎥⎦ ±1, both of which have magnitude 1. By part (iii) of the definition of an Anosov automorphism, this matrix is not the matrix of an Anosov automorphism. ⎡3 2 ⎤ are The entries of the matrix ⎢ ⎣1 1 ⎥⎦ integers, its determinant is 1; and neither of its eigenvalues, 2 ± 3, has magnitude 1. Consequently, this is the matrix of an Anosov automorphism. ⎡1 0 ⎤ The eigenvalues of the matrix ⎢ are ⎣0 1 ⎥⎦ both equal to 1, and so both have magnitude 1. By part (iii) of the definition, this is not the matrix of an Anosov automorphism. ⎡5 7 ⎤ The entries of the matrix ⎢ are ⎣ 2 3⎥⎦ integers; its determinant is 1; and neither of its eigenvalues, 4 ± 15, has magnitude 1. Consequently, this is the matrix of an Anosov automorphism. ⎡6 2 ⎤ The determinant of the matrix ⎢ is 2, ⎣5 2 ⎥⎦

249



⎡1 ⎢⎣1 ⎡1 ⎢⎣1 ⎡1 ⎢⎣1

1 , otherwise we obtain the 2-cycle 2 ⎡1⎤ ⎡0⎤ ⎡1⎤ ⎢2⎥ → ⎢1⎥ → ⎢2⎥. ⎣0⎦ ⎣2⎦ ⎣0⎦ Similarly, starting with a point of the form ⎡0⎤ ⎢⎣ y ⎥⎦ with 0 < y < 1, we obtain a 4-cycle ⎡0⎤ ⎡ y⎤ ⎡ 0 ⎤ ⎡1 − y ⎤ ⎡ 0 ⎤ ⎢⎣ y ⎥⎦ → ⎢⎣ 0 ⎥⎦ → ⎢⎣1 − y ⎥⎦ → ⎢⎣ 0 ⎥⎦ → ⎢⎣ y ⎥⎦ if 1 y ≠ , otherwise we obtain the 2-cycle 2 ⎡0⎤ ⎡1⎤ ⎡0⎤ ⎢ 1 ⎥ → ⎢ 2 ⎥ → ⎢ 1 ⎥ . Thus every point not in ⎣2⎦ ⎣0⎦ ⎣2⎦ the interior of S is a periodic point with 1, 2, or 4. Finally because no point in S can have a dense set of iterates, it follows that the mapping cannot be chaotic. x≠

1

( 2 , 1) III'

(0, 0)

(1, 1)

I'

IV'

( 12 , 0)

1 ⎤ ⎡ x2 ⎤ ⎡ a ⎤ ⎡ 12 ⎤ + = , 2 ⎥⎦ ⎢⎣ y2 ⎥⎦ ⎢⎣ b ⎥⎦ ⎢⎣ 1 ⎥⎦ 1 ⎤ ⎡ x3 ⎤ ⎡ a ⎤ ⎡1⎤ . + = 2 ⎥⎦ ⎢⎣ y3 ⎥⎦ ⎢⎣ b ⎥⎦ ⎢⎣1⎥⎦

⎡1 1 ⎤ ⎡ 2 −1⎤ The inverse of the matrix ⎢ is ⎢ . ⎥ 1 2 ⎣ ⎦ ⎣ −1 1⎥⎦ We multiply the above three matrix equations by ⎡ c ⎤ ⎡ 2 −1⎤ ⎡ a ⎤ this inverse and set ⎢ ⎥ = ⎢ . Notice ⎣ d ⎦ ⎣ −1 1⎥⎦ ⎢⎣ b ⎥⎦ that c and d must be integers. This leads to ⎡ x1 ⎤ ⎡ 2 −1⎤ ⎡0 ⎤ ⎡ c ⎤ ⎡c⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢0 ⎥ − ⎢ d ⎥ = − ⎢ d ⎥ , ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 1⎦ ⎣ 1 0⎤ ⎡ c ⎤ ⎡ ⎤ ⎡ x ⎡ 2 ⎤ ⎡ 2 −1⎤ 2 ⎡ c ⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢ ⎥ − ⎢ d ⎥ = ⎢ 1 ⎥ − ⎢ d ⎥ , ⎦ ⎣1⎦ ⎣ ⎦ ⎣ 2 ⎦ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎡ x3 ⎤ ⎡ 2 −1⎤ ⎡1⎤ ⎡ c ⎤ ⎡1 ⎤ ⎡ c ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ − ⎢ ⎥. ⎣ y3 ⎦ ⎣ −1 1⎦ ⎣1⎦ ⎣ d ⎦ ⎣0 ⎦ ⎣ d ⎦ The only integer values of c and d that will give values of xi and yi in the interval [0, 1] are c = d = 0. This then gives a = b = 0 and the ⎡ x ⎤ ⎡1 1 ⎤ ⎡ x ⎤ mapping ⎢ ⎥ → ⎢ maps the three ⎣ y ⎦ ⎣1 2 ⎥⎦ ⎢⎣ y ⎥⎦ ⎛ 1⎞ points (0, 0), ⎜ 0, ⎟ , and (1, 0) to the three ⎝ 2⎠ ⎛1 ⎞ points (0, 0), ⎜ , 1⎟ , and (1, 1), respectively. ⎝2 ⎠ ⎛ 1⎞ The three points (0, 0), ⎜ 0, ⎟ , and (1, 0) define ⎝ 2⎠ the triangular region labeled I in the diagram below, which then maps onto the region I ′. (0, 1) (1, 1)

9. As per the hint, we wish to find the regions in S that map onto the four indicated regions in the figure below.

(0, 1)

1 ⎤ ⎡ x1 ⎤ ⎡ a ⎤ ⎡0 ⎤ + = , 2 ⎥⎦ ⎢⎣ y2 ⎥⎦ ⎢⎣ b ⎥⎦ ⎢⎣0 ⎥⎦

II' (1, 0)

We first consider region I ′ with vertices (0, 0), ⎛1 ⎞ ⎜ , 1⎟ , and (1, 1). We seek points ( x1 , y1 ), ⎝2 ⎠ ( x2 , y2 ), and ( x3 , y3 ), with entries that lie in [0, 1], that map onto these three points under the ⎡ x ⎤ ⎡1 1 ⎤ ⎡ x ⎤ ⎡ a ⎤ for certain mapping ⎢ ⎥ → ⎢ + ⎣ y ⎦ ⎣1 2 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ b ⎥⎦ integer values of a and b to be determined. This leads to the three equations

IV 1

(0, 2 )

II

1

III

(1, 2 )

I (0, 0)

(1, 0)

For region II ′, the calculations are as follows:

250


Section 10.14

⎡1 1 ⎤ ⎡ x1 ⎤ ⎡ a ⎤ ⎡ 12 ⎤ ⎢⎣1 2⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢ 0 ⎥ , ⎣ 1⎦ ⎣ ⎦

⎡1 1 ⎤ ⎡ x2 ⎤ ⎡ a ⎤ ⎡1⎤ ⎢⎣1 2 ⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢⎣1⎥⎦ , ⎣ 2⎦ ⎡1 1 ⎤ ⎡ x3 ⎤ ⎡ a ⎤ ⎡ 12 ⎤ ⎢⎣1 2 ⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢ 0 ⎥ ; ⎣ 3⎦ ⎣ ⎦ ⎡ x1 ⎤ ⎡ 2 −1⎤ ⎡0 ⎤ ⎡ c ⎤ ⎡c⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢0 ⎥ − ⎢ d ⎥ = − ⎢ d ⎥ , ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 1⎦ ⎣

⎡1 1 ⎤ ⎡ x2 ⎤ ⎡ a ⎤ ⎡1⎤ ⎢⎣1 2 ⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢⎣1⎥⎦ , ⎣ 2⎦ ⎡1 1 ⎤ ⎡ x3 ⎤ ⎡ a ⎤ ⎡1 ⎤ ⎢⎣1 2⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢⎣0 ⎥⎦ ; ⎣ 3⎦ ⎡ x1 ⎤ ⎡ 2 −1⎤ ⎡ 12 ⎤ ⎡ c ⎤ ⎡ 1 ⎤ ⎡ c ⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢ ⎥ − ⎢ d ⎥ = ⎢ − 1 ⎥ − ⎢ d ⎥ , ⎦ ⎣0⎦ ⎣ ⎦ ⎣ 2 ⎦ ⎣ ⎦ ⎣ 1⎦ ⎣ ⎡ x2 ⎤ ⎡ 2 −1⎤ ⎡1⎤ ⎡ c ⎤ ⎡1 ⎤ ⎡ c ⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢1⎥ − ⎢ d ⎥ = ⎢ 0⎥ − ⎢ d ⎥ , ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 2⎦ ⎣

⎡ x2 ⎤ ⎡ 2 −1⎤ ⎡1⎤ ⎡ c ⎤ ⎡1 ⎤ ⎡ c ⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢1⎥ − ⎢ d ⎥ = ⎢ 0⎥ − ⎢ d ⎥ , ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎡ x3 ⎤ ⎡ 2 −1⎤ ⎡ 12 ⎤ ⎡ c ⎤ ⎡ 1 ⎤ ⎡ c ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ − ⎢ ⎥ = ⎢ 1⎥ − ⎢ ⎥. ⎣ y3 ⎦ ⎣ −1 1⎦ ⎣ 0 ⎦ ⎣ d ⎦ ⎣ − 2 ⎦ ⎣ d ⎦ Only c = 0 and d = −1 will work. This leads to a = −1, b = −2 and the mapping ⎡ x ⎤ ⎡1 1 ⎤ ⎡ x ⎤ ⎡ −1⎤ ⎢⎣ y ⎥⎦ → ⎢⎣1 2 ⎥⎦ ⎢⎣ y ⎥⎦ + ⎢⎣ −2 ⎥⎦ maps region IV with ⎛ 1⎞ vertices (0, 1), (1, 1), and ⎜1, ⎟ onto region ⎝ 2⎠ IV ′.

⎡ x3 ⎤ ⎡ 2 −1⎤ ⎡1 ⎤ ⎡ c ⎤ ⎡ 2⎤ ⎡ c ⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢0 ⎥ − ⎢ d ⎥ = ⎢ −1⎥ − ⎢ d ⎥ . ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 3⎦ ⎣ Only c = 1 and d = −1 will work. This leads to a = 0, b = −1 and the mapping ⎡ x ⎤ ⎡1 1⎤ ⎡ x ⎤ ⎡ 0 ⎤ ⎢⎣ y ⎥⎦ → ⎢⎣1 2 ⎥⎦ ⎢⎣ y ⎥⎦ + ⎢⎣ −1⎥⎦ maps region II with ⎛ 1⎞ vertices ⎜ 0, ⎟ , (0, 1), and (1, 0) onto region ⎝ 2⎠ ′ II . For region III ′, the calculations are as follows: ⎡1 1 ⎤ ⎡ x1 ⎤ ⎡ a ⎤ ⎡0 ⎤ ⎢⎣1 2⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢⎣0 ⎥⎦ , ⎣ 1⎦

12. As per the hint, we want to find all solutions of ⎡ x ⎤ ⎡ 2 3⎤ ⎡ x0 ⎤ ⎡ r ⎤ the matrix equation ⎢ 0 ⎥ = ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎣ y0 ⎦ ⎣ 3 5⎦ ⎣ y0 ⎦ ⎣ s ⎦ where 0 ≤ x0 < 1, 0 ≤ y0 < 1, and r and s are nonnegative integers. This equation can be ⎡1 3 ⎤ ⎡ x0 ⎤ ⎡ r ⎤ rewritten as ⎢ , which has the ⎢ ⎥= ⎣3 4 ⎥⎦ ⎣ y0 ⎦ ⎢⎣ s ⎥⎦ −4r + 3s 3r − s and y0 = . First solution x0 = 5 5 trying r = 0 and s = 0, 1, 2, ..., then r = 1 and s = 0, 1, 2, ..., etc., we find that the only values of r and s that yield values of x0 and y0 lying in [0, 1) are: 2 1 r = 1 and s = 2, which give x0 = and y0 = ; 5 5 1 3 r = 2 and s = 3, which give x0 = and y0 = ; 5 5 4 2 r = 2 and s = 4, which give x0 = and y0 = ; 5 5 3 4 r = 3 and s = 5, which give x0 = and y0 = . 5 5 ⎛2 1⎞ ⎛3 4⎞ We can then check that ⎜ , ⎟ and ⎜ , ⎟ ⎝5 5⎠ ⎝5 5⎠ ⎛1 3⎞ ⎛4 2⎞ form one 2-cycle and ⎜ , ⎟ and ⎜ , ⎟ form ⎝5 5⎠ ⎝5 5⎠ another 2-cycle.

⎡1 1 ⎤ ⎡ x2 ⎤ ⎡ a ⎤ ⎡ 12 ⎤ ⎢⎣1 2 ⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢ 1 ⎥ , ⎣ 2⎦ ⎣ ⎦ ⎡1 1 ⎤ ⎡ x3 ⎤ ⎡ a ⎤ ⎡0 ⎤ ⎢⎣1 2⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢⎣1 ⎥⎦ ; ⎣ 3⎦ ⎡ x1 ⎤ ⎡ 2 −1⎤ ⎡0 ⎤ ⎡ c ⎤ ⎡c⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢0 ⎥ − ⎢ d ⎥ = − ⎢ d ⎥ , ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 1⎦ ⎣ 1 ⎡ x2 ⎤ ⎡ 2 −1⎤ ⎡ 2 ⎤ ⎡ c ⎤ ⎡ 0 ⎤ ⎡ c ⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢ ⎥ − ⎢ d ⎥ = ⎢ 1 ⎥ − ⎢ d ⎥ , ⎦ ⎣1⎦ ⎣ ⎦ ⎣ 2 ⎦ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎡ x3 ⎤ ⎡ 2 −1⎤ ⎡0 ⎤ ⎡ c ⎤ ⎡ −1⎤ ⎡ c ⎤ ⎢ y ⎥ = ⎢ −1 1⎥ ⎢1 ⎥ − ⎢ d ⎥ = ⎢ 1⎥ − ⎢ d ⎥ . ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 3⎦ ⎣ Only c = −1 and d = 0 will work. This leads to a = −1, b = −1 and the mapping ⎡ x ⎤ ⎡1 1 ⎤ ⎡ x ⎤ ⎡ −1⎤ ⎢⎣ y ⎥⎦ → ⎢⎣1 2 ⎥⎦ ⎢⎣ y ⎥⎦ + ⎢⎣ −1⎥⎦ maps region III with ⎛ 1⎞ vertices (1, 0), ⎜1, ⎟ , and (0, 1) onto region ⎝ 2⎠ III ′. For region IV ′, the calculations are as follows: ⎡1 1 ⎤ ⎡ x1 ⎤ ⎡ a ⎤ ⎡0 ⎤ ⎢⎣1 2⎥⎦ ⎢ y ⎥ + ⎢⎣ b ⎥⎦ = ⎢⎣0 ⎥⎦ , ⎣ 1⎦ 251



14. Begin with a 101 × 101 array of white pixels and add the letter ‘A’ in black pixels to it. Apply the mapping T to this image, which will scatter the black pixels throughout the image. Then superimpose the letter ‘B’ in black pixels onto this image. Apply the mapping T again and then superimpose the letter ‘C’ in black pixels onto the resulting image. Repeat this procedure with the letter ‘D’ and ‘E’. The next application of the mapping will return you to the letter ‘A’ with the pixels for the letters ‘B’ through ‘E’ scattered in the background.

⎡9 1 ⎤ , det(A) = 18 − 7 = 11, 2. (a) For A = ⎢ ⎣ 7 2 ⎥⎦ which is not divisible by 2 or 13. Therefore by Corollary 10.15.3, A is invertible. From Eqation (2): ⎡ 2 −1⎤ A−1 = (11)−1 ⎢ ⎣ −7 9 ⎥⎦ ⎡ 2 −1⎤ = 19 ⎢ ⎣ −7 9 ⎥⎦ ⎡ 38 −19⎤ =⎢ ⎣ −133 171⎥⎦ ⎡12 7 ⎤ (mod 26). =⎢ ⎣ 23 15⎥⎦ Checking: ⎡ 9 1⎤ ⎡12 7 ⎤ AA−1 = ⎢ ⎣ 7 2⎥⎦ ⎢⎣ 23 15⎥⎦ ⎡131 78⎤ =⎢ ⎣130 79 ⎥⎦ ⎡ 1 0⎤ (mod 26) =⎢ ⎣ 0 1⎥⎦

Section 10.15 Exercise Set 10.15 1. First group the plaintext into pairs, add the dummy letter T, and get the numerial equivalents from Table 1. DA RK NI GH TT 4 1 18 11 14 9 7 8 20 20

⎡ 1 3⎤ (a) For the enciphering matrix A = ⎢ , ⎣ 2 1⎥⎦ reducing everything mod 26, we have ⎡ 1 3⎤ ⎡ 4 ⎤ ⎡ 7 ⎤ G ⎢⎣ 2 1⎥⎦ ⎢⎣ 1 ⎥⎦ = ⎢⎣9 ⎥⎦ I ⎡1 3⎤ ⎡18⎤ ⎡ 51⎤ ⎡ 25⎤ Y ⎢⎣ 2 1⎥⎦ ⎢⎣11⎥⎦ = ⎢⎣ 47 ⎥⎦ = ⎢⎣ 21⎥⎦ U O ⎡1 3⎤ ⎡14 ⎤ ⎡ 41⎤ ⎡15⎤ ⎢⎣ 2 1⎥⎦ ⎢⎣ 9 ⎥⎦ = ⎢⎣37 ⎥⎦ = ⎢⎣11⎥⎦ K ⎡ 1 3⎤ ⎡ 7 ⎤ ⎡ 31⎤ ⎡ 5⎤ E ⎢⎣ 2 1⎥⎦ ⎢⎣ 8 ⎥⎦ = ⎢⎣ 22 ⎥⎦ = ⎢⎣ 22 ⎥⎦ V ⎡1 3⎤ ⎡ 20 ⎤ ⎡80 ⎤ ⎡ 2 ⎤ B ⎢⎣ 2 1⎥⎦ ⎢⎣ 20 ⎥⎦ = ⎢⎣60 ⎥⎦ = ⎢⎣ 8 ⎥⎦ H The Hill cipher is GIYUOKEVBH.

⎡12 7 ⎤ ⎡ 9 1⎤ A−1 A = ⎢ ⎣ 23 15⎥⎦ ⎢⎣ 7 2 ⎥⎦ ⎡157 26 ⎤ =⎢ ⎣312 53⎥⎦ ⎡ 1 0⎤ (mod 26). =⎢ ⎣0 1⎥⎦ ⎡3 1⎤ (b) For A = ⎢ , det(A) = 9 − 5 = 4, which ⎣5 3⎥⎦ is divisible by 2. Therefore by Corollary 10.15.3, A is not invertible. ⎡8 11⎤ (c) For A = ⎢ , ⎣ 1 9 ⎥⎦ det(A) = 72 − 11 = 61 = 9 (mod 26), which is not divisible by 2 or 13. Therefore by Corollary 10.15.3, A is invertible. From (2): ⎡ 9 −11⎤ A−1 = (9) −1 ⎢ 8⎥⎦ ⎣ −1 ⎡ 9 −11⎤ = 3⎢ 8⎥⎦ ⎣ −1 ⎡ 27 −33⎤ =⎢ ⎣ −3 24 ⎥⎦ ⎡ 1 19 ⎤ (mod 26). =⎢ ⎣ 23 24 ⎥⎦ Checking:

⎡4 3⎤ (b) For the enciphering matrix A = ⎢ , ⎣ 1 2 ⎥⎦ reducing everything mod 26, we have ⎡ 4 3 ⎤ ⎡ 4 ⎤ ⎡19 ⎤ S ⎢⎣ 1 2 ⎥⎦ ⎢⎣ 1⎥⎦ = ⎢⎣ 6 ⎥⎦ F ⎡ 4 3⎤ ⎡18⎤ ⎡105⎤ ⎡ 1⎤ A ⎢⎣1 2⎥⎦ ⎢⎣11⎥⎦ = ⎢⎣ 40 ⎥⎦ = ⎢⎣14 ⎥⎦ N E ⎡ 4 3 ⎤ ⎡14 ⎤ ⎡ 83⎤ ⎡ 5⎤ ⎢⎣1 2 ⎥⎦ ⎢⎣ 9 ⎥⎦ = ⎢⎣32⎥⎦ = ⎢⎣ 6 ⎥⎦ F Z ⎡ 4 3 ⎤ ⎡ 7 ⎤ ⎡52⎤ ⎡ 0 ⎤ ⎢⎣ 1 2 ⎥⎦ ⎢⎣ 8⎥⎦ = ⎢⎣ 23⎥⎦ = ⎢⎣ 23⎥⎦ W J ⎡ 4 3 ⎤ ⎡ 20 ⎤ ⎡140 ⎤ ⎡10 ⎤ ⎢⎣1 2 ⎥⎦ ⎢⎣ 20 ⎥⎦ = ⎢⎣ 60 ⎥⎦ = ⎢⎣ 8⎥⎦ H The Hill cipher is SFANEFZWJH. 252


Section 10.15

⎡8 11⎤ ⎡ 1 19 ⎤ AA−1 = ⎢ ⎣ 1 9⎥⎦ ⎢⎣ 23 24 ⎥⎦ ⎡ 261 416 ⎤ =⎢ ⎣ 208 235⎥⎦ ⎡ 1 0⎤ (mod 26) =⎢ ⎣ 0 1⎥⎦

⎡1 8⎤ ⎡15 12⎤ AA−1 = ⎢ ⎣1 3⎥⎦ ⎢⎣ 21 5⎥⎦ ⎡183 52 ⎤ =⎢ ⎣ 78 27 ⎥⎦ ⎡ 1 0⎤ (mod 26) =⎢ ⎣ 0 1⎥⎦

⎡ 1 19 ⎤ ⎡8 11⎤ A−1 A = ⎢ ⎣ 23 24 ⎥⎦ ⎢⎣ 1 9 ⎥⎦ ⎡ 27 182 ⎤ =⎢ ⎣ 208 469 ⎥⎦ ⎡ 1 0⎤ (mod 26). =⎢ ⎣0 1⎥⎦

⎡15 A−1 A = ⎢ ⎣ 21 ⎡ 27 =⎢ ⎣ 26 ⎡1 =⎢ ⎣0

⎡ 2 1⎤ (d) For A = ⎢ , det(A) = 14 − 1 = 13, ⎣ 1 7 ⎥⎦ which is divisible by 13. Therefore by Corollary 10.15.4, A is not invertible.

12 ⎤ ⎡1 8⎤ 5⎥⎦ ⎢⎣1 3⎥⎦ 156⎤ 183⎥⎦ 0⎤ (mod 26). 1⎥⎦

3. From Table 1 the numerical equivalent of this ciphertext is 19 1 11 14 15 24 1 15 10 24 ⎡ 4 1⎤ Now we have to find the inverse of A = ⎢ . ⎣ 3 2 ⎥⎦ Since det(A) = 8 − 3 = 5, we have by (2): ⎡ 2 −1⎤ A−1 = (5)−1 ⎢ ⎣ −3 4 ⎥⎦ ⎡ 2 −1⎤ = 21 ⎢ ⎣ −3 4 ⎥⎦ ⎡ 42 −21⎤ =⎢ ⎣ −63 84⎥⎦ ⎡16 5⎤ (mod 26). =⎢ ⎣15 6 ⎥⎦ To obtain the plaintext, multiply each ciphertext

⎡ 3 1⎤ (e) For A = ⎢ , det(A) = 6 − 6 = 0, so that ⎣6 2 ⎥⎦ A is not invertible by Corollary 10.15.4. ⎡1 8⎤ (f) For A = ⎢ , det(A) = 3 − 8 = −5 = 21 ⎣1 3⎥⎦ (mod 26), which is not divisible by 2 or 13. Therefore by Corollary 10.15.4, A is invertible. From (2): ⎡ 3 −8 ⎤ A−1 = (21)−1 ⎢ ⎣ −1 1⎥⎦ ⎡ 3 −8 ⎤ = 5⎢ ⎣ −1 1⎥⎦ ⎡ 15 −40 ⎤ =⎢ 5⎥⎦ ⎣ −5 ⎡15 12 ⎤ (mod 26). =⎢ ⎣ 21 5⎥⎦ Checking:

vector by A−1 and reduce the results modulo 26. ⎡16 5⎤ ⎡19 ⎤ ⎡309 ⎤ ⎡ 23⎤ W ⎢⎣15 6 ⎥⎦ ⎢⎣ 1⎥⎦ = ⎢⎣ 291⎥⎦ = ⎢⎣ 5⎥⎦ E L ⎡16 5 ⎤ ⎡11⎤ ⎡ 246 ⎤ ⎡12⎤ ⎢⎣15 6 ⎥⎦ ⎢⎣14 ⎥⎦ = ⎢⎣ 249 ⎥⎦ = ⎢⎣15⎥⎦ O ⎡16 5 ⎤ ⎡ 15⎤ ⎡360⎤ ⎡ 22⎤ V ⎢⎣15 6 ⎥⎦ ⎢⎣ 24 ⎥⎦ = ⎢⎣369⎥⎦ = ⎢⎣ 5⎥⎦ E ⎡16 5 ⎤ ⎡ 1⎤ ⎡ 91⎤ ⎡13⎤ M ⎢⎣15 6 ⎥⎦ ⎢⎣15⎥⎦ = ⎢⎣105⎥⎦ = ⎢⎣ 1⎥⎦ A ⎡16 5 ⎤ ⎡ 10 ⎤ ⎡ 280 ⎤ ⎡ 20 ⎤ T ⎢⎣15 6 ⎥⎦ ⎢⎣ 24 ⎥⎦ = ⎢⎣ 294 ⎥⎦ = ⎢⎣ 8⎥⎦ H The plaintext is thus WE LOVE MATH. 4. From Table 1 the numerical equivalent of the known plaintext is AR MY 1 18 13 25 and the numerical equivalent of the corresponding ciphertext is 253



Since det( A−1 ) = 35 − 90 = −55 = 23(mod 26),

SL HK 19 12 8 11 so the corresponding plaintext and ciphertext vectors are ⎡ 1⎤ ⎡19 ⎤ p1 = ⎢ ⎥ ↔ c1 = ⎢ ⎥ 18 ⎣ ⎦ ⎣12 ⎦ ⎡ 13⎤ ⎡ 8⎤ p 2 = ⎢ ⎥ ↔ c2 = ⎢ ⎥ 25 ⎣ ⎦ ⎣11⎦

A = ( A−1 )−1 ⎡ 5 −15⎤ = 23−1 ⎢ 7 ⎥⎦ ⎣ −6 ⎡ 5 −15⎤ = 17 ⎢ 7 ⎥⎦ ⎣ −6 ⎡ 85 −255⎤ =⎢ ⎣ −102 119 ⎥⎦ ⎡7 5⎤ (mod 26) =⎢ ⎣ 2 15⎥⎦ is the enciphering matrix.

⎡cT ⎤ ⎡19 12 ⎤ We want to reduce C = ⎢ 1T ⎥ = ⎢ ⎥ to I by ⎢⎣c2 ⎥⎦ ⎣ 8 11⎦ elementary row operations and simultaneously ⎡pT ⎤ ⎡ 1 18⎤ apply these operations to P = ⎢ 1T ⎥ = ⎢ ⎥. ⎢⎣p 2 ⎥⎦ ⎣13 25⎦ The calculations are as follows: ⎡19 12 1 18⎤ Form the matrix ⎢⎣ 8 11 13 25⎥⎦ [C P ] . ⎡ 1 132 11 198⎤ ⎢⎣8 11 13 25⎥⎦

5. From Table 1 the numerical equivalent of the known plaintext is AT OM 1 20 15 13 and the numerical equivalent of the corresponding ciphertext is JY QO 10 25 17 15 The corresponding plaintext and ciphertext vectors are: ⎡ 1⎤ ⎡ 10 ⎤ p1 = ⎢ ⎥ ↔ c1 = ⎢ ⎥ ⎣ 20 ⎦ ⎣ 25⎦ ⎡15⎤ ⎡17 ⎤ p2 = ⎢ ⎥ ↔ c2 = ⎢ ⎥ ⎣13⎦ ⎣15⎦ ⎡10 25⎤ We want to reduce C = ⎢ to I by ⎣17 15 ⎥⎦ elementary row operations and simultaneously ⎡ 1 20 ⎤ apply these operations to P = ⎢ . ⎣15 13⎥⎦ The calculations are as follows: ⎡10 25 1 20⎤ Form the matrix ⎢⎣17 15 15 13⎥⎦ [C P ] .

Multiply the first row by 19−1 = 11 (mod 26).

⎡ 1 2 11 16 ⎤ ⎢⎣8 11 13 25⎥⎦

Replace 132 and 198 by their residues modulo 26.

11 16 ⎤ ⎡1 2 ⎢⎣0 −5 −75 −103⎥⎦ −8 times the first row to the second. ⎡ 1 2 11 16 ⎤ Replace the entries in ⎢⎣0 21 3 1⎥⎦ the second row by their residues modulo 26. 1 2 11 16 ⎡ ⎤ Multiply the second row ⎢⎣0 1 15 5⎥⎦

⎡ 27 40 16 33⎤ ⎢⎣ 17 15 15 13⎥⎦

by 21−1 = 5 (mod 26). ⎡ 1 0 −19 6 ⎤ ⎢⎣0 1 15 5⎥⎦

the first (since 10−1 does not exist mod 26).


⎡ 1 0 7 6⎤ ⎢⎣0 1 15 5⎥⎦

Add the second row to

⎡ 1 14 16 7 ⎤ ⎢⎣17 15 15 13⎥⎦

Replace −19 by its

Replace the entries in

the first row by their residues modulo 26. 1 14 16 7 ⎡ ⎤ ⎢⎣0 −223 −257 −106 ⎥⎦ Add −17 times the first row to the second.

residue modulo 26. ⎡ 7 6⎤ so the deciphering Thus ( A−1 )T = ⎢ ⎣15 5⎥⎦ ⎡7 15⎤ (mod 26). matrix is A−1 = ⎢ ⎣ 6 5⎥⎦ 254


⎡ 1 14 16 7 ⎤ ⎢⎣0 11 3 24 ⎥⎦

Section 10.15

which yields the message THEY SPLIT THE ATOM.

Replace the entries in the second row by their residues modulo 26.

7⎤ ⎡ 1 14 16 ⎢⎣0 1 57 456 ⎥⎦

6. Since we want a Hill 3-cipher, we will group the letters in triples. From Table 1 the numerical equivalents of the known plaintext are I H A V E C O M E 9 8 1 22 5 3 15 13 5 and the numerical equivalent of the corresponding ciphertext are H P A F Q G G D U 8 16 1 6 17 7 7 4 21 The corresponding plaintext and ciphertext vectors are ⎡9 ⎤ ⎡ 8⎤ p1 = ⎢ 8⎥ ↔ c1 = ⎢16 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1⎦ ⎣ 1⎦ ⎡ 22 ⎤ ⎡ 6⎤ p 2 = ⎢ 5⎥ ↔ c2 = ⎢17 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 3⎦ ⎣ 7⎦ ⎡15⎤ ⎡ 7⎤ p3 = ⎢13⎥ ↔ c3 = ⎢ 4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 5⎦ ⎣ 21⎦ ⎡ 8 16 1⎤ We want to reduce C = ⎢ 6 17 7 ⎥ to I by ⎢ ⎥ ⎣7 4 21⎦ elementary row operations and simultaneously ⎡ 9 8 1⎤ apply these operations to P = ⎢ 22 5 3⎥ . ⎢ ⎥ ⎣ 15 13 5⎦ The calculations are as follows: ⎡ 8 16 1 9 8 1⎤ ⎢ 6 17 7 22 5 3⎥ Form the matrix ⎢ ⎥ ⎣7 4 21 15 13 5⎦

Multiply the second row by 11−1 = 19 (mod 26).

⎡ 1 14 16 7 ⎤ ⎢⎣0 1 5 14 ⎥⎦

Replace the entries in the second row by their residues modulo 26.

⎡ 1 0 −54 −189 ⎤ ⎢⎣0 1 5 14 ⎥⎦


⎡ 1 0 24 19 ⎤ ⎢⎣0 1 5 14 ⎥⎦

Replace −54 and −189

by their residues modulo 26. ⎡ 24 19 ⎤ , and so the Thus ( A−1 )T = ⎢ ⎣ 5 14 ⎥⎦ ⎡ 24 5⎤ . deciphering matrix is A−1 = ⎢ ⎣ 19 14 ⎥⎦ From Table 1 the numerical equivalent of the given ciphertext is LN GI HG YB VR EN JY 12 14 7 9 8 7 25 2 22 18 5 14 10 25 QO 17 15 To obtain the plaintext pairs, we multiply each ciphertext vector by A−1: ⎡ 24 5⎤ ⎡12 ⎤ ⎡ 358⎤ ⎡ 20 ⎤ ⎢⎣ 19 14 ⎥⎦ ⎢⎣14 ⎥⎦ = ⎢⎣ 424 ⎥⎦ = ⎢⎣ 8⎥⎦ ⎡ 24 5⎤ ⎡7 ⎤ ⎡ 213⎤ ⎡ 5⎤ ⎢⎣ 19 14 ⎥⎦ ⎢⎣ 9 ⎥⎦ = ⎢⎣ 259 ⎥⎦ = ⎢⎣ 25⎥⎦ ⎡ 24 5⎤ ⎡ 8⎤ ⎡ 227 ⎤ ⎡19 ⎤ ⎢⎣ 19 14 ⎥⎦ ⎢⎣7 ⎥⎦ = ⎢⎣ 250 ⎥⎦ = ⎢⎣16 ⎥⎦ ⎡ 24 5⎤ ⎡ 25⎤ ⎡610 ⎤ ⎡12 ⎤ ⎢⎣ 19 14 ⎥⎦ ⎢⎣ 2 ⎥⎦ = ⎢⎣ 503⎥⎦ = ⎢⎣ 9 ⎥⎦ ⎡ 24 5⎤ ⎡ 22 ⎤ ⎡ 618⎤ ⎡ 20 ⎤ ⎢⎣ 19 14 ⎥⎦ ⎢⎣ 18⎥⎦ = ⎢⎣ 670 ⎥⎦ = ⎢⎣ 20 ⎥⎦ ⎡ 24 5⎤ ⎡ 5⎤ ⎡190⎤ ⎡8⎤ ⎢⎣ 19 14 ⎥⎦ ⎢⎣14 ⎥⎦ = ⎢⎣ 291⎥⎦ = ⎢⎣5⎥⎦ ⎡ 24 5⎤ ⎡ 10 ⎤ ⎡365⎤ ⎡ 1⎤ ⎢⎣ 19 14 ⎥⎦ ⎢⎣ 25⎥⎦ = ⎢⎣540 ⎥⎦ = ⎢⎣ 20⎥⎦ ⎡ 24 5⎤ ⎡17 ⎤ ⎡ 483⎤ ⎡15⎤ ⎢⎣ 19 14 ⎥⎦ ⎢⎣15⎥⎦ = ⎢⎣ 533⎥⎦ = ⎢⎣13⎥⎦

T H E Y S P L (mod 26) I T T H E A T O M

[C

P].

⎡15 20 22 24 21 6 ⎤ ⎢ 6 17 7 22 5 3⎥ ⎢ ⎥ ⎣ 7 4 21 15 13 5⎦ Add the third row to the first since 8−1 does not exist modulo 26. ⎡ 1 140 154 168 147 42⎤ ⎢ 6 17 7 22 5 3⎥ ⎢ ⎥ 4 21 15 13 5⎦ ⎣7 Multiply the first row by 15−1 = 7 (mod 26).

255



⎡ 1 10 24 12 17 16⎤ ⎢ 6 17 7 22 5 3⎥ ⎢ ⎥ ⎣7 4 21 15 13 5⎦ Replace the entries in the first row by their residues modulo 26. 24 12 17 16⎤ ⎡ 1 10 ⎢0 −43 −137 −50 −97 −93⎥ ⎢ ⎥ ⎣0 −66 −147 −69 −106 −107 ⎦

⎡ 1 10 24 12 17 ⎢0 9 19 2 7 ⎢ ⎣0 12 9 9 24

⎡ 1 10 24 12 17 ⎢0 1 57 6 21 ⎢ ⎣0 12 9 9 24

⎡ 1 0 0 4 15 ⎢0 1 0 9 17 ⎢ ⎣0 0 1 15 6

⎡ 4 Thus, ( A−1 )T = ⎢ 9 ⎢ ⎣15

Add −6 times the first row to the second and −7 times the first row to the third. 16 ⎤ 11⎥ ⎥ 23⎦ Replace the entries in the second and third rows by their residues modulo 26. 16 ⎤ 33⎥ ⎥ 23⎦ Multiply the second row

⎡ 4 9 15⎤ deciphering matrix is A−1 = ⎢ 15 17 6 ⎥ . ⎢ ⎥ ⎣ 24 0 17 ⎦ From Table 1 the numerical equivalent of the given ciphertext is H P A F Q G G D U G G D 8 16 1 6 17 7 7 4 21 7 4 4 H P G O D Y N O R 8 16 7 15 4 25 14 15 18 To obtain the plaintext triples, we multiply each

ciphertext vector by A−1 : ⎡ 4 9 15⎤ ⎡ 8⎤ ⎡ 191⎤ ⎡9 ⎤ ⎢ 15 17 6 ⎥ ⎢16 ⎥ = ⎢ 398⎥ = ⎢8⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 24 0 17 ⎦ ⎣ 1⎦ ⎣ 209 ⎦ ⎣ 1⎦

−1

⎡ 1 10 24 ⎢0 1 5 ⎢ ⎣0 12 9

⎡ 1 0 −26 ⎢0 1 5 ⎢ 0 0 − 51 ⎣

⎡1 0 0 ⎢0 1 5 ⎢ ⎣0 0 1

⎡1 0 0 ⎢0 1 0 ⎢ ⎣0 0 1

Add −5 times the third row to the second. 24 ⎤ 0⎥ ⎥ 17 ⎦ Replace the entries in the second row by their residues modulo 26. 15 24⎤ 17 0⎥ and so the ⎥ 6 17 ⎦

by 9 = 3 (modulo 26). 12 17 16 ⎤ 6 21 7 ⎥ ⎥ 9 24 23⎦ Replace the entries in the second row by their residues modulo 26. −48 −193 −54 ⎤ 6 21 7⎥ ⎥ −63 −228 −61⎦

⎡ 4 9 15⎤ ⎡ 6 ⎤ ⎡ 282 ⎤ ⎡ 22 ⎤ ⎢ 15 17 6 ⎥ ⎢17 ⎥ = ⎢ 421⎥ = ⎢ 5⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 24 0 17 ⎦ ⎣ 7 ⎦ ⎣ 263⎦ ⎣ 3⎦ ⎡ 4 9 15⎤ ⎡ 7 ⎤ ⎡ 379⎤ ⎡15⎤ ⎢ 15 17 6 ⎥ ⎢ 4 ⎥ = ⎢ 299⎥ = ⎢13⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 24 0 17 ⎦ ⎣ 21⎦ ⎣ 525⎦ ⎣ 5⎦

I H A V E C O M E

⎡ 4 9 15⎤ ⎡7 ⎤ ⎡ 124 ⎤ ⎡ 20 ⎤ T ⎢ 15 17 6 ⎥ ⎢ 4⎥ = ⎢ 197 ⎥ = ⎢ 15⎥ O ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 24 0 17 ⎦ ⎣ 4⎦ ⎣ 236 ⎦ ⎣ 2 ⎦ B

Add −10 times the second row to the first and −12 times the second row to the third. 4 15 24⎤ 6 21 7 ⎥ ⎥ 15 6 17 ⎦ Replace the entries in the first and second row by their residues modulo 26. 4 15 24 ⎤ −69 −9 −78⎥ ⎥ 15 6 17 ⎦

⎡ 4 9 15⎤ ⎡ 8⎤ ⎡ 281⎤ ⎡ 21⎤ U ⎢ 15 17 6 ⎥ ⎢16 ⎥ = ⎢ 434 ⎥ = ⎢ 18⎥ R ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 24 0 17 ⎦ ⎣ 7 ⎦ ⎣ 311⎦ ⎣ 25⎦ Y ⎡ 4 9 15⎤ ⎡ 15⎤ ⎡ 471⎤ ⎡3⎤ C ⎢ 15 17 6 ⎥ ⎢ 4 ⎥ = ⎢ 443⎥ = ⎢ 1⎥ A ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 24 0 17 ⎦ ⎣ 25⎦ ⎣785⎦ ⎣5⎦ E ⎡ 4 9 15⎤ ⎡14 ⎤ ⎡ 461⎤ ⎡19 ⎤ S ⎢ 15 17 6 ⎥ ⎢15⎥ = ⎢ 573⎥ = ⎢ 1⎥ A ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 24 0 17 ⎦ ⎣18⎦ ⎣642 ⎦ ⎣18⎦ R Finally, the message is I HAVE COME TO BURY CAESAR.

256


Section 10.16

⎡0 1 1⎤ Thus A−1 = ⎢ 1 1 1⎥ . ⎢ ⎥ ⎣ 1 0 1⎦ ⎡0 1 1⎤ ⎡ 0 ⎤ ⎡1 ⎤ ⎢ 1 1 1⎥ ⎢1 ⎥ = ⎢1 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 1 0 1⎦ ⎣ 0 ⎦ ⎣0 ⎦

7. (a) Multiply each of the triples of the message ⎡ 1 1 0⎤ by A = ⎢ 0 1 1⎥ and reduce the results ⎢ ⎥ ⎣ 1 1 1⎦ modulo 2. ⎡ 1 1 0 ⎤ ⎡ 1⎤ ⎡ 2 ⎤ ⎡0 ⎤ ⎢0 1 1⎥ ⎢ 1⎥ = ⎢ 1⎥ = ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1 1 1⎦ ⎣ 0 ⎦ ⎣ 2 ⎦ ⎣0 ⎦ ⎡ 1 1 0 ⎤ ⎡ 1⎤ ⎡ 1⎤ ⎡ 1⎤ ⎢0 1 1⎥ ⎢ 0 ⎥ = ⎢ 1⎥ = ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1 1 1⎦ ⎣ 1⎦ ⎣ 2 ⎦ ⎣0 ⎦ ⎡ 1 1 0 ⎤ ⎡1⎤ ⎡ 2 ⎤ ⎡ 0 ⎤ ⎢0 1 1⎥ ⎢1⎥ = ⎢ 2 ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1 1 1⎦ ⎣1⎦ ⎣ 3⎦ ⎣ 1⎦ The encoded message is 010110001. (b) Reduce [ A I ] to ⎡⎣ I ⎡ 1 1 0 1 0 0⎤ ⎢0 1 1 0 1 0 ⎥ ⎢ ⎥ ⎣ 1 1 1 0 0 1⎦ ⎡ 1 1 0 1 0 0⎤ ⎢0 1 1 0 1 0⎥ ⎢ ⎥ ⎣ 2 2 1 1 0 1⎦

⎡0 1 1⎤ ⎡1 ⎤ ⎡1 ⎤ ⎡1 ⎤ ⎢1 1 1⎥ ⎢1 ⎥ = ⎢ 2 ⎥ = ⎢0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣1 0 1⎦ ⎣ 0 ⎦ ⎣1 ⎦ ⎣1 ⎦ ⎡0 1 1⎤ ⎡ 0⎤ ⎡1⎤ ⎢ 1 1 1⎥ ⎢ 0⎥ = ⎢1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 1 0 1⎦ ⎣1 ⎦ ⎣1⎦ The decoded message is 110101111, which is the original message.

8. Since 29 is a prime number, by Corollary 10.15.2 a matrix A with entries in Z 29 is invertible if and only if det(A) ≠ 0 (mod 29).

−1 ⎤

A ⎦ modulo 2.

Form the matrix

Section 10.16

[A

Exercise Set 10.16

I ].

1. Use induction on n, the case n = 1 being already given. If the result is true for n − 1, then

Add the first row to

M n = M n −1M

= ( PD n −1 P −1 )( PDP −1 )

the third row. ⎡ 1 1 0 1 0 0⎤ ⎢0 1 1 0 1 0 ⎥ ⎢ ⎥ ⎣0 0 1 1 0 1⎦

= PD n −1 ( P −1P) DP −1

Replace 2 by its

= PD n −1DP −1

= PD n P −1 , proving the result.

residue modulo 2. ⎡ 1 1 0 1 0 0⎤ ⎢0 1 2 1 1 1⎥ ⎢ ⎥ ⎣0 0 1 1 0 1⎦

Add the third row

2. Using Table 1 and notations of Example 1, we derive the following equations: 1 1 an = an −1 + bn −1 2 4 1 1 1 bn = an −1 + bn −1 + cn −1 2 2 2 1 1 cn = bn −1 + cn −1. 4 2 ⎡ 1 1 0⎤ ⎢2 4 ⎥ The transition matrix is thus M = ⎢ 12 12 12 ⎥ . ⎢ 1 1⎥ ⎢⎣ 0 4 2 ⎥⎦ The characteristic polynomial of M is

to the second row. ⎡ 1 1 0 1 0 0⎤ ⎢0 1 0 1 1 1⎥ ⎢ ⎥ ⎣0 0 1 1 0 1⎦

Replace 2 by its residue modulo 2.

⎡ 1 2 0 2 1 1⎤ ⎢0 1 0 1 1 1⎥ ⎢ ⎥ ⎣0 0 1 1 0 1⎦

Add the second row to the first row.

⎡ 1 0 0 0 1 1⎤ ⎢0 1 0 1 1 1⎥ ⎢ ⎥ ⎣0 0 1 1 0 1⎦

Replace 2 by its residue modulo 2.

257



⎛3⎞ ⎛1⎞ det(λI − M ) = λ3 − ⎜ ⎟ λ 2 + ⎜ ⎟ λ ⎝2⎠ ⎝2⎠ 1⎞ ⎛ = λ(λ − 1) ⎜ λ − ⎟ , ⎝ 2⎠

x( 2 n) = ( M 2 M1 )n x( 0) and x( 2 n +1) = M1 ( M 2 M1 )n x( 0) . We have

0 ⎤ ⎡1 1 0 ⎤ ⎡ 12 83 14 ⎤ 2 ⎥ ⎥ ⎥ ⎢ 1⎥ ⎢ 1 1⎥ = ⎢ 1 1 1 ⎥ . 0 ⎢ 2 2 2 2 2 1 ⎥ ⎢0 0 0 ⎥ ⎢ 0 1 1 ⎥ ⎦ ⎢⎣ 2 ⎦⎥ ⎣ 8 4 ⎦⎥ The characteristic polynomial of this matrix is 5 1 λ3 − λ 2 + λ, so the eigenvalues are λ1 = 1, 4 4 1 λ 2 = , λ3 = 0. Corresponding eigenvectors 4 ⎡1 ⎢2 M 2 M1 = ⎢ 12 ⎢ ⎣⎢ 0

1 so the eigenvalues of M are λ = 1, λ 2 = , and 2 λ3 = 0. Corresponding eigenvectors (found by solving (λI − M)x = 0) are e1 = [1 2 1] , T

e2 = [1 0 −1] , and e3 = [1 −2 1] . Thus T

T

Mn

= PD n P −1

are e1 = [5 6 1]T , e2 = [ −1 0 1]T , and

1 1⎤ 1 0 0 ⎤ ⎡ 14 4 4⎥ ⎡ 1 1 1⎤ ⎢⎡ ⎢ ⎥ 1 n 0 − 12 ⎥ . = ⎢ 2 0 −2 ⎥ ⎢0 1 ⎢2 0 ⎥ 2 ⎢ ⎥ ⎥⎢ 1⎥ ⎣ 1 −1 1⎦ ⎢⎣0 0 0 ⎦ ⎢⎣ 14 − 14 4 ⎥⎦ This yields ⎡ an ⎤ x( n) = ⎢ bn ⎥ ⎢ ⎥ ⎢⎣ cn ⎥⎦ ⎡ 1 1 n +1 1 1 1 n +1 ⎤ − 2 ⎢4 + 2 ⎥ ⎡ a0 ⎤ 4 4 ⎢ ⎥⎢ ⎥ 1 1 1 =⎢ 2 2 2 ⎥ ⎢ b0 ⎥ . ⎢ 1 1 n +1 1 1 1 n +1 ⎥ ⎣⎢ c0 ⎦⎥ + 2 ⎢⎣ 4 − 2 ⎥⎦ 4 4 Remembering that a0 + b0 + c0 = 1, we obtain

e3 = [1 −2 1]T . Thus,

()

()

an =

( M 2 M1 ) n

= PD n P −1

1 1 1⎤ 1 0 0 ⎤ ⎡ 12 12 12 ⎥ ⎡ 5 −1 1⎤ ⎢⎡ ⎢ ⎥ n 1 1 2⎥. = ⎢ 6 0 −2 ⎥ ⎢ 0 1 0⎥ ⎢− 3 6 3 4 ⎢ ⎥⎢ ⎢ ⎥ ⎥ ⎣ 1 1 1⎦ ⎣0 0 0 ⎦ ⎢⎣ 14 − 14 14 ⎥⎦ Using the notation of Example 1 (recall that a0 + b0 + c0 = 1), we obtain

()

()

()

1 1 1 ⎛1⎞ a0 + b0 + c0 + ⎜ ⎟ 4 4 4 ⎝2⎠ n +1

⎛1⎞ a0 − ⎜ ⎟ ⎝2⎠

b2 n

n +1

c0

c2 n

1 ⎛1⎞ + ⎜ ⎟ (a0 − c0 ) 4 ⎝2⎠ 1 1 1 1 bn = a0 + b0 + c0 = 2 2 2 2 =

cn = =

and 2 1 + (2a0 − b0 − 4c0 ) 3 6 ⋅ 4n 1 1 b2 n +1 = − (2a0 − b0 − 4c0 ) 3 6 ⋅ 4n c2 n +1 = 0. a2n +1 =

n +1 n +1 1 1 1 ⎛1⎞ ⎛1⎞ a0 + b0 + c0 − ⎜ ⎟ a0 + ⎜ ⎟ c0 4 4 2 ⎝2⎠ ⎝2⎠

1 ⎛1⎞ −⎜ ⎟ 4 ⎝2⎠

⎛1⎞ Since ⎜ ⎟ ⎝2⎠

n +1

n +1

5 1 + (2a0 − b0 − 4c0 ) 12 6 ⋅ 4n 1 = 2 1 1 = − (2a0 − b0 − 4c0 ) 12 6 ⋅ 4n

a2 n =

()

n +1

1 4 1 2 1 4

(a0 − c0 ).

4. The characteristic polynomial of M is 1⎞ ⎛ (λ − 1) ⎜ λ − ⎟ , so the eigenvalues are λ1 = 1 2⎠ ⎝ 1 and λ 2 = . Corresponding eigenvectors are 2

approaches zero as n → ∞, we

1 1 1 obtain an → , bn → , and cn → as 4 2 4 n → ∞.

easily found to be e1 = [1 0]T and e2 = [1 −1]T . From this point, the verification of Equation (7) is in the text.

3. Call M1 the matrix of Example 1, and M 2 the matrix of Exercise 2. Then

258


1 5. From Equation (9), if b0 = .25 = , we get b1 = 4

Section 10.16

1 4 9 8

2 = , then b2 = 9

2 9 10 9

1 = , b3 = 5

1 5 11 10

=

2 , and, in general, 11

2 2 . We will reach = .10 in 12 generations. According to Equation (8), under the controlled program 8+ n 20 1 1 the percentage would be in 12 generations, or = .00006 = .006%. 14 16 , 384 2 bn =

1 ⎡ ⎤ 2 ⎢ ⎥ 1 ⎢ ⎥ 2 ⎢ ⎥ 0 ⎥ 6. P −1x(0) = ⎢ 0 ⎢ ⎥ ⎢1 ( ⎥ ) ⎢ 12 1 + 5 ⎥ ⎢ 1 (1 − 5 ) ⎥ ⎣ 12 ⎦ 1 ⎡ ⎤ 2 ⎢ ⎥ 1 ⎢ ⎥ 2 ⎢ ⎥ 0 ⎥ n −1 (0) ⎢ D P x =⎢ 0 ⎥ ⎢1 1 n +1 ⎥ ⎢ 3 ⎡⎣ 4 (1 + 5 ) ⎤⎦ ⎥ ⎢ n +1 ⎥ ⎢ 1 ⎡ 1 (1 − 5 ) ⎤ ⎥ ⎦ ⎦ ⎣3 ⎣4 n +1 n +1 ⎡1 + 1 ⋅ 1 ⎡ + ( −3 + 5 )(1 − 5 ) ⎤⎦ ⎤ ⎢ 2 3 4n + 2 ⎣( −3 − 5 )(1 + 5 ) ⎥ ⎢ ⎥ n +1 n +1 ⎤ 1⋅ 1 ⎡ + (1 − 5 ) ⎦ 3 4n +1 ⎣(1 + 5 ) ⎢ ⎥ ⎢ ⎥ n n⎤ 1⋅ 1 ⎡ ⎥ 3 4n +1 ⎣(1 + 5 ) + (1 − 5 ) ⎦ n −1 (0) ⎢ PD P x = ⎢ ⎥ n n 1⋅ 1 ⎡ ⎤ ⎢ ⎥ 3 4n +1 ⎣(1 + 5 ) + (1 − 5 ) ⎦ ⎢ ⎥ n +1 n +1 1⋅ 1 ⎡ ⎢ ⎥ + (1 − 5 ) ⎤⎦ 3 4n +1 ⎣(1 + 5 ) ⎢ ⎥ ⎢ 1 + 1 ⋅ n1+ 2 ⎡( −3 − 5 )(1 + 5 )n +1 + ( −3 + 5 )(1 − 5 )n +1 ⎤ ⎥ ⎦⎦ ⎣2 3 4 ⎣

⎡1⎤ ⎢2⎥ ⎢0⎥ ⎢0⎥ 1 1 [As n tends to infinity, and approach 0, so x( n) approaches ⎢ ⎥ . n+ 2 n +1 0 4 4 ⎢ ⎥ ⎢0⎥ ⎢1⎥ ⎣2⎦

7. From (13) we have that the probability that the limiting sibling-pairs will be type (A, AA) is 2 1 2 1 a0 + b0 + c0 + d0 + e0 . 3 3 3 3 The proportion of A genes in the population at the outset is as follows: all the type

259



⎡857 ⎤ (c) x(6) = Lx(5) = ⎢ ⎣ 285⎥⎦ ⎡ 855⎤ x(6) λ1x(5) = ⎢ ⎣ 287 ⎥⎦

2 1 of the type (A, Aa) genes, 3 3 the type (A, aa) genes, etc. ...yielding 2 1 2 1 a0 + b0 + c0 + d0 + e0 . 3 3 3 3 (A, AA) genes,

5. a1 is the average number of offspring produced

8. From an (A, AA) pair we get only (A, AA) pairs and similarly for (a, aa). From either (A, aa) or (a, AA) pairs we must get an Aa female, who will not mature. Thus no offspring will come from such pairs. The transition matrix is then ⎡1 0 0 0 ⎤ ⎢0 0 0 0⎥ ⎢ ⎥. ⎢0 0 0 0⎥ ⎣⎢0 0 0 1 ⎦⎥

in the first age period. a2b1 is the number of offspring produced in the second period times the probability that the female will live into the second period, i.e., it is the expected number of offspring per female during the second period, and so on for all the periods. Thus, the sum of these, which is the net reproduction rate, is the expected number of offspring produced by a given female during her expected lifetime. ⎛ 1 ⎞ ⎛ 1 ⎞⎛ 1 ⎞ 19 7. R = 0 + 4 ⎜ ⎟ + 3 ⎜ ⎟⎜ ⎟ = = 2.375 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 4 ⎠ 8

9. For the first column of M we realize that parents of type (A, AA) can produce offspring only of that type, and similarly for the last column. The fifth column is like the second column, and follows the analysis in the text. For the middle two columns, say the third, note that male offspring from (A, aa) must be of type a, and females are of type Aa, because of the way the genes are inherited.

8. R = 0 + (.00024)(.99651) + " + (.00240)(.99651)" (.987) = 1.49611. Section 10.18 Exercise Set 10.18

Section 10.17

1. (a) The characteristic polynomial of L is 3 ⎛ 1⎤ 3⎞⎡ ⎛ 3⎞ λ3 − 2λ − = ⎜ λ − ⎟ ⎢λ 2 + ⎜ ⎟ λ + ⎥ , so 8 ⎝ 4⎦ ⎝ 2⎠ 2⎠⎣

Exercise Set 10.17 1. (a) The characteristic polynomial of L is 3 λ 2 − λ − , so the eigenvalues of L are 4 3 1 3 λ = and λ = − , thus λ1 = and 2 2 2 ⎡1⎤ ⎢ 1 ⎥ ⎡1 ⎤ x1 = ⎢ 2 ⎥ = ⎢ 1 ⎥ is the corresponding ⎢ 3 ⎥ ⎣3⎦ ⎣2⎦ eigenvector.

3 λ1 = . Thus h, the fraction harvested of 2 2 1 each age group, is 1 − = so the yield is 3 3 1 33 % of the population. The eigenvector 3 ⎡1⎤ ⎢1⎥ 3 corresponding to λ = is ⎢ 3 ⎥ ; this is the 2 ⎢1⎥ ⎣ 18 ⎦ age distribution vector after each harvest.

⎡100 ⎤ ⎡175⎤ (b) x(1) = Lx(0) = ⎢ ⎥ , x( 2) = Lx(1) = ⎢ ⎥ , 50 ⎣ ⎦ ⎣ 50 ⎦ ⎡ 250⎤ x(3) = Lx( 2) = ⎢ , ⎣ 88⎥⎦ ⎡382 ⎤ 5 ⎡570 ⎤ x( 4) = Lx(3) = ⎢ , x = Lx ( 4 ) = ⎢ ⎣125⎥⎦ ⎣ 191⎥⎦

(b) From Equation (10), the age distribution

vector x1 is ⎡1 ⎣ us that h1 = 1 −

1 2

1 19 8

=

1⎤ 8⎦

T

. Equation (9) tells

11 11 , so we harvest 19 19

or 57.9% of the youngest age class. Since Lx1 = ⎡ 19 ⎣8 260

1 2

1⎤ 8⎦

T

, the youngest class


Section 10.19

contains 79.2% of the population. Thus the yield is 57.9% of 79.2%, or 45.8% of the population. 2. The Leslie matrix of Example 1 has b1 = .845, b2 = .975, b3 = .965, etc. This, together with the harvesting data from Equations (13) and the formula of Equation (5) yields x1 = [1 .845 .824 .795 .755 .699 .626 .5323 0 0 0 0]T Lx1 = [ 2.090 .845 .824 .795 .755 .699 .626 .532 .418 0 0 0]T . The total of the entries of Lx1 is 7.584. The proportion of sheep harvested is h1 ( Lx1 )2 + h9 ( Lx1 )9 = 1.51, or 19.9% of the population. 4. In this situation we have hI ≠ 0, and h1 = h2 = " = hI −1 = hI +1 = " = hn = 0. Equation (4) then takes the form a1 + a2 b1 + a3b1b2 + " + aI b1b2 " bI −1 (1 − hI ) + aI +1b1b2 " bI (1 − hI ) + " + an b1b2 " bn −1 (1 − hI ) = 1.

(1 − hI )[ aI b1b2 " bI −1 + aI +1b1b2 " bI + " + an b1b2 " bn −1 ] = 1 − a1 − a2 b1 − " − aI −1b1b2 " bI − 2 .

a + a b + " + aI −1b1b2 " bI − 2 − 1 So hI = 1 2 1 +1 aI b1b2 " bI −1 + " + an b1b2 " bn −1 a + a b + " + aI −1b1b2 " bI − 2 − 1 + aI b1b2 " bI −1 + " + an b1b2 " bn −1 = 1 21 aI b1b2 " bI −1 + " + an b1b2 " bn −1 R −1 = aI b1b2 " bI −1 + " + an b1b2 " bn −1 5. Here hJ = 1, hI ≠ 0, and all the other hk’s are zero. Then Equation (4) becomes

a1 + a2 b1 + " + aI −1b1b2 " bI − 2 + (1 − hI )[ aI b1b2 " bI −1 + " + aJ −1b1b2 " bJ − 2 ] = 1.

a1 + a2b1 + " + aI −1b1b2 " bI − 2 − 1 +1 aI b1b2 " bI −1 + " + aJ −1b1b2 " bJ −2 a + a b + " + aJ −1b1b2 " bJ − 2 − 1 . = 1 21 aI b1b2 " bI −1 + " + aJ −1b1b2 " bJ −2

We solve for hI to obtain hI =

Section 10.19 Exercise Set 10.19 1. From Theorem 10.19.1, we compute a0 =

bk =

π2 3

1

2π

π ∫0

1

2π

π ∫0

2 1 (t − π )2 dt = π 2 , ak = π 3

2π

∫0

(t − π )2 cos kt dt =

4 k2

, and

(t − π )2 sin kt dt = 0. So the least-squares trigonometric polynomial of order 3 is

4 + 4 cos t + cos 2t + cos 3t . 9

2. From Theorem 10.19.2, we compute a0 =

2 T 2 2 2 2 T 2 2 kπ t T2 , t dt = T = = and a t cos dt k 3 T ∫0 T ∫0 T k 2π 2

2 T 2 2 kπ t T2 = − t sin dt . T ∫0 T kπ So the least-squares trigonometric polynomial of order 4 is bk =

2 T2 T2 ⎛ 2π t 1 4π t 1 6π t 1 8π t ⎞ T ⎛ 2π t 1 4π t 1 6π t 1 8π t ⎞ + − cos cos cos cos + + + + sin + sin + sin ⎜ ⎟ ⎜ sin ⎟. 3 π2 ⎝ T T T 16 T ⎠ π ⎝ T T T T ⎠ 4 9 2 3 4

261


3. From Theorem 10.19.2, a0 =

ak =

1

2π

π ∫0


f (t )dt =

1 π

π

2

∫0 sin t dt = π . (Note the upper limit on the second integral),

1 π

π

∫0 sin t cos kt dt π

1⎛ 1 ⎞ = ⎜ 2 [ k sin kt sin t + cos kt cos t ⎟ π ⎝ k −1 ⎠0 1⎛ 1 ⎞ = ⎜ 2 [0 + (−1)k −1 − 1] ⎟ π ⎝ k −1 ⎠ 0 if is odd k ⎧ ⎪ 2 =⎨ if k is even. − ⎪ π (k 2 − 1) ⎩ ⎧1 1 π ⎪ if k = 1 bk = ∫ sin kt sin t dt = ⎨ 2 . π 0 ⎪⎩0 if k > 1

So the least-squares trigonometric polynomial of order 4 is


ak =

1 2π

π ∫0

=−

1 2π

π ∫0

1

π

1 2 2 cos 2t − cos 4t . + sin t − 2 3π 15π

1 4 sin t dt = . 2 π

1 sin t cos kt dt 2 4

π (4k 2 − 1)

4 , π (2k − 1)(2k + 1) 1 2π 1 bk = ∫ sin t sin kt dt = 0. 0 2 π =−

So the least-square trigonometric polynomial of order n is

2

π

−

4 ⎛ cos t

cos 2t

2 T 2 1T 2 T T f (t )dt = ∫ 2 t dt + ∫ 1 (T − t )dt = . ∫ 2 T 0 T 0 T 2T 1T 2 2kπ t 2 T 2 kπ t ak = ∫ 2 t cos dt + ∫ 1 (T − t ) cos dt T 0 T T 2T T 4T ((−1)k − 1) = 4k 2π 2 if k is even ⎧0 ⎪ 8T . =⎨ if k is odd ⎪ ( 2k ) 2 π 2 ⎩ So the least-squares trigonometric polynomial of order n is: 2π t 1 6π t 1 10π t 1 2π nt ⎞ T 8T ⎛ 1 + cos + +" + cos cos cos − ⎜ 2 2 2 2 T T 10 T T ⎟⎠ 4 π2 ⎝2 6 ( 2 n)


if n is even; the last term involves n − 1 if n is odd.

262

cos 3t

cos nt

⎞

+ + +" + . 3⋅5 5⋅7 (2n − 1)(2n + 1) ⎟⎠ π ⎜⎝ 1 ⋅ 3


6. (a)

(b)

1 =

2π

∫0

cos kt = =

Section 10.20

(c) As in part (a) the system for c1 , c2 , and c3 ⎡3 −2 3⎤ ⎡ c1 ⎤ ⎡0 ⎤ is ⎢3 −2 0 ⎥ ⎢c2 ⎥ = ⎢0 ⎥ which has the ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 1 1 1⎦ ⎣ c3 ⎦ ⎣1 ⎦

dt = 2π 2π

∫0

cos2 kt dt

2 3 unique solution c1 = , c2 = , and 5 5 c3 = 0. Because these coefficients are all nonnegative, it follows that v is a convex combination of the vectors v1 , v 2 , and

2π ⎛ 1 + cos 2t ⎞

∫0

⎜ ⎝

2

⎟ dt ⎠

= π. (c)

2π

sin kt =

∫0

=

∫0

sin 2 kt dt

v3 .

2π ⎛ 1 − cos 2t ⎞

⎜ ⎝

2

⎟ dt ⎠

(d) As in part (a) the system for c1 , c2 , and c3

⎡3 −2 3⎤ ⎡ c1 ⎤ ⎡1 ⎤ is ⎢3 −2 0 ⎥ ⎢c2 ⎥ = ⎢0 ⎥ which has the ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 1 1 1⎦ ⎣ c3 ⎦ ⎣1 ⎦ 4 6 unique solution c1 = , c2 = , and 15 15 5 c3 = . Because these coefficients are all 15 nonnegative, it follows that v is a convex combination of the vectors v1 , v 2 , and

= π Section 10.20 Exercise Set 10.20

⎡1⎤ ⎡3⎤ ⎡ 4 ⎤ ⎡3⎤ 1. (a) Equation (2) is c1 ⎢ ⎥ + c2 ⎢ ⎥ + c3 ⎢ ⎥ = ⎢ ⎥ ⎣1⎦ ⎣5⎦ ⎣ 2 ⎦ ⎣3⎦ and Equation (3) is c1 + c2 + c3 = 1. These equations can be written in combined matrix ⎡1 3 4 ⎤ ⎡ c1 ⎤ ⎡3⎤ form as ⎢1 5 2 ⎥ ⎢c2 ⎥ = ⎢3⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣1 1 1 ⎦ ⎣ c3 ⎦ ⎣1⎦

v3 . 2. For both triangulations the number of triangles, m, is equal to 7; the number of vertex points, n, is equal to 7; and the number of boundary vertex points, k, is equal to 5. Equation (7), m = 2n − 2 − k, becomes 7 = 2(7) − 2 − 5, or 7 = 7.

1 This system has the unique solution c1 = , 5 2 2 c2 = , and c3 = . Because these 5 5 coefficients are all nonnegative, it follows that v is a convex combination of the vectors v1 , v 2 , and v3 .

3. Combining everything that is given in the statement of the problem, we obtain: w = Mv + b = M (c1 v1 + c2 v 2 + c3 v3 ) + (c1 + c2 + c3 )b = c1 ( Mv1 + b) + c2 ( Mv 2 + b) + c3 ( Mv3 + b) = c1w1 + c2 w 2 + c3 w 3 .

(b) As in part (a) the system for c1 , c2 , and c3 ⎡1 3 4 ⎤ ⎡ c1 ⎤ ⎡ 2 ⎤ is ⎢1 5 2 ⎥ ⎢c2 ⎥ = ⎢ 4 ⎥ which has the ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣1 1 1 ⎦ ⎣ c3 ⎦ ⎣1 ⎦

v1

4. (a)

2 4 unique solution c1 = , c2 = , and 5 5 1 c3 = − . Because one of these coefficients 5 is negative, it follows that v is not a convex combination of the vectors v1 , v 2 , and

v2 v3

v4 v5 v6

v3 .

263

v7


(b)

v1


v2

Solving these two linear systems leads to ⎡3 −1⎤ ⎡0 ⎤ and b = ⎢ ⎥ . M =⎢ ⎣1 ⎦ ⎣ 1 1⎥⎦

v3 v4

(c) As in part (a), we are led to the following two linear systems: ⎡ −2 1 1⎤ ⎡ m11 ⎤ ⎡ 0⎤ ⎢ 3 5 1⎥ ⎢ m12 ⎥ = ⎢ 5⎥ and ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎣ 1 0 1⎦ ⎣ b1 ⎦ ⎣ 3⎦

v5 v6

v7

⎡ −2 1 1⎤ ⎡ m21 ⎤ ⎡ −2 ⎤ ⎢ 3 5 1⎥ ⎢ m22 ⎥ = ⎢ 2 ⎥ . ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎣ 1 0 1⎦ ⎣ b2 ⎦ ⎣ −3⎦ Solving these two linear systems leads to ⎡1 0 ⎤ ⎡ 2⎤ M =⎢ and b = ⎢ ⎥ . ⎥ 0 1 ⎣ ⎦ ⎣ −3⎦

m12 ⎤ ⎡m ⎡ b1 ⎤ 5. (a) Let M = ⎢ 11 ⎥ and b = ⎢b ⎥ . Then m m ⎣ 21 ⎣ 2⎦ 22 ⎦ the three matrix equations Mvi + b = w i , i = 1, 2, 3, can be written as the six scalar equations m11 + m12 + b1 = 4 m21 + m22 + b2 = 3

(d) As in part (a), we are led to the following two linear systems: ⎡ 5⎤ ⎡ 0 2 1⎤ ⎡ m11 ⎤ ⎢ 2 ⎥ ⎢ 2 2 1⎥ ⎢ m12 ⎥ = ⎢ 7 ⎥ and ⎥ ⎢ 2⎥ ⎢ ⎥⎢ ⎣ −4 −2 1⎦ ⎣ b1 ⎦ ⎢ − 7 ⎥ ⎣ 2⎦ ⎡ 0 2 1⎤ ⎡ m21 ⎤ ⎡ −1⎤ ⎢ 2 2 1⎥ ⎢ m22 ⎥ = ⎢ 3⎥ . ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎣ −4 −2 1⎦ ⎣ b2 ⎦ ⎣ −9 ⎦

2m11 + 3m12 + b1 = 9 2m21 + 3m22 + b2 = 5 2m11 + m12 + b1 = 5 2m21 + m22 + b2 = 3 The first, third, and fifth equations can be written in matrix form as ⎡1 1 1⎤ ⎡ m11 ⎤ ⎡ 4 ⎤ ⎢ 2 3 1⎥ ⎢ m12 ⎥ = ⎢9 ⎥ and the second, ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎣ 2 1 1⎦ ⎣ b1 ⎦ ⎣ 5 ⎦ fourth, and sixth equations as ⎡1 1 1⎤ ⎡ m21 ⎤ ⎡3⎤ ⎢ 2 3 1⎥ ⎢ m22 ⎥ = ⎢5⎥ . ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎣ 2 1 1⎦ ⎣ b2 ⎦ ⎣3⎦ The first system has the solution m11 = 1, m12 = 2, b1 = 1 and the second system has the solution m21 = 0, m22 = 1, b2 = 2.

Solving these two linear systems leads to ⎡ 1 1⎤ ⎡ 1⎤ M = ⎢2 ⎥ and b = ⎢ 2 ⎥ . ⎣ 2 0⎦ ⎣ −1⎦ 7. (a) The vertices v1 , v 2 , and v3 of a triangle can be written as the convex combinations v1 = (1) v1 + (0) v 2 + (0) v3 , v 2 = (0) v1 + (1) v 2 + (0) v3 , and v3 = (0) v1 + (0) v 2 + (1) v3 . In each of these cases, precisely two of the coefficients are zero and one coefficient is one.

⎡1 2 ⎤ ⎡1 ⎤ and b = ⎢ ⎥ . Thus we obtain M = ⎢ ⎣2⎦ ⎣0 1 ⎥⎦

(b) If, for example, v lies on the side of the triangle determined by the vectors v1 and

(b) As in part (a), we are led to the following two linear systems: ⎡ −2 2 1⎤ ⎡ m11 ⎤ ⎡ −8⎤ ⎢ 0 0 1⎥ ⎢ m12 ⎥ = ⎢ 0⎥ and ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎣ 2 1 1⎦ ⎣ b1 ⎦ ⎣ 5⎦

v 2 then from Exercise 6(a) we must have

that v = c1 v1 + c2 v 2 + (0) v3 where c1 + c2 = 1. Thus at least one of the coefficients, in this example c3 , must equal zero.

⎡ −2 2 1⎤ ⎡ m21 ⎤ ⎡1 ⎤ ⎢ 0 0 1⎥ ⎢ m22 ⎥ = ⎢1 ⎥ . ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎣ 2 1 1⎦ ⎣ b2 ⎦ ⎣ 4 ⎦

264


Section 10.20

(c) From part (b), if at least one of the coefficients in the convex combination is zero, then the vector must lie on one of the sides of the triangle. Consequently, none of the coefficients can be zero if the vector lies in the interior of the triangle. 8. (a) Consider the vertex v1 of the triangle and its opposite side determined by the vectors v 2 and v3 . The midpoint v m of this

( v 2 + v3 ) and the point on 2 the line segment from v1 to v m that is twothirds of the distance to v m is given by opposite side is

1 2 1 2 ⎛ v + v3 ⎞ v1 + v m = v1 + ⎜ 2 ⎟ 3 3 3 3⎝ 2 ⎠ 1 1 1 = v1 + v 2 + v3 . 3 3 3 (b)

1 ⎡ 2 ⎤ 1 ⎡ 5 ⎤ 1 ⎡1⎤ 1 ⎡8 ⎤ ⎡ 83 ⎤ + + = =⎢ ⎥ 3 ⎢⎣ 3 ⎥⎦ 3 ⎢⎣ 2 ⎥⎦ 3 ⎢⎣1⎥⎦ 3 ⎢⎣6 ⎥⎦ ⎣⎢ 2 ⎦⎥

265

Chapter 2 Determinants 4 −1 6 (b) M 23 = 4 1 14 4 1 2 1 14 4 14 4 1 =4 − (−1) +6 1 2 4 2 4 1 = 4(2 − 14) + (8 − 56) + 6(4 − 4) = −96


7 −1 1. M11 = = 28 − (−1) = 29 1 4 C11 = (−1)1+1 M11 = 29 M12 =

6 −1 = 24 − 3 = 21 −3 4

C23 = (−1)2 +3 M 23 = 96

C12 = (−1)1+ 2 M12 = −21 M13 =

4 1 6 (c) M 22 = 4 0 14 4 3 2 1 6 4 6 4 1 = −4 +0 − 14 3 2 4 2 4 3 = −4(2 − 18) − 14(12 − 4) = −48

6 7 = 6 − (−21) = 27 −3 1

C13 = (−1)1+3 M13 = 27 M 21 =

−2 3 = −8 − 3 = −11 1 4

C22 = (−1) 2+ 2 M 22 = −48

C21 = (−1)2 +1 M 21 = 11 M 22 =

1 3 = 4 − (−9) = 13 −3 4

−1 1 6 (d) M 21 = 1 0 14 1 3 2 1 6 −1 6 −1 1 = −1 +0 − 14 3 2 1 2 1 3 = −(2 − 18) − 14(−3 − 1) = 72

C22 = (−1) 2+ 2 M 22 = 13 M 23 =

1 −2 = 1 − 6 = −5 −3 1

C23 = (−1)2 +3 M 23 = 5 M 31 =

−2 3 = 2 − 21 = −19 7 −1

C21 = (−1)2 +1 M 21 = −72

C31 = (−1)3+1 M 31 = −19 M 32

5.

1 3 = = −1 − 18 = −19 6 −1

C32 = (−1) M 33 =

3+ 2

⎡ 3 5⎤ ⎢⎣ −2 4⎥⎦

M 32 = 19

1 −2 = 7 − (−12) = 19 6 7 7.

C33 = (−1)3+3 M 33 = 19

3. (a) M13

3 5 = 12 − (−10) = 22 −2 4 −1

C13 = (−1)1+3 M13 = 0

44

5 2 1 ⎡ 4 −5⎤ ⎡ 11 − 22 ⎤ ⎢ ⎥ = 3 ⎥ 3⎥⎦ ⎢ 1 22 ⎢⎣ 2 22 ⎦ ⎣ 11

−5 7 = 10 − (−49) = 59 −7 −2 ⎡ −5 7 ⎤ ⎢⎣ −7 −2 ⎥⎦

0 0 3 4 1 = 4 1 14 = 3 = 3(4 − 4) = 0 4 1 4 1 2

=

−1

=

2 1 ⎡ −2 −7 ⎤ ⎡ − 59 ⎢ = 59 ⎢⎣ 7 −5⎥⎦ ⎢ 7 ⎣ 59

7 ⎤ − 59 ⎥ 5⎥ − 59 ⎦


9.

Section 2.1

a −3 5 = (a − 3)(a − 2) − (5)(−3) −3 a − 2 = a 2 − 5a + 6 + 15 = a 2 − 5a + 21

11.

13.

−2 1 4 −2 1 4 −2 1 3 5 −7 = 3 5 −7 3 5 1 6 2 1 6 2 1 6 = [(−2)(5)(2) + (1)(−7)(1) + (4)(3)(6)] − [(4)(5)(1) + (−2)(−7)(6) + (1)(3)(2)] = [ −20 − 7 + 72] − [ 20 + 84 + 6] = −65 3 0 0 3 0 03 0 2 −1 5 = 2 −1 5 2 −1 1 9 −4 1 9 −4 1 9 = [(3)(−1)(−4) + (0)(5)(1) + (0)(2)(9)] − [0(−1)(1) + (3)(5)(9) + (0)(2)(−4)] = 12 − 135 = −123

15. det( A) = (λ − 2)(λ + 4) − (−5)

= λ 2 + 2λ − 8 + 5 = λ 2 + 2λ − 3 = (λ − 1)(λ + 3) det(A) = 0 for λ = 1 or −3. 17. det(A) = (λ − 1)(λ + 1) − 0 = (λ − 1)(λ + 1) det(A) = 0 for λ = 1 or −1.

3 0 0 −1 5 2 5 2 −1 19. (a) 2 −1 5 = 3 −0 +0 9 −4 1 −4 1 9 1 9 −4 = 3(4 − 45) = −123 3 0 0 −1 5 0 0 0 0 (b) 2 −1 5 = 3 −2 +1 9 −4 9 −4 −1 5 1 9 −4 = 3(4 − 45) − 2(0 − 0) + (0 − 0) = −123

(c)

3 0 0 0 0 3 0 3 0 + (−1) −5 2 −1 5 = −2 9 −4 1 −4 1 9 1 9 −4 = −2(0 − 0) − (−12 − 0) − 5(27 − 0) = 12 − 135 = −123

45

Chapter 2: Determinants


3 0 0 2 5 3 0 3 0 (d) 2 −1 5 = −0 + (−1) −9 1 −4 1 −4 2 5 1 9 −4 = −(−12 − 0) − 9(15 − 0) = 12 − 135 = −123

(e)

(f)

3 0 0 0 0 3 0 3 0 −9 + (−4) 2 −1 5 = 1 −1 5 2 5 2 −1 1 9 −4 = (0 − 0) − 9(15 − 0) − 4(−3 − 0) = −135 + 12 = −123 3 0 0 2 −1 3 0 3 0 2 −1 5 = 0 −5 + (−4) 1 9 1 9 2 −1 1 9 −4 = −5(27 − 0) − 4(−3 − 0) = −135 + 12 = −123

21. Expand along the second column. −3 7 det( A) = 5 = 5[ −15 − (−7)] = −40 −1 5 23. Expand along the first column.

det( A) = 1

k

k2

k

k2

−1

k

k2

k

k2

+1

k

k2

k

k2

= (k 3 − k 3 ) − (k 3 − k 3 ) + (k 3 − k 3 ) =0 25. Expand along the third column, then along the first rows of the 3 × 3 matrices. 3 3 5 3 3 5 det( A) = −3 2 2 −2 − 3 2 2 −2 2 10 2 4 1 0

⎛ 2 −2 2 −2 2 2 ⎞ ⎛ 2 −2 2 −2 2 2⎞ = 3⎜ 3 −3 +5 ⎟ − 3⎜ 3 1 0 − 3 4 0 + 5 4 1 ⎟ 10 2 2 2 2 10 ⎝ ⎠ ⎝ ⎠ = −3[3(24) − 3(8) + 5(16)] − 3[3(2) − 3(8) + 5(−6)] = −3(128) − 3(−48) = −240

27.

1 0 0 0 −1 0 = (1)(−1)(1) = −1 0 0 1

46


0 1 29. 0 1

0 2 4 2

0 0 3 3

1 0 31. 0 0

2 7 −3 1 −4 1 = (1)(1)(2)(3) = 6 0 2 7 0 0 3

Section 2.1

0 0 = (0)(2)(3)(8) = 0 0 8

33. Expand along the third column. sin(θ ) cos(θ ) 0 sin(θ ) cos(θ ) sin(θ ) 0 = − cos(θ ) − cos(θ ) sin(θ ) sin(θ ) − cos(θ ) sin(θ ) + cos(θ ) 1

= sin 2 (θ ) − (− cos2 (θ )) = sin 2 (θ ) + cos2 (θ ) =1 1 35. M11 = 0

f = 1 in both matrices, so expanding along the first row or column gives that d 2 = d1 + λ. 1

True/False 2.1 (a) False; the determinant is ad − bc.

(b) False;

1 0 0 1 0 = 0 1 0 =1 0 1 0 0 1

(c) True; if i + j is even then (−a)i + j = 1 and Cij = (−1)i + j M ij = M ij . Similarly, if Cij = (−1)i + j M ij = M ij , then

(−1)i + j = 1 so i + j must be even. ⎡a b (d) True; let A = ⎢ b d ⎢ ⎣c e

c⎤ b e ⎥ , then C12 = (−1) c ⎥ f⎦

e b = −(bf − ce) and C21 = (−1) f e

c = −(bf − ce). Similar f

arguments show that C23 = C32 and C13 = C31 . (e) True; the value is det(A) regardless of the row or column chosen. (f) True (g) False; the determinant would be the product, not the sum.

c 0 ⎡1 0 ⎤ = c 2 . In general, if A is an n × n matrix, then (h) False; let A = ⎢ , then det(A) = 1, but det(cA) = ⎥ 0 c 0 1 ⎣ ⎦ det(cA) = c n A.

47



⎡1 0 ⎤ ⎡ −1 0 ⎤ (i) False; let A = ⎢ and B = ⎢ , then ⎣ 0 −1⎥⎦ ⎣0 1 ⎥⎦ det(A) = det(B) = 1, but det(A + B) = 0. 2 ⎡a b⎤ 2 ⎡ a + bc ab + bd ⎤ = A (j) True; let A = ⎢ so that ⎢ ⎥ and ⎣ c d ⎥⎦ ⎢⎣ ac + cd bc + d 2 ⎦⎥

det( A2 ) = (a 2 + bc)(bc + d 2 ) − (ab + bd )(ac + cd ) = a 2 bc + a 2 d 2 + bcd 2 + b 2 c 2 − (a 2bc + 2abcd + bcd 2 ) = a 2 d 2 − 2abcd + b 2 c 2 = (ad − bc)2 = (det( A))2 Section 2.2 Exercise Set 2.2 1. det(A) = (−2)(4) − (3)(1) = −8 − 3 = −11 ⎡ −2 1⎤ AT = ⎢ ⎣ 3 4 ⎥⎦

det( AT ) = (−2)(4) − (1)(3) = −8 − 3 = −11 3. det( A) = [(2)(2)(6) + (−1)(4)(5) + (3)(1)(−3)] − [(3)(2)(5) + (2)(4)(−3) + (−1)(1)(6)] = [ 24 − 20 − 9] − [30 − 24 − 6] = −5

det( AT ) = [(2)(2)(6) + (1)(−3)(3) + (5)(−1)(4)] − [(5)(2)(3) + (2)(−3)(4) + (1)(−1)(6)] = [ 24 − 9 − 20] − [30 − 24 − 6] = −5 5. The matrix is I 4 with the third row multiplied by −5. 1 0 0 0 0 1 0 0 = −5 0 0 −5 0 0 0 0 1 7. The matrix is I 4 with the second and third rows interchanged. 1 0 0 0 0 0 1 0 = −1 0 1 0 0 0 0 0 1 9. The matrix is I 4 with −9 times the fourth row added to the second row. 1 0 0 0 0 1 0 −9 =1 0 0 1 0 0 0 0 1

48


11.

13.

Section 2.2

0 3 1 1 1 2 1 1 2 =−0 3 1 3 2 4 3 2 4 1 1 2 1 =−0 3 0 −1 −2 1 1 2 = 0 −1 −2 0 3 1 1 1 2 =−0 1 2 0 3 1 1 1 2 =−0 1 2 0 0 −5 = −(−5) =5

15.

3 −6 9 1 −2 3 2 7 2 3 2 7 2 − − = − − 0 1 5 0 1 5 1 −2 3 3 4 =30 0 1 5 1 −2 3 1 5 = −3 0 0 3 4 1 −2 3 1 5 = −3 0 0 0 −11 = −3(−11) = 33

17.

2 1 0 0

1 0 2 1

3 1 1 2

1 1 =− 0 3

1 0 1 1 2 1 3 1 0 2 1 0 0 1 2 3 1 0 1 1 0 1 1 −1 =− 0 2 1 0 0 1 2 3 1 0 1 1 0 1 1 −1 =− 0 0 −1 2 0 0 1 4 1 0 1 1 0 1 1 −1 = 0 0 1 −2 0 0 1 4 1 0 1 1 0 1 1 −1 = 0 0 1 −2 0 0 0 6 =6

1 3 1 5 3 1 3 −2 −7 0 −4 2 0 −1 0 0 1 0 1= 0 0 0 0 2 1 1 0 0 0 0 0 1 1 0 0 1 3 0 1 =−0 0 0 0 0 0 1 0 =−0 0 0 1 0 =−0 0 0 = −2

49

1 2 1 2 0 1 −2 1 2 0

5 6 0 1 1

3 8 1 1 1 5 3 −6 −8 0 1 1 1 1 1

3 1 5 1 −2 −6 0 1 0 0 0 1 0 0 1 3 1 5 1 −2 −6 0 1 0 0 0 1 0 0 0

3 −8 1 −1 1 3 −8 1 −1 2



19. Exercise 14: First add multiples of the first row to the remaining rows, then expand by cofactors along the first column. Repeat these steps with the 3 × 3 determinant, then evaluate the 2 × 2 determinant. 1 −2 3 1 1 −2 3 1 5 −9 6 3 0 1 −9 −2 = −1 2 −6 −2 0 0 −3 −1 2 8 6 1 0 12 0 −1 1 −9 −2 = 0 −3 −1 12 0 −1 1 −9 −2 = 0 −3 −1 0 108 23 −3 −1 = 108 23 = (−3)(23) − (−1)(108) = 39 Exercise 15: Add −2 times the second row to the first row, then expand by cofactors along the first column. For the 3 × 3 determinant, add multiples of the first row to the remaining row, then expand by cofactors of the first column. Finally, evaluate the 2 × 2 determinant. 2 1 3 1 0 1 1 −1 1 0 1 1 1 0 1 1 = 0 2 1 0 0 2 1 0 0 1 2 2 0 1 2 3 1 1 −1 = (−1) 2 1 0 1 2 3 1 1 −1 = − 0 −1 2 0 1 4 −1 2 =− 1 4 = −(−4 − 2) =6 1 Exercise 16: Add − times the first row to the 2 second row, then expand by cofactors along the fourth column. For the 3 × 3 determinant, add −2 times the second row to the third row, then expand by cofactors along the second column. Finally, evaluate the 2 × 2 determinant.

0 1 2 2 3 − 13

1 1 1

1 2 1 3 2 3

1 3

1

0

1 2

1 2 2 3 − 13

0

0 0

=

1 0

1 1 0

1 2 1 3

1 0 3 2 0 0 3 1 0 1 2 2 = (−1) 23 13 13 − 13 32 0 1 0 1 2 2 1 = − 23 13 3 − 53 0 − 23 1 1 ⎛1⎞ 2 = − ⎜ ⎟ 25 ⎝ 3 ⎠ − 3 − 23

1 ⎡ 1 ⎛ 5 ⎞⎤ = − ⎢− − ⎜ − ⎟⎥ 3 ⎣ 3 ⎝ 6 ⎠⎦ 1 =− 6 Exercise 17: Add 2 times the first row to the second row, then expand by cofactors along the first column two times. For the 3 × 3 determinant, add −2 times the first row to the second row, then expand by cofactors along the first column. Finally, evaluate the 2 × 2 determinant. 1 3 1 5 3 1 3 1 5 3 −2 −7 0 −4 2 0 −1 2 6 8 0 0 1 0 1= 0 0 1 0 1 0 0 2 1 1 0 0 2 1 1 0 0 0 1 1 0 0 0 1 1 −1 2 6 8 0 1 0 1 = 0 2 1 1 0 0 1 1 1 0 1 = (−1) 2 1 1 0 1 1 1 0 1 = − 0 1 −1 0 1 1 1 −1 =− 1 1 = −[1 − (−1)] = −2

50


Section 2.3

21. Interchange the second and third rows, then the first and second rows to arrive at the determinant whose value is given. d e f d e f a b c g h i = − a b c = d e f = −6 a b c g h i g h i

29.

1 a

1 b

1 c

a2

b2

c2

1 = 0

23. Remove the common factors of 3 from the first row, −1 from the second row, and 4 from the third row to arrive at the determinant whose value is given. 3a 3b 3c a b c − d −e − f = 3 − d −e − f 4 g 4h 4i 4 g 4h 4i a b c e f = −3 d 4 g 4h 4i a b c = −12 d e f g h i = −12(−6) = 72

=

1 b−a

1 c−a

0 b2 − a 2 c2 − a 2 b−a c−a b2 − a2

c2 − a2

= (b − a)(c 2 − a 2 ) − (c − a)(b 2 − a 2 ) = (b − a)(c + a)(c − a) − (c − a)(b + a)(b − a) = (b − a)(c − a)[c + a − (b + a)] = (b − a)(c − a)(c − b) True/False 2.2 (a) True; interchanging two rows of a matrix causes the determinant to be multiplied by −1, so two such interchanges result in a multiplication by (−1)(−1) = 1. (b) True; consider the transposes of the matrices so that the column operations performed on A to get

B are row operations on AT .

25. Add −1 times the third row to the first row to arrive at the matrix whose value is given. a+ g b+h c+i a b c d e f = d e f = −6 g h i g h i

(c) False; adding a multiple of one row to another does not change the determinant. In this case, det(B) = det(A). (d) False; multiplying each row by its row number causes the determinant to be multiplied by the row number, so det(B) = n! det(A).

27. Remove the common factor of −3 from the first row, then add 4 times the second row to the third to arrive at the matrix whose value is given. −3a −3b −3c d e f g − 4 d h − 4e i − 4 f a b c 3 d e f =− g − 4d h − 4e i − 4 f a b c = −3 d e f g h i = −3(−6) = 18

(e) True; since the columns are identical, they are proportional, so det(A) = 0 by Theorem 2.2.5. (f) True; −1 times the second and fourth rows can be added to the sixth row without changing det(A). Since this introduces a row of zeros, det(A) = 0. Section 2.3 Exercise Set 2.3 1. det(A) = −4 − 6 = −10 −2 4 det(2 A) = = −16 − 24 = −40 = 22 (−10) 6 8

51



3. det(A) = [20 − 1 + 36] − [6 + 8 − 15] = 56 −4 2 −6 det(−2 A) = −6 −4 −2 −2 −8 −10 = [ −160 + 8 − 288] − [ −48 − 64 + 120] = −448

13. Expand by cofactors of the first row. 1 0 det( A) = 2 = 2(6) = 12 3 6

Since det(A) ≠ 0, A is invertible. 15. det( A) = (k − 3)(k − 2) − 4

= k 2 − 5k + 6 − 4

= (−2)3 (56)

= k 2 − 5k + 2 Use the quadratic formula to solve

9 −1 8 5. det( AB) = 31 1 17 10 0 2 = [18 − 170 − 0] − [80 + 0 − 62] = −170 −1 −3 6 det( BA) = 17 11 4 10 5 2 = [ −22 − 120 + 510] − [660 − 20 − 102] = −170 2 1 det( A) = 2 = 2(8 − 3) = 10 3 4 det( B) = [1 − 10 + 0] − [15 + 0 − 7] = −17

k 2 − 5k + 2 = 0.

k=

−(−5) ± (−5)2 − 4(1)(2) 5 ± 17 = 2(1) 2

A is invertible for k ≠

5 ± 17 . 2

17. det( A) = [ 2 + 12k + 36] − [ 4k + 18 + 12] = 8k + 8 A is invertible for k ≠ −1. 19. A is the same matrix as in Exercise 7, which was shown to be invertible (det(A) = −1). The cofactors of A are C11 = −3, C12 = −(−3) = 3,

3 0 3 det( A + B) = 10 5 2 5 0 3 3 3 =5 5 3 = 5(9 − 15) = −30 ≠ det( A) + det( B)

C13 = −4 + 2 = −2, C21 = −(15 − 20) = 5, C22 = 6 − 10 = −4, C23 = −(8 − 10) = 2, C31 = 0 + 5 = 5, C32 = −(0 + 5) = −5, and C33 = −2 + 5 = 3.

⎡ −3 3 −2 ⎤ The matrix of cofactors is ⎢ 5 −4 2 ⎥ . ⎢ ⎥ 3⎦ ⎣ 5 −5 ⎡ −3 5 5⎤ 1 1 ⎢ A−1 = adj( A) = 3 −4 −5⎥ ⎥ −1 ⎢ −2 2 det( A) 3⎦ ⎣ ⎡ 3 −5 −5 ⎤ = ⎢ −3 4 5⎥ ⎢ ⎥ ⎣ 2 −2 −3⎦

7. Add multiples of the second row to the first and third row to simplify evaluating the determinant. 0 3 5 3 5 det( A) = −1 −1 0 = −(−1) = 9 − 10 = −1 2 3 0 2 3

Since det(A) ≠ 0, A is invertible. 9. Expand by cofactors of the first column. 1 −3 det( A) = 2 = 2(2) = 4 0 2

21. A is the same matrix as in Exercise 9, which was shown to be invertible (det(A) = 4). The cofactors of A are C11 = 2, C12 = 0, C13 = 0,

Since det(A) ≠ 0, A is invertible.

C21 = −(−6) = 6, C22 = 4, C23 = 0, C31 = 9 − 5 = 4, C32 = −(−6) = 6, and C33 = 2.

11. The third column of A is twice the first column, so det(A) = 0 and A is not invertible.

52


Section 2.3

⎡2 0 0⎤ The matrix of cofactors is ⎢ 6 4 0 ⎥ . ⎢ ⎥ ⎣4 6 2⎦ ⎡2 6 4⎤ 1 1⎢ −1 adj( A) = 0 4 6 ⎥ A = 4 ⎢0 0 2⎥ det( A) ⎣ ⎦ ⎡1 ⎢2 = ⎢0 ⎢ ⎢⎣ 0

1 3 1 C23 = (−1) 1 3 9 = 0 1 3 2 1 3 1 C24 = 1 3 8 = 0 1 3 2

1⎤ ⎥ 1 32 ⎥ ⎥ 0 12 ⎥ ⎦ 3 2

3 1 1 C31 = 5 2 2 = 0 3 2 2 1 1 1 C32 = (−1) 2 2 2 = 0 1 2 2

23. Use row operations and cofactor expansion to simplify the determinant. 1 3 1 1 0 −1 0 0 det( A) = 0 0 7 8 0 0 1 1 −1 0 0 = 0 7 8 0 1 1 7 8 = (−1) 1 1 = −(7 − 8) =1 Since det(A) ≠ 0, A is invertible. The cofactors of A are 5 2 2 C11 = 3 8 9 = −4 3 2 2

1 3 1 C33 = 2 5 2 = −1 1 3 2 1 3 1 C34 = (−1) 2 5 2 = −(−1) = 1 1 3 2 3 1 1 C41 = (−1) 5 2 2 = −(1) = −1 3 8 9 1 1 1 C42 = 2 2 2 = 0 1 8 9 1 3 1 C43 = (−1) 2 5 2 = −(−8) = 8 1 3 9

2 2 2 C12 = (−1) 1 8 9 = −(−2) = 2 1 2 2

1 3 1 C44 = 2 5 2 = −7 1 3 8

2 5 2 C13 = 1 3 9 = −7 1 3 2

6⎤ ⎡ −4 2 −7 ⎢ 3 −1 0 0⎥ The matrix of cofactors is ⎢ ⎥. ⎢ 0 0 −1 1⎥ 8 −7 ⎦⎥ ⎣⎢ −1 0

2 5 2 C14 = (−1) 1 3 8 = −(−6) = 6 1 3 2

A

3 1 1 C21 = (−1) 3 8 9 = −(−3) = 3 3 2 2 1 1 1 C22 = 1 8 9 = −1 1 2 2

53

−1

⎡ −4 1 1⎢ 2 adj( A) = ⎢ = det( A) 1 ⎢ −7 ⎢⎣ 6 ⎡ −4 ⎢ 2 =⎢ ⎢ −7 ⎢⎣ 6

3 0 −1⎤ −1 0 0 ⎥ ⎥ 0 −1 8⎥ 0 1 −7 ⎥⎦ 3 0 −1⎤ 0⎥ −1 0 ⎥ 0 −1 8 ⎥ 0 1 −7 ⎥⎦



1⎤ ⎡ 4 −3 A1 = ⎢ −2 −1 0 ⎥ ⎢ ⎥ ⎣ 0 0 −3⎦ 4 −3 det( A1 ) = (−3) = −3(−4 − 6) = 30 −2 −1

⎡ 4 5 0⎤ 25. A = ⎢11 1 2⎥ ⎢ ⎥ ⎣ 1 5 2⎦ 4 5 4 5 det( A) = −2 +2 1 5 11 1 = −2(20 − 5) + 2(4 − 55) = −132 ⎡ 2 5 0⎤ Ax = ⎢ 3 1 2⎥ ⎢ ⎥ ⎣ 1 5 2⎦

1⎤ ⎡1 4 A2 = ⎢ 2 −2 0 ⎥ ⎢ ⎥ ⎣ 4 0 −3⎦ 2 −2 1 4 det( A2 ) = + (−3) 4 0 2 −2 = (0 + 8) − 3(−2 − 8) = 38 ⎡ 1 −3 4 ⎤ A3 = ⎢ 2 −1 −2⎥ ⎢ ⎥ ⎣ 4 0 0⎦

2 5 2 5 +2 1 5 3 1 = −2(10 − 5) + 2(2 − 15) = −36 ⎡ 4 2 0⎤ Ay = ⎢11 3 2 ⎥ ⎢ ⎥ ⎣ 1 1 2⎦

det( Ax ) = −2

det( A3 ) = 4

4 2 4 2 +2 1 1 11 3 = −2(4 − 2) + 2(12 − 22) = −24 ⎡ 4 5 2⎤ Az = ⎢11 1 3⎥ ⎢ ⎥ ⎣ 1 5 1⎦

det( Ay ) = −2

−3 4 = 4(6 + 4) = 40 −1 −2

det( A1 ) 30 30 = =− det( A) −11 11 det( A2 ) 38 38 x2 = = =− det( A) −11 11 det( A3 ) 40 40 x3 = = =− det( A) −11 11 30 38 The solution is x1 = − , x2 = − , 11 11 40 x3 = − . 11 x1 =

1 3 5 2 5 2 − 11 + 5 1 5 1 1 3 = 4(1 − 15) − 11(5 − 10) + (15 − 2) = 12 det( Ax ) −36 3 = = x= det( A) −132 11 det( Ay ) −24 2 y= = = det( A) −132 11 det( Az ) 12 1 z= = =− det( A) −132 11 3 2 1 The solution is x = , y = , z = − . 11 11 11 det( Az ) = 4

⎡ 3 −1 1⎤ 29. A = ⎢ −1 7 −2 ⎥ ⎢ ⎥ ⎣ 2 6 −1⎦ Note that the third row of A is the sum of the first and second rows. Thus, det(A) = 0 and Cramer’s rule does not apply. ⎡ 4 1 1 1⎤ ⎢ 3 7 −1 1⎥ 31. A = ⎢ ⎥ ⎢ 7 3 −5 8⎥ ⎢⎣ 1 1 1 2 ⎥⎦ Use row operations and cofactor expansion to simplify the determinant.

1⎤ ⎡ 1 −3 27. A = ⎢ 2 −1 0⎥ ⎢ ⎥ ⎣ 4 0 −3⎦ 2 −1 1 −3 + (−3) 4 0 2 −1 = (0 + 4) − 3(−1 + 6) = −11

det( A) =

54


Section 2.3

8 ⎛ 1⎞ (c) det(2 A−1 ) = 23 det( A−1 ) = 8 ⎜ − ⎟ = − 7 7 ⎝ ⎠

0 −3 −3 −7 0 4 −4 −5 det( A) = 0 −4 −12 −6 1 1 1 2 −3 −3 −7 = −(1) 4 −4 −5 −4 −12 −6 −3 −3 −7 = − 4 −4 −5 0 −16 −11

(d) det((2 A)−1 ) =

=

23 det( A) 1 = 8(−7) 1 =− 56

⎛ −4 −5 −3 −7 ⎞ = − ⎜ −3 −4 − − − 16 11 16 −11 ⎟⎠ ⎝ = 3(44 − 80) + 4(33 − 112) = −424 1 1⎤ ⎡4 6 ⎢3 1 −1 1⎥ Ay = ⎢ ⎥ ⎢ 7 −3 −5 8 ⎥ 1 2 ⎦⎥ ⎣⎢ 1 3

⎡a (e) det ⎢ b ⎢ ⎣c

0 −6 −3 7 0 −8 −4 −5 det( Ay ) = 0 −24 −12 −6 1 3 1 2 −6 −3 7 = −(1) −8 −4 −5 −24 −12 −6 −6 −3 7 = − −8 −4 −5 0 0 −34 −6 −3 = −(−34) −8 −4 = −(−34)(24 − 24) =0 det( Ay ) 0 = =0 y= det( A) −424

in A

(b) det( A−1 ) =

d⎤ ⎡a b e ⎥ = det ⎢ g h ⎥ ⎢ f⎦ ⎣d e ⎡a b = − det ⎢ d e ⎢ ⎣g h = − det( A) =7

c⎤ i⎥ ⎥ f⎦ c⎤ f⎥ ⎥ i⎦

1 1 = det( A) 7

⎛1⎞ 8 (c) det(2 A−1 ) = 23 det( A−1 ) = 8 ⎜ ⎟ = ⎝7⎠ 7 (d) det((2 A)−1 ) =

=

1 det(2 A) 1

23 det( A) 1 = 8(7) 1 = 56 True/False 2.3 (a) False; det(2 A) = 23 det( A) if A is a 3 × 3 matrix.

are integers.

⎡ −1 0 ⎤ ⎡ 1 0⎤ (b) False; consider A = ⎢ and B = ⎢ . ⎣0 1⎥⎦ ⎣ 0 −1⎥⎦ Then det(A) = det(B) = 1, but det(A + B) = 0 ≠ det(2A) = 4.

35. (a) det(3 A) = 33 det( A) = 27(−7) = −189 (b) det( A−1 ) =

g h i

37. (a) det(3 A) = 33 det( A) = 27(7) = 189

33. If all the entries in A are integers, then all the cofactors of A are integers, so all the entries in adj(A) are integers. Since det(A) = 1 and 1 adj( A) = adj( A), then all the entries A−1 = det( A) −1

1 det(2 A) 1

1 1 1 = =− det( A) −7 7

55



(c) True; since A is invertible, then 1 det( A−1 ) = and det( A) 1 det( A−1BA) = ⋅ det( B) ⋅ det( A) = det( B). det( A)

(b)

(d) False; a square matrix A is invertible if and only if det(A) ≠ 0. (e) True; since adj(A) is the transpose of the matrix

of cofactors of A, and ( BT )T = B for any matrix B.

3. (a)

(f) True; the ij entry of A ⋅ adj(A) is ai1C j1 + ai 2 C j 2 + " + ain C jn . If i ≠ j then this

sum is 0, and when i = j then this sum is det(A). Thus, A ⋅ adj(A) is a diagonal matrix with diagonal entries all equal to det(A) or (det( A)) I n .

(b)

(g) True; if det(A) were not 0, the system Ax = 0 would have exactly one solution (the trivial solution). (h) True; if the reduced row echelon form of A were I n , then there could be no such matrix b.

5. (a) (i) True; since E is invertible. (j) True; if A is invertible, then so is A−1 and so

both Ax = 0 and A−1x = 0 have only the trivial solution. (b)

(k) True; if A is invertible, then det(A) ≠ 0 and A−1

is invertible, so det( A) ⋅ A−1 = adj( A) is invertible.

(l) False; if the kth row of A is a row of zeros, then every cofactor Cij = 0 for i ≠ k, hence the

−1 5 2 2 −1 5 2 0 2 −1 = −1 + (−3) 1 1 2 −1 −3 1 1 = −(2 + 1) − 3(−5 − 4) = 24 5 2 −1 5 2 −1 0 2 −1 = 0 2 −1 0 −14 −5 −3 1 1 2 −1 5 = 0 2 −1 0 0 −12 = (−1)(2)(−12) = 24 3 0 −1 1 1 0 −1 1 1 1 =3 − 4 2 4 2 0 4 2 = 3(2 − 4) − (0 + 4) = −10 3 0 −1 1 1 1 1 1=−3 0 0 4 2 0 4 1 1 = − 0 −3 0 4

1 −1 2 1 −4 2

1 1 1 = − 0 −3 −4 0 0 − 10 3

cofactor matrix has a row of zeros and the corresponding column (not row) of adj(A) is a column of zeros.

⎛ 10 ⎞ = −(1)(−3) ⎜ − ⎟ ⎝ 3⎠ = −10

Chapter 2 Supplementary Exercises 1. (a)

3 3 −4 2 =− 3 3 −4 2 1 1 = −3 −4 2 1 1 = −3 0 6 = −3(1)(6) = −18

−4 2 = −4(3) − 2(3) = −18 3 3

56



3 −2 7. (a) 1 −9

6 0 3 1 0 −1 2 −2

1 −2 −2 3 3 1 4 1 4 1 4 = 3 0 −1 1 − 6 1 −1 1 − 1 1 0 −1 1 −9 −2 2 −9 2 −2 2 −2 2 2 ⎛ −1 1 −1 1 −2 3 ⎞ 1 4⎞ ⎛ 1 1 1 −1 ⎞ ⎛ 3 1 = 3⎜ 3 +2 ⎟ − 6 ⎜ −2 −2 2 − 1 −9 2 + 4 −9 −2 ⎟ − ⎜ −1 2 −2 − (−1) −9 2 ⎟ − − 2 2 1 1 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 3[3(−2 + 2) + 2(1 + 4)] − 6[ −2(−2 + 2) − (2 + 9) + 4(−2 − 9)] − [ −(−6 − 2) + (−4 + 27)] = 30 + 330 − 31 = 329

3 −2 (b) 1 −9

6 0 3 1 0 −1 2 −2

1 4 =− 1 2

1 0 −2 3 3 6 −9 2 1 0 0 3 =− 0 6 0 2 1 0 0 3 =−0 0 0 0

−1 1 0 −2 −1 −1 3 −11 −1 −1 5 31 −3

1 4 1 2 1 6 −2 11 1 6 −14 7

0 −1 1 3 −1 6 0 5 −14 0 0 − 329 15 ⎛ 329 ⎞ = −(1)(3)(5) ⎜ − ⎟ ⎝ 15 ⎠ = 329 1 0 =−0 0

9. Exercise 3: −1 5 2 −1 5 2 −1 5 0 2 −1 = 0 2 −1 0 2 −3 1 1 −3 1 1 −3 1 = [(−1)(2)(1) + (5)(−1)(−3) + (2)(0)(1)] − [(2)(2)(−3) + (−1)(−1)(1) + (5)(0)(1)] = 13 − (−11) = 24 Exercise 4: −1 −2 −3 −1 −2 −3 −1 −2 −4 −5 −6 = −4 −5 −6 −4 −5 −7 −8 −9 −7 −8 −9 −7 −8 = [(−1)(−5)(−9) + (−2)(−6)(−7) + (−3)(−4)(−8)] − [(−3)(−5)(−7) + (−1)(−6)(−8) + (−2)(−4)(−9)] = −225 − (−225) =0

57



Exercise 5: 3 0 −1 3 0 −1 3 0 1 1 1= 1 1 11 1 0 4 2 0 4 20 4 = [(3)(1)(2) + (0)(1)(0) + (−1)(1)(4)] − [(−1)(1)(0) + (3)(1)(4) + (0)(1)(2)] = 2 − 12 = −10 Exercise 6: −5 1 4 −5 1 4 −5 1 3 0 2 = 3 0 2 3 0 1 −2 2 1 −2 2 1 −2 = [(−5)(0)(2) + (1)(2)(1) + (4)(3)(−2)] − [(4)(0)(1) + (−5)(2)(−2) + (1)(3)(2)] = −22 − 26 = −48 11. The determinants for Exercises 1, 3, and 4 were evaluated in Exercises 1, 3, and 9. Exercise 2: 7 −1 = 7(−6) − (−1)(−2) = −44 −2 −6

The determinants of the matrices in Exercises 1−3 are nonzero, so those matrices are invertible. The matrix in Exercise 4 is not invertible, since its determinant is zero. 13.

5 b−3 = 5(−3) − (b − 3)(b − 2) b − 2 −3 = −15 − b2 + 5b − 6 = −b 2 + 5b − 21

0 0 15. 0 0 5

0 0 0 −3 0 0 −4 0 0 −1 0 0 = (5)(2)(−1)(−4)(−3) 2 0 0 0 0 0 0 0 = −120

⎡ −4 2 ⎤ 17. Let A = ⎢ . Then det(A) = −18. The cofactors of A are C11 = 3, C12 = −3, C21 = −2, and C22 = −4. The ⎣ 3 3⎥⎦ ⎡ 3 −3⎤ matrix of cofactors is ⎢ . ⎣ −2 −4 ⎥⎦

A−1 =

1 1 1 ⎡ 3 −2 ⎤ ⎡ − 6 ⎢ = adj( A) = −18 ⎢⎣ −3 −4 ⎥⎦ ⎢ 1 det( A) ⎣ 6

1⎤ 9⎥ 2⎥ 9⎦

⎡ −1 5 2 ⎤ 19. Let A = ⎢ 0 2 −1⎥ . Then det(A) = 24. ⎢ ⎥ ⎣ −3 1 1⎦ The cofactors of A are C11 = 2 + 1 = 3, C12 = −(0 − 3) = 3, C13 = 0 + 6 = 6, C21 = −(5 − 2) = −3, C22 = −1 + 6 = 5,

C23 = −(−1 + 15) = −14, C31 = −5 − 4 = −9, C32 = −(1 − 0) = −1, and C33 = −2 − 0 = −2.

58



6⎤ ⎡ 3 3 The matrix of cofactors is ⎢ −3 5 −14⎥ . ⎢ ⎥ ⎣ −9 −1 −2⎦ A−1 =

9 16 + =1 25 25 ⎡x − 4⎤ 5⎥ Ax′ = ⎢ 3⎥ ⎢⎣ y 5⎦ 3 4 det( Ax′ ) = x + y 5 5 3 ⎡ x⎤ ⎥ Ay′ = ⎢ 5 ⎢⎣ 54 y ⎥⎦ 3 4 4 3 det( Ay′ ) = y − x = − x + y 5 5 5 5 3 4 4 3 The solution is x ′ = x + y, y ′ = − x + y. 5 5 5 5 det( A) =

⎡ 3 −3 −9 ⎤ 1 1 ⎢ adj( A) = 3 5 −1⎥ det( A) 24 ⎢6 −14 −2 ⎥ ⎣ ⎦ ⎡ 1 − 1 − 3⎤ 8 8⎥ ⎢8 5 1 ⎥ = ⎢ 18 − 24 24 ⎢1 ⎥ 7 1 ⎢⎣ 4 − 12 − 12 ⎥⎦

⎡ 3 6 0 1⎤ ⎢ −2 3 1 4⎥ 23. Let A = ⎢ ⎥. ⎢ 1 0 −1 1⎥ ⎣⎢ −9 2 −2 2⎦⎥ Then det(A) = 329. The cofactors of A are C11 = 10, C12 = 55, C13 = −21, C14 = −31,

1 α⎤ 1 β⎥ ⎥ β 1⎦

⎡1 27. A = ⎢ 1 ⎢ ⎣α

C21 = −2, C22 = −11, C23 = 70, C24 = 72,

α β −α 0 β −α 1−α 2 β −α 0 = β − α 1− α 2

C31 = 52, C32 = −43, C33 = −175, C34 = 102,

1 det( A) = 0

C41 = −27, C42 = 16, C43 = −42, and C44 = −15. The matrix of cofactors is ⎡ 10 55 −21 −31⎤ ⎢ −2 −11 70 72 ⎥ ⎢ ⎥. ⎢ 52 −43 −175 102 ⎥ ⎣⎢ −27 16 −42 −15⎦⎥

1 0

= 0 − (β − α )2 = −( β − α ) 2 det(A) = 0 if and only if α = β , and the system has a nontrivial solution if and only if det(A) = 0.

1 adj( A) det( A) 52 −27 ⎤ ⎡ 10 −2 1 ⎢ 55 −11 −43 16 ⎥ = ⎢ ⎥ 329 ⎢ −21 70 −175 −42 ⎥ ⎢⎣ −31 72 102 −15⎥⎦ 10 52 27 ⎤ 2 ⎡ − 329 − 329 329 ⎢ 329 ⎥ 55 43 16 ⎥ 11 − 329 − 329 ⎢ 329 329 =⎢ ⎥ 3 10 25 6 − − ⎢ − 47 47 47 47 ⎥ ⎢ 31 72 102 − 15 ⎥ 329 329 329 ⎦⎥ ⎣⎢ − 329

A−1 =

29. (a) Draw the perpendicular from the vertex of angle γ to side c as shown. b α c1

γ

c

a c2

β

c1 c and cos β = 2 , so b a c = c1 + c2 = a cos β + b cos α . A similar construction gives the other two equations in the system. The matrix form of the system is ⎡0 c b ⎤ ⎡ cos α ⎤ ⎡ a ⎤ ⎢ c 0 a ⎥ ⎢ cos β ⎥ = ⎢ b ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣b a 0 ⎦ ⎣ cos γ ⎦ ⎣ c ⎦ Then cos α =

25. In matrix form, the system is ⎡ 3 − 4 ⎤ x′ 5 ⎥ ⎡ ⎤ = ⎡ x⎤ . ⎢5 3 ⎥ ⎣⎢ y ′⎦⎥ ⎣⎢ y ⎦⎥ ⎢ 45 5⎦ ⎣ ⎡3 − 4⎤ 5⎥ A = ⎢5 3⎥ ⎢ 45 5⎦ ⎣

59



31. If A is invertible then A−1 is also invertible and

0 c b det( A) = c 0 a b a 0 c a c 0 = −c +b b 0 b a = −c(0 − ab) + b(ac − 0) = 2abc ⎡a c b⎤ Aα = ⎢ b 0 a ⎥ ⎢ ⎥ ⎣ c a 0⎦ det( Aα ) = −c

det(A) ≠ 0, so det( A) A−1 is invertible. Thus, adj( A) = det( A) A−1 is invertible and (adj( A))−1 = (det( A) A−1 )−1 1 = ( A−1 )−1 det( A) 1 = A det( A) = adj(A −1 )

b a a b −a c 0 b a

since adj( A−1 ) = det( A−1 )( A−1 )−1 =

= −c(0 − ac) − a(a 2 − b 2 )

33. Let A be an n × n matrix for which the sum of the entries in each row is zero and let x1 be the

= a (c 2 + b 2 − a 2 ) det( Aα ) cos α = det( A) a(c 2 + b 2 − a 2 ) = 2abc c 2 + b2 − a 2 = 2bc

n × 1 column matrix whose entries are all one. ⎡ 0⎤ ⎢ 0⎥ Then Ax1 = ⎢ ⎥ because each entry in the ⎢# ⎥ ⎣⎢ 0⎦⎥ product is the sum of the entries in a row of A. That is, Ax = 0 has a nontrivial solution, so det(A) = 0.

⎡0 a b ⎤ (b) Aβ = ⎢ c b a ⎥ ⎢ ⎥ ⎣b c 0 ⎦ c a c b det( Aβ ) = −a +b b 0 b c

35. Add 10,000 times the first column, 1000 times the second column, 100 times the third column, and 10 times the fourth column to the fifth column. The entries in the fifth column are now 21,375, 38,798, 34,162, 40,223, and 79,154, respectively. Evaluating the determinant by expanding by cofactors of the fifth column, gives 21, 375C15 + 38, 798C25 + 34,162C35 + 40, 223C45 + 79,154C55 .

= −a(0 − ab) + b(c 2 − b2 ) cos β =

= b( a 2 + c 2 − b 2 ) det( Aβ )

det( A) b( a 2 + c 2 − b 2 ) = 2abc 2 a + c 2 − b2 = 2ac ⎡0 c a ⎤ Aγ = ⎢ c 0 b ⎥ ⎢ ⎥ ⎣b a c ⎦

det( Aγ ) = −c

Since each coefficient of Ci5 is divisible by 19, this sum must also be divisible by 19. Note that the column operations do not change the value of the determinant⎯consider them as row operations on the transpose.

c b c 0 +a b c b a

= −c(c 2 − b 2 ) + a(ac − 0) = c( a 2 + b 2 − c 2 ) cos γ =

c( a 2 + b 2 − c 2 ) det( A) 2abc a 2 + b2 − c 2 = 2ab

det( Aγ )

1 A. det( A)

=

60

Chapter 3 Euclidean Vector Spaces Section 3.1

(f)

z

y

Exercise Set 3.1 x

z

1. (a)

(–3, 4, –5) (3, 4, 5)

y

3. (a)

y

x

x z

(b)

(–3, 4, 5)

y

(b)

x

y x

z

(c) (3, –4, 5)

(c)

y

(d)

z

x

y

x

z

(d)

y

y x (3, 4, –5)

x

(e) (–3, –4, 5)

z

x

y

61

Chapter 3: Euclidean Vector Spaces


z

(e)

9. (a) Let the terminal point be B(b1 , b2 ). Then (b1 − 1, b2 − 1) = (1, 2) so b1 = 2 and b2 = 3. The terminal point is B(2, 3). (b) Let the initial point be A(a1, a2 , a3 ). Then (−1 − a1 , − 1 − a2 , 2 − a3 ) = (1, 1, 3), so

y

a1 = −2, a2 = −2, and a3 = −1. The initial

point is A(−2, −2, −1).

x z

(f)

11. (a) One possibility is u = v, then the initial point (x, y, z) satisfies (3 − x, 0 − y, −5 − z) = (4, −2, −1), so that x = −1, y = 2, and z = −4. One possibility is (−1, 2, −4).

y

(b) One possibility is u = −v, then the initial point (x, y, z) satisfies (3 − x, 0 − y, −5 − z) = (−4, 2, 1), so that x = 7, y = −2, and z = −6. One possibility is (7, −2, −6).

x

→ 5. (a) P1 P2 = (3 − 4, 7 − 8) = (−1, − 1) y x

13. (a) u + w = (4, −1) + (−3, −3) = (1, −4) (b) v − 3u = (0, 5) − (12, −3) = (−12, 8)

→ (b) P1 P2 = (−4 − 3, − 7 − (−5)) = (−7, − 2)

(c) 2(u − 5w ) = 2((4, − 1) − (−15, − 15)) = 2(19, 14) = (38, 28)

y x

(d) 3v − 2(u + 2w ) = (0, 15) − 2((4, − 1) + (−6, − 6)) = (0, 15) − 2(−2, − 7) = (0, 15) + (4, 14) = (4, 29)

→ (c) P1 P2 = (−2 − 3, 5 − (−7), − 4 − 2) = (−5, 12, − 6) z

(e) −3(w − 2u + v) = −3((−3, − 3) − (8, − 2) + (0, 5)) = −3(−11, 4) = (33, − 12)

y

x

(f) (−2u − v) − 5(v + 3w) = ((−8, 2) − (0, 5)) − 5((0, 5) + (−9, −9)) = (−8, −3) − 5(−9, −4) = (−8, −3) + (45, 20) = (37, 17)

→ 7. (a) P1 P2 = (2 − 3, 8 − 5) = (−1, 3)

15. (a) v − w = (4, 7, − 3, 2) − (5, − 2, 8, 1) = (−1, 9, − 11, 1)

→ (b) P1 P2 = (2 − 5, 4 − (−2), 2 − 1) = (−3, 6, 1)

(b) 2u + 7 v = (−6, 4, 2, 0) + (28, 49, − 21, 14) = (22, 53, − 19, 14) 62


Section 3.1

(c) −u + (v − 4w) = (3, −2, −1, 0) + ((4, 7, −3, 2) − (20, −8, 32, 4)) = (3, −2, −1, 0) + (−16, 15, −35, −2) = (−13, 13, −36, −2) (d) 6(u − 3v) = 6((−3, 2, 1, 0) − (12, 21, − 9, 6)) = 6(−15, − 19, 10, − 6) = (−90, − 114, 60, − 36) (e) − v − w = (−4, − 7, 3, − 2) − (5, − 2, 8, 1) = (−9, − 5, − 5, − 3) (f) (6v − w) − (4u + v) = ((24, 42, −18, 12) − (5, −2, 8, 1)) − ((−12, 8, 4, 0) + (4, 7, −3, 2)) = (19, 44, −26, 11) − (−8, 15, 1, 2) = (27, 29, −27, 9) 17. (a) w − u = (−4, 2, − 3, − 5, 2) − (5, − 1, 0, 3, − 3) = (−9, 3, − 3, − 8, 5) (b) 2 v + 3u = (−2, − 2, 14, 4, 0) + (15, − 3, 0, 9, − 9) = (13, − 5, 14, 13, − 9) (c) −w + 3(v − u) = (4, −2, 3, 5, −2) + 3((−1, −1, 7, 2, 0) − (5, −1, 0, 3, −3)) = (4, −2, 3, 5, −2) + 3(−6, 0, 7, −1, 3) = (4, −2, 3, 5, −2) + (−18, 0, 21, −3, 9) = (−14, −2, 24, 2, 7) (d) 5(−v + 4u − w) = 5((1, 1, −7, −2, 0) + (20, −4, 0, 12, −12) − (−4, 2, −3, −5, 2)) = 5(25, −5, −4, 15, −14) = (125, −25, −20, 75, −70) (e) −2(3w + v) + (2u + w) = −2((−12, 6, − 9, − 15, 6) + (−1, − 1, 7, 2, 0)) + ((10, − 2, 0, 6, − 6) + (−4, 2, − 3, − 5, 2)) = −2(−13, 5, − 2, − 13, 6) + (6, 0, − 3, 1, − 4) = (26, − 10, 4, 26, − 12) + (6, 0, − 3, 1, − 4) = (32, − 10, 1, 27, − 16) (f)

1 1 (w − 5v + 2u) + v = ((−4, 2, − 3, − 5, 2) − (−5, − 5, 35, 10, 0) + (10, − 2, 0, 6, − 6)) + (−1, − 1, 7, 2, 0) 2 2 1 = (11, 5, − 38, − 9, − 4) + (−1, − 1, 7, 2, 0) 2 9 ⎛ 11 5 ⎞ = ⎜ , , − 19, − , − 2 ⎟ + (−1, − 1, 7, 2, 0) 2 ⎝2 2 ⎠ 5 ⎛9 3 ⎞ = ⎜ , , − 12, − , − 2 ⎟ 2 ⎝2 2 ⎠

19. (a) v − w = (4, 0, − 8, 1, 2) − (6, − 1, − 4, 3, − 5) = (−2, 1, − 4, − 2, 7) (b) 6u + 2 v = (−18, 6, 12, 24, 24) + (8, 0, − 16, 2, 4) = (−10, 6, − 4, 26, 28)

63



(c) (2u − 7w ) − (8v + u) = ((−6, 2, 4, 8, 8) − (42, − 7, − 28, 21, − 35)) − ((32, 0, − 64, 8, 16) + (−3, 1, 2, 4, 4)) = (−48, 9, 32, − 13, 43) − (29, 1, − 62, 12, 20) = (−77, 8, 94, − 25, 23) 21. Let x = ( x1, x2 , x3 , x4 , x5 ).

2u − v + x = (−6, 2, 4, 8, 8) − (4, 0, − 8, 1, 2) + ( x1 , x2 , x3 , x4 , x5 ) = (−10 + x1, 2 + x2 , 12 + x3 , 7 + x4 , 6 + x5 ) 7x + w = (7 x1, 7 x2 , 7 x3 , 7 x4 , 7 x5 ) + (6, − 1, − 4, 3, − 5) = (7 x1 + 6, 7 x2 − 1, 7 x3 − 4, 7 x4 + 3, 7 x5 − 5) Equate components. 8 −10 + x1 = 7 x1 + 6 ⇒ x1 = − 3 1 2 + x2 = 7 x2 − 1 ⇒ x2 = 2 8 12 + x3 = 7 x3 − 4 ⇒ x3 = 3 2 7 + x4 = 7 x4 + 3 ⇒ x4 = 3 11 6 + x5 = 7 x5 − 5 ⇒ x5 = 6 ⎛ 8 1 8 2 11 ⎞ x = ⎜− , , , , ⎟ ⎝ 3 2 3 3 6⎠ 23. (a) There is no scalar k such that ku is the given vector, so the given vector is not parallel to u. (b) −2u = (4, −2, 0, −6, −10, −2) The given vector is parallel to u. (c) The given vector is 0, which is parallel to all vectors. 25. au + bv = (a, − a, 3a, 5a) + (2b, b, 0, − 3b) = (a + 2b, − a + b, 3a, 5a − 3b) = (1, − 4, 9, 18) Equating the third components give 3a = 9 or a = 3. Equating any other components gives b = −1. The scalars are a = 3, b = −1. 27. Equating components gives the following system of equations. c1 + 3c2 = −1 −c1 + 2c2 + c3 = 1 c2 + 4c3 = 19

⎡ 1 0 0 2⎤ ⎡ 1 3 0 −1⎤ The reduced row echelon form of ⎢ −1 2 1 1⎥ is ⎢0 1 0 −1⎥ . ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 1 5⎦ ⎣ 0 1 4 19⎦ The scalars are c1 = 2, c2 = −1, c3 = 5.

64


Section 3.2

29. Equating components gives the following system of equations. −c1 + 2c2 + 7c3 + 6c4 = 0 3c1 + c3 + 3c4 = 5 2c1 + 4c2 + c3 + c4 = 6 −c2 + 4c3 + 2c4 = −3 The reduced row echelon form of ⎡ 1 0 0 0 1⎤ ⎡ −1 2 7 6 0 ⎤ ⎢0 1 0 0 1⎥ ⎢ 3 0 1 3 5⎥ ⎥. ⎢ ⎥ is ⎢ 2 4 1 1 6 ⎢0 0 1 0 −1⎥ ⎢ ⎥ ⎣⎢0 0 0 1 1⎦⎥ ⎣⎢ 0 −1 4 2 −3⎦⎥

True/False 3.1 (a) False; vector equivalence is determined solely by length and direction, not location. (b) False; (a, b) is a vector in 2-space, while (a, b, 0) is a vector in 3-space. Alternatively, equivalent vectors must have the same number of components. (c) False; v and kv are parallel for all values of k. (d) True; apply Theorem 3.1.1 parts (a), (b), and (a) again: v + (u + w) = v + (w + u) = ( v + w) + u = (w + v) + u.

The scalars are c1 = 1, c2 = 1, c3 = −1, c4 = 1. 31. Equating components gives the following system of equations. −2c1 − 3c2 + c3 = 0 9c1 + 2c2 + 7c3 = 5 6c1 + c2 + 5c3 = 4 The reduced row echelon form of ⎡ 1 0 1 0⎤ ⎡ −2 −3 1 0 ⎤ ⎢ 9 2 7 5⎥ is ⎢0 1 −1 0 ⎥ . ⎢ ⎥ ⎢ ⎥ 1 5 4⎦ ⎣0 0 0 1⎦ ⎣ 6

(e) True; the vector −u can be added to both sides of the equation. (f) False; a and b must be nonzero scalars for the statement to be true. (g) False; the vectors v and −v have the same length and can be placed to be collinear, but are not equal.

From the bottom row, the system has no solution, so no such scalars exist. 33. (a)

(h) True; vector addition is defined component-wise.

1 → 1 PQ = (7 − 2, − 4 − 3, 1 − (−2)) 2 2 1 = (5, − 7, 3) 2 7 3⎞ ⎛5 =⎜ , − , ⎟ 2 2⎠ ⎝2

(i) False; (k + m)(u + v) = (k + m)u + (k + m)v. 1 (j) True; x = (5v + 4w ) 8 (k) False; for example, if v 2 = − v1 then

a1 v1 + a2 v 2 = b1 v1 + b2 v 2 as long as

1 → PQ with its initial point at P 2 gives the midpoint. ⎛ 5 3⎞ ⎛9 1 1⎞ ⎛ 7⎞ ⎜ 2 + , 3 + ⎜ − ⎟, − 2 + ⎟ = ⎜ , − , − ⎟ 2 2⎠ ⎝2 2 2⎠ ⎝ 2⎠ ⎝

a1 − a2 = b1 − b2 .

Locating

(b)


3 → 3 21 9 ⎞ ⎛ 15 PQ = (5, − 7, 3) = ⎜ , − , ⎟ 4 4 4 4 4⎠ ⎝

1. (a)

v = 42 + (−3)2 = 25 = 5

v 1 ⎛ 4 3⎞ = (4, − 3) = ⎜ , − ⎟ has the v 5 ⎝5 5⎠ same direction as v. v 1 ⎛ 4 3⎞ = − (4, − 3) = ⎜ − , ⎟ has the u2 = − 5 v ⎝ 5 5⎠ direction opposite v. u1 =

3 → Locating PQ with its initial point at P 4 gives the desired point. ⎛ 15 9⎞ ⎛ 21 ⎞ ⎜ 2 + , 3 + ⎜ − ⎟, − 2 + ⎟ 4 4⎠ ⎝ 4⎠ ⎝ ⎛ 23 9 1 ⎞ =⎜ , − , ⎟ 4 4⎠ ⎝ 4

65


(b)


v = 22 + 22 + 22 = 12 = 2 3 ⎛ 1 1 1 ⎞ v 1 = (2, 2, 2) = ⎜ , , ⎟ has the same direction as v. v 2 3 ⎝ 3 3 3⎠ v u2 = − v 1 =− (2, 2, 2) 2 3 ⎛ 1 1 1 ⎞ ,− ,− = ⎜− ⎟ 3 3 3⎠ ⎝ has the direction opposite v. u1 =

(c)

v = 12 + 02 + 22 + 12 + 32 = 15 u1 = =

v v 1

(1, 0, 2, 1, 3) 15 ⎛ 1 2 1 3 ⎞ , 0, , , =⎜ ⎟ 15 15 15 ⎠ ⎝ 15 has the same direction as v. v u2 = − v 1 =− (1, 0, 2, 1, 3) 15 ⎛ 1 2 1 3 ⎞ = ⎜− , 0, − ,− ,− ⎟ 15 15 15 ⎠ ⎝ 15 has the direction opposite v.

3. (a)

u + v = (2, − 2, 3) + (1, − 3, 4) = (3, − 5, 7) = 32 + (−5)2 + 72 = 83

(b)

u + v = (2, − 2, 3) + (1, − 3, 4) = 22 + (−2)2 + 32 + 12 + (−3)2 + 42 = 17 + 26

(c)

−2u + 2 v = (−4, 4, − 6) + (2, − 6, 8) = (−2, − 2, 2) = (−2)2 + (−2) 2 + 22 = 12 =2 3

66


(d)

Section 3.2

3u − 5v + w = (6, − 6, 9) − (5, − 15, 20) + (3, 6, − 4) = (4, 15, − 15) = 42 + 152 + (−15)2 = 466

5. (a)

3u − 5v + w = (−6, − 3, 12, 15) − (15, 5, − 25, 35) + (−6, 2, 1, 1) = (−27, − 6, 38, − 19) = (−27)2 + (−6)2 + 382 + (−19)2 = 2570

(b)

3u − 5 v + w = (−6, − 3, 12, 15) − 5 (3, 1, − 5, 7) + (−6, 2, 1, 1) = (−6)2 + (−3)2 + 122 + 152 − 5 32 + 12 + (−5) 2 + 7 2 + (−6)2 + 22 + 12 + 12 = 414 − 5 84 + 42 = 3 46 − 10 21 + 42

(c)

− u v = − (−2)2 + (−1)2 + 42 + 52 (3, 1, − 5, 7) = − 46 (3, 1, − 5, 7) = 46 32 + 12 + (−5) 2 + 7 2 = 46 84 = 2 966

7.

kv = k (−2, 3, 0, 6)

= k

(−2)2 + 32 + 0 + 62

= k 49 =7 k

5 5 5 7 k = 5 if k = , so k = or k = − . 7 7 7 9. (a) u ⋅ v = (3)(2) + (1)(2) + (4)(−4) = −8

u ⋅u = u

2

= 32 + 12 + 42 = 26

v⋅v = v

2

= 22 + 22 + (−4)2 = 24

(b) u ⋅ v = (1)(2) + (1)(−2) + (4)(3) + (6)(−2) =0 u ⋅u = u

2

= 12 + 12 + 42 + 62 = 54

v⋅v = v

2

= 22 + (−2)2 + 32 + (−2)2 = 21

67



11. (a) d (u, v) = u − v = (3 − 1)2 + (3 − 0)2 + (3 − 4)2 = 22 + 32 + (−1)2 = 14 (b) d (u, v) = u − v = (0 − (−3))2 + (−2 − 2)2 + (−1 − 4)2 + (1 − 4)2 = 32 + (−4)2 + (−5)2 + (−3)2 = 59 (c) d (u, v) = u − v = (3 − (−4))2 + (−3 − 1)2 + (−2 − (−1)) 2 + (0 − 5) 2 + (−3 − 0)2 + (13 − (−11))2 + (5 − 4)2 = 72 + (−4)2 + (−1)2 + (−5)2 + (−3)2 + 242 + 12 = 677 13. (a) cos θ = =

u⋅v u v (3)(1) + (3)(0) + (3)(4) 32 + 32 + 32 12 + 02 + 42 15

=

27 17 Since cos θ is positive, θ is acute. (b) cos θ =

u⋅v u v (0)(−3) + (−2)(2) + (−1)(4) + (1)(4)

=

2

0 + (−2)2 + (−1) 2 + 12 (−3)2 + 22 + 42 + 42 −4

= =−

6 45 4

6 45 Since cos θ is negative, θ is obtuse. (c) cos θ =

u⋅v u v

= = =−

(3)(−4) + (−3)(1) + (−2)(−1) + (0)(5) + (−3)(0) + (13)(−11) + (5)(4) 32 + (−3)2 + (−2) 2 + 02 + (−3)2 + 132 + 52 (−4)2 + 12 + (−1)2 + 52 + 02 + (−11) 2 + 42 −136 225 180 136

225 180 Since cos θ is negative, θ is obtuse.

68


Section 3.2

15. The angle θ between the vectors is 30°. 3 a ⋅ b = a b cosθ = 9 ⋅ 5 cos 30° = 45 2

23. (a) cos θ = =

17. (a) u ⋅ (v ⋅ w) does not make sense because v ⋅ w is a scalar.

=−

u ⋅ v does not make sense because u ⋅ v is a scalar.

(b) cos θ =

(d) (u ⋅ v) − u makes sense since both u ⋅ v =

and u are scalars.

19. (a)

(b)

(c)

(−4, − 3) (−4, − 3) = (−4, − 3) (−4)2 + (−3)2 (−4, − 3) = 25 3⎞ ⎛ 4 = ⎜− , − ⎟ 5 5 ⎝ ⎠

=−

=

( −3, 2, 3 ) 2 (−3)2 + 22 + ( 3 ) ( −3, 2, 3 ) =

(d) cos θ =

(d)

=

=0

42 27

(−2)2 + 22 + 32 12 + 72 + (−4)2 0 17 66

25. (a) u ⋅ v = (3)(4) + (2)(−1) = 10 u v = 32 + 22 42 + (−1) 2 = 13 17 ≈ 14.866

21. Divide v by v to get a unit vector that has the same direction as v, then multiply by m: m

10

u⋅v u v (−2)(1) + (2)(7) + (3)(−4)

=

(1, 2, 3, 4, 5) (1, 2, 3, 4, 5) (1, 2, 3, 4, 5) = 2 1 + 22 + 32 + 42 + 52 (1, 2, 3, 4, 5) = 55 ⎛ 1 2 3 4 5 ⎞ =⎜ , , , , ⎟ 55 55 55 55 ⎠ ⎝ 55

8 10 3

12 + (−5)2 + 42 32 + 32 + 32 0

=0

16 ⎛ 3 1 3⎞ = ⎜− , , ⎜ 4 2 4 ⎟⎟ ⎝ ⎠

40 16 24

u⋅ v u v (1)(3) + (−5)(3) + (4)(3)

=

=

962

(−6)2 + (−2)2 42 + 02 −24

=−

(c) cos θ =

13 74 11

u⋅v u v (−6)(4) + (−2)(0)

=

(1, 7) (1, 7) (1, 7) ⎛ 1 7 ⎞ = = =⎜ , ⎟ 2 2 (1, 7) 50 ⎝ 5 2 5 2 ⎠ 1 +7

( −3, 2, 3 ) ( −3, 2, 3 )

22 + 32 52 + (−7) 2 −11

=

(b) u ⋅ (v + w) makes sense. (c)

u⋅ v u v (2)(5) + (3)(−7)

v . v 69



(b) u ⋅ v = (−3)(2) + (1)(−1) + (0)(3) = −7 = 7

Section 3.3

u v = (−3)2 + 12 + 02 22 + (−1)2 + 32

Exercise Set 3.3

= 10 14 ≈ 11.832 (c)

1. (a) u ⋅ v = (6)(2) + (1)(0) + (4)(−3) = 0, so u and v are orthogonal.

u ⋅ v = (0)(1) + (2)(1) + (2)(1) + (1)(1) = 5

(b) u ⋅ v = (0)(1) + (0)(1) + (−1)(1) = −1 ≠ 0, so u and v are not orthogonal.

u v = 02 + 22 + 22 + 12 12 + 12 + 12 + 12 = 9 4 =6

(c) u ⋅ v = (−6)(3) + (0)(1) + (4)(6) = 6 ≠ 0, so u and v are not orthogonal. (d) u ⋅ v = (2)(5) + (4)(3) + (−8)(7) = −34 ≠ 0, so u and v are not orthogonal.

27. It is a sphere of radius 1 centered at ( x0 , y0 , z0 ).

3. (a) v1 ⋅ v 2 = (2)(3) + (3)(2) = 12 ≠ 0 The vectors do not form an orthogonal set.

True/False 3.2

(b) v1 ⋅ v 2 = (−1)(1) + (1)(1) = 0 The vectors form an orthogonal set.

(a) True; 2 v = 2 v . (b) True

(c) v1 ⋅ v 2 = (−2)(1) + (1)(0) + (1)(2) = 0 v1 ⋅ v3 = (−2)(−2) + (1)(−5) + (1)(1) = 0

(c) False; 0 = 0. (d) True; the vectors are

(e) True; cos

π 3

=

v 2 ⋅ v3 = (1)(−2) + (0)(−5) + (2)(1) = 0 The vectors form an orthogonal set.

v v and − . v v

(d) v1 ⋅ v 2 = (−3)(1) + (4)(2) + (−1)(5) = 0 v1 ⋅ v3 = (−3)(4) + (4)(−3) + (−1)(0) = −24

1 u⋅v . = 2 u v

Since v1 ⋅ v3 ≠ 0, the vectors do not form an orthogonal set.

(f) False; (u ⋅ v) + w does not make sense because u ⋅ v is a scalar. u ⋅ (v + w) is a meaningful expression.

5. Let x = ( x1, x2 , x3 ). u ⋅ x = (1)( x1 ) + (0)( x2 ) + (1)( x3 ) = x1 + x3 v ⋅ x = (0)( x1 ) + (1)( x2 ) + (1)( x3 ) = x2 + x3

(g) False; for example, if u = (1, 1), v = (−1, 1), and w = (1, −1), then u ⋅ v = u ⋅ w = 0.

Thus x1 = x2 = − x3 . One vector orthogonal to both u and v is x = (1, 1, −1). x (1, 1, − 1) = x 12 + 12 + (−1)2

(h) False; let u = (1, 1) and v = (−1, 1), then u ⋅ v = (1)(−1) + (1)(1) = 0. u ⋅ v = 0 indicates that the angle between u and v is 90°.

=

(i) True; if u = (u1 , u2 ) and v = (v1 , v2 ) then

(1, 1, − 1)

3 ⎛ 1 1 1 ⎞ =⎜ , ,− ⎟ 3⎠ ⎝ 3 3

u1 , u2 > 0 and v1, v2 < 0 so u1v1 , u2 v2 < 0, hence u1v1 + u2 v2 < 0.

⎛ 1 1 1 ⎞ The possible vectors are ± ⎜ , ,− ⎟. 3 3 3⎠ ⎝

(j) True; use the triangle inequality twice. u + v + w = (u + v) + w ≤ u + v + w u+ v + w ≤ u + v + w

70


Section 3.3

→

7. AB = (−2 − 1, 0 − 1, 3 − 1) = (−3, − 1, 2)

proja u =

19. (a)

→

BC = (−3 − (−2), − 1 − 0, 1 − 3) = (−1, − 1, − 2) =

→

CA = (1 − (−3), 1 − (−1), 1 − 1) = (4, 2, 0) →

→

=

AB ⋅ BC = (−3)(−1) + (−1)(−1) + (2)(−2) = 0 →

→

→

→

Since AB ⋅ BC = 0, AB and BC are orthogonal and the points form the vertices of a right triangle.

=

u ⋅a u ⋅a 2

a a

a u ⋅a a (1)(−4) + (−2)(−3) (−4) 2 + (−3)2 2

=

9. −2(x − (−1)) + 1(y − 3) + (−1)(z − (−2)) = 0 −2(x + 1) + (y − 3) − (z + 2) = 0

2

a

25

2 = 5

11. 0(x − 2) + 0(y − 0) + 2(z − 0) = 0 2z = 0

proja u =

(b) 13. A normal to 4x − y + 2z = 5 is n1 = (4, − 1, 2). A normal to 7x − 3y + 4z = 8 is n 2 = (7, − 3, 4).

=

Since n1 and n 2 are not parallel, the planes are not parallel.

=

15. 2y = 8x − 4z + 5 ⇒ 8x − 2y − 4z = −5 A normal to the plane is n1 = (8, − 2, − 4).

u ⋅a a (3)(2) + (0)(3) + (4)(3)

22 + 32 + 32 18 22

21. u ⋅ a = (6)(3) + (2)(−9) = 0 The vector component of u along a is u⋅a 0 proja u = (3, − 9) = (0, 0). a= 2 2 a a

1 1 1 1 z+ y ⇒ x− y− z =0 2 4 4 2 1 1⎞ ⎛ A normal to the plane is n 2 = ⎜1, − , − ⎟ . 4 2⎠ ⎝ Since n1 = 8n 2 , the planes are parallel. x=

The vector component of u orthogonal to a is u − proja u = (6, 2) − (0, 0) = (6, 2). 23. u ⋅ a = (3)(1) + (1)(0) + (−7)(5) = −32

17. A normal to 3x − y + z − 4 = 0 is n1 = (3, − 1, 1) and a normal to x + 2z = −1 is n 2 = (1, 0, 2).

a

n1 ⋅ n 2 = (3)(1) + (−1)(0) + (1)(2) = 5

2

= 12 + 02 + 52 = 26

The vector component of u along a is u⋅a proja u = a 2 a −32 (1, 0, 5) = 26 80 ⎞ ⎛ 16 = ⎜ − , 0, − ⎟ . 13 ⎠ ⎝ 13 The vector component of u orthogonal to a is 80 ⎞ ⎛ 16 u − proja u = (3, 1, − 7) − ⎜ − , 0, − ⎟ 13 ⎠ ⎝ 13 11 ⎞ ⎛ 55 = ⎜ , 1, − ⎟ . 13 ⎠ ⎝ 13

Since n1 and n 2 are not orthogonal, the planes are not perpendicular.

71



25. u ⋅ a = (1)(0) + (1)(2) + (1)(−1) = 1 a

2

2

2

31. The line is 4x + y − 2 = 0. ax + by0 + c D= 0 a 2 + b2 4(2) + (1)(−5) − 2 = 42 + 12 1 = 17

2

= 0 + 2 + (−1) = 5

The vector component of u along a is u⋅a proja u = a 2 a 1 = (0, 2, − 1) 5 ⎛ 2 1⎞ = ⎜ 0, , − ⎟ . 5⎠ ⎝ 5 The vector component of u orthogonal to a is ⎛ 2 1⎞ u − proja u = (1, 1, 1) − ⎜ 0, , − ⎟ ⎝ 5 5⎠ ⎛ 3 6⎞ = ⎜ 1, , ⎟ . ⎝ 5 5⎠

33. The plane is x + 2y − 2z − 4 = 0. ax + by0 + cz0 + d D= 0 a2 + b2 + c2 (1)(3) + 2(1) − 2(−2) − 4 = 12 + 22 + (−2)2 5 = 9 5 = 3

27. u ⋅ a = (2)(4) + (1)(−4) + (1)(2) + (2)(−2) = 2 a

2

= 42 + (−4)2 + 22 + (−2)2 = 40

The vector component of u along a is u⋅a proja u = a 2 a 2 = (4, − 4, 2, − 2) 40 1⎞ ⎛1 1 1 = ⎜ , − , , − ⎟. 5 5 10 10 ⎝ ⎠ The vector component of u orthogonal to a is 1⎞ ⎛1 1 1 u − proja u = (2, 1, 1, 2) − ⎜ , − , , − ⎟ ⎝ 5 5 10 10 ⎠ ⎛ 9 6 9 21 ⎞ = ⎜ , , , ⎟. ⎝ 5 5 10 10 ⎠ 29. D =

=

35. The plane is 2x + 3y − 4z − 1 = 0. ax + by0 + cz0 + d D= 0 a2 + b2 + c2 2(−1) + 3(2) − 4(1) − 1 = 22 + 32 + (−4)2 −1 = 29 1 = 29 37. A point on 2x − y − z = 5 is P0 (0, − 5, 0). The

ax0 + by0 + c

distance between P0 and the plane

a2 + b2 4(−3) + 3(1) + 4

−4x + 2y + 2z = 12 is −4(0) + 2(−5) + 2(0) − 12 D= (−4)2 + 22 + 22 −22 = 24 22 = 2 6 11 . = 6

= =1

−5

42 + 32

25

39. Note that the equation of the second plane is −2 times the equation of the first plane, so the planes coincide. The distance between the planes is 0. 72


Section 3.4

41. (a) α is the angle between v and i, so v ⋅i cos α = v i (a)(1) + (b)(0) + (c)(0) = v ⋅1 a = . v (b) Similar to part (a), cos β =

cos γ =

(c)

Section 3.4 Exercise Set 3.4 1. The vector equation is x = x0 + tv. (x, y) = (−4, 1) + t(0, −8) Equating components gives the parametric equations. x = −4, y = 1 − 8t

b and v

3. Since the given point is the origin, the vector equation is x = tv. (x, y, z) = t(−3, 0, 1) Equating components gives the parametric equations. x = −3t, y = 0, z = t

c . v

v ⎛ a b c ⎞ =⎜ , , ⎟ = (cos α , cos β , cos γ ) v ⎜⎝ v v v ⎟⎠

(d) Since 1, so

5. For t = 0, the point is (3, −6). The coefficients of t give the parallel vector (−5, −1).

v is a unit vector, its magnitude is v

7. For t = 0, the point is (4, 6). x = (4 − 4t, 6 − 6t) + (−2t, 0) = (4 − 6t, 6 − 6t) The coefficients of t give the parallel vector (−6, −6).

cos2 α + cos2 β + cos2 γ = 1 or

cos2 α + cos2 β + cos2 γ = 1.

9. The vector equation is x = x0 + t1v1 + t2 v 2 . ( x, y, z ) = (−3, 1, 0) + t1 (0, − 3, 6) + t2 (−5, 1, 2) Equating components gives the parametric equations. x = −3 − 5t2 , y = 1 − 3t1 + t2 , z = 6t1 + 2t2

43. v ⋅ (k1w1 + k2 w 2 ) = v ⋅ (k1w1 ) + v ⋅ (k2 w 2 ) = k1 ( v ⋅ w1 ) + k2 ( v ⋅ w 2 ) = k1 (0) + k2 (0) =0

11. The vector equation is x = x0 + t1v1 + t2 v 2 .

True/False 3.3

( x, y, z ) = (−1, 1, 4) + t1 (6, − 1, 0) + t2 (−1, 3, 1) Equating components gives the parametric equations. x = −1 + 6t1 − t2 , y = 1 − t1 + 3t2 , z = 4 + t2

(a) True; the zero vector is orthogonal to all vectors. (b) True; (ku) ⋅ (mv) = km(u ⋅ v) = km(0) = 0. (c) True; the orthogonal projection of u on a has the same direction as a while the vector component of u orthogonal to a is orthogonal (perpendicular) to a.

13. One vector orthogonal to v is w = (3, 2). Using w, the vector equation is (x, y) = t(3, 2) and the parametric equations are x = 3t, y = 2t.

(d) True; since projb u has the same direction as b.

15. Two possible vectors that are orthogonal to v are v1 = (0, 1, 0) and v 2 = (5, 0, 4). Using v1 and v 2 , the vector equation is

(e) True; if v has the same direction as a, then proja v = v.

( x, y, z ) = t1 (0, 1, 0) + t2 (5, 0, 4) and the parametric equations are x = 5t2 , y = t1,

(f) False; for a = (1, 0), u = (1, 10) and v = (1, 7), proja u = proja v.

z = 4t2 .

(g) False; u + v ≤ u + v .

73



⎡ 1 1 1 0⎤ 17. The augmented matrix of the system is ⎢ 2 2 2 0 ⎥ which has reduced row echelon form ⎢ ⎥ ⎣ 3 3 3 0⎦ The solution of the system is x1 = −r − s, x2 = r , x3 = s, or x = (−r − s, r, s) in vector form.

⎡ 1 1 1 0⎤ ⎢0 0 0 0⎥ . ⎢ ⎥ ⎣0 0 0 0⎦

⎡ 1 1 1⎤ Since the row vectors of ⎢ 2 2 2 ⎥ are all multiples of (1, 1, 1), only r1 needs to be considered. ⎢ ⎥ ⎣ 3 3 3⎦ r1 ⋅ x = (1, 1, 1) ⋅ (− r − s, r , s) = 1(−r − s) + 1(r ) + 1( s) =0 ⎡1 5 1 2 −1 0 ⎤ which has reduced row echelon form 19. The augmented matrix of the system is ⎢ ⎣1 −2 −1 3 2 0 ⎥⎦ 8 0⎤ ⎡ 1 0 − 3 19 7 7 7 ⎢ ⎥. 2 − 1 − 3 0⎥ ⎢0 1 7 7 7 ⎣ ⎦ 3 19 8 2 1 3 The solution of the system is x1 = r − s − t , x2 = − r + s + t , x3 = r , x4 = s, x5 = t , or 7 7 7 7 7 7 19 8 2 1 3 ⎛3 ⎞ x = ⎜ r − s − t , − r + s + t , r , s, t ⎟ in vector form. 7 7 7 7 7 7 ⎝ ⎠ For this system, r1 = (1, 5, 1, 2, − 1) and r2 = (1, − 2, − 1, 3, 2). 19 8 ⎞ ⎛ 2 1 3 ⎞ ⎛3 r1 ⋅ x = 1⎜ r − s − t ⎟ + 5 ⎜ − r + s + t ⎟ + 1(r ) + 2( s) − 1(t ) 7 7 7 7 7 7 ⎝ ⎠ ⎝ ⎠ 3 19 8 10 5 15 = r − s − t − r + s + t + r + 2s − t 7 7 7 7 7 7 =0 19 8 ⎞ ⎛ 2 1 3 ⎞ ⎛3 r2 ⋅ x = 1⎜ r − s − t ⎟ − 2 ⎜ − r + s + t ⎟ − 1(r ) + 3( s) + 2(t ) 7 7 ⎠ ⎝ 7 7 7 ⎠ ⎝7 3 19 8 4 2 6 = r − s − t + r − s − t − r + 3s + 2t 7 7 7 7 7 7 =0 21. (a) A particular solution of x + y + z = 1 is (1, 0, 0). The general solution of x + y + z = 0 is x = −s − t, y = s, z = t, which is s(−1, 1, 0) + t(−1, 0, 1) in vector form. The equation can be represented by (1, 0, 0) + s(−1, 1, 0) + t(−1, 0, 1). (b) Geometrically, the form of the equation indicates that it is a plane in R3 , passing through the point P(1, 0, 0) and parallel to the vectors (−1, 1, 0) and (−1, 0, 1). 23. (a) If x = (x, y, z) is orthogonal to a and b, then a ⋅ x = x + y + z = 0 and b ⋅ x = −2x + 3y = 0. x + y+z =0. The system is −2 x + 3 y =0 ⎡1 0 ⎡ 1 1 1 0⎤ ⎢ (b) The reduced row echelon form of ⎢ is ⎢⎣0 1 ⎣ −2 3 0 0⎥⎦

3 5 2 5

0⎤ ⎥ so the solution will require one 0 ⎥⎦

parameter. Since only one parameter is required, the solution space is a line through the origin in R3 . 74


Section 3.5 x2 = s, x3 = t , x4 = 0 and a particular solution

(c) From part (b), the solution of the system is 3 2 x = − t , y = − t , z = t, or 5 5 2 ⎞ ⎛ 3 x = ⎜ − t , − t , t ⎟ in vector form. Here, 5 ⎠ ⎝ 5 r1 = (1, 1, 1) and r2 = (−2, 3, 0).

1 of the nonhomogeneous system is x1 = , 3 x2 = 0, x3 = 0, x4 = 1. True/False 3.4

⎛ 3 ⎞ ⎛ 2 ⎞ r1 ⋅ x = 1⎜ − t ⎟ + 1⎜ − t ⎟ + 1(t ) = 0 ⎝ 5 ⎠ ⎝ 5 ⎠ ⎛ 3 ⎞ ⎛ 2 ⎞ r2 ⋅ x = −2 ⎜ − t ⎟ + 3 ⎜ − t ⎟ + 0(t ) = 0 ⎝ 5 ⎠ ⎝ 5 ⎠

(a) True; the vector equation is then x = x0 + tv where x0 is the given point and v is the given vector. (b) False; two non collinear vectors that are parallel to the plane are needed.

25. (a) The reduced row echelon form of ⎡ 1 2 − 1 0⎤ ⎡ 3 2 −1 0 ⎤ 3 3 ⎥ ⎢ 6 4 −2 0 ⎥ is ⎢⎢0 0 0 0⎥ . ⎢ ⎥ ⎢0 0 1 0⎦ 0 0⎥ ⎣ −3 −2 ⎣ ⎦ The solution of the system is 2 1 x1 = − s + t , x2 = s, x3 = t . 3 3

(c) True; the equation of the line is x = tv where v is any nonzero vector on the line. (d) True; if b = 0, then the statement is true by Theorem 3.4.3. If b ≠ 0, so that there is a nonzero entry, say bi of b, and if x0 is a solution vector of the system, then by matrix multiplication, ri ⋅ x0 = bi where ri is the ith row vector of A.

(b) Since the second and third rows of ⎡ 3 2 −1 2 ⎤ ⎢ 6 4 −2 4 ⎥ are scalar multiples of ⎢ ⎥ 1 −2 ⎦ ⎣ −3 −2

(e) False; a particular solution of Ax = b must be used to obtain the general solution of Ax = b from the general solution of Ax = 0.

the first, it suffices to show that x1 = 1, x2 = 0, x3 = 1 is a solution of 3x1 + 2 x2 − x3 = 2. 3(1) + 2(0) − 1 = 3 − 1 = 2.

(f) True; A(x1 − x2 ) = Ax1 − Ax2 = b − b = 0. Section 3.5 Exercise Set 3.5

(c) In vector form, add (1, 0, 1) to 1 ⎛ 2 ⎞ ⎜ − s + t , s, t ⎟ to get 3 3 ⎝ ⎠ 2 1 x1 = 1 − s + t , x2 = s, x3 = 1 + t . 3 3

1. (a) v × w = (0, 2, − 3) × (2, 6, 7) ⎛ 2 −3 0 −3 0 2 ⎞ =⎜ ,− , 6 7 2 7 2 6 ⎟⎠ ⎝ = (14 + 18, − (0 + 6), 0 − 4) = (32, − 6, − 4)

27. The reduced row echelon form of ⎡1 4 1 0 1⎤ ⎡ 3 4 1 2 3⎤ 3 3 3 ⎢6 8 2 5 7 ⎥ is ⎢⎢0 0 0 1 1⎥⎥ . ⎢ ⎥ ⎢0 0 0 0 0⎥ ⎣9 12 3 10 13⎦ ⎣ ⎦ A general solution of the system is 1 4 1 x1 = − s − t , x2 = s, x3 = t , x4 = 1. 3 3 3 From the general solution of the nonhomogeneous system, a general solution of 4 1 the homogeneous system is x1 = − s − t , 3 3

(b) u × ( v × w ) = (3, 2, − 1) × (32, − 6, − 4) ⎛ 2 −6 3 −1 3 2 ⎞ =⎜ ,− , 1 4 32 − − −4 32 −6 ⎟⎠ ⎝ = (−8 − 6, − (−12 + 32), − 18 − 64) = (−14, − 20, − 82)

75



(c) u × v = (3, 2, − 1) × (0, 2, − 3) ⎛ 2 −1 3 −1 3 2 ⎞ =⎜ ,− , 2 3 0 − −3 0 2 ⎟⎠ ⎝ = (−6 + 2, − (−9 − 0), 6 − 0) = (−4, 9, 6) (u × v) × w = (−4, 9, 6)× (2, 6, 7) ⎛9 6 −4 6 −4 9 ⎞ =⎜ ,− , 6 7 2 7 2 6 ⎟⎠ ⎝ = (63 − 36, − (−28 − 12), − 24 − 18) = (27, 40, − 42)

→

P1 P4 = (3, 1) The parallelogram can be considered in R3 as being determined by u = (3, 2, 0) and v = (3, 1, 0). ⎛2 0 3 0 3 2⎞ ,− , u× v = ⎜ 3 0 3 1 ⎟⎠ ⎝1 0 = (0, 0, − 3) u × v = 02 + 02 + (−3) 2 = 3 The area is 3. →

The triangle can be considered in R3 as being half of the parallelogram formed by u = (1, 4, 0) and v = (−3, 2, 0). ⎛4 0 1 0 1 4⎞ ,− , u× v = ⎜ −3 0 −3 2 ⎟⎠ ⎝2 0 = (0, 0, 14)

5. u × v is orthogonal to both u and v. u × v = (−2, 1, 5) × (3, 0, − 3) ⎛1 5 −2 5 −2 1 ⎞ =⎜ ,− , 0 3 3 −3 3 0 ⎟⎠ − ⎝ = (−3 − 0, − (6 − 15), 0 − 3) = (−3, 9, − 3)

u × v = 02 + 02 + 142 = 14 The area of the triangle is →

→

v = P1P3 = (2, 0, 3). ⎛ −5 2 −1 2 −1 −5 ⎞ ,− , u× v = ⎜ 0 3 2 3 2 0 ⎟⎠ ⎝ = (−15, 7, 10)

u × v = (−7)2 + (−1) 2 + 32 = 59

u × v = (−15)2 + 7 2 + 102 = 374

59.

The area of the triangle is

9. u × v = (2, 3, 0) × (−1, 2, − 2) ⎛3 0 2 0 2 3⎞ =⎜ ,− , 2 2 1 −2 −1 2 ⎟⎠ − − ⎝ = (−6 − 0, − (−4 − 0), 4 + 3) = (−6, 4, 7)

→

1 374 u× v = . 2 2

2 −6 2 17. u ⋅ ( v × w) = 0 4 −2 2 2 −4 4 −2 −6 2 =2 +2 2 −4 4 −2 = 2(−12) + 2(4) = −16 The volume of the parallelepiped is −16 = 16.

u × v = (−6)2 + 42 + 72 = 101 The area is

1 u × v = 7. 2

15. Let u = P1 P2 = (−1, − 5, 2) and

7. u × v = (1, − 1, 2) × (0, 3, 1) ⎛ −1 2 1 2 1 −1 ⎞ =⎜ ,− , 0 1 0 3 ⎟⎠ ⎝ 3 1 = (−1 − 6, − (1 − 0), 3 − 0) = (−7, − 1, 3)

The area is

→

13. AB = (1, 4) and AC = (−3, 2)

3. u × v is orthogonal to both u and v. u × v = (−6, 4, 2) × (3, 1, 5) ⎛4 2 −6 2 −6 4 ⎞ =⎜ ,− , 1 5 3 5 3 1 ⎟⎠ ⎝ = (20 − 2, − (−30 − 6), − 6 − 12) = (18, 36, − 18)

101. →

11. P1 P2 = (3, 2) and P3 P4 = (−3, − 2) so the side determined by P1 and P2 is parallel to the side determined by P3 and P4 , but the direction is opposite, thus P1 P2 and P1 P4 are adjacent sides. 76


Section 3.5

1 −1 −2 19. u ⋅ ( v × w) = 3 0 −2 5 −4 0 3 0 −1 −2 = +2 5 −4 5 −4 = −12 + 2(14) = 16 The vectors do not lie in the same plane.

(b)

−2 0 6 1 −3 1 −5 −1 1 1 −3 −3 1 = −2 +6 −1 1 −5 −1 = −2(−2) + 6(−16) = −92

→ = (−1)2 + 22 + 22 = 3 AB Let x be the length of the altitude from vertex C to side AB, then 1 → 26 x = 2 AB 2 26 26 x= = → 3 AB

21. u ⋅ ( v × w) =

29. (u + v) × (u − v) = (u + v ) × u − (u + v ) × v = (u × u) + ( v × u) − ((u × v) + ( v × v)) = 0 + ( v × u ) − (u × v ) − 0 = ( v × u) + ( v × u) = 2( v × u)

a 0 0 23. u ⋅ ( v × w) = 0 b 0 = abc 0 0 c

37. (a) a = PQ = (3, − 1, − 3)

→

→

b = PR = (2, − 1, 1) →

25. Since u ⋅ (v × w) = 3, then ⎡ u1 u2 u3 ⎤ det ⎢ v1 v2 v3 ⎥ = 3. ⎢ ⎥ ⎢⎣ w1 w2 w3 ⎥⎦

c = PS = (4, − 4, 3) 3 −1 −3 a ⋅ (b × c) = 2 −1 1 4 −4 3 2 1 2 −1 −1 1 =3 +1 −3 4 3 4 −4 −4 3 = 3(1) + 1(2) − 3(−4) = 17 1 17 The volume is a ⋅ (b × c) = . 6 6

⎡ u1 u2 u3 ⎤ (a) u ⋅ (w × v) = det ⎢ w1 w2 w3 ⎥ = −3 ⎢ ⎥ ⎣⎢ v1 v2 v3 ⎦⎥ (Rows 2 and 3 were interchanged.) (b) (v × w) ⋅ u = u ⋅ (v × w) = 3 ⎡ w1 w2 w3 ⎤ (c) w ⋅ (u × v) = det ⎢ u1 u2 u3 ⎥ = 3 ⎢ ⎥ ⎣⎢ v1 v2 v3 ⎦⎥ (Rows 2 and 3 were interchanged, then rows 1 and 2 were interchanged.)

→

(b) a = PQ = (1, 2, − 1) →

b = PR = (3, 4, 0) →

c = PS = (−1, − 3, 4) 1 2 −1 a ⋅ (b × c) = 3 4 0 −1 −3 4 3 4 1 = −1 +4 3 −1 −3 = −(−5) + 4(−2) = −3 1 The volume is a ⋅ (b × c) = 6

→

27. (a) Let u = AB = (−1, 2, 2) and →

v = AC = (1, 1, − 1). ⎛2 2 −1 2 −1 2 ⎞ ,− , u× v = ⎜ 1 1 1 −1 1 1 ⎟⎠ − ⎝ = (−4, 1, − 3) u × v = (−4) 2 + 12 + (−3) 2 = 26 The area of the triangle is

1 26 u× v = . 2 2 77

2 4

3 1 = . 6 2


SSM: Elementary Linear Algebra (d) u ⋅ w = (−2)(2) + (0)(−5) + (4)(−5) = −24

True/False 3.5

w

(a) True; for nonzero vectors u and v, u × v = u v sin θ will only be 0 if θ = 0, i.e., if u and v are parallel.

= 22 + (−5)2 + (−5)2 = 54

projw u =

u⋅w

w 2 w −24 (2, − 5, − 5) = 54 12 = − (2, − 5, − 5) 27

(b) True; the cross product of two nonzero and non collinear vectors will be perpendicular to both vectors, hence normal to the plane containing the vectors.

−2 0 4 (e) u ⋅ ( v × w) = 3 −1 6 2 −5 −5 3 −1 −1 6 = −2 +4 2 −5 −5 −5 = −2(35) + 4(−13) = −122

(c) False; the scalar triple product is a scalar, not a vector. (d) True (e) False; for example (i × i) × j = 0 × j = 0 i × (i × j) = i × k = −j

(f) −5v + w = (−15, 5, − 30) + (2, − 5, − 5) = (−13, 0, − 35) (u ⋅ v)w = ((−2)(3) + (0)(−1) + (4)(6))(2, − 5, − 5) = 18(2, − 5, − 5) = (36, − 90, − 90) (−5 v + w ) × (u ⋅ v)w ⎛ 0 −35 −13 −35 −13 0⎞ =⎜ , − , 36 −90 36 −90 ⎟⎠ ⎝ −90 −90 = (−3150, − 2430, 1170)

(f) False; for example, if v and w are distinct vectors that are both parallel to u, then u × v = u × w = 0, but v ≠ w. Chapter 3 Supplementary Exercises 1. (a) 3v − 2u = (9, − 3, 18) − (−4, 0, 8) = (13, − 3, 10) (b)

2

u+v+w = (−2 + 3 + 2, 0 − 1 − 5, 4 + 6 − 5) = (3, − 6, 5)

3. (a) 3v − 2u = (−9, 0, 24, 0) − (−4, 12, 4, 2) = (−5, − 12, 20, − 2)

= 32 + (−6)2 + 52 = 70

(b)

(c) The distance between −3u and v + 5w is −3u − ( v + 5w ) = −3u − v − 5w . −3u − v − 5w = (6, 0, − 12) − (3, − 1, 6) − (10, − 25, − 25) = (−7, 26, 7)

u+ v+w = (−2 − 3 + 9, 6 + 0 + 1, 2 + 8 − 6, 1 + 0 − 6) = (4, 7, 4, − 5) = 42 + 7 2 + 42 + (−5) 2 = 106

= (−7)2 + 262 + 7 2 = 774

78



(c) The distance between −3u and v + 5w is −3u − ( v + 5w ) = −3u − v − 5w . −3u − v − 5w = (6, − 18, − 6, − 3) − (−3, 0, 8, 0) − (45, 5, − 30, − 30) = (−36, − 23, 16, 27) = (−36) 2 + (−23)2 + 162 + 272 = 2810

(d) u ⋅ w = (−2)(9) + (6)(1) + (2)(−6) + (1)(−6) = −30 w

2

= 92 + 12 + (−6)2 + (−6)2 = 154

projw u =

u⋅w

w 2 w −30 (9, 1, − 6, − 6) = 154 15 = − (9, 1, − 6, − 6) 77

5. u = (−32, −1, 19), v = (3, −1, 5), w = (1, 6, 2) u ⋅ v = (−32)(3) + (−1)(−1) + (19)(5) = 0 u ⋅ w = (−32)(1) + (−1)(6) + (19)(2) = 0 v ⋅ w = (3)(1) + (−1)(6) + (5)(2) = 7 Since v ⋅ w ≠ 0, the vectors do not form an orthogonal set. 7. (a) The set of all such vectors is the line through the origin which is perpendicular to the given vector. (b) The set of all such vectors is the plane through the origin which is perpendicular to the given vector. (c) The only vector in R 2 that can be orthogonal to two non collinear vectors is 0; the set is {0}, the origin. (d) The set of all such vectors is the line through the origin which is perpendicular to the plane containing the two given vectors. 9. True; u + v

2

= (u + v) ⋅ (u + v) = u ⋅u + u ⋅ v + v ⋅u + v ⋅ v = u

Thus, if u + v

2

2

= u

+ u ⋅ v + v ⋅u + v 2

2

+ v 2 , u ⋅ v = v ⋅ u = 0, so u and v are orthogonal. →

→

11. Let S be S (−1, s2 , s3 ). Then u = PQ = (3, 1, − 2) and v = RS = (−6, s2 − 1, s3 − 1). u and v are parallel if u × v = 0. ⎛ 1 3 3 1 ⎞ −2 −2 ,− , u× v = ⎜ 1 1 6 1 6 s s s s − − − − − − 1 ⎟⎠ 2 3 3 2 ⎝ = ( s3 − 1 + 2( s2 − 1), − (3( s3 − 1) − 12), 3( s2 − 1) + 6) = (2s2 + s3 − 3, − 3s3 + 15, 3s2 + 3) If u × v = 0, then from the second and third components, s2 = −1 and s3 = 5, which also causes the first component to be 0. The point is S(−1, −1, 5).

79


→

SSM: Elementary Linear Algebra 21. One point on the line is (0, −5). Since the slope 3 of the line is m = , the vector (1, 3) is parallel 1 to the line. Using this point and vector gives the vector equation (x, y) = (0, −5) + t(1, 3). Equating components gives the parametric equations x = t, y = −5 + 3t.

→

13. u = PQ = (3, 1, − 2), v = PR = (2, 2, − 3) cos θ = = =

u⋅v u v (3)(2) + (1)(2) + (−2)(−3) 32 + 12 + (−2)2 22 + 22 + (−3)2 14

23. From the vector equation, (−1, 5, 6) is a point on the plane and (0, −1, 3) × (2, −1, 0) will be a normal to the plane. (0, − 1, 3) × (2, − 1, 0) ⎛ −1 3 0 3 0 −1 ⎞ =⎜ ,− , 2 0 2 −1 ⎟⎠ ⎝ −1 0 = (3, 6, 2) A point-normal equation for the plane is 3(x + 1) + 6(y − 5) + 2(z − 6) = 0.

14 17 14 = 17 15. The plane is 5x − 3y + z + 4 = 0. 5(−3) − 3(1) + 3 + 4 11 D= = 35 52 + (−3) 2 + 12 →

17. The plane will contain u = PQ = (1, − 2, − 2)

25. Two vectors in the plane are →

→

u = PQ = (−10, 4, − 1) and

and v = PR = (5, − 1, − 5).

→

Using P(−2, 1, 3), the vector equation is ( x, y, z ) = (−2, 1, 3) + t1 (1, − 2, − 2) + t2 (5, − 1, − 5). Equating components gives the parametric equations x = −2 + t1 + 5t2 , y = 1 − 2t1 − t2 ,

v = PR = (−9, 6, − 6). A normal to the plane is u × v. ⎛ 4 −1 −10 −1 −10 4 ⎞ u× v = ⎜ ,− , −9 −6 −9 6 ⎟⎠ ⎝ 6 −6 = (−18, − 51, − 24) Using point P(9, 0, 4), a point-normal equation for the plane is −18(x − 9) − 51y − 24(z − 4) = 0.

z = 3 − 2t1 − 5t2 . 19. The vector equation is (x, y) = (0, −3) + t(8, −1). Equating components gives the parametric equations x = 8t, y = −3 − t.

29. The equation represents a plane perpendicular to the xy-plane which intersects the xy-plane along the line Ax + By = 0.

80

Chapter 4 General Vector Spaces Section 4.1 Exercise Set 4.1 1. (a) u + v = (−1, 2) + (3, 4) = (−1 + 3, 2 + 4) = (2, 6)

3u = 3(−1, 2) = (0, 6) (b) The sum of any two real numbers is a real number. The product of two real numbers is a real number and 0 is a real number. (c) Axioms 1−5 hold in V because they also hold in R 2 . (e) Let u = (u1 , u2 ) with u1 ≠ 0. Then 1u = 1(u1 , u2 ) = (0, u2 ) ≠ u. 3. The set is a vector space with the given operations. 5. The set is not a vector space. Axiom 5 fails to hold because of the restriction that x ≥ 0. Axiom 6 fails to hold for k < 0. 7. The set is not a vector space. Axiom 8 fails to hold because (k + m)2 ≠ k 2 + m 2 . 9. The set is a vector space with the given operations. 11. The set is a vector space with the given operations. 23.

u+w = v+w (u + w) + (− w) = ( v + w) + (− w) u + [ w + (−w )] = v + [ w + (−w)] u+0 = v+0 u=v

Hypothesis Add − w to both sides. Axiom 3 Axiom 5 Axiom 4

25. (1) Axiom 7 (2) Axiom 4 (3) Axiom 5 (4) Axiom 1 (5) Axiom 3 (6) Axiom 5 (7) Axiom 4 True/False 4.1 (a) False; vectors are not restricted to being directed line segments. (b) False; vectors are not restricted to being n-tuples of real numbers. (c) True

81

Chapter 4: General Vector Spaces


(d) False; if a vector space V had exactly two elements, one of them would necessarily be the zero vector 0. Call the other vector u. Then 0 + 0 = 0, 0 + u = u, and u + 0 = u. Since −u must exist and −u ≠ 0, then −u = u and 1 1 u + u = 0. Consider the scalar . Since u 2 2 1 1 would be an element of V, u = 0 or u = u. If 2 2 1 u = 0, then 2 1 1 ⎛1 1⎞ u = 1u = ⎜ + ⎟ u = u + u = 0 + 0 = 0. 2 2 ⎝2 2⎠ 1 If u = u, then 2 1 1 ⎛1 1⎞ u = 1u = ⎜ + ⎟ u = u + u = u + u = 0. 2 2 ⎝2 2⎠ Either way we get u = 0 which contradicts our assumption that we had exactly two elements.

(d) This is a subspace of P3 . 5. (a) This is a subspace of R ∞ . (b) This is not a subspace of R ∞ , since for k ≠ 1, kv is not in the set. (c) This is a subspace of R ∞ . (d) This is a subspace of R ∞ . 7. Consider k1u + k2 v = (a, b, c). Thin (0k1 + k2 , − 2k1 + 3k2 , 2k1 − k2 ) = (a, b, c). Equating components gives the system k2 = a ⎡ 0 1⎤ −2k1 + 3k2 = b or Ax = b where A = ⎢ −2 3⎥ , ⎢ ⎥ 2k1 − k2 = c ⎣ 2 −1⎦ ⎡a ⎤ ⎡k ⎤ x = ⎢ 1 ⎥ , and b = ⎢ b ⎥ . ⎢ ⎥ ⎣ k2 ⎦ ⎣c ⎦ The system can be solved simultaneously by ⎡ 0 1 2 3 0 0⎤ reducing the matrix ⎢ −2 3 2 1 4 0 ⎥ , ⎢ ⎥ ⎣ 2 −1 2 5 5 0 ⎦ ⎡ 1 0 2 4 0 0⎤ which reduces to ⎢0 1 2 3 0 0 ⎥ . The ⎢ ⎥ ⎣0 0 0 0 1 0 ⎦ results can be read from the reduced matrix.

(e) False; the zero vector would be 0 = 0 + 0x which does not have degree exactly 1. Section 4.2 Exercise Set 4.2 1. (a) This is a subspace of R3 . (b) This is not a subspace of R3 , since (a1, 1, 1) + (a2 , 1, 1) = (a1 + a2 , 2, 2) which is not in the set.

(a) (2, 2, 2) = 2u + 2v is a linear combination of u and v.

(c) This is a subspace of R3 .

(b) (3, 1, 5) = 4u + 3v is a linear combination of u and v.

(d) This is not a subspace of R3 , since for k ≠ 1 k (a + c + 1) ≠ ka + kc + 1, so k(a, b, c) is not in the set.

(c) (0, 4, 5) is not a linear combination of u and v. (d) (0, 0, 0) = 0u + 0v is a linear combination of u and v.

3

(e) This is a subspace of R .

9. Similar to the process in Exercise 7, determining whether the given matrices are linear combinations of A, B, and C can be accomplished by reducing the matrix ⎡ 4 1 0 6 0 6 −1⎤ ⎢ 0 −1 2 −8 0 0 5⎥ ⎢ ⎥. ⎢ −2 2 1 −1 0 3 7 ⎥ ⎣⎢ −2 3 4 −8 0 8 1⎦⎥

3. (a) This is a subspace of P3 . (b) This is a subspace of P3 . (c) This is not a subspace of P3 since

k (a0 + a1 x + a2 x 2 + a3 x3 ) is not in the set for all noninteger values of k.

82


⎡1 ⎢0 The matrix reduces to ⎢ ⎢0 ⎢⎣0

0 1 0 0

Section 4.2

0 1 0 2 1 −3 0 0

0 1 0⎤ 0 2 0⎥ ⎥. 0 1 0⎥ 0 0 1⎥⎦

⎡ 6 −8 ⎤ (a) ⎢ = 1A + 2 B − 3C is a linear combination of A, B, and C. ⎣ −1 −8⎥⎦ ⎡0 0⎤ = 0 A + 0 B + 0C is a linear combination of A, B, and C. (b) ⎢ ⎣0 0⎥⎦ ⎡6 0⎤ = 1A + 2 B + 1C is a linear combination of A, B, and C. (c) ⎢ ⎣ 3 8⎥⎦ ⎡ −1 5 ⎤ is not a linear combination of A, B, and C. (d) ⎢ ⎣ 7 1⎥⎦

11. Let b = (b1 , b2 , b3 ) be an arbitrary vector in R3 . 2k1 = b1 (a) If k1v1 + k2 v 2 + k3 v3 = b, then (2k1 , 2k1 + k3 , 2k1 + 3k2 + k3 ) = (b1 , b2 , b3 ) or 2k1 + k3 = b2 . 2k1 + 3k2 + k3 = b3 ⎡2 0 0⎤ Since det ⎢ 2 0 1⎥ = −6 ≠ 0, the system is consistent for all values of b1 , b2 , and b3 so the given vectors ⎢ ⎥ ⎣ 2 3 1⎦ span R3 .

(b) If k1v1 + k2 v 2 + k3 v3 = b, then (2k1 + 4k2 + 8k3 , − k1 + k2 − k3 , 3k1 + 2k2 + 8k3 ) = (b1 , b2 , b3 ) or 2k1 + 4k2 + 8k3 = b1 − k1 + k2 − k3 = b2 . 3k1 + 2k2 + 8k3 = b3 ⎡ 2 4 8⎤ Since det ⎢ −1 1 −1⎥ = 0, the system is not consistent for all values of b1 , b2 , and b3 so the given vectors ⎢ ⎥ ⎣ 3 2 8⎦ do not span R3 .

(c) If k1v1 + k2 v 2 + k3 v3 + k4 v 4 = b, then (3k1 + 2k2 + 5k3 + k4 , k1 − 3k2 − 2k3 + 4k4 , 4k1 + 5k2 + 9k3 − k4 ) = (b1, b2 , b3 ) or 3k1 + 2k2 + 5k3 + k4 = b1 k1 − 3k2 − 2k3 + 4k4 = b2 . 4k1 + 5k2 + 9k3 − k4 = b3 5 1 b1 ⎤ ⎡3 2 Reducing the matrix ⎢ 1 −3 −2 4 b2 ⎥ leads to ⎢ ⎥ ⎣ 4 5 9 −1 b3 ⎦ The system has a solution only if −

⎡1 −3 −2 4 b2 ⎤ ⎢ ⎥ 3 1 −1 11 b1 − 11 b2 ⎢0 1 1 ⎥. ⎢0 0 0 0 − 17 b + 7 b + b ⎥ 3⎦ 11 1 11 2 ⎣

17 7 b1 + b2 + b3 = 0 or 17b1 = 7b2 + 11b3 . This restriction means that the 11 11

span of v1 , v 2 , v3 , and v 4 is not all of R3 , i.e., the given vectors do not span R3 . 83



(d) If k1v1 + k2 v 2 + k3 v3 + k4 v 4 = b, then (k1 + 3k2 + 4k3 + 3k4 , 2k1 + 4k2 + 3k3 + 3k4 , 6k1 + k2 + k3 + k4 ) = (b1, b2 , b3 ) or k1 + 3k2 + 4k3 + 3k4 = b1 2k1 + 4k2 + 3k3 + 3k4 = b2 . 6k1 + k2 + k3 + k4 = b3 ⎡1 0 ⎢ ⎢0 1 ⎢ ⎢⎣0 0 Thus, for any values of b1 , b2 , and b3 , values of ⎡1 3 4 3 b1 ⎤ The matrix ⎢ 2 4 3 3 b2 ⎥ reduces to ⎢ ⎥ ⎣6 1 1 1 b3 ⎦

1 b − 1 b + 7 b ⎤ − 39 1 39 2 39 3 ⎥ 16 23 b − 5 b ⎥ . − 39 b1 + 39 0 2 39 3 ⎥ 22 b − 17 b + 2 b 1 39 1 39 2 39 3 ⎥⎦ k1 , k2 , k3 , and k4 can be found. The given vectors span

0

1 39 16 39 17 39

R3 .

13. Let p = a0 + a1 x + a2 x 2 be an arbitrary polynomial in P2 . If k1p1 + k2p 2 + k3p3 + k4 p 4 = p, then (k1 + 3k2 + 5k3 − 2k4 ) + (−k1 + k2 − k3 − 2k4 ) x + (2k1 + 4k3 + 2k4 ) x 2 = a0 + a1 x + a2 x 2 or k1 + 3k2 + 5k3 − 2k4 = a0 − k1 + k2 − k3 − 2k4 = a1 . 2k1 + 4k3 + 2k4 = a2

⎡ 1 3 5 −2 a0 ⎤ Reducing the matrix ⎢ −1 1 −1 −2 a1 ⎥ leads to ⎢ ⎥ ⎣ 2 0 4 2 a2 ⎦

a0 ⎡ 1 3 5 −2 ⎤ ⎢ ⎥ 1 1a − + 0 1 1 1 a ⎢ ⎥. 4 0 4 1 ⎢0 0 0 0 − 1 a + 3 a + a ⎥ 2⎦ ⎣ 2 0 2 1

1 3 The system has a solution only if − a0 + a1 + a2 = 0 or a0 = 3a1 + 2a2 . This restriction means that the span of 2 2 p1 , p 2 , p3 , and p 4 is not all of P3 , i.e., they do not span P3 .

⎡ −1 1 1 0 ⎤ 15. (a) Since det(A) = 0, reduce the matrix ⎢ 3 −1 0 0 ⎥ . The matrix reduces to ⎢ ⎥ ⎣ 2 −4 −5 0 ⎦

⎡ 1 0 1 0⎤ 2 ⎢ ⎥ ⎢0 1 32 0 ⎥ . The solution ⎢ ⎥ ⎣0 0 0 0 ⎦

1 3 set is x = − t , y = − t , z = t which is a line through the origin. 2 2 3 0⎤ ⎡ 1 −2 (b) Since det(A) = 0, reduce the matrix ⎢ −3 6 9 0 ⎥ . The matrix reduces to ⎢ ⎥ ⎣ −2 4 −6 0 ⎦ set is x = 2t, y = t, z = 0, which is a line through the origin.

⎡ 1 −2 0 0 ⎤ ⎢0 0 1 0 ⎥ . The solution ⎢ ⎥ ⎣0 0 0 0 ⎦

(c) Since det(A) = −1 ≠ 0, the system has only the trivial solution, i.e., the origin. (d) Since det(A) = 8 ≠ 0, the system has only the trivial solution, i.e., the origin. ⎡ 1 −1 1 0 ⎤ (e) Since det(A) = 0, reduce the matrix ⎢ 2 −1 4 0⎥ . The matrix reduces to ⎢ ⎥ ⎣ 3 1 11 0⎦ is x = −3t, y = −2t, z = t, which is a line through the origin.

84

⎡ 1 0 3 0⎤ ⎢0 1 2 0⎥ . The solution set ⎢ ⎥ ⎣0 0 0 0⎦


Section 4.3

⎡ 1 −3 1 0 ⎤ (f) Since det(A) = 0, reduce the matrix ⎢ 2 −6 2 0 ⎥ . The matrix reduces to ⎢ ⎥ ⎣ 3 −9 3 0 ⎦ is x − 3y + z = 0, which is a plane through the origin.

⎡ 1 −3 1 0 ⎤ ⎢0 0 0 0 ⎥ . The solution set ⎢ ⎥ ⎣0 0 0 0 ⎦

17. Let f = f(x) and g = g(x) be elements of the set. Then b

f + g = ∫ ( f ( x) + g ( x))dx a b

b

a

a

= ∫ f ( x)dx + ∫ g ( x)dx =0

b

b

a

a

and f + g is in the set. Let k be any scalar. Then kf = ∫ kf ( x)dx = k ∫ f ( x)dx = k ⋅ 0 = 0 so kf is in the set. Thus the set is a subspace of C[a, b]. True/False 4.2 (a) True; this is part of the definition of a subspace. (b) True; since a vector space is a subset of itself. (c) False; for instance, the set W in Example 4 contains the origin (zero vector), but is not a subspace of R 2 . (d) False; R 2 is not a subset of R3 . (e) False; if b ≠ 0, i.e., the system is not homogeneous, then the solution set does not contain the zero vector. (f) True; this is by the definition of the span of a set of vectors. (g) True; by Theorem 4.2.2. (h) False; consider the subspaces W1 = {( x, y) y = x} and W2 = {( x, y) y = − x} in R 2 . v1 = (1, 1) is in W1 and v 2 = (1, − 1) is in W2 , but v1 + v 2 = (2, 0) is not in W1 ∪ W2 = {( x, y) y = ± x} .

(i) False; let v1 = (1, 0), v 2 = (0, 1), u1 = (−1, 0), and u 2 = (0, − 1). Then {v1 , v 2 } and {u1, u 2 } both span R 2 but {u1, u 2 } ≠ {v1 , v 2 }. (j) True; the sum of two upper triangular matrices is upper triangular and the scalar multiple of an upper triangular matrix is upper triangular. (k) False; the span of x − 1, ( x − 1)2 , and ( x − 1)3 will only contain polynomials for which p(1) = 0, i.e., not all of P3 . Section 4.3 Exercise Set 4.3 1. (a) u 2 = −5u1 , i.e., u 2 is a scalar multiple of u1 . (b) Since there are 3 vectors and 3 > 2, the set is linearly dependent by Theorem 4.3.3. (c) p 2 = 2p1 , i.e., p 2 is a scalar multiple of p1 .

85



(d) B = −A, i.e., B is a scalar multiple of A. 3. (a) The equation k1 (3, 8, 7, − 3) + k2 (1, 5, 3, − 1) + k3 (2, − 1, 2, 6) + k4 (1, 4, 0, 3) = (0, 0, 0, 0) generates the homogeneous system. 3k1 + k2 + 2k3 + k4 = 0 8k1 + 5k2 − k3 + 4k4 = 0 7k1 + 3k2 + 2k3 =0 −3k1 − k2 + 6k3 + 3k4 = 0

⎡ 3 1 2 ⎢ 8 5 −1 Since det ⎢ ⎢ 7 3 2 ⎢⎣ −3 −1 6 independent.

1⎤ 4⎥ ⎥ = 128 ≠ 0, the system has only the trivial solution and the vectors are linearly 0⎥ 3⎥⎦

(b) The equation k1 (0, 0, 2, 2) + k2 (3, 3, 0, 0) + k3 (1, 1, 0, − 1) = (0, 0, 0, 0) generates the homogeneous system 3k2 + k3 = 0 3k2 + k3 = 0 2k1 =0 2k1 − k3 = 0 The third equation gives k1 = 0, which gives k3 = 0 in the fourth equation. k3 = 0 gives k2 = 0 in the first two equations. Since the system has only the trivial solution, the vectors are linearly independent. (c) The equation k1 (0, 3, − 3, − 6) + k2 (−2, 0, 0, − 6) + k3 (0, − 4, − 2, − 2) + k4 (0, − 8, 4, − 4) = (0, 0, 0, 0) generates the homogeneous system −2 k 2 =0 3k1 − 4 k3 − 8 k 4 = 0 . −3k1 − 2k3 + 4k4 = 0 −6k1 − 6k2 − 2k3 − 4k4 = 0

⎡ 0 −2 0 0 ⎤ ⎢ 3 0 −4 −8 ⎥ Since det ⎢ ⎥ = 480 ≠ 0, the system has only the trivial solution and the vectors are linearly ⎢ −3 0 −2 4 ⎥ ⎣⎢ −6 −6 −2 −4 ⎦⎥ independent. (d) The equation k1 (3, 0, − 3, 6) + k2 (0, 2, 3, 1) + k3 (0, − 2, − 2, 0) + k4 (−2, 1, 2, 1) = (0, 0, 0, 0) generates the homogeneous system 3k1 − 2 k4 = 0 2k2 − 2k3 + k4 = 0 . −3k1 + 3k2 − 2k3 + 2k4 = 0 6k1 + k2 + k4 = 0

⎡ 3 ⎢ 0 Since det ⎢ ⎢ −3 ⎢⎣ 6 independent.

0 0 −2 ⎤ 2 −2 1⎥ ⎥ = 36 ≠ 0, the system has only the trivial solution and the vectors are linearly 3 −2 2⎥ 1 0 1⎥⎦

86


Section 4.3

5. If the vectors lie in a plane, then they are linearly dependent.

3 7 3 t = − : v 2 = v1 + v3 2 2 2 7 2 t = 1: v3 = − v1 + v 2 3 3

(a) The equation k1v1 + k2 v 2 + k3 v3 = 0 generates the homogeneous system 2k1 + 6k2 + 2k3 = 0 −2k1 + k2 = 0. 4 k 2 − 4 k3 = 0

9. The equation k1v1 + k2 v 2 + k3 v3 = 0 generates the homogeneous system 1 1 λk1 − k2 − k3 = 0 2 2 1 1 − k1 + λk2 − k3 = 0 . 2 2 1 1 − k1 − k2 + λk3 = 0 2 2 ⎡ λ − 1 − 1⎤ 2 2⎥ ⎢ 1 det ⎢ − 12 λ − 12 ⎥ = (4λ3 − 3λ − 1) ⎢ 1 ⎥ 4 1 λ ⎥⎦ ⎢⎣ − 2 − 2 1 = (λ − 1)(2λ + 1)2 4 1 For λ = 1 and λ = − , the determinant is zero 2 and the vectors form a linearly dependent set.

⎡ 2 6 2⎤ Since det ⎢ −2 1 0⎥ = −72 ≠ 0, the ⎢ ⎥ ⎣ 0 4 −4 ⎦ vectors are linearly independent, so they do not lie in a plane. (b) The equation k1v1 + k2 v 2 + k3 v3 = 0 generates the homogeneous system −6k1 + 3k2 + 4k3 = 0 7k1 + 2k2 − k3 = 0 . 2k1 + 4k2 + 2k3 = 0

⎡ −6 3 4 ⎤ Since det ⎢ 7 2 −1⎥ = 0, the vectors are ⎢ ⎥ ⎣ 2 4 2⎦ linearly dependent, so they do lie in a plane.

11. Let {v a , vb , … , v n } be a (nonempty) subset of S. If this set were linearly dependent, then there would be a nonzero solution (ka , kb , … , kn ) to

7. (a) The equation k1v1 + k2 v 2 + k3 v3 = 0 generates the homogeneous system 6 k 2 + 4 k3 = 0 3k1 − 7 k3 = 0 . k1 + 5k2 + k3 = 0 − k1 + k2 + 3k3 = 0

⎡ 0 ⎢ 3 The matrix ⎢ ⎢ 1 ⎣⎢ −1

ka v a + kb vb + + kn v n = 0. This can be expanded to a nonzero solution of k1v1 + k2 v 2 + + kr v r = 0 by taking all other coefficients as 0. This contradicts the linear independence of S, so the subset must be linearly independent.

6 4 0⎤ 0 −7 0 ⎥ ⎥ reduces to 5 1 0⎥ 1 3 0 ⎥⎦

13. If S is linearly dependent, then there is a nonzero solution (k1 , k2 , …, kr ) to

⎡ 1 0 − 7 0⎤ 3 ⎢ ⎥ 2 0⎥ ⎢0 1 . Since the system has a 3 ⎢ ⎥ 0 0 0 0 ⎢ ⎥ 0 0 ⎦⎥ ⎣⎢0 0 nontrivial solution, the vectors are linearly dependent. (b) The solution of the system is k1 =

k1v1 + k2 v 2 +

+ kr v r = 0. Thus

(k1 , k2 , … , kr , 0, 0, … , 0) is a nonzero solution to k1v1 + k2 v 2 + + kr v r + kr +1 v r +1 + + kn v n =0 so the set {v1 , v 2 , … , v r , v r +1 , … , v n } is linearly dependent.

7 t, 3

15. If {v1, v 2 , v3} were linearly dependent, there would be a nonzero solution to k1v1 + k2 v 2 + k3 v3 = 0 with k3 ≠ 0 (since v1

2 k 2 = − t , k3 = t . 3 3 2 3 t = : v1 = v 2 − v3 7 7 7

and v 2 are linearly independent). Solving for

87



determinant was simplified by adding the first row to the third row.

k k v3 gives v3 = − 1 v1 − 2 v 2 which contradicts k3 k3 that v3 is not in span {v1 , v 2 }. Thus,

True/False 4.3

{v1, v 2 , v3} is linearly independent.

(a) False; the set {0} is linearly dependent.

19. (a) If v1 , v 2 , and v3 are placed with their initial points at the origin, they will not lie in the same plane, so they are linearly independent.

(b) True (c) False; the set {v1 , v 2 } in R 2 where v1 = (0, 1) and v 2 = (0, 2) is linearly dependent.

(b) If v1 , v 2 , and v3 are placed with their initial points at the origin, they will lie in the same plane, so they are linearly dependent. Alternatively, note that v1 + v 2 = v3 by

(d) True; suppose {kv1, kv 2 , kv3} were linearly dependent, then there would be a nontrivial solution to k1 (kv1 ) + k2 (kv 2 ) + k3 (kv3 ) = 0 which would give a nontrivial solution to k1v1 + k2 v 2 + k3 v3 = 0.

placing v1 along the z-axis with its terminal point at the origin. 21. The Wronskian is x cos x W ( x) = = − x sin x − cos x. 1 − sin x

(e) True; consider the sets {v1 , v 2 }, {v1 , v 2 , v3},

..., {v1 , v 2 , … , v n }. If {v1 , v 2 } is linearly dependent, then v 2 = kv1 for some k and the

W(0) = −cos 0 = −1, so the functions are linearly independent.

condition is met. If v 2 ≠ kv1 then there is some 2 < k ≤ n for which the set {v1 , v 2 , …, v k −1} is linearly independent but {v1, v 2 , … , v k } is linearly dependent. This means that there is a nontrivial solution of a1 v1 + a2 v 2 + + ak −1 v k −1 + ak v k = 0 with

1 x ex 23. (a) W ( x) = 0

x 1 ex = e

0 0 ex Since W ( x) = e x is nonzero for all x, the vectors are linearly independent.

ak ≠ 0 and the equation can be solved to give v k as a linear combination of v1 , v 2 , ..., v k −1 .

1 x x2 (b) W ( x) = 0 1 2 x = 2 0 0 2 Since W(x) = 2 is nonzero for all x, the vectors are linearly independent.

(f) False; the set is ⎧ ⎡1 1 ⎤ ⎡1 0 ⎤ ⎡1 0⎤ ⎡ 0 1 ⎤ ⎡ 0 1⎤ ⎨ ⎢ 0 0 ⎥ , ⎢1 0 ⎥ , ⎢0 1 ⎥ , ⎢1 0 ⎥ , ⎢ 0 1⎥ , ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎩⎣

⎡0 0⎤ ⎫ ⎢⎣1 1 ⎥⎦ ⎬ and ⎭ ⎡1 1 ⎤ ⎡0 0 ⎤ ⎡1 0⎤ ⎡0 1⎤ 1⎢ + 1⎢ + (−1) ⎢ + (−1) ⎢ ⎥ ⎥ ⎥ 0 0 1 1 1 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣0 1⎥⎦ ⎡0 0⎤ =⎢ ⎣ 0 0 ⎥⎦ so the set is linearly dependent.

x cos x sin x cos x 25. W ( x) = cos x − sin x cos x − x sin x − sin x − cos x − x cos x − 2 sin x x cos x sin x cos x = cos x − sin x cos x − x sin x −2 sin x 0 0 sin x cos x = −2 sin x cos x − sin x 2

(g) True; the first polynomial has a nonzero constant term, so it cannot be expressed as a linear combination of the others, and since the two others are not scalar multiples of one another, the three polynomials are linearly independent.

2

= −2 sin x(− sin x − cos x) = 2 sin x Since 2sin x is not identically zero, the vectors are linearly independent, hence span a threedimensional subspace of F(−∞, ∞). Note that the 88


Section 4.4

(h) False; the functions f1 and f 2 are linearly dependent if there are scalars k1 and k2 such that k1 f1 ( x) + k2 f 2 ( x) = 0 for all real numbers x. Section 4.4 Exercise Set 4.4 1. (a) The set S = {u1, u 2 , u3} is not linearly independent; u3 = 2u1 + u 2 . (b) The set S = {u1, u 2 } spans a plane in R3 , not all of R3 . (c) The set S = {p1, p 2 } does not span P2 ; x 2 is not a linear combination of p1 and p 2 . (d) The set S = {A, B, C, D, E} is not linearly independent; E can be written as linear combination of A, B, C, and D.

⎡ 1 2 3⎤ 3. (a) As in Example 3, it is sufficient to consider det ⎢ 0 2 3⎥ = 6 ≠ 0. ⎢ ⎥ ⎣ 0 0 3⎦ Since this determinant is nonzero, the set of vectors is a basis for R3 .

⎡ 3 2 1⎤ (b) It is sufficient to consider det ⎢ 1 5 4 ⎥ = 26 ≠ 0. Since this determinant is nonzero, the set of vectors is a ⎢ ⎥ ⎣ −4 6 8⎦ basis for R3 . 0⎤ ⎡ 2 4 (c) It is sufficient to consider det ⎢ −3 1 −7 ⎥ = 0. Since this determinant is zero, the set of vectors is not ⎢ ⎥ 1⎦ ⎣ 1 1 linearly independent.

⎡ 1 2 −1⎤ (d) It is sufficient to consider det ⎢ 6 4 2⎥ = 0. Since this determinant is zero, the set of vectors is not ⎢ ⎥ ⎣ 4 −1 5⎦ linearly independent. ⎡3 6 ⎤ ⎡ 0 −1⎤ ⎡ 0 −8 ⎤ ⎡ 1 0 ⎤ ⎡ 0 0⎤ 5. The equations c1 ⎢ + c2 ⎢ + c3 ⎢ + c4 ⎢ = and ⎥ ⎥ ⎥ 3 6 1 0 12 4 − − − − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ −1 2 ⎥⎦ ⎢⎣ 0 0⎥⎦ ⎡3 6 ⎤ ⎡ 0 −1⎤ ⎡ 0 −8 ⎤ ⎡ 1 0⎤ ⎡ a b ⎤ c1 ⎢ + c2 ⎢ + c3 ⎢ + c4 ⎢ = generate the systems ⎥ ⎥ ⎥ 3 6 1 0 12 4 − − − − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ −1 2 ⎥⎦ ⎢⎣ c d ⎥⎦ 3c1 + c4 = 0 3c1 + c4 = a 6c1 − c2 − 8c3 =0 6c1 − c2 − 8c3 =b and 3c1 − c2 − 12c3 − c4 = 0 3c1 − c2 − 12c3 − c4 = c −6c1 − 4c3 + 2c4 = 0 −6c1 − 4c3 + 2c4 = d

89


SSM: Elementary Linear Algebra c1 − 4c2 + 7c3 = 5 2c1 + 5c2 − 8c3 = −12 . The matrix 3c1 + 6c2 + 9c3 = 3

which have the same coefficient matrix. Since 0 1⎤ ⎡ 3 0 ⎢ 6 −1 −8 0 ⎥ det ⎢ ⎥ = 48 ≠ 0 the matrices ⎢ 3 −1 −12 −1⎥ ⎢⎣ −6 0 −4 2 ⎥⎦ are linearly independent and also span M 22 , so they are a basis.

5⎤ ⎡ 1 −4 7 ⎢2 5 −8 −12⎥ reduces to ⎢ ⎥ 3⎦ ⎣3 6 9 ⎡ 1 0 0 −2 ⎤ ⎢0 1 0 0 ⎥ . ⎢ ⎥ ⎣0 0 1 1⎦

7. (a) Since S is the standard basis for R 2 , (w) S = (3, − 7).

The solution of the system is c1 = −2, c2 = 0, c3 = 1 and ( v) S = (−2, 0, 1).

(b) Solve c1u1 + c2 u 2 = w, or in terms of

components, (2c1 + 3c2 , − 4c1 + 8c2 ) = (1, 1). 2c1 + 3c2 = 1 Equating components gives . −4c1 + 8c2 = 1

11. Solve c1 A1 + c2 A2 + c3 A3 + c4 A4 = A. By inspection, c3 = −1 and c4 = 3. Thus it remains

−c1 + c2 = 2 which can be c1 + c2 = 0

⎡ 2 3 1⎤ The matrix ⎢ reduces to ⎣ −4 8 1⎥⎦

to solve the system

⎡1 0 5 ⎤ 3⎞ 28 ⎥ . Thus, ( w) = ⎛ 5 ⎢ ⎜ , ⎟. S 3 ⎝ 28 14 ⎠ ⎢⎣0 1 14 ⎥⎦

from which c1 = −1. Thus, ( A) S = (−1, 1, − 1, 3).

solved by adding the equations to get c2 = 1,

⎡0 0 ⎤ , 13. Solving c1 A1 + c2 A2 + c3 A3 + c4 A4 = ⎢ ⎣0 0 ⎥⎦ ⎡a b ⎤ c1 A1 + c2 A2 + c3 A3 + c4 A4 = ⎢ , and ⎣ c d ⎥⎦ c1 A1 + c2 A2 + c3 A3 + c4 A4 = A gives the systems

(c) Solve c1u1 + c2 u 2 = w, or in terms of

components, (c1 , c1 + 2c2 ) = (a, b). It is clear that c1 = a and solving a + 2c2 = b for c2 gives c2 =

b−a . Thus, 2

⎛ b−a⎞ ( w ) S = ⎜ a, ⎟. ⎝ 2 ⎠ 9. (a) Solve c1v1 + c2 v 2 + c3 v3 = v, or in terms of components, (c1 + 2c2 + 3c3 , 2c2 + 3c3 , 3c3 ) = (2, − 1, 3). Equating components gives c1 + 2c2 + 3c3 = 2 2c2 + 3c3 = −1 which can easily be 3c3 = 3

c1 c1 + c2 c1 + c2 + c3 c1 + c2 + c3 + c4

=0 =0 , =0 =0

c1 c1 + c2 c1 + c2 + c3 c1 + c2 + c3 + c4

=1 =0 . =1 =0

c1 c1 + c2 c1 + c2 + c3 c1 + c2 + c3 + c4

⎡1 ⎢1 It is easy to see that det ⎢ ⎢1 ⎣⎢1

solved by back-substitution to get c1 = 3, c2 = −2, c3 = 1. Thus, ( v) S = (3, − 2, 1).

0 1 1 1

0 0 1 1

=a =b , and =c =d

0⎤ 0⎥ ⎥ = 1 ≠ 0, so 0⎥ 1⎦⎥

{A1, A2 , A3 , A4 } spans M 22 . The third system is easy to solve by inspection to get c1 = 1,

(b) Solve c1v1 + c2 v 2 + c3 v3 = v, or in terms of components, (c1 − 4c2 + 7c3 , 2c1 + 5c2 − 8c3 , 3c1 + 6c2 + 9c3 )

c2 = −1, c3 = 1, c4 = −1, thus, A = A1 − A2 + A3 − A4 .

= (5, −12, 3). Equating components gives

15. Solving the systems c1p1 + c2p 2 + c3p3 = 0, c1p1 + c2p 2 + c3p3 = a + bx + cx 2 , and c1p1 + c2 p 2 + c3p3 = p gives the systems 90


c1 c1 + c2 c1 + c2 + c3 c1 c1 + c2 c1 + c2 + c3

Section 4.5

= 0 c1 =a = 0 , c1 + c2 = b , and = 0 c1 + c2 + c3 = c =7 = −1 . =2

True/False 4.4 (a) False; {v1 , … , v n } must also be linearly independent to be a basis. (b) False; the span of the set must also be V for the set to be a basis.

⎡1 0 0 ⎤ It is easy to see that det ⎢1 1 0 ⎥ = 1 ≠ 0, so ⎢ ⎥ ⎣1 1 1 ⎦ {p1, p 2 , p3} spans P2 . The third system is easy

(c) True; a basis must span the vector space. (d) True; the standard basis is used for the coordinate vectors in R n .

to solve by inspection to get c1 = 7, c2 = −8,

(e) False; the set {1 + x + x 2 + x3 + x 4 ,

c3 = 3, thus, p = 7p1 − 8p 2 + 3p3 .

x + x 2 + x3 + x 4 , x 2 + x3 + x 4 , x3 + x 4 , x 4 } is a basis for P4 .

17. From the diagram, j = u 2 and u1 = (cos 30°)i + (sin 30°) j =

3 1 i + j. 2 2

Section 4.5

3 1 i + u 2 for i in terms of u1 2 2 2 1 and u 2 gives i = u1 − u2 . 3 3

Exercise Set 4.5

Solving u1 =

(a)

(

3, 1) is

⎡ 1 1 −1 0 ⎤ 1. The matrix ⎢ −2 −1 2 0 ⎥ reduces to ⎢ ⎥ ⎣ −1 0 1 0 ⎦ ⎡ 1 0 −1 0 ⎤ ⎢0 1 0 0 ⎥ . The solution of the system is ⎢ ⎥ ⎣0 0 0 0 ⎦ x1 = t , x2 = 0, x3 = t , which can be written as ( x1 , x2 , x3 ) = (t , 0, t ) or ( x1 , x2 , x3 ) = t (1, 0, 1). Thus, the solution space has dimension 1 and a basis is (1, 0, 1).

3i + j.

1 ⎛ 2 ⎞ u1 − u 2 ⎟ + u2 3i + j = 3 ⎜ 3 ⎠ ⎝ 3 = 2u1 − u 2 + u 2 = 2u1 The x ′y ′-coordinates are (2, 0). (b) (1, 0) is i which has x ′y ′-coordinates

⎡ 1 −4 3 −1 0 ⎤ 3. The matrix ⎢ reduces to ⎣ 2 −8 6 −2 0⎥⎦ ⎡ 1 −4 3 −1 0 ⎤ ⎢⎣0 0 0 0 0 ⎥⎦ . The solution of the system is x1 = 4r − 3s + t , x2 = r , x3 = s, x4 = t , which can be written as ( x1 , x2 , x3 , x4 ) = (4r − 3s + t , r , s, t ) or

1 ⎞ ⎛ 2 ,− ⎜ ⎟. 3⎠ ⎝ 3 (c) Since (0, 1) = j = u 2 , the x ′y ′-coordinates are (0, 1). (d) (a, b) is ai + bj. 1 ⎛ 2 ⎞ u1 − u 2 ⎟ + bu 2 ai + bj = a ⎜ 3 ⎠ ⎝ 3 a ⎞ 2 ⎛ = au1 + ⎜ b − ⎟ u2 3⎠ 3 ⎝ a ⎞ ⎛ 2 a, b − The x ′y ′-coordinates are ⎜ ⎟. 3⎠ ⎝ 3

( x1 , x2 , x3 , x4 ) = r (4, 1, 0, 0) + s(−3, 0, 1, 0) + t (1, 0, 0, 1). Thus, the solution space has dimension 3 and a basis is (4, 1, 0, 0), (−3, 0, 1, 0), (1, 0, 0, 1).

91



n(n + 1) entries on 2 or above the main diagonal, so the space has n(n + 1) . dimension 2

⎡2 1 3 0⎤ 5. The matrix ⎢ 1 0 5 0 ⎥ reduces to ⎢ ⎥ ⎣0 1 1 0⎦ ⎡ 1 0 0 0⎤ ⎢0 1 0 0 ⎥ , so the system has only the trivial ⎢ ⎥ ⎣0 0 1 0 ⎦ solution, which has dimension 0 and no basis.

(c) As in part (b), there are

11. (a) Let p1 = p1 ( x) and p 2 = p2 ( x) be elements of W. Then ( p1 + p2 )(1) = p1 (1) + p2 (1) = 0 so p1 + p 2 is in W. (kp1 )(1) = k ( p1 (1)) = k ⋅ 0 = 0 so kp1 is in W. Thus, W is a subspace of P2 .

7. (a) Solving the equation for x gives 2 5 x = y − z, so parametric equations are 3 3 2 5 x = r − s, y = r, z = s, which can be 3 3 written in vector form as 5 ⎛2 ⎞ ( x, y, z ) = ⎜ r − s, r , s ⎟ 3 ⎝3 ⎠ ⎛2 ⎞ ⎛ 5 ⎞ = r ⎜ , 1, 0 ⎟ + s ⎜ − , 0, 1⎟ . ⎝3 ⎠ ⎝ 3 ⎠ ⎛2 ⎞ ⎛ 5 ⎞ A basis is ⎜ , 1, 0 ⎟ , ⎜ − , 0, 1⎟ . ⎝3 ⎠ ⎝ 3 ⎠

(b), (c) A basis for W is {−1 + x, − x + x 2 } so the dimension is 2. 13. Let u1 = (1, 0, 0, 0), u 2 = (0, 1, 0, 0), u3 = (0, 0, 1, 0), u 4 = (0, 0, 0, 1). Then the set {v1 , v 2 , u1, u 2 , u3 , u 4} clearly spans R 4 . The equation c1v1 + c2 v 2 + k1u1 + k2 u 2 + k3u3 + k4 u 4 = 0 leads to the system c1 − 3c2 + k1 =0 −4c1 + 8c2 + k2 =0 . 2c1 − 4c2 + k3 =0 −3c1 + 6c2 + k4 = 0

(b) Solving the equation for x gives x = y, so parametric equations are x = r, y = r, z = s, which can be written in vector form as ( x, y, z ) = (r , r , s) = r (1, 1, 0) + s(0, 0, 1). A basis is (1, 1, 0), (0, 0, 1).

⎡ 1 −3 ⎢ −4 8 The matrix ⎢ − 2 4 ⎢ ⎢⎣ −3 6 ⎡ 1 0 −2 0 0 ⎢ 0 1 −1 0 0 ⎢ to ⎢ 0 0 0 1 0 ⎢ ⎢0 0 0 0 1 ⎣

(c) In vector form, the line is (x, y, z) = (2t, −t, 4t) = t(2, −1, 4). A basis is (2, −1, 4). (d) The vectors can be parametrized as a = r, b = r + s, c = s or (a, b, c) = (r , r + s, s) = r (1, 1, 0) + s(0, 1, 1). A basis is (1, 1, 0), (0, 1, 1). 9. (a) The dimension is n, since a basis is the set {A1, A2 , … , An } where the only nonzero entry in Ai is aii = 1.

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0⎤ 0⎥ ⎥ reduces 0⎥ 0 ⎥⎦

−1 0 ⎤ − 13 0⎥ ⎥ . − 43 0⎥ ⎥ 2 0⎥ 3 ⎦

From the reduced matrix it is clear that u1 is a linear combination of v1 and v 2 so it will not be in the basis, and that {v1, v 2 , u 2 , u3} is linearly independent so it is one possible basis. Similarly, u 4 and either u 2 or u3 will produce a linearly independent set. Thus, any two of the vectors (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1) can be used.

(b) In a symmetric matrix, the elements above the main diagonal determine the elements below the main diagonal, so the entries on and above the main diagonal determine all the entries. There are n(n + 1) entries on or n + (n − 1) + + 1 = 2 above the main diagonal, so the space has n(n + 1) . dimension 2 92


Section 4.6

15. Let v3 = (a, b, c). The equation c1v1 + c2 v 2 + c3 v3 = 0 leads to the system c1 + ac3 = 0 −2c1 + 5c2 + bc3 = 0 . 3c1 − 3c2 + cc3 = 0

(i) True; the set of n × n matrices has dimension 2 n 2 , while the set {I , A, A2 , …, An } contains

n 2 + 1 matrices. (j) False; since P2 has dimension three, then by Theorem 4.5.6(c), the only three-dimensional subspace of P2 is P2 itself.

⎡ 1 0 a 0⎤ The matrix ⎢ −2 5 b 0⎥ reduces to ⎢ ⎥ ⎣ 3 −3 c 0⎦ 0⎤ a ⎡1 0 ⎢0 1 2a+ 1b 0 ⎥ and the 5 5 ⎢ ⎥ ⎢0 0 − 9 a + 3 b + c 0 ⎥ 5 5 ⎣ ⎦ homogeneous system has only the trivial 9 3 solution if − a + b + c ≠ 0. Thus, adding any 5 5 vector v3 = (a, b, c) with 9a − 3b − 5c ≠ 0 will

Section 4.6 Exercise Set 4.6 1. (a) Since {u1, u 2 } is the standard basis B for ⎡ 3⎤ R 2 , [ w ]s = ⎢ ⎥ . ⎣ −7 ⎦ ⎡ 2 (b) ⎢ ⎣ −4 ⎡1 ⎢ ⎢⎣0 and

3

create a basis for R . True/False 4.5 (a) True; by definition.

0 1

2 7 1 7

3⎤ ⎡2 − 28 ⎥ . Thus PB→ S = ⎢ 7 1⎥ ⎢⎣ 17 14 ⎦

⎡2 [ w ]S = PB→ S [ w ]B = ⎢ 7 ⎢⎣ 17

(b) True; any basis of R17 will contain 17 linearly independent vectors. (c) False; to span R17 , at least 17 vectors are required.

⎡1 0 1 0 ⎤ (c) ⎢ reduces to ⎣1 2 0 1⎥⎦

(d) True; since the dimension for R5 is 5, then linear independence is sufficient for 5 vectors to be a basis.

(f) True; if the set contains n vectors, it is a basis, while if it contains more than n vectors, it can be reduced to a basis.

1 0⎤ ⎡1 0 ⎢0 1 − 1 1 ⎥ . 2 2⎦ ⎣

a ⎡ ⎤ =⎢ 1 ⎥. 1 ⎣− 2 a + 2 b ⎦

(g) True; if the set contains n vectors, it is a basis, while if it contains fewer than n vectors, it can be enlarged to a basis.

3. (a) Since S is the standard basis B for P2 , ⎡ 4⎤ (p) S = (4, − 3, 1) or [p]S = ⎢ −3⎥ . ⎢ ⎥ ⎣ 1⎦

⎡0 1 ⎤ ⎡ 0 1⎤ ⎫ ⎢⎣1 0 ⎥⎦ , ⎢⎣ −1 0 ⎥⎦ ⎬ form a ⎭

93

3⎤ − 28 ⎥ 1⎥ 14 ⎦

3⎤ 5 − 28 1 ⎡ ⎤ ⎥ ⎡⎢ ⎤⎥ = ⎢ 28 ⎥ . 1 ⎥ ⎣1⎦ ⎢ 3 ⎥ 14 ⎦ ⎣ 14 ⎦

⎡ 1 0⎤ Thus PB→ S = ⎢ 1 1 ⎥ and ⎣− 2 2 ⎦ [ w ]S = PB → S [ w ]B ⎡ 1 0⎤ ⎡ a ⎤ = ⎢ 1 1⎥ ⎢ ⎥ ⎣− 2 2 ⎦ ⎣b ⎦

(e) True; since the dimension of R5 is 5, then spanning is sufficient for 5 vectors to be a basis.

(h) True; the set ⎧ ⎡1 0 ⎤ ⎡ 1 0 ⎤ ⎨ ⎢ 0 1 ⎥ , ⎢0 −1⎥ , ⎦ ⎣ ⎦ ⎩⎣ basis for M 22 .

3 1 0⎤ reduces to 8 0 1⎥⎦


⎡1 (b) ⎢ 1 ⎢ ⎣0 ⎡1 ⎢ ⎢0 ⎢ ⎣⎢0


⎡ −1 ⎢ 1 [ B]E = PS → E [ B ]S = ⎢ ⎢ 0 ⎢⎣ 0 ⎡ 15⎤ ⎢ −1⎥ =⎢ ⎥ ⎢ 6⎥ ⎣⎢ 3⎦⎥

1 0 1 0 0⎤ 0 1 0 1 0 ⎥ reduces to ⎥ 1 1 0 0 1⎦ 1 1 − 1⎤ 0 0 2 2 2⎥ 1 −1 1 ⎥ . Thus, 1 0 2 2 2 1 1⎥ 0 1 − 12 2 2 ⎦⎥

1 − 1⎤ ⎡ 1 2 2⎥ ⎢ 2 1 ⎥ and PB→ S = ⎢ 12 − 12 2 ⎢ 1 1 1⎥ ⎢⎣ − 2 2 2 ⎥⎦ [p]S = PB → S [p]B 1 − 1⎤ ⎡ 1 2 2 ⎥ ⎡ 2⎤ ⎢ 2 1 ⎥ ⎢ −1⎥ = ⎢ 12 − 12 2 ⎢ ⎥ ⎢ 1 1 1 ⎥ ⎣ 1⎦ − 2 2 ⎦⎥ ⎣⎢ 2 ⎡ 0⎤ = ⎢ 2⎥ ⎢ ⎥ ⎣ −1⎦

0 0 1 0

0 ⎤ ⎡ −8 ⎤ 0⎥ ⎢ 7⎥ ⎥⎢ ⎥ 0⎥ ⎢ 6⎥ 1⎥⎦ ⎢⎣ 3⎥⎦

⎡15 −1⎤ and B = ⎢ . ⎣ 6 3⎥⎦ ⎡ 2 4 1 −1⎤ 7. (a) ⎢ reduces to ⎣ 2 −1 3 −1⎥⎦ ⎡ 1 0 13 10 ⎢ ⎢⎣0 1 − 25

⎡ 13 − 12 ⎤ ⎥ so PB′→ B = ⎢ 10 ⎢⎣ − 52 0 ⎥⎦

− 12 ⎤ ⎥. 0⎥⎦

⎡ 1 −1 2 4 ⎤ (b) ⎢ reduces to ⎣3 −1 2 −1⎥⎦ ⎡1 0 0 − 5⎤ ⎡ 0 − 5⎤ 2 ⎥ so P 2⎥. ⎢ = B → B′ ⎢ 13 ⎥ − − 2 ⎢⎣0 1 −2 − 13 ⎥ ⎢ 2⎦ 2⎦ ⎣

or (p) S = (0, 2, − 1).

⎡ 1 2 3⎤ 5. (a) From Theorem 4.6.2, PS → B = ⎢0 2 3⎥ ⎢ ⎥ ⎣ 0 0 3⎦

⎡2 4 1 0⎤ (c) ⎢ reduces to ⎣ 2 −1 0 1⎥⎦

where B is the standard basis for R3 . Thus, ⎡ 1 2 3⎤ ⎡ 6 ⎤ ⎡16 ⎤ [ w ]B = PS → B [ w ]S = ⎢0 2 3⎥ ⎢ −1⎥ = ⎢10 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 3⎦ ⎣ 4 ⎦ ⎣12 ⎦ and w = (16, 10, 12).

⎡1 0 1 10 ⎢ ⎢⎣0 1 15

2⎤ 5⎥ − 15 ⎥ ⎦

⎡1 so PE → B = ⎢ 10 ⎢⎣ 15

2⎤ 5⎥ − 15 ⎥ ⎦

where E is the standard basis for R 2 . [ w ]B = PE → B [ w ]E 2⎤ ⎡1 10 5 ⎥ ⎡ 3⎤ ⎢ = 1 ⎢⎣ 5 − 51 ⎥⎦ ⎢⎣ −5⎥⎦ ⎡ − 17 ⎤ = ⎢ 10 ⎥ ⎢ 8⎥ ⎣ 5⎦ [ w ]B′ = PB→ B′ [ w ]B ⎡ 0 − 5 ⎤ ⎡ − 17 ⎤ 2 ⎥ ⎢ 10 ⎥ =⎢ ⎢⎣ −2 − 13 ⎥⎢ 8 ⎥ 2 ⎦⎣ 5 ⎦ ⎡ −4 ⎤ =⎢ ⎥ ⎣ −7 ⎦

(b) The basis in Exercise 3(a) is the standard ⎡3⎤ basis B for P2 , so [q]B = [q]S = ⎢ 0 ⎥ and ⎢ ⎥ ⎣4⎦ q = 3 + 4 x2 . (c) From Theorem 4.6.2, ⎡ −1 1 0 ⎢ 1 1 0 PS → E = ⎢ ⎢ 0 0 1 ⎣⎢ 0 0 0

1 1 0 0

0⎤ 0⎥ ⎥ where E is the 0⎥ 1⎦⎥

standard basis for M 22 . Thus,

94


⎡ 1 −1 1 0 ⎤ (d) ⎢ reduces to ⎣3 −1 0 1⎥⎦

Section 4.6

⎡1 0 − 1 2 ⎢ ⎢⎣0 1 − 32

[ w ]B′ = PB → B′ [ w ]B 5⎤ ⎡ 3 2 2 ⎥ ⎡ 9⎤ ⎢ = ⎢ −2 −3 − 1 ⎥ ⎢ −9 ⎥ 2 ⎢ ⎥ ⎢ ⎥ 1 6 ⎦ ⎣ −5⎦ ⎣ 5

1⎤ 2⎥ 1⎥ 2⎦

⎡− 1 1 ⎤ so PE → B′ = ⎢ 2 2 ⎥ . ⎢⎣ − 23 12 ⎥⎦ [ w ]B′ = PE → B′ [ w ]E ⎡ − 1 1 ⎤ ⎡ 3⎤ = ⎢ 23 2 ⎥ ⎢ ⎥ ⎢⎣ − 2 12 ⎥⎦ ⎣ −5⎦ ⎡ −4 ⎤ =⎢ ⎥ ⎣ −7 ⎦

⎡− 7 ⎤ ⎢ 2⎥ = ⎢ 23 ⎥ ⎢ 2⎥ ⎣ 6⎦ ⎡ 3 (c) ⎢ 1 ⎢ ⎣ −5 ⎡1 ⎢ ⎢0 ⎢ ⎣0

1 −1 2 2 1⎤ 1 0 1 −1 2 ⎥ reduces to ⎥ −3 2 1 1 1⎦ 5⎤ 0 0 3 2 2⎥ 1 0 −2 −3 − 12 ⎥ so ⎥ 0 1 5 1 6⎦ 5⎤ ⎡ 3 2 2⎥ ⎢ PB→ B′ = ⎢ −2 −3 − 1 ⎥ . 2 ⎢ ⎥ 1 6⎦ ⎣ 5

⎡ 3 9. (a) ⎢ 1 ⎢ ⎣ −5 ⎡1 ⎢ ⎢0 ⎢ ⎣0

1 −1 1 0 0 ⎤ 1 0 0 1 0 ⎥ reduces to ⎥ −3 2 0 0 1⎦ 1⎤ 0 0 1 12 2⎥ 1 0 −1 12 − 12 ⎥ so ⎥ 0 1 1 2 1⎦

⎡ 1 ⎢ PE → B′ = ⎢ −1 ⎢ ⎣ 1

1 2 1 2

2

1⎤ 2⎥ − 12 ⎥ .

⎥ 1⎦

[ w ]B′ = PE → B′ [ w ]E 1⎤ ⎡ 1 1 2 2 ⎥ ⎡ −5 ⎤ ⎢ = ⎢ −1 1 − 1 ⎥ ⎢ 8 ⎥ 2 2 ⎢ ⎥ ⎢ ⎥ 1⎦ ⎣ −5⎦ ⎣ 1 2

⎡2 2 1 1 0 0⎤ (b) ⎢ 1 −1 2 0 1 0 ⎥ reduces to ⎢ ⎥ ⎣ 1 1 1 0 0 1⎦ 3 1 − 5⎤ ⎡1 0 0 2 2 2⎥ ⎢ 3 ⎥ so 1 1 ⎢0 1 0 − 2 − 2 2 ⎢ ⎥ 0 2⎦ ⎣0 0 1 −1 1 − 5⎤ ⎡ 3 2 2⎥ ⎢ 2 3 ⎥ where E is the PE → B = ⎢ − 1 − 1 2 2 2 ⎢ ⎥ 0 2⎦ ⎣ −1

⎡− 7 ⎤ ⎢ 2⎥ = ⎢ 23 ⎥ ⎢ 2⎥ ⎣ 6⎦ 11. (a) The span of f1 and f2 is the set of all linear combinations af1 + bf 2 = a sin x + b cos x and this vector can be represented by (a, b). Since g1 = 2f1 + f 2 and g 2 = 3f2 , it is ⎡ 2 1⎤ sufficient to compute det ⎢ = 6. Since ⎣ 0 3⎥⎦ this determinant is nonzero, g1 and g 2 form a basis for V.

standard basis for R3 . [ w ]B = PE → B [ w ]E 1 − 5⎤ ⎡ 3 2 2 ⎥ ⎡ −5 ⎤ ⎢ 2 3 1 1 = ⎢− −2 ⎥ ⎢ 8⎥ 2 2 ⎢ ⎥ ⎢ ⎥ 0 2 ⎦ ⎣ −5 ⎦ ⎣ −1 ⎡ 9⎤ = ⎢ −9 ⎥ ⎢ ⎥ ⎣ −5⎦

(b) Since B can be represented as {(1, 0), (0, 1)} ⎡ 2 0⎤ PB′→ B = ⎢ . ⎣ 1 3⎥⎦ ⎡ 2 0 1 0⎤ (c) ⎢ reduces to ⎣ 1 3 0 1⎥⎦ ⎡ 1 so PB→ B′ = ⎢ 2 ⎢⎣ − 16 95

0⎤ ⎥. 1⎥ 3⎦

1 ⎡1 0 2 ⎢ ⎢⎣0 1 − 16

0⎤ ⎥ 1⎥ 3⎦



⎡ 2⎤ (d) [h]B = ⎢ ⎥ since B is the standard basis ⎣ −5 ⎦ for V. ⎡ 1 0 ⎤ ⎡ 2 ⎤ ⎡ 1⎤ ⎥ [h]B′ = PB → B′ [h ]B = ⎢ 2 = ⎢⎣ − 16 13 ⎥⎦ ⎢⎣ −5⎥⎦ ⎢⎣ −2 ⎥⎦

⎡ 1 2 1 1⎤ ⎡ 1 0 3 5⎤ 15. (a) ⎢ reduces to ⎢ ⎣0 1 −1 −2 ⎥⎦ ⎣ 2 3 3 4⎥⎦ ⎡ 3 5⎤ . so PB2 → B1 = ⎢ ⎣ −1 −2 ⎥⎦ ⎡ 1 1 1 2⎤ ⎡ 1 0 2 5⎤ reduces to ⎢ (b) ⎢ ⎣0 1 −1 −3⎥⎦ ⎣3 4 2 3⎥⎦ ⎡ 2 5⎤ so PB1 → B2 = ⎢ . ⎣ −1 −3⎥⎦

⎡ 1 2 3⎤ 13. (a) PB→ S = ⎢ 2 5 3⎥ ⎢ ⎥ ⎣ 1 0 8⎦

⎡ 1 2 1 0⎤ ⎡ 1 0 −3 2 ⎤ reduces to ⎢ (d) ⎢ ⎣0 1 2 −1⎥⎦ ⎣ 2 3 0 1⎥⎦ ⎡ −3 2 ⎤ so PS → B1 = ⎢ where S is the ⎣ 2 −1⎥⎦

⎡ 1 2 3 1 0 0⎤ (b) ⎢ 2 5 3 0 1 0⎥ reduces to ⎢ ⎥ ⎣ 1 0 8 0 0 1⎦ ⎡ 1 0 0 −40 16 9 ⎤ ⎢0 1 0 13 −5 −3⎥ so ⎢ ⎥ 0 0 1 5 −2 −1⎦ ⎣ ⎡ −40 16 9 ⎤ PS → B = ⎢ 13 −5 −3⎥ . ⎢ ⎥ ⎣ 5 −2 −1⎦

standard basis {(1, 0), (0, 1)} for R 2 . Since w is the standard basis vector (0, 1) for R 2 , ⎡ 2⎤ [ w ]B1 = ⎢ ⎥ . ⎣ −1⎦ [ w ]B2 = PB1 → B2 [ w ]B1

⎡ 2 5⎤ ⎡ 2 ⎤ =⎢ ⎣ −1 −3⎥⎦ ⎢⎣ −1⎥⎦ ⎡ −1⎤ =⎢ ⎥ ⎣ 1⎦

(d) [ w ]B = PS → B [ w ]S ⎡ −40 16 9 ⎤ ⎡ 5⎤ = ⎢ 13 −5 −3⎥ ⎢ −3⎥ ⎢ ⎥⎢ ⎥ ⎣ 5 −2 −1⎦ ⎣ 1⎦ ⎡ −239⎤ = ⎢ 77 ⎥ ⎢ ⎥ ⎣ 30⎦ [ w ]S = PB →S [ w ]B ⎡ 1 2 3⎤ ⎡ −239 ⎤ = ⎢ 2 5 3⎥ ⎢ 77 ⎥ ⎢ ⎥⎢ ⎥ ⎣ 1 0 8⎦ ⎣ 30 ⎦ ⎡ 5⎤ = ⎢ −3⎥ ⎢ ⎥ ⎣ 1⎦

⎡ 1 1 1 0⎤ ⎡ 1 0 4 −1⎤ reduces to ⎢ (e) ⎢ ⎥ 3 4 0 1 ⎣ ⎦ ⎣0 1 −3 1⎥⎦ ⎡ 4 −1⎤ . so PS → B2 = ⎢ ⎣ −3 1⎥⎦ ⎡ 4 −1⎤ ⎡ 2 ⎤ ⎡ 3⎤ [ w ]B2 = PS → B2 [ w ]S = ⎢ = ⎣ −3 1⎥⎦ ⎢⎣ 5⎥⎦ ⎢⎣ −1⎥⎦ [ w ]B = PB 1

2 → B1

[ w ]B

2

⎡ 3 5 ⎤ ⎡ 3⎤ =⎢ ⎣ −1 −2⎥⎦ ⎢⎣ −1⎥⎦ ⎡ 4⎤ =⎢ ⎥ ⎣ −1⎦

⎡ 3⎤ (e) [ w ]S = ⎢ −5⎥ ⎢ ⎥ ⎣ 0⎦ [ w ]B = PS → B [ w ]S ⎡ −40 16 9 ⎤ ⎡ 3⎤ = ⎢ 13 −5 −3⎥ ⎢ −5⎥ ⎢ ⎥⎢ ⎥ ⎣ 5 −2 −1⎦ ⎣ 0 ⎦ ⎡ −200⎤ = ⎢ 64⎥ ⎢ ⎥ ⎣ 25⎦

96


⎡ 3 17. (a) ⎢ 1 ⎢ ⎣ −5 ⎡1 ⎢ ⎢0 ⎢ ⎣0

Section 4.6

1 −1 2 2 1⎤ 1 0 1 −1 2 ⎥ reduces to ⎥ −3 2 1 1 1⎦ 5⎤ 0 0 3 2 2⎥ 1 0 −2 −3 − 12 ⎥ so ⎥ 0 1 5 1 6⎦

PB → B 1

2

⎡ 3 (c) ⎢ 1 ⎢ ⎣ −5 ⎡1 ⎢ ⎢0 ⎢ ⎣0

5⎤ ⎡ 3 2 2⎥ ⎢ = ⎢ −2 −3 − 1 ⎥ . 2 ⎢ ⎥ 1 6⎦ ⎣ 5

1 −1 1 0 0 ⎤ 1 0 0 1 0 ⎥ reduces to ⎥ −3 2 0 0 1⎦ 1⎤ 0 0 1 12 2⎥ 1 0 −1 12 − 12 ⎥ so ⎥ 0 1 1 2 1⎦

1⎤ ⎡ 1 1 2 2⎥ ⎢ PS → B = ⎢ −1 1 − 1 ⎥ . 2 2 2 ⎢ ⎥ 1⎦ ⎣ 1 2 [ w ]B2 = PS → B2 [ w ]S

⎡2 (b) ⎢ 1 ⎢ ⎣1 ⎡1 ⎢ ⎢0 ⎢ ⎣0

2 1 1 0 0⎤ −1 2 0 1 0 ⎥ reduces to ⎥ 1 1 0 0 1⎦ 3 1 − 5⎤ 0 0 2 2 2⎥ 3 ⎥ , so 1 0 − 12 − 12 2 ⎥ 0 1 −1 0 2⎦ 1 − 5⎤ ⎡ 3 2 2⎥ ⎢ 2 3 ⎥ where S is the PS → B = ⎢ − 1 − 1 1 2 2 ⎢ 2 ⎥ 0 2⎦ ⎣ −1

1⎤ ⎡ 1 1 2 2 ⎥ ⎡ −5 ⎤ ⎢ = ⎢ −1 1 − 1 ⎥ ⎢ 8 ⎥ 2 2 ⎢ ⎥ ⎢ ⎥ 1⎦ ⎣ −5⎦ ⎣ 1 2 ⎡− 7 ⎤ ⎢ 2⎥ = ⎢ 23 ⎥ ⎢ 2⎥ ⎣ 6⎦

19. (a)

y v1

standard basis for R3 . [ w ]B1 = PS → B1 [ w ]S

[ w ]B2

θ e1

1 − 5⎤ ⎡ 3 2 2 ⎥ ⎡ −5 ⎤ ⎢ 2 3 1 1 = ⎢− −2 ⎥ ⎢ 8⎥ 2 2 ⎢ ⎥ ⎢ ⎥ 0 2⎦ ⎣ −5⎦ ⎣ −1 ⎡ 9⎤ = ⎢ −9 ⎥ ⎢ ⎥ ⎣ −5 ⎦ = PB1 → B2 [ w ]B1

x

From the diagram, it is clear that [ v1 ]S = (cos 2θ , sin 2θ ). y e2

θ

x

v2

From the diagram, the angle between the positive y-axis and v 2 is

5⎤ ⎡ 3 2 2 ⎥ ⎡ 9⎤ ⎢ = ⎢ −2 −3 − 1 ⎥ ⎢ −9 ⎥ 2 ⎢ ⎥ ⎢ ⎥ 1 6 ⎦ ⎣ −5 ⎦ ⎣ 5 ⎡− 7 ⎤ ⎢ 2⎥ = ⎢ 23 ⎥ ⎢ 2⎥ ⎣ 6⎦

⎛π ⎞ 2 ⎜ − θ ⎟ = π − 2θ , so the angle between ⎝2 ⎠ v 2 and the positive x-axis is π π⎞ ⎛ − ⎜ π − 2θ − ⎟ = 2θ − and 2 2⎠ ⎝ ⎛ π⎞ π ⎞⎞ ⎛ ⎛ [ v 2 ]S = ⎜ cos ⎜ 2θ − ⎟ , sin ⎜ 2θ − ⎟ ⎟ 2 2 ⎠⎠ ⎝ ⎠ ⎝ ⎝ = (sin 2θ , − cos 2θ ) Thus by Theorem 4.6.2, sin 2θ ⎤ ⎡cos 2θ PB→ S = ⎢ sin 2 cos 2θ ⎥⎦ θ − ⎣

97



Section 4.7

21. If w is a vector in the space and [ w ]B′ is the coordinate vector of w relative to B ′, then [ w ]B = P[ w ]B′ and [ w ]C = Q[ w ]B , so

Exercise Set 4.7 1. Row vectors: r1 = [ 2 −1 0 1],

[ w ]C = Q( P[ w ]B′ ) = QP[ w ]B′ . The transition matrix from B′ to C is QP. The transition matrix from C to B′ is (QP)−1 = P −1Q −1.

r2 = [3 5 7 −1], r3 = [1 4 2 7]

⎡2⎤ ⎡ −1⎤ Column vectors: c1 = ⎢ 3⎥ , c2 = ⎢ 5⎥ , ⎢ ⎥ ⎢ ⎥ ⎣ 1⎦ ⎣ 4⎦ ⎡0⎤ ⎡ 1⎤ c3 = ⎢ 7 ⎥ , c4 = ⎢ −1⎥ ⎢ ⎥ ⎢ ⎥ ⎣2⎦ ⎣ 7⎦

23. (a) By Theorem 4.6.2, P is the transition matrix from B = {(1, 1, 0), (1, 0, 2), (0, 2, 1)} to S.

⎡ 1 1 0 1 0 0⎤ (b) ⎢ 1 0 2 0 1 0 ⎥ reduces to ⎢ ⎥ ⎣0 2 1 0 0 1⎦ 4 ⎡1 0 0 5 ⎢ 1 ⎢0 1 0 5 ⎢ 2 0 0 1 − 5 ⎣⎢

1 5 − 51 2 5

⎡ 1 3 −2 ⎤ ⎡ 1 0 1⎤ 3. (a) ⎢ reduces to ⎢ , so ⎥ 4 6 10 − ⎣ ⎦ ⎣0 1 −1⎥⎦ ⎡ −2 ⎤ ⎡ 1⎤ ⎡ 3⎤ ⎢⎣ 10 ⎥⎦ = ⎢⎣ 4 ⎥⎦ − ⎢⎣ −6⎥⎦ .

− 52 ⎤ ⎥ 2⎥. 5 1⎥ 5 ⎦⎥

Thus, P is the transition matrix from S to ⎧⎛ 4 1 2⎞ ⎛1 1 2⎞ B = ⎨⎜ , , − ⎟ , ⎜ , − , ⎟ , 5⎠ ⎝5 5 5⎠ ⎩⎝ 5 5 ⎛ 2 2 1 ⎞⎫ ⎜ − , , ⎟⎬ ⎝ 5 5 5 ⎠⎭

⎡ 1 0 1 0⎤ ⎡ 1 1 2 −1⎤ (b) ⎢ 1 0 1 0⎥ reduces to ⎢0 1 1 0⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 2 1 3 2⎦ ⎣0 0 0 1⎦ so the system Ax = b is inconsistent and b is not in the column space of A. ⎡ 1 −1 1 5 ⎤ (c) ⎢9 3 1 1⎥ reduces to ⎢ ⎥ ⎣ 1 1 1 −1⎦

27. B must be the standard basis for R n , since, in particular [ei ]B = ei for the standard basis

vectors e1 , e2 , … , e n .

1⎤ ⎡1 0 0 ⎢0 1 0 −3⎥ , so ⎢ ⎥ ⎣0 0 1 1⎦ ⎡ 5⎤ ⎡ 1⎤ ⎡ −1⎤ ⎡1⎤ ⎢ 1⎥ = ⎢9 ⎥ − 3 ⎢ 3⎥ + ⎢1⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎣ −1⎦ ⎣ 1⎦ ⎣ 1⎦ ⎣1⎦

True/False 4.6 (a) True (b) True; by Theorem 4.6.1. (c) True

⎡ 1 −1 1 2 ⎤ (d) ⎢ 1 1 −1 0 ⎥ reduces to ⎢ ⎥ ⎣ −1 −1 1 0 ⎦ ⎡ x1 ⎤ ⎡ 1 0 0 1⎤ ⎢0 1 −1 −1⎥ , so the system A ⎢ x2 ⎥ = b ⎢ ⎥ ⎢ ⎥ ⎣0 0 0 0 ⎦ ⎣ x3 ⎦ has the solution x1 = 1, x2 = t − 1, x3 = t .

(d) True (e) False; consider the bases S = {e1 , e2 , e3} and

B = {2e3 , 3e1 , 5e 2} for R3 , then ⎡0 3 0⎤ PB→ S = ⎢ 0 0 5⎥ which is not a diagonal ⎢ ⎥ ⎣2 0 0⎦ matrix.

⎡ 2 ⎤ ⎡ 1⎤ ⎡ −1⎤ ⎡ 1⎤ Thus ⎢ 0 ⎥ = ⎢ 1⎥ + (t − 1) ⎢ 1⎥ + t ⎢ −1⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 ⎦ ⎣ −1⎦ ⎣ −1⎦ ⎣ 1⎦

(f) False; A would have to be invertible, not just square.

98


⎡1 ⎢0 (e) ⎢ ⎢1 ⎢⎣0

2 1 2 1

0 2 1 2

1 1 3 2

⎡1 ⎢0 ⎢ ⎢0 ⎢⎣0

0 1 0 0

0 0 1 0

0 −26 ⎤ 0 13⎥ ⎥ , so 0 −7 ⎥ 1 4 ⎥⎦

Section 4.7 x3 = s, x4 = t , or in vector form

4⎤ 3⎥ ⎥ reduces to 5⎥ 7 ⎥⎦

⎡ −1⎤ ⎡2⎤ ⎡ −1⎤ ⎡ −2 ⎤ ⎢ 0⎥ ⎢ 1⎥ ⎢ 0⎥ ⎢ 0⎥ x = ⎢ ⎥ + r ⎢ ⎥ + s⎢ ⎥ +t ⎢ ⎥. ⎢ 0⎥ ⎢0⎥ ⎢ 1⎥ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦ The general form of the solution of Ax = 0 is ⎡2⎤ ⎡ −1⎤ ⎡ −2 ⎤ ⎢ 1⎥ ⎢ 0⎥ ⎢ 0⎥ x = r ⎢ ⎥ + s ⎢ ⎥ +t ⎢ ⎥. 0 ⎢ ⎥ ⎢ 1⎥ ⎢ 0 ⎥ ⎣⎢ 0 ⎦⎥ ⎣⎢ 0 ⎦⎥ ⎣⎢ 1⎦⎥

⎡4⎤ ⎡ 1⎤ ⎡ 2⎤ ⎡ 0⎤ ⎡ 1⎤ ⎢ 3⎥ ⎢0⎥ ⎢ 1⎥ ⎢ 2⎥ ⎢ 1⎥ ⎢ ⎥ = −26 ⎢ ⎥ + 13 ⎢ ⎥ − 7 ⎢ ⎥ + 4 ⎢ ⎥ . ⎢ 5⎥ ⎢ 1⎥ ⎢ 2⎥ ⎢ 1⎥ ⎢ 3⎥ ⎣⎢7 ⎦⎥ ⎣⎢ 0 ⎦⎥ ⎣⎢ 1⎦⎥ ⎣⎢ 2 ⎦⎥ ⎣⎢ 2 ⎦⎥

1 4⎤ ⎡ 1 2 −3 ⎢ −2 1 2 1 −1⎥ (d) ⎢ ⎥ reduces to 3⎥ ⎢ −1 3 −1 2 ⎣⎢ 4 −7 0 −5 −5⎦⎥

⎡ 1 −3 1⎤ ⎡ 1 −3 1⎤ 5. (a) ⎢ reduces to ⎢ , and ⎥ ⎣ 2 −6 2 ⎦ ⎣0 0 0 ⎥⎦ the solution of Ax = b is x1 = 1 + 3t , x2 = t ,

⎡1 ⎢ ⎢0 ⎢ ⎢0 ⎢⎣0

⎡ 1⎤ ⎡3⎤ or in vector form, x = ⎢ ⎥ + t ⎢ ⎥ . ⎣ 0⎦ ⎣ 1⎦ The general form of the solution of Ax = 0 is ⎡3⎤ x = t ⎢ ⎥. ⎣ 1⎦

0 − 75 1 − 45

− 15 3 5

6⎤ 5⎥ 7⎥ 5

and the solution of ⎥ 0 0⎥ 0 0 ⎥⎦ 6 7 1 Ax = b is x1 = + s + t , 5 5 5 7 4 3 x2 = + s − t , x3 = s, x4 = t , or in 5 5 5

⎡ 1 1 2 5⎤ (b) ⎢ 1 0 1 −2 ⎥ reduces to ⎢ ⎥ ⎣ 2 1 3 3⎦ ⎡ 1 0 1 −2 ⎤ ⎢0 1 1 7 ⎥ , and the solution of ⎢ ⎥ ⎣0 0 0 0 ⎦ Ax = b is x1 = −2 − t , x2 = 7 − t , x3 = t , or

0 0

0 0

⎡6⎤ ⎡7⎤ ⎡ 1⎤ ⎢5⎥ ⎢5⎥ ⎢ 5⎥ 7⎥ 3 4 ⎢ vector form, x = 5 + s ⎢ 5 ⎥ + t ⎢ − 5 ⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0⎥ ⎢ 1⎥ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣ 0⎥⎦ The general form of the solution of Ax = 0 is ⎡7⎤ ⎡ 1⎤ ⎢5⎥ ⎢ 5⎥ 3 4 x = s ⎢ 5 ⎥ + t ⎢− 5 ⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ 1⎥ ⎢ 0 ⎥ ⎢⎣ 0⎥⎦ ⎢⎣ 1⎥⎦

⎡ −2 ⎤ ⎡ −1⎤ in vector form, x = ⎢ 7 ⎥ + t ⎢ −1⎥ . The ⎢ ⎥ ⎢ ⎥ ⎣ 0 ⎦ ⎣ 1⎦ general form of the solution of Ax = 0 is ⎡ −1⎤ x = t ⎢ −1⎥ . ⎢ ⎥ ⎣ 1⎦

7. (a) A basis for the row space is r1 = [1 0 2], r2 = [0 0 1]. A basis for the column

⎡ 1 −2 1 2 −1⎤ ⎢ 2 −4 2 4 −2 ⎥ (c) ⎢ ⎥ reduces to 1⎥ ⎢ −1 2 −1 −2 ⎢⎣ 3 −6 3 6 −3⎥⎦

⎡ 1⎤ ⎡2⎤ space is c1 = ⎢0 ⎥ , c2 = ⎢ 1⎥ . ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣0⎦

⎡ 1 −2 1 2 −1⎤ ⎢0 0 0 0 0 ⎥ ⎢ ⎥ , and the solution of ⎢0 0 0 0 0 ⎥ ⎣⎢0 0 0 0 0 ⎦⎥ Ax = b is x1 = −1 + 2r − s − 2t , x2 = r , 99



(b) A basis for the row space is r1 = [1 −3 0 0], r2 = [0 1 0 0].

(b) A basis for the row space is r1 = [1 −3 0 0], r2 = [0 1 0 0].

⎡ 1⎤ ⎢0 ⎥ A basis for the column space is c1 = ⎢ ⎥ , ⎢0 ⎥ ⎣⎢0 ⎦⎥


⎡ −3⎤ ⎢ 1⎥ c2 = ⎢ ⎥ . ⎢ 0⎥ ⎣⎢ 0 ⎦⎥

⎡ −3⎤ ⎢ 1⎥ c2 = ⎢ ⎥ . ⎢ 0⎥ ⎣⎢ 0 ⎦⎥

(c) A basis for the row space is r1 = [1 2 4 5], r2 = [0 1 −3 0],

(c) A basis for the row space is r1 = [1 2 4 5], r2 = [0 1 −3 0],

r3 = [0 0 1 −3], r4 = [0 0 0 1].

r3 = [0 0 1 −3], r4 = [0 0 0 1].

⎡ 1⎤ ⎢0 ⎥ ⎢ ⎥ A basis for the column space is c1 = ⎢0 ⎥ , ⎢0 ⎥ ⎢⎣0 ⎥⎦ ⎡2⎤ ⎡ 4⎤ ⎡ 5⎤ ⎢ 1⎥ ⎢ −3⎥ ⎢ 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ c2 = ⎢ 0 ⎥ , c3 = ⎢ 1⎥ , c4 = ⎢ −3⎥ . ⎢0⎥ ⎢ 0⎥ ⎢ 1⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0⎥⎦ ⎢⎣ 0 ⎥⎦

⎡ 1⎤ ⎢0 ⎥ ⎢ ⎥ A basis for the column space is c1 = ⎢0 ⎥ , ⎢0 ⎥ ⎢⎣0 ⎥⎦ ⎡2⎤ ⎡ 4⎤ ⎡ 5⎤ ⎢ 1⎥ ⎢ −3⎥ ⎢ 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ c2 = ⎢ 0 ⎥ , c3 = ⎢ 1⎥ , c4 = ⎢ −3⎥ . ⎢0⎥ ⎢ 0⎥ ⎢ 1⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0⎥⎦ ⎢⎣ 0 ⎥⎦

(d) A basis for the row space is r1 = [1 2 −1 5], r2 = [0 1 4 3],

(d) A basis for the row space is r1 = [1 2 −1 5], r2 = [0 1 4 3],

r3 = [0 0 1 −7], r4 = [0 0 0 1].

r3 = [0 0 1 −7], r4 = [0 0 0 1].



⎡ −1⎤ ⎡2⎤ ⎡ 5⎤ ⎢ 4⎥ ⎢ 1⎥ ⎢ 3⎥ c 2 = ⎢ ⎥ , c3 = ⎢ ⎥ , c 4 = ⎢ ⎥ . ⎢ 1⎥ ⎢0⎥ ⎢ −7 ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦

⎡2⎤ ⎡ −1⎤ ⎡ 5⎤ ⎢ 1⎥ ⎢ 4⎥ ⎢ 3⎥ c 2 = ⎢ ⎥ , c3 = ⎢ ⎥ , c 4 = ⎢ ⎥ . ⎢0⎥ ⎢ 1⎥ ⎢ −7 ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦

9. (a) A basis for the row space is r1 = [1 0 2],

⎡ 1 1 −4 −3⎤ 11. (a) A row echelon form of ⎢ 2 0 2 −2 ⎥ ⎢ ⎥ ⎣ 2 −1 3 2 ⎦ ⎡ 1 1 −4 −3⎤ is ⎢0 1 −5 −2 ⎥ . Thus a basis for the ⎢ ⎥ 1 − 12 ⎦⎥ ⎣⎢0 0

r2 = [0 0 1]. A basis for the column ⎡ 1⎤ ⎡2⎤ ⎢ ⎥ space is c1 = 0 , c2 = ⎢ 1⎥ . ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣0⎦

subspace is (1, 1, −4, −3), (0, 1, −5, −2), 1⎞ ⎛ ⎜ 0, 0, 1, − ⎟ . 2⎠ ⎝ 100


Section 4.7

⎡ −1 1 −2 0 ⎤ (b) A row echelon form of ⎢ 3 3 6 0 ⎥ is ⎢ ⎥ ⎣ 9 0 0 3⎦ 0⎤ ⎡ 1 −1 2 ⎢0 1 0 0 ⎥ . Thus a basis for the ⎢ 1⎥ ⎢⎣0 0 1 − 6 ⎥⎦

(b) det(A) = det(B) = 5 A and B are invertible so the systems Ax = 0 and Bx = 0 have only the trivial solution. That is, the null spaces of A and B are both the origin. The second row of C is a multiple of the first and it is clear that if 3x + y = 0, then (x, y) is in the null space of C, i.e., the null space of C is the line 3x + y = 0. ⎡x⎤ The equation Dx = 0 has all x = ⎢ ⎥ as ⎣ y⎦ solutions, so the null space of D is the entire xy-plane.

subspace is (1, −1, 2, 0), (0, 1, 0, 0), 1⎞ ⎛ ⎜ 0, 0, 1, − ⎟ . 6⎠ ⎝ ⎡ 1 1 ⎢ 0 0 (c) A row echelon form of ⎢ ⎢ −2 0 ⎢⎣ 0 −3

0 1 2 0

0⎤ 1⎥ ⎥ is 2⎥ 3⎥⎦

True/False 4.7 (a) True; by the definition of column space.

⎡ 1 1 0 0⎤ ⎢0 1 1 1⎥ ⎢ ⎥ . Thus a basis for the ⎢0 0 1 1⎥ ⎣⎢0 0 0 1⎦⎥ subspace is (1, 1, 0, 0), (0, 1, 1, 1), (0, 0, 1, 1), (0, 0, 0, 1).

(b) False; the system Ax = b may not be consistent. (c) False; the column vectors of A that correspond to the column vectors of R containing the leading 1’s form a basis for the column space of A. (d) False; the row vectors of A may be linearly dependent if A is not in row echelon form.

⎡ 1 0 0⎤ 15. (a) The row echelon form of A is ⎢0 1 0 ⎥ , ⎢ ⎥ ⎣0 0 0 ⎦ so the general solution of the system Ax = 0 ⎡0 ⎤ is x = 0, y = 0, z = t or t ⎢0 ⎥ . Thus the null ⎢ ⎥ ⎣ 1⎦ space of A is the z-axis, and the column ⎡ 1⎤ ⎡0 ⎤ space is the span of c1 = ⎢ 1⎥ , c2 = ⎢0 ⎥ , ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣0 ⎦ which is all linear combinations of y and x, i.e., the xy-plane.

⎡1 3⎤ ⎡1 3 ⎤ (e) False; the matrices A = ⎢ and B = ⎢ ⎣0 0 ⎥⎦ ⎣ 2 6 ⎥⎦ have the same row space but different column spaces. (f) True; elementary row operations do not change the null space of a matrix. (g) True; elementary row operations do not change the row space of a matrix. (h) False; see (e) for an example. (i) True; this is the contrapositive of Theorem 4.7.1.

(b) Reversing the reasoning in part (a), it is ⎡0 0 0 ⎤ clear that one such matrix is ⎢0 1 0 ⎥ . ⎢ ⎥ ⎣0 0 1⎦

(j) False; assuming that A and B are both n × n matrices, the row space of A will have n vectors in its basis since being invertible means that A is row equivalent to I n . However, since B is

17. (a) If the null space of a matrix A is the given line, then the equation Ax = 0 is equivalent ⎡ 3 −5⎤ ⎡ x ⎤ ⎡0 ⎤ to ⎢ = and a general form ⎣0 0 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣0 ⎥⎦ ⎡3a −5a ⎤ where a and b are real for A is ⎢ ⎣ 3b −5b ⎥⎦ numbers, not both of which are zero.

singular, it is not row equivalent to I n , so in reduced row echelon form it has at least one row of zeros. Thus the row space of B will have fewer than n vectors in its basis.

101


SSM: Elementary Linear Algebra 7. (a) Since rank(A) = rank [ A b ] , the system is

Section 4.8

consistent, and since nullity(A) = 3 − 3 = 0, there are 0 parameters in the general solution.

Exercise Set 4.8 1. The reduced row echelon form of A is ⎡1 0 − 6 − 4⎤ 7 7⎥ ⎢ 2 ⎥ , so rank (A) = 2. ⎢0 1 17 7 7 ⎢ ⎥ 0 0⎦ ⎣0 0 ⎡ 1 −3 −2 ⎤ ⎢2 1 3⎥ AT = ⎢ ⎥ and the reduced row 4 5 9⎥ ⎢ ⎣⎢ 0 2 2 ⎦⎥

⎡1 ⎢0 echelon form of A is ⎢ ⎢0 ⎣⎢0 T

0 1 0 0

(b) Since rank( A) < rank ⎡⎣ A b ⎤⎦ , the system is

inconsistent. (c) Since rank( A) = rank ⎡⎣ A b ⎤⎦ , the system is


1⎤ 1⎥ ⎥ , so 0⎥ 0 ⎦⎥

(d) Since rank( A) = rank ⎡⎣ A b ⎤⎦ , the system is


rank( AT ) = rank( A) = 2.

(e) Since rank( A) < rank ⎡⎣ A b ⎤⎦ , the system is

3. (a) Since rank (A) = 2, there are 2 leading variables. Since nullity(A) = 3 − 2 = 1, there is 1 parameter.

inconsistent. (f) Since rank( A = rank ⎡⎣ A b ⎤⎦ , the system is


(b) Since rank(A) = 1, there is 1 leading variable. Since nullity(A) = 3 − 1 = 2, there are 2 parameters.

(g) Since rank( A) = rank ⎡⎣ A b ⎤⎦ , the system is

(c) Since rank(A) = 2, there are 2 leading variables. Since nullity(A) = 4 − 2 = 2, there are 2 parameters.


(d) Since rank(A) = 2, there are 2 leading variables. Since nullity(A) = 5 − 2 = 3, there are 3 parameters.

⎡1 −3 b1 ⎤ ⎢1 −2 b ⎥ 2⎥ ⎢ 9. The augmented matrix is ⎢1 1 b3 ⎥ , which is ⎢1 −4 b4 ⎥ ⎢1 5 b ⎥ 5⎦ ⎣ 3b2 − 2b1 ⎤ ⎡1 0 ⎢0 1 b2 − b1 ⎥ ⎢ ⎥ row equivalent to ⎢0 0 b3 − 4b2 + 3b1 ⎥ . Thus, ⎢0 0 b4 + b2 − 2b1 ⎥ ⎢0 0 b − 8b + 7b ⎥ 5 2 1⎦ ⎣ the system is consistent if and only if =0 3b1 − 4b2 + b3 −2b1 + b2 + b4 =0 + b5 = 0 7b1 − 8b2

(e) Since rank(A) = 3, there are 3 leading variables. Since nullity(A) = 5 − 3 = 2, there are 2 parameters. 5. (a) Since A is 4 × 4, rank(A) + nullity(A) = 4, and the largest possible value of rank(A) is 4, so the smallest possible value of nullity(A) is 0. (b) Since A is 3 × 5, rank(A) + nullity(A) = 5, and the largest possible value of rank(A) is 3, so the smallest possible value of nullity(A) is 2. (c) Since A is 5 × 3, rank(A) + nullity(A) = 3, and the largest possible value of rank(A) is 3, so the smallest possible value of nullity(A) is 0.

which has the general solution b1 = r , b2 = s, b3 = −3r + 4s, b4 = 2r − s, b5 = −7 r + 8s. 102


Section 4.9

11. No, since rank(A) + nullity(A) = 3 and nullity(A) = 1, the row space and column space must both be 2-dimensional, i.e., planes in 3-space.

(e) True; if the rows are linearly dependent, then the rank is at least 1 less than the number of rows, so since the matrix is square, its nullity is at least 1.

13. The rank is 2 if r = 2 and s = 1. Since the third column will always have a nonzero entry, the rank will never be 1.

(f) False; if the nullity of A is zero, then Ax = b is consistent for every vector b. (g) False; the dimension of the row space is always equal to the dimension of the column space.

17. (a) The number of leading 1’s is at most 3, since a matrix cannot have more leading 1’s than rows.

(h) False; rank(A) = rank(AT ) for all matrices.

(b) The number of parameters is at most 5, since the solution cannot have more parameters than A has columns.

(i) True; the row space and null space cannot both have dimension 1 since dim(row space) + dim(null space) = 3.

(c) The number of leading 1’s is at most 3, since the row space of A has the same dimension as the column space, which is at most 3.

(j) False; a vector w in W ⊥ need not be orthogonal

to every vector in V; the relationship is that V ⊥ is a subspace of W ⊥ .

(d) The number of parameters is at most 3 since the solution cannot have more parameters than A has columns.


⎡0 1⎤ ⎡ 1 2⎤ 19. Let A = ⎢ and B = ⎢ . Then ⎥ 0 0 ⎣ ⎦ ⎣ 2 4 ⎥⎦

1. (a) Since A has size 3 × 2, the domain of TA is R 2 and the codomain of TA is R3 .

⎡0 0⎤ rank(A) = rank(B) = 1, but since A = ⎢ ⎣ 0 0 ⎥⎦ ⎡ 5 10⎤ and B 2 = ⎢ , ⎣10 20⎥⎦

(b) Since A has size 2 × 3, the domain of TA is

rank( A2 ) = 0 ≠ rank( B 2 ) = 1.

(c) Since A has size 3 × 3, the domain of TA is

2

R3 and the codomain of TA is R 2 .

R3 and the codomain of TA is R3 .

True/False 4.8

(d) Since A has size 1 × 6, the domain of TA is

(a) False; the row vectors are not necessarily linearly independent, and if they are not, then neither are the column vectors, since dim(row space) = dim(column space).

R 6 and the codomain of TA is R1 . 3. The domain of T is R 2 and the codomain of T is

(b) True; if the row vectors are linearly independent then nullity(A) = 0 and rank(A) = n = the number of rows. But, since rank(A) + nullity(A) = the number of columns, A must be square.

R3 . T(1, −2) = (1 − 2, −(−2), 3(1)) = (−1, 2, 3) 5. (a) The transformation is linear, with domain

(c) False; the nullity of a nonzero m × n matrix is at most n − 1.

R3 and codomain R 2 . (b) The transformation is nonlinear, with

(d) False; if the added column is linearly dependent on the previous columns, adding it does not affect the rank.

domain R 2 and codomain R3 . (c) The transformation is linear, with domain R3 and codomain R3 .

103



⎡0 ⎢1 ⎢ (d) The matrix is ⎢0 ⎢0 ⎢⎣ 1

(d) The transformation is nonlinear, with

domain R 4 and codomain R 2 . 7. (a) T is a matrix transformation; ⎡0 0 0⎤ . TA = ⎢ ⎣ 0 0 0 ⎥⎦

(c) T is a matrix transformation; ⎡ 3 −4 0 ⎤ . TA = ⎢ ⎣ 2 0 −5⎥⎦

⎡ 2 −1 1⎤ ⎡ 2 ⎤ ⎡ 0 ⎤ (b) AT x = ⎢ 0 1 1⎥ ⎢ 1⎥ = ⎢ −2 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 0 0 ⎦ ⎣ −3⎦ ⎣ 0 ⎦ T(2, 1, −3) = (2(2) − (1) + (−3), 1 + (−3), 0) = (0, −2, 0)

(d) T is not a matrix transformation, since for k ≠ 1,

T (k ( x, y, z )) = (k 2 y 2 , kz ) ≠ kT ( x, y, z )

⎡ 1 0 0⎤ ⎡ 2 ⎤ ⎡ 2 ⎤ 15. (a) ⎢0 1 0⎥ ⎢ −5⎥ = ⎢ −5⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 −1⎦ ⎣ 3⎦ ⎣ −3⎦

= (ky 2 , kz ).

The result is (2, −5, −3).

(e) T is not a matrix transformation, since T(0, 0, 0) = (−1, 0) ≠ 0.

⎡ 1 0 0⎤ ⎡ 2⎤ ⎡2⎤ (b) ⎢0 −1 0 ⎥ ⎢ −5⎥ = ⎢ 5⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 1⎦ ⎣ 3⎦ ⎣ 3⎦ The result is (2, 5, 3).

⎡ 3 5 −1⎤ 9. By inspection, T = ⎢ 4 −1 1⎥ . ⎢ ⎥ ⎣ 3 2 −1⎦ For (x, y, z) = (−1, 2, 4), w1 = 3(−1) + 5(2) − 4 = 3

⎡ −1 0 0⎤ ⎡ 2 ⎤ ⎡ −2 ⎤ (c) ⎢ 0 1 0⎥ ⎢ −5⎥ = ⎢ −5⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 0 1⎦ ⎣ 3⎦ ⎣ 3⎦ The result is (−2, −5, 3).

w2 = 4(−1) − (2) + 4 = −2 w3 = 3(−1) + 2(2) − (4) = −3 so T(−1, 2, 4) = (3, −2, −3). ⎡ 3 5 −1⎤ ⎡ −1⎤ ⎡ 3⎤ Also, ⎢ 4 −1 1⎥ ⎢ 2 ⎥ = ⎢ −2 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 3 2 −1⎦ ⎣ 4 ⎦ ⎣ −3⎦

⎡ 1 0 0 ⎤ ⎡ −2 ⎤ ⎡ −2 ⎤ 17. (a) ⎢0 1 0 ⎥ ⎢ 1⎥ = ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 0 0 ⎦ ⎣ 3⎦ ⎣ 0 ⎦ The result is (−2, 1, 0).

⎡ 0 1⎤ ⎢ −1 0 ⎥ 11. (a) The matrix is ⎢ ⎥. ⎢ 1 3⎥ ⎣⎢ 1 −1⎦⎥

⎡ 1 0 0 ⎤ ⎡ −2 ⎤ ⎡ −2 ⎤ (b) ⎢0 0 0 ⎥ ⎢ 1⎥ = ⎢ 0⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 1⎦ ⎣ 3⎦ ⎣ 3⎦ The result is (−2, 0, 3).

⎡ 7 2 −1 1⎤ (b) The matrix is ⎢ 0 1 1 0 ⎥ . ⎢ ⎥ ⎣ −1 0 0 0 ⎦

0 0 0 0 0

1⎤ 0⎥ ⎥ 0⎥ . 0⎥ 0 ⎥⎦

⎡ −1 1⎤ ⎡ −1⎤ ⎡ 5 ⎤ 13. (a) AT x = ⎢ = ⎣ 0 1⎥⎦ ⎢⎣ 4⎥⎦ ⎢⎣ 4 ⎥⎦ T(−1, 4) = (−(−1) + 4, 4) = (5, 4)

(b) T is not a matrix transformation since T(0, 0, 0) ≠ 0.

⎡0 ⎢0 ⎢ (c) The matrix is ⎢0 ⎢0 ⎢⎣0

0 0 0 0 0 1 1 0 0 −1

⎡ 0 0 0 ⎤ ⎡ −2 ⎤ ⎡ 0 ⎤ (c) ⎢0 1 0 ⎥ ⎢ 1⎥ = ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 1⎦ ⎣ 3⎦ ⎣ 3⎦ The result is (0, 1, 3).

0⎤ 0⎥ ⎥ 0⎥ . 0⎥ 0 ⎥⎦ 104


Section 4.9

0 ⎤ ⎡1 0 0 0 ⎤ ⎡ −2 ⎤ ⎢ ⎡1 ⎡ −2 ⎤ 3 1⎥ 19. (a) ⎢0 cos 30° − sin 30°⎥ ⎢ 1⎥ = ⎢ 0 2 − 2 ⎥ ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢ ⎥ 3 ⎥ ⎣ 2⎦ ⎣0 sin 30° cos 30° ⎦ ⎣ 2 ⎦ ⎢ 0 1 ⎥ ⎣ 2 2 ⎦ ⎡ −2 ⎤ ⎢ 3 −2 ⎥ =⎢ 2 ⎥ ⎢ 1+ 2 3 ⎥ ⎣⎢ 2 ⎦⎥ ⎛ The result is ⎜ −2, ⎝

3 − 2 1+ 2 3 ⎞. , ⎟ 2 2 ⎠

⎡ 1 0 ⎡ cos 45° 0 sin 45° ⎤ ⎡ −2 ⎤ ⎢ 2 ⎢ ⎥ ⎢ ⎥ (b) 0 1 0 1 =⎢ 0 1 ⎢ ⎥⎢ ⎥ ⎢ 1 − ° ° sin cos 45 0 45 2 0 ⎣ ⎦ ⎣ ⎦ ⎢− ⎣ 2 ⎡ 0 ⎤ =⎢ 1 ⎥ ⎢ ⎥ ⎢⎣ 2 2 ⎥⎦

1 ⎤ 2 ⎥ ⎡ −2 ⎤

0 ⎥ ⎢ 1⎥ 1 ⎥ ⎢⎣ 2 ⎥⎦ 2 ⎥⎦

The result is ( 0, 1, 2 2 ) .

⎡cos 90° − sin 90° 0 ⎤ ⎡ −2 ⎤ ⎡ 0 −1 0 ⎤ ⎡ −2 ⎤ (c) ⎢ sin 90° cos 90° 0⎥ ⎢ 1⎥ = ⎢ 1 0 0 ⎥ ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ 0 1 ⎦ ⎣ 2 ⎦ ⎣ 0 0 1⎦ ⎣ 2 ⎦ ⎣ 0 ⎡ −1⎤ = ⎢ −2 ⎥ ⎢ ⎥ ⎣ 2⎦ The result is (−1, −2, 2). 0⎤ ⎡1 0 0 0 ⎡1 ⎤ ⎡ −2 ⎤ ⎢ ⎡ −2 ⎤ 3 1 ⎥⎢ ⎥ 21. (a) ⎢0 cos(−30°) − sin(−30°) ⎥ ⎢ 1⎥ = ⎢0 2 ⎥ 1 2 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ 2⎥ − ° − ° 0 30 30 2 sin( ) cos( ) 3 1 ⎣ ⎦ ⎣ ⎦ ⎢0 − ⎣ ⎦ 2 2 ⎥⎦ ⎣ ⎡ −2 ⎤ ⎢ 3+2 ⎥ =⎢ 2 ⎥ ⎢ −1+ 2 3 ⎥ ⎢⎣ 2 ⎥⎦ ⎛ ⎞ The result is ⎜ −2, 3 + 2 , −1 + 2 3 ⎟ . ⎝ 2 2 ⎠

105


⎡ 1 ⎡ cos( −45°) 0 sin(−45°) ⎤ ⎡ −2 ⎤ ⎢ 2 ⎥ ⎢ 1⎥ = ⎢ 0 (b) ⎢ 0 1 0 ⎢ ⎥⎢ ⎥ ⎢ ⎣ − sin(−45°) 0 cos(−45°) ⎦ ⎣ 2 ⎦ ⎢ 1 ⎣ 2 ⎡ −2 =⎢ 1 ⎢ ⎣⎢ 0


0 − 1 0

1 ⎤ 2 ⎥ ⎡ −2 ⎤

0 ⎥ ⎢ 1⎥ 1 ⎥ ⎢⎣ 2 ⎥⎦ 2 ⎥⎦

2⎤ ⎥ ⎥ ⎦⎥

The result is ( −2 2 , 1, 0 ) . ⎡cos(−90°) − sin(−90°) 0⎤ ⎡ −2 ⎤ ⎡ 0 1 0 ⎤ ⎡ −2 ⎤ (c) ⎢ sin(−90°) cos(−90°) 0⎥ ⎢ 1⎥ = ⎢ −1 0 0 ⎥ ⎢ 1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ 0 0 1 ⎦ ⎣ 2 ⎦ ⎣ 0 0 1⎦ ⎣ 2 ⎦ ⎣ ⎡1 ⎤ = ⎢2⎥ ⎢ ⎥ ⎣2⎦ The result is (1, 2, 2). 25.

⎛ 2 2 1⎞ = ⎜ , , ⎟ = (a, b, c) ⎝ 3 3 3⎠ 2 + 2 +1 cos 180° = −1, sin 180° = 0 The matrix is ⎡ 2 2 2 2 (1 − (−1)) − 1 (0) ⎢ 3 (1 − (−1)) + (−1) 3 3 3 ⎢ 2 2 (1 − ( −1)) + ( −1) ⎢ 23 23 (1 − (−1)) + 13 (0) 3 ⎢ ⎢ 2 1 (1 − (−1)) − 2 (0) 2 1 (1 − (−1)) + 2 (0) 3 3 3 3 ⎣ 3 3 v v

=

() ( )( ) ( )( )

(2, 2, 1)

2

2

2

( )( ) () ( )( )

( 23 )( 13 ) (1 − (−1)) + ( 23 ) (0)⎤⎥ ⎡ 94 (2) − 1 ⎥ ⎢ ( 23 )( 13 ) (1 − (−1)) − ( 32 ) (0) ⎥⎥ = ⎢⎢ 94 (2) 2 2 1 (1 − ( −1)) + ( −1) ⎥ ⎢⎣ 9 ( 2) (3) ⎦ ⎡− 1 ⎢ 9 = ⎢ 89 ⎢ 4 ⎢⎣ 9

8 9 − 19 4 9

⎤ ⎥ 4 ( 2) − 1 ⎥ 9 ⎥ 2 ( 2) 1 ( 2) − 1 ⎥⎦ 9 9 4 ( 2) 9

2 ( 2) 9 2 ( 2) 9

4⎤ 9⎥ 4⎥ 9 ⎥ − 79 ⎥ ⎦

29. (a) The result is twice the orthogonal projection of x on the x-axis. (b) The result is twice the reflection of x about the x-axis. ⎡cos 2θ 31. Since cos 2θ = cos 2 θ − sin 2 θ and sin 2θ = 2 sin θ cos θ, the matrix can be written as ⎢ ⎣ sin 2θ has the effect of rotating x through the angle 2θ.

− sin 2θ ⎤ which cos 2θ ⎥⎦

33. The result of the transformation is rotation through the angle θ followed by translation by x0 . It is not a matrix transformation since T (0) = x0 ≠ 0. 35. The image of the line under the operator T is the line in R n which has vector representation x = T (x0 ) + tT ( v), unless T is the zero operator in which case the image is 0.

106


Section 4.10

⎡1 1⎤ 3. (a) The standard matrix for T1 is ⎢ and ⎣1 −1⎥⎦

True/False 4.9 (a) False; if A is 2 × 3, then the domain of TA is

⎡ 3 0⎤ the standard matrix for T2 is ⎢ . ⎣ 2 4 ⎥⎦

R3 . (b) False; if A is m × n, then the codomain of TA is

(b) The standard matrix for T2 T1 is

Rm .

⎡ 3 0 ⎤ ⎡1 1⎤ ⎡ 3 3⎤ ⎢⎣ 2 4 ⎥⎦ ⎢⎣1 −1⎥⎦ = ⎢⎣ 6 −2 ⎥⎦ . The standard matrix for T1 T2 is

(c) False; T(0) = 0 is not a sufficient condition. For

instance if T: R n → R1 is given by T ( v) = v then T(0) = 0, but T is not a matrix transformation.

⎡1 1⎤ ⎡ 3 0 ⎤ ⎡5 4⎤ ⎢⎣1 −1⎥⎦ ⎢⎣ 2 4 ⎥⎦ = ⎢⎣ 1 −4⎥⎦ .

(d) True

(c) T1 (T2 ( x1 , x2 )) = (5 x1 + 4 x2 , x1 − 4 x2 )

T2 (T1 ( x1 , x2 )) = (3x1 + 3x2 , 6 x1 − 2 x2 )

(e) False; every matrix transformation will have this property by the homogeneity property.

5. (a) The standard matrix for a rotation of 90° is ⎡cos 90° − sin 90°⎤ ⎡ 0 −1⎤ ⎢⎣ sin 90° cos 90°⎥⎦ = ⎢⎣ 1 0⎥⎦ and the standard matrix for reflection about the line ⎡0 1⎤ y = x is ⎢ . ⎣ 1 0 ⎥⎦ ⎡0 1⎤ ⎡0 −1⎤ ⎡ 1 0 ⎤ ⎢⎣ 1 0 ⎥⎦ ⎢⎣ 1 0 ⎥⎦ = ⎢⎣0 −1⎥⎦

(f) True; the only matrix transformation with this property is the zero transformation (consider x = y ≠ 0). (g) False; since b ≠ 0, then T(0) = 0 + b = b ≠ 0, so T cannot be a matrix operator. (h) False; there is no angle θ for which 1 cos θ = sin θ = . 2

(b) The standard matrix for orthogonal ⎡0 0 ⎤ projection on the y-axis is ⎢ and the ⎣0 1⎥⎦ standard matrix for a contraction with factor ⎡ 1 0⎤ 1 ⎥. k = is ⎢ 2 2 ⎢⎣ 0 12 ⎥⎦

(i) True; when a = 1 it is the matrix for reflection about the x-axis and when a = −1 it is the matrix for reflection about the y-axis. Section 4.10 Exercise Set 4.10

⎡1 ⎢2 ⎢⎣ 0

1. The standard matrix for TB TA is

3⎤ ⎡ 1 −2 1⎥ ⎢ 4 1 ⎥⎢ 7⎦ ⎣5 2 21⎤ 4⎥ . ⎥ 25⎦ The standard matrix for TA ⎡2 BA = ⎢ 5 ⎢ ⎣6 ⎡ 5 = ⎢10 ⎢ ⎣ 45

⎡1 AB = ⎢ 4 ⎢ ⎣5 ⎡ −8 = ⎢ −5 ⎢ ⎣ 44

−3 0 1 −1 −8 3

0⎤ −3⎥ ⎥ 4⎦

0⎤ ⎡ 0 0 ⎤ ⎡0 ⎥ = 1 ⎥ ⎢⎣ 0 1⎥⎦ ⎢ 0 ⎣ 2⎦

0⎤ 1⎥ 2⎦

(c) The standard matrix for reflection about the ⎡ 1 0⎤ x-axis is ⎢ and the standard matrix ⎣0 −1⎥⎦ ⎡ 3 0⎤ for a dilation with factor k = 3 is ⎢ . ⎣0 3⎥⎦ ⎡ 3 0⎤ ⎡ 1 0⎤ ⎡ 3 0⎤ ⎢⎣0 3⎥⎦ ⎢⎣ 0 −1⎥⎦ = ⎢⎣ 0 −3⎥⎦

TB is

−2 0 ⎤ ⎡ 2 −3 3⎤ 1 −3⎥ ⎢ 5 0 1⎥ ⎥⎢ ⎥ 2 4⎦ ⎣ 6 1 7⎦ −3 1⎤ −15 −8⎥ . ⎥ −11 45⎦ 107



⎡ −1 0 0 ⎤ 7. (a) The standard matrix for reflection about the yz-plane is ⎢ 0 1 0 ⎥ and the standard matrix for orthogonal ⎢ ⎥ ⎣ 0 0 1⎦ ⎡ 1 0 0⎤ projection on the xz-plane is ⎢0 0 0 ⎥ . ⎢ ⎥ ⎣0 0 1⎦ ⎡ 1 0 0 ⎤ ⎡ −1 0 0⎤ ⎡ −1 0 0 ⎤ ⎢0 0 0 ⎥ ⎢ 0 1 0⎥ = ⎢ 0 0 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 1⎦ ⎣ 0 0 1⎦ ⎣ 0 0 1⎦ ⎡ 1 ⎡ cos 45° 0 sin 45° ⎤ ⎢ 2 (b) The standard matrix for a rotation of 45° about the y-axis is ⎢ 0 1 0 ⎥=⎢ 0 ⎢ ⎥ ⎢ ⎣ − sin 45° 0 cos 45° ⎦ ⎢ − 1 ⎣ 2

⎡ 2 ⎢ the standard matrix for a dilation with factor k = 2 is ⎢ 0 ⎢⎣ 0 ⎡ 2 ⎢ ⎢ 0 ⎣⎢ 0

0 2 0

⎡ 1 0 ⎤⎢ 2 ⎥ 0 ⎥⎢ 0 ⎢ 2 ⎦⎥ ⎢ − 1 ⎣ 2

0 1 0

1 ⎤ 2⎥

⎡1 0 ⎥=⎢0 ⎢ 1 ⎥ ⎢ −1 ⎣ 2 ⎦⎥

0 2 0

0 2 0

0 1 0

1 ⎤ 2⎥

0 ⎥ and 1 ⎥ 2 ⎦⎥

0 ⎤ ⎥ 0 ⎥. 2 ⎥⎦

1⎤ 0 ⎥⎥ 1 ⎥⎦

⎡ 1 0 0⎤ (c) The standard matrix for orthogonal projection on the xy-plane is ⎢0 1 0 ⎥ and the standard matrix for ⎢ ⎥ ⎣0 0 0 ⎦ ⎡ −1 0 0 ⎤ reflection about the yz-plane is ⎢ 0 1 0 ⎥ . ⎢ ⎥ ⎣ 0 0 1⎦ ⎡ −1 0 0 ⎤ ⎡ 1 0 0 ⎤ ⎡ −1 0 0 ⎤ ⎢ 0 1 0⎥ ⎢ 0 1 0⎥ = ⎢ 0 1 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 0 1⎦ ⎣ 0 0 0⎦ ⎣ 0 0 0 ⎦ ⎡1 0 ⎤ ⎡0 0⎤ . 9. (a) [T1 ] = ⎢ and [T2 ] = ⎢ ⎣0 0 ⎥⎦ ⎣ 0 1 ⎥⎦ ⎡0 0 ⎤ so T1 T2 = T2 T1. [T1 ][T2 ] = [T2 ][T1 ] = ⎢ ⎣0 0 ⎥⎦

108


Section 4.10

⎡ cos(θ1 + θ 2 ) − sin(θ1 + θ 2 ) ⎤ (b) From Example 1, [T2 T1 ] = ⎢ ⎥ ⎣ sin(θ1 + θ 2 ) cos(θ1 + θ 2 ) ⎦ [T1 T2 ] = [T1 ][T2 ] ⎡cos θ1 − sin θ1 ⎤ ⎡cos θ 2 − sin θ 2 ⎤ =⎢ cos θ1 ⎥⎦ ⎢⎣ sin θ 2 cos θ 2 ⎥⎦ ⎣ sin θ1 ⎡ cos θ1 cos θ 2 − sin θ1 sin θ 2 − cos θ1 sin θ 2 − sin θ1 cos θ 2 ⎤ =⎢ ⎥ ⎣sin θ1 cos θ 2 + cos θ1 sin θ 2 − sin θ1 sin θ 2 + cos θ1 cos θ 2 ⎦ ⎡cos(θ1 + θ 2 ) − sin(θ1 + θ 2 ) ⎤ =⎢ ⎥ ⎣ sin(θ1 + θ 2 ) cos(θ1 + θ 2 ) ⎦ = [T2 T1 ] Thus T1 T2 = T2 T1. ⎡ 1 0⎤ ⎡ cos θ − sin θ ⎤ and [T2 ] = ⎢ . (c) [T1 ] = ⎢ ⎥ cos θ ⎥⎦ ⎣0 0 ⎦ ⎣ sin θ ⎡ 1 0 ⎤ ⎡cos θ − sin θ ⎤ [T1 ][T2 ] = ⎢ cos θ ⎥⎦ ⎣0 0 ⎥⎦ ⎢⎣ sin θ ⎡cos θ − sin θ ⎤ =⎢ 0 ⎥⎦ ⎣ 0 ⎡cos θ − sin θ ⎤ ⎡ 1 0 ⎤ [T2 ][T1 ] = ⎢ cos θ ⎥⎦ ⎢⎣0 0 ⎥⎦ ⎣ sin θ ⎡cos θ 0 ⎤ =⎢ ⎣ sin θ 0 ⎥⎦ T1 T2 ≠ T2 T1 11. (a) Not one-to-one because the second row (column) is all 0’s, so the determinant is 0. (b) One-to-one because the determinant is −1. (c) One-to-one because the determinant is −1. (d) One-to-one because the determinant is k 2 ≠ 0. (e) One-to-one because the determinant is 1. (f) One-to-one because the determinant is −1. (g) One-to-one because the determinant is k 3 ≠ 0.

⎡ 1 2⎤ 13. (a) [T ] = ⎢ ⎣ −1 1⎥⎦ det[T] = 1 + 2 = 3 ≠ 0, so T is one-to-one. 1 2 1 ⎡1 −2 ⎤ ⎡ 3 − 3 ⎤ ⎢ ⎥ [T −1 ] = ⎢ = 1⎥ 3 ⎣1 1⎥⎦ ⎢ 1 3⎦ ⎣3 2 1 1 ⎞ ⎛1 T −1 ( w1 , w2 ) = ⎜ w1 − w2 , w1 + w2 ⎟ 3 3 3 ⎠ ⎝3

109



⎡ 4 −6 ⎤ (b) [T ] = ⎢ 3⎥⎦ ⎣ −2 det[T] = 12 − 12 = 0, so T is not one-to-one. ⎡ 0 −1⎤ (c) [T ] = ⎢ ⎣ −1 0 ⎥⎦ det[T] = 0 − 1 = −1 ≠ 0, so T is one-to-one. 1 ⎡0 1⎤ ⎡ 0 −1⎤ [T −1 ] = = −1 ⎢⎣ 1 0 ⎥⎦ ⎢⎣ −1 0 ⎥⎦ T −1 ( w1 , w2 ) = (− w2 , − w1 )

⎡ 3 0⎤ (d) [T ] = ⎢ ⎣ −5 0 ⎥⎦ det[T] = 3(0) − (−5)(0) = 0, so T is not one-to-one. 15. (a) The inverse is (another) reflection about the x-axis in R 2 . (b) The inverse is rotation through the angle −

(c) The inverse is contraction by a factor of

π in R 2 . 4

1 in R 2 . 3

(d) The inverse is (another) reflection about the yz-plane in R3 . (e) The inverse is dilation by a factor of 5 in R3 . 17. Let u = ( x1 , y1 ), v = ( x2 , y2 ), and k be a scalar. (a) T (u + v) = T ( x1 + x2 , y1 + y2 ) = (2( x1 + x2 ) + ( y1 + y2 ), ( x1 + x2 ) − ( y1 + y2 )) = ((2 x1 + y1 ) + (2 x2 + y2 ), ( x1 − y1 ) + ( x2 − y2 )) = T (u ) + T ( v ) T (ku) = T (kx1 , ky1 ) = (2kx1 + ky1, kx1 − ky1 ) = k (2 x1 + y1 , x1 − y1 ) = kT (u) T is a matrix operator.

(b) T (ku) = T (kx1 , ky1 ) = (kx1 + 1, ky1 ) ≠ k ( x1 + 1, y1 ) T is not a matrix operator, which can also be shown by considering T(u + v).

110


Section 4.10

(c) T (u + v) = T ( x1 + x2 , y1 + y2 ) = ( y1 + y2 , y1 + y2 ) = T (u) + T ( v) T (ku) = T (kx1 , ky1 ) = (ky1 , ky1 ) = k ( y1 , y1 ) = kT (u) T is a matrix operator. (d) T (u + v) = T ( x1 + x2 , y1 + y2 )

( 3 x1 + x2 , 3 y1 + y2 ) ≠ ( 3 x1 + 3 x2 , 3 y1 + 3 y2 ) =

T is not a matrix operator, which can also be shown by considering T(ku). 19. Let u = ( x1 , y1 , z1 ), v = ( x2 , y2 , z2 ), and k be a scalar. (a) T (u + v) = T ( x1 + x2 , y1 + y2 , z1 + z2 ) = (0, 0) = T (u) + T ( v) T (ku) = T (kx1, ky1, kz1 ) = (0, 0) = k (0, 0) = kT (u) T is a matrix transformation. (b) T (u + v) = T ( x1 + x2 , y1 + y2 , z1 + z2 ) = (3( x1 + x2 ) − 4( y1 + y2 ), 2( x1 + x2 ) − 5( z1 + z2 )) = ((3x1 − 4 y1 ) + (3x2 − 4 y2 ), (2 x1 − 5 z1 ) + (2 x2 − 5 z2 )) = T (u ) + T ( v ) T (ku) = T (kx1 , ky1 , kz1 ) = (3kx1 − 4ky1 , 2kx1 − 5kz1 ) = k (3x1 − 4 y1, 2 x1 − 5 z1 ) = kT (u) T is a matrix operator.

⎡ −1 0 ⎤ ⎡ 1 0 ⎤ ⎡ −1 0 ⎤ 21. (a) ⎢ = ⎣ 0 1⎥⎦ ⎢⎣0 0 ⎥⎦ ⎢⎣ 0 0 ⎥⎦ ⎡ 1 0 ⎤ ⎡0 1⎤ ⎡ 0 1⎤ (b) ⎢ = ⎣0 −1⎥⎦ ⎢⎣ 1 0 ⎥⎦ ⎢⎣ −1 0 ⎥⎦ ⎡0 0 ⎤ ⎡0 1⎤ ⎡ 3 0 ⎤ ⎡ 0 (c) ⎢ = ⎣0 1⎥⎦ ⎢⎣ 1 0 ⎥⎦ ⎢⎣0 3⎥⎦ ⎢⎣ 1 = ⎡0 ⎢⎣ 3

0⎤ ⎡ 3 0⎤ 0 ⎥⎦ ⎢⎣ 0 3⎥⎦ 0⎤ 0⎥⎦

23. (a) TA (e1 ) = (−1, 2, 4), TA (e2 ) = (3, 1, 5), and TA (e3 ) = (0, 2, − 3)

111



(b) TA (e1 + e2 + e3 ) = (−1 + 3 + 0, 2 + 1 + 2, 4 + 5 − 3) = (2, 5, 6)

(b) True; use c1 = c2 = 1 to get the additivity

(c) TA (7e3 ) = 7(0, 2, − 3) = (0, 14, − 21)

(c) True; if T is a one-to-one matrix transformation with matrix A, then Ax = 0 has only the trivial solution. That is T(x − y) = 0 ⇒ A(x − y) = 0 ⇒ x = y, so there are no such distinct vectors.

property and c1 = k , c2 = 0 to get the homogeneity property.

25. (a) Yes; if u and v are distinct vectors in the domain of T1 which is one-to-one, then T1 (u) and T1 ( v) are distinct. If T1 (u) and

(d) False; the zero transformation is one example.

T1 ( v) are in the domain of T2 which is also one-to-one, then T2 (T1 (u)) and T2 (T1 ( v))

(e) False; the zero transformation is one example.

are distinct. Thus T2 T1 is one-to-one.

(f) False; the zero transformation is one example.

(b) Yes; if T1 ( x, y) = ( x, y, 0) and

Section 4.11

T2 ( x, y, z ) = ( x, y), then T1 is one-to-one

Exercise Set 4.11

but T2 is not one-to-one and the

1. (a) Reflection about the line y = −x maps (1, 0) to (0, −1) and (0, 1) to (−1, 0), so the ⎡ 0 −1⎤ standard matrix is ⎢ . ⎣ −1 0 ⎥⎦

composition T2 T1 is the identity transformation on R 2 which is one-to-one. However, when T1 is not one-to-one and T2 is one-to-one, the composition T2 T1 is not one-to-one, since T1 (u) = T1 ( v) for some

(b) Reflection through the origin maps (1, 0) to (−1, 0) and (0, 1) to (0, −1), so the standard ⎡ −1 0 ⎤ matrix is ⎢ . ⎣ 0 −1⎥⎦

distinct vectors u and v in the domain of T1 and consequently T2 (T1 (u)) = T2 (T1 ( v)) for those same vectors. 27. (a) The product of any m × n matrix and the n × 1 matrix of all zeros (the zero vector in

(c) Orthogonal projection on the x-axis maps (1, 0) to (1, 0) and (0, 1) to (0, 0), so the ⎡ 1 0⎤ standard matrix is ⎢ . ⎣0 0 ⎥⎦

R n ) is the m × 1 matrix of all zero, which is the zero vector in R m .

(d) Orthogonal projection on the y-axis maps (1, 0) to (0, 0) and (0, 1) to (0, 1), so the ⎡0 0 ⎤ standard matrix is ⎢ . ⎣0 1⎥⎦

(b) One example is T ( x1, x2 ) = ( x12 + x22 , x1 x2 ) 29. (a) The range of T must be a proper subset of R n , i.e., not all of R n . Since det(A) = 0,

3. (a) A reflection through the xy-plane maps (1, 0, 0) to (1, 0, 0), (0, 1, 0) to (0, 1, 0), and (0, 0, 1) to (0, 0, −1), so the standard matrix ⎡ 1 0 0⎤ is ⎢0 1 0 ⎥ . ⎢ ⎥ ⎣0 0 −1⎦

there is some b1 in R n such that Ax = b1 is inconsistent, hence b1 is not in the range of T. (b) T must map infinitely many vectors to 0. True/False 4.10 2

(b) A reflection through the xz-plane maps (1, 0, 0) to (1, 0, 0), (0, 1, 0) to (0, −1, 0), and (0, 0, 1) to (0, 0, 1), so the standard ⎡ 1 0 0⎤ matrix is ⎢0 −1 0 ⎥ . ⎢ ⎥ ⎣0 0 1⎦

2

(a) False; for example T: R → R given by T ( x1, x2 ) = ( x12 + x22 , x1 x2 ) is not a matrix transformation, although T(0) = T(0, 0) = (0, 0) = 0. 112


Section 4.11

(b) The geometric effect is expansion by a factor of 5 in the y-direction and reflection about the x-axis.

(c) A reflection through the yz-plane maps (1, 0, 0) to (−1, 0, 0), (0, 1, 0) to (0, 1, 0), and (0, 0, 1) to (0, 0, 1), so the standard ⎡ −1 0 0 ⎤ matrix is ⎢ 0 1 0 ⎥ . ⎢ ⎥ ⎣ 0 0 1⎦

(c) The geometric effect is shearing by a factor of 4 in the x-direction.

⎡ 1 0⎤ ⎡ 12 0 ⎤ ⎡ 12 0 ⎤ 13. (a) ⎢ ⎢ ⎥=⎢ ⎥ ⎣0 5⎥⎦ ⎣ 0 1⎦ ⎣ 0 5⎦

5. (a) Rotation of 90° about the z-axis maps (1, 0, 0) to (0, 1, 0), (0, 1, 0) to (−1, 0, 0), and (0, 0, 1) to (0, 0, 1), so the standard ⎡0 −1 0 ⎤ matrix is ⎢ 1 0 0 ⎥ . ⎢ ⎥ ⎣0 0 1⎦

⎡ 1 0⎤ ⎡ 1 0⎤ ⎡ 1 0⎤ = (b) ⎢ ⎣ 2 1⎥⎦ ⎢⎣ 0 5⎥⎦ ⎢⎣ 2 5⎥⎦ ⎡cos180° − sin 180°⎤ ⎡ 0 1⎤ (c) ⎢ ⎣ sin 180° cos180°⎥⎦ ⎢⎣ 1 0 ⎥⎦ ⎡ −1 0⎤ ⎡ 0 1⎤ =⎢ ⎣ 0 −1⎥⎦ ⎢⎣ 1 0 ⎥⎦ ⎡ 0 −1⎤ =⎢ ⎣ −1 0⎥⎦

(b) Rotation of 90° about the x-axis maps (1, 0, 0) to (1, 0, 0), (0, 1, 0) to (0, 0, 1), and (0, 0, 1) to (0, −1, 0), so the standard matrix ⎡ 1 0 0⎤ is ⎢0 0 −1⎥ . ⎢ ⎥ ⎣0 1 0 ⎦

−1

⎡0 1⎤ ⎡0 1⎤ , the inverse =⎢ 15. (a) Since ⎢ ⎥ 1 0 ⎣ ⎦ ⎣ 1 0 ⎥⎦ transformation is reflection about y = x.

(c) Rotation of 90° about the y-axis maps (1, 0, 0) to (0, 0, −1), (0, 1, 0) to (0, 1, 0), and (0, 0, 1) to (1, 0, 0), so the standard ⎡ 0 0 1⎤ matrix is ⎢ 0 1 0 ⎥ . ⎢ ⎥ ⎣ −1 0 0 ⎦

⎡k 0⎤ (b) Since for 0 < k < 1, ⎢ ⎣ 0 1⎥⎦

−1

⎡ 1 0⎤ = ⎢k ⎥ and ⎣ 0 1⎦

−1 ⎡ 1 0 ⎤ ⎡ 1 0⎤ ⎢⎣0 k ⎥⎦ = ⎢ 0 1 ⎥ , the inverse k⎦ ⎣ transformation is an expansion along the same axis.

⎡ − 3 0 ⎤ ⎡ 0 ⎤ ⎡ 0 ⎤ ⎡ − 3 0 ⎤ ⎡ 1 ⎤ ⎡ − 3⎤ 7. ⎢ = , , = ⎣ 0 1⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 0 1⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡ −3 0 ⎤ ⎡0 ⎤ ⎡0 ⎤ ⎡ −3 0 ⎤ ⎡1⎤ ⎡ −3⎤ ⎢⎣ 0 1⎥⎦ ⎢⎣1 ⎥⎦ = ⎢⎣1 ⎥⎦ , ⎢⎣ 0 1⎥⎦ ⎢⎣1⎥⎦ = ⎢⎣ 1⎥⎦ The image is a rectangle with vertices at (0, 0), (−3, 0), (0, 1), and (−3, 1).

⎡ −1 0 ⎤ (c) Since ⎢ ⎣ 0 1⎥⎦

y

−1

⎡ −1 0 ⎤ =⎢ and ⎣ 0 1⎥⎦

−1

⎡ 1 0⎤ ⎡ 1 0⎤ ⎢⎣0 −1⎥⎦ = ⎢⎣ 0 −1⎥⎦ the inverse transformation is a reflection about the same coordinate axis.

x

9. (a) Shear by a factor of k = 4 in the y-direction ⎡ 1 0⎤ has matrix ⎢ . ⎣ 4 1⎥⎦

⎡1 k⎤ (d) Since k ≠ 0, ⎢ ⎣0 1⎥⎦ −1

−1

⎡ 1 −k ⎤ =⎢ and 1⎥⎦ ⎣0

⎡ 1 0⎤ ⎡ 1 0⎤ ⎢⎣ k 1⎥⎦ = ⎢⎣ −k 1⎥⎦ , the inverse transformation is a shear (in the opposite direction) along the same axis.

(b) Shear by a factor of k = −2 in the x-direction ⎡ 1 −2 ⎤ has matrix ⎢ . 1⎥⎦ ⎣0 11. (a) The geometric effect is expansion by a factor of 3 in the x-direction. 113



⎡ t⎤ 17. The line y = 2x can be represented by ⎢ ⎥ . ⎣ 2t ⎦

23. (a) To map (x, y, z) to (x + kz, y + kz, z), the ⎡1 0 k⎤ standard matrix is ⎢0 1 k ⎥ . ⎢ ⎥ ⎣0 0 1⎦

⎡ 1 3⎤ ⎡ t ⎤ ⎡ t + 6t ⎤ ⎡ 7t ⎤ = = (a) ⎢ ⎣0 1⎥⎦ ⎢⎣ 2t ⎥⎦ ⎢⎣ 0 + 2t ⎥⎦ ⎢⎣ 2t ⎥⎦ The image is x = 7t, y = 2t, and using 1 2 t = x, this is the line y = x. 7 7

(b) Shear in the xz-direction with factor k maps (x, y, z) to (x + ky, y, z + ky). The standard ⎡ 1 k 0⎤ matrix is ⎢0 1 0 ⎥ . ⎢ ⎥ ⎣0 k 1⎦ Shear in the yz-direction with factor k maps (x, y, z) to (x, y + kx, z + kx). ⎡ 1 0 0⎤ The standard matrix is ⎢ k 1 0⎥ . ⎢ ⎥ ⎣ k 0 1⎦

⎡ 1 0 ⎤ ⎡ t ⎤ ⎡t ⎤ (b) ⎢ 1⎥⎢ ⎥ = ⎢ ⎥ ⎣0 2 ⎦ ⎣ 2t ⎦ ⎣t ⎦ The image is x = t, y = t which is the line y = x. ⎡0 1⎤ ⎡ t ⎤ ⎡ 2t ⎤ (c) ⎢ = ⎣ 1 0⎥⎦ ⎢⎣ 2t ⎥⎦ ⎢⎣ t ⎥⎦ The image is x = 2t, y = t, which is the line 1 y = x. 2

True/False 4.11 (a) False; the image will be a parallelogram, but it will not necessarily be a square. (b) True; an invertible matrix operator can be expressed as a product of elementary matrices, each of which has geometric effect among those listed.

⎡ −1 0 ⎤ ⎡ t ⎤ ⎡ −t ⎤ (d) ⎢ = ⎣ 0 1⎥⎦ ⎢⎣ 2t ⎥⎦ ⎢⎣ 2t ⎥⎦ The image is x = −t, y = 2t, which is the line y = −2x.

(c) True; by Theorem 4.11.3.

3⎤ ⎡1 ⎡cos 60° − sin 60°⎤ ⎡ t ⎤ ⎢ 2 − 2 ⎥ ⎡ t ⎤ (e) ⎢ = 1 ⎥ ⎢⎣ 2t ⎥⎦ ⎣ sin 60° cos 60°⎥⎦ ⎢⎣ 2t ⎥⎦ ⎢ 3 2 ⎦ ⎣2 1− 2 3 ⎤ ⎡ t = ⎢ 2 ⎥ ⎢ 3+2 t ⎥ ⎣ 2 ⎦

⎡ 1 0⎤ (d) True; ⎢ ⎣0 −1⎥⎦ ⎡ −1 0 ⎤ ⎢⎣ 0 1⎥⎦

3+2 1− 2 3

x or y = −

⎡ 1 0⎤ =⎢ , ⎣0 −1⎥⎦

⎡ −1 0 ⎤ ⎡0 1⎤ =⎢ , and ⎢ ⎥ 0 1 ⎣ 1 0⎥⎦ ⎣ ⎦

−1

⎡0 1⎤ =⎢ . ⎣ 1 0 ⎥⎦

⎡1 1⎤ (e) False; ⎢ maps the unit square to the ⎣1 −1⎥⎦

1− 2 3 3+2 t, y = t, 2 2 2 x, this is the line and using t = 1− 2 3 The image is x =

y=

−1

−1

parallelogram with vertices (0, 0), (1, 1), (1, −1), and (2, 0) which is not a reflection about a line.

⎡ 1 −2 ⎤ (f) False; ⎢ maps the unit square to the 1⎥⎦ ⎣2 parallelogram with vertices (0, 0), (1, 2), (−2, 1), and (−1, 3) which is not a shear in either the xor y-direction.

8+5 3 x. 11

⎡ 3 1⎤ ⎡ x ⎤ ⎡ 3x + y ⎤ ⎡ x′ ⎤ 19. (a) ⎢ = = ⎣6 2 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣6 x + 2 y ⎥⎦ ⎢⎣ y ′⎥⎦ Thus, the image ( x′, y ′) satisfies y ′ = 2 x′ for every (x, y) in the plane.

(g) True; this is an expansion by a factor of 3 in the y-direction.

(b) No, since the matrix is not invertible.

Section 4.12 Exercise Set 4.12 1. (a) A is a stochastic matrix. 114


Section 4.12

(b) A is not a stochastic matrix since the sums of the entries in each column are not 1.

9 . The steady-state vector is 17 ⎡8 9 ⎤ ⎡ 8 ⎤ q = ⎢ 9 17 ⎥ = ⎢ 17 ⎥ . ⎢ 9 ⎥ ⎢9⎥ ⎣ 17 ⎦ ⎣ 17 ⎦

so s =

( )

(c) A is a stochastic matrix. (d) A is not a stochastic matrix because the third column is not a probability vector due to the negative entry.

⎡0.5 3. x1 = Px0 = ⎢ ⎣0.5 ⎡ 0.5 x 2 = Px1 = ⎢ ⎣ 0.5

9. Each column of P is a probability vector and ⎡3 1 1⎤ ⎢8 2 6⎥ 7 ⎥ , so P is a regular stochastic P 2 = ⎢ 13 83 18 ⎢7 1 4⎥ ⎢⎣ 24 8 9 ⎥⎦ matrix. The system (I − P)q = 0 is ⎡ 1 −1 0⎤ ⎡ q ⎤ 2 ⎡ 0⎤ ⎢ 2 ⎥ 1 1 − 1 ⎥ ⎢q ⎥ = ⎢ ⎥ . ⎢ − 14 0 2 2 3 ⎢ ⎥ ⎢ ⎥ ⎢ 1 ⎥ q 0⎦ 1 ⎣ ⎣ 3⎦ 0 ⎢⎣ − 4 3 ⎥⎦ The reduced row echelon form of I − P is ⎡1 0 − 4⎤ 3⎥ ⎢ ⎢0 1 − 43 ⎥ , so the general solution of the ⎢ ⎥ 0⎦ ⎣0 0 4 4 system is q1 = s, q2 = s, q3 = s. Since q 3 3 must be a probability vector, 11 3 1 = q1 + q2 + q3 = s, so s = . 11 3 ⎡4 3 ⎤ ⎡ 4 ⎤ ⎢ 3 11 ⎥ ⎢ 11 ⎥ 4 ⎥. The steady-state vector is q = ⎢ 4 3 ⎥ = ⎢ 11 ⎢ 3 11 ⎥ ⎢ ⎥ 3 ⎢ 3 ⎥ ⎣⎢ 11 ⎦⎥ ⎣ 11 ⎦

0.6 ⎤ ⎡0.5⎤ ⎡0.55⎤ = 0.4 ⎥⎦ ⎢⎣0.5⎥⎦ ⎢⎣0.45⎥⎦ 0.6 ⎤ ⎡0.55⎤ ⎡ 0.545⎤ = 0.4 ⎥⎦ ⎢⎣0.45⎥⎦ ⎢⎣ 0.455⎥⎦

⎡0.5 0.6 ⎤ ⎡ 0.545⎤ ⎡0.5455⎤ x 3 = Px 2 = ⎢ = ⎣0.5 0.4 ⎥⎦ ⎢⎣ 0.455⎥⎦ ⎢⎣0.4545⎥⎦ ⎡0.5 0.6 ⎤ ⎡0.5455⎤ ⎡ 0.54545⎤ x 4 = Px 3 = ⎢ = ⎣0.5 0.4 ⎥⎦ ⎢⎣0.4545⎥⎦ ⎢⎣ 0.45455⎥⎦ ⎡ 0.5455 0.5454 ⎤ ⎡0.5⎤ x 4 = P 4 x0 = ⎢ ⎣ 0.4545 0.4546 ⎥⎦ ⎢⎣0.5⎥⎦ ⎡ 0.54545⎤ =⎢ ⎣ 0.45455⎥⎦

5. (a) P is a regular stochastic matrix. (b) P is not a regular stochastic matrix, since the entry in row 1, column 2 of P k will be 0 for all k ≥ 1.

( ) ( )

(c) P is a regular stochastic matrix since ⎡ 21 1 ⎤ P 2 = ⎢ 25 5 ⎥ . 4 4⎥ ⎢⎣ 25 5⎦ 7. Since P is a stochastic matrix with all positive entries, P is a regular stochastic matrix. The system (I − P)q = 0 is ⎡ 3 − 2⎤ ⎡ q ⎤ 3 ⎥ 1 = ⎡0⎤ . ⎢ 4 3 2 ⎥ ⎢ q ⎥ ⎢⎣ 0 ⎥⎦ ⎢− ⎣ 2⎦ 3⎦ ⎣ 4 ⎡ 3 − 2 0⎤ ⎡ 1 − 8 0⎤ 3 9 ⎢ 4 ⎥ reduces to ⎢ ⎥ so the 2 0⎥ ⎢ − 34 0 0 0 ⎢ ⎣ ⎦⎥ 3 ⎣ ⎦ 8 general solution is q1 = s, q2 = s. Since q 9 17 must be a probability vector 1 = q1 + q2 = s 9

11. (a) The entry 0.2 = p11 represents the probability that something in state 1 stays in state 1. (b) The entry 0.1 = p12 represents the probability that something in state 2 moves to state 1. ⎡0.2 0.1⎤ ⎡1 ⎤ ⎡0.2 ⎤ (c) ⎢ = ⎣ 0.8 0.9 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 0.8 ⎥⎦ The probability is 0.8. ⎡0.2 0.1⎤ ⎡0.5⎤ ⎡ 0.15⎤ (d) ⎢ = ⎣ 0.8 0.9 ⎥⎦ ⎢⎣0.5⎥⎦ ⎢⎣ 0.85⎥⎦ The probability is 0.85.

115



⎡ 0.95 0.55⎤ 13. (a) Let state 1 be good air. The transition matrix is P = ⎢ . ⎣ 0.05 0.45⎥⎦ ⎡1 ⎤ ⎡ 0.93 0.77 ⎤ ⎡1 ⎤ ⎡ 0.93⎤ = (b) P 2 ⎢ ⎥ = ⎢ ⎣ 0⎦ ⎣ 0.07 0.23⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣0.07 ⎥⎦ The probability is 0.93 that the air will be good. ⎡0 ⎤ ⎡ 0.922 0.858⎤ ⎡ 0 ⎤ ⎡ 0.858 ⎤ (c) P3 ⎢ ⎥ = ⎢ = ⎣1 ⎦ ⎣ 0.078 0.142 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣0.142 ⎥⎦ The probability is 0.142 that the air will be bad. ⎡0.2 ⎤ ⎡0.95 0.55⎤ ⎡0.2 ⎤ ⎡ 0.63⎤ (d) P ⎢ ⎥ = ⎢ = ⎣ 0.8⎦ ⎣0.05 0.45⎥⎦ ⎢⎣ 0.8 ⎥⎦ ⎢⎣0.37 ⎥⎦ The probability is 0.63 that the air will be good. ⎡ 0.95 0.03⎤ 15. Let state 1 be living in the city. Then the transition matrix is P = ⎢ . The initial state vector is ⎣ 0.05 0.97 ⎥⎦ ⎡100 ⎤ x0 = ⎢ ⎥ (in thousands). ⎣ 25⎦ ⎡0.95 0.03⎤ ⎡100 ⎤ ⎡95.75 ⎤ = (a) Px0 = ⎢ ⎣0.05 0.97 ⎥⎦ ⎢⎣ 25⎥⎦ ⎢⎣ 29.25⎥⎦ 2

⎡0.95 0.03⎤ ⎡100⎤ ⎡91.84 ⎤ P 2 x0 = ⎢ = ⎣0.05 0.97 ⎥⎦ ⎢⎣ 25⎥⎦ ⎢⎣33.16 ⎥⎦ 3

⎡0.95 0.03⎤ ⎡100 ⎤ ⎡ 88.243 ⎤ P3x0 = ⎢ ≈ ⎣0.05 0.97 ⎥⎦ ⎢⎣ 25⎥⎦ ⎢⎣36.757 ⎥⎦ 4

⎡0.95 0.03⎤ ⎡100⎤ ⎡ 84.933 ⎤ P 4 x0 = ⎢ ≈ ⎣0.05 0.97 ⎥⎦ ⎢⎣ 25⎥⎦ ⎢⎣ 40.067 ⎥⎦ 5

⎡0.95 0.03⎤ ⎡100⎤ ⎡81.889⎤ P5 x0 = ⎢ ≈ ⎣0.05 0.97 ⎥⎦ ⎢⎣ 25⎥⎦ ⎢⎣ 43.111⎥⎦ Year

1

2

3

4

5

City

95,750

91,840

88,243

84,933

81,889

Suburbs

29,250

33,160

36,757

40,067

43,111

⎡ 0.05 −0.03⎤ ⎡ q1 ⎤ ⎡0 ⎤ . (b) The system (I − P)q = 0 is ⎢ ⎢ ⎥= ⎣ −0.05 0.03⎥⎦ ⎣ q2 ⎦ ⎢⎣0 ⎥⎦ ⎡1 −0.6 ⎤ so the general solution is q1 = 0.6s, q2 = s. Since The reduced row echelon form of I − P is ⎢ 0 ⎥⎦ ⎣0 1 ⎡ 0.6(0.625) ⎤ ⎡0.375⎤ 1 = q1 + q2 = 1.6s, s = = 0.625 and the steady-state vector is q = ⎢ = . 1.6 ⎣ 0.625 ⎥⎦ ⎢⎣0.625⎥⎦ Over the long term, the population of 125,000 will be distributed with 0.375(125,000) = 46,875 in the city and 0.625(125,000) = 78,125 in the suburbs.

116


⎡1 ⎢ 10 17. The transition matrix is P = ⎢ 4 ⎢ 51 ⎣⎢ 10 ⎡ 23 ⎡1 ⎤ ⎢ 100 (a) P 2 ⎢ 0⎥ = ⎢ 17 ⎢ ⎥ ⎢ 50 ⎣ 0⎦ ⎢ 43 ⎣ 100

19 50 7 20 27 100

The probability is

11 ⎤ 50 ⎥ ⎡1 ⎤ 29 ⎥ ⎢ ⎥ 0 50 ⎢ ⎥ ⎥ 1 ⎣0 ⎦ 5 ⎥⎦


1 5 3 10 1 2

3⎤ 5⎥ 1⎥. 5 ⎥ 1 5 ⎦⎥

1 3 1 Column 3: 1 − − = 5 10 2 ⎡7 1 1⎤ ⎢ 10 10 5 ⎥ 3 1⎥ P = ⎢ 15 10 ⎢ 1 3 32 ⎥ ⎢⎣ 10 5 10 ⎥⎦ The system (I − P)q = 0 is ⎡ 3 − 1 − 1⎤ 10 5 ⎥ ⎡ q1 ⎤ ⎡ 0 ⎤ ⎢ 10 7 − 1 ⎥ ⎢q ⎥ = ⎢ ⎥ . ⎢ − 15 0 10 2 ⎢ 2⎥ ⎢ ⎥ ⎢ 1 ⎥ q 0 3 7 ⎢⎣ − 10 − 5 10 ⎥⎦ ⎣ 3 ⎦ ⎣ ⎦ The reduced row echelon form of I − P is ⎡ 1 0 −1⎤ ⎢0 1 −1⎥ , so the general solution is q1 = s, ⎢ ⎥ ⎣0 0 0 ⎦

⎡ 23 ⎤ ⎢ 100 ⎥ = ⎢ 17 ⎥ ⎢ 50 ⎥ 43 ⎢⎣ 100 ⎥⎦

23 . 100

(b) The system (I − P)q = 0 is ⎡ 9 − 1 − 3⎤ 5 5 ⎥ ⎡ q1 ⎤ ⎡ 0 ⎤ ⎢ 10 7 − 1 ⎥ ⎢ q ⎥ = ⎢0 ⎥ . ⎢ −4 5 ⎢ 2⎥ ⎢ ⎥ ⎢ 15 101 ⎥ q 4 ⎣ 3 ⎦ ⎣0 ⎦ ⎢⎣ − 10 − 2 5 ⎥⎦ The reduced row echelon form of I − P is ⎡ 1 0 − 46 ⎤ 47 ⎥ ⎢ ⎢0 1 − 66 ⎥ , so the general solution is 47 ⎢ ⎥ 0⎦ ⎣0 0 46 66 q1 = s , q2 = s, q3 = s. 47 47 159 47 Since 1 = q1 + q2 + q3 = s, s = and 47 159 the steady-state vector is ⎡ 46 47 ⎤ ⎡ 46 ⎤ ⎢ 47 159 ⎥ ⎢ 159 ⎥ q = ⎢ 66 47 ⎥ = ⎢ 22 ⎥. ⎢ 47 159 ⎥ ⎢ 53 ⎥ ⎢ 47 ⎥ ⎢ 47 ⎥ ⎣ 159 ⎦ ⎣ 159 ⎦

q2 = s, q3 = s. Since 1 = q1 + q2 + q3 = 3s, ⎡1 ⎤ ⎢3⎥ 1 s = and the steady-state vector is q = ⎢ 13 ⎥ . 3 ⎢1 ⎥ ⎢⎣ 3 ⎥⎦

21. Since q is the steady-state vector for P, Pq = q, hence P k q = P k −1 ( Pq) = P k −1q = P k −2 ( Pq) =

=q

and P k q = q for every positive integer k.

( ) ( )

True/False 4.12 (a) True; the sum of the entries is 1 and all entries are nonnegative. (b) True; the column vectors are probability vectors 2

⎡0.2 1⎤ ⎡0.84 0.2 ⎤ . =⎢ and ⎢ ⎥ 0 . 8 0 ⎣ ⎦ ⎣0.16 0.8⎥⎦

⎡ 0.2893⎤ ⎡35⎤ (c) 120q ≈ 120 ⎢ 0.4151⎥ ≈ ⎢50⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0.2956 ⎦ ⎣35⎦ Over the long term, the probability that a car ends up at location 1 is about 0.2893, location 2 is about 0.4151, and location 3 is about 0.2956. Thus the vector 120q gives the long term distribution of 120 cars. The locations should have 35, 50, and 35 parking spaces, respectively.

(c) True; this is part of the definition of a transition matrix. (d) False; the steady-state vector is a solution which is also a probability vector. (e) True Chapter 4 Supplementary Exercises

7 1 1 19. Column 1: 1 − − = 10 10 5 3 3 1 Column 2: 1 − − = 10 5 10

1. (a) u + v = (3 + 1, −2 + 5, 4 + (−2)) = (4, 3, 2) (−1)u = ((−1)3, 0, 0) = ( −3, 0, 0)

117



⎡ 1 0 1⎤ ⎡ 1 0 1⎤ 9. (a) A = ⎢ 0 1 0⎥ reduces to ⎢0 1 0 ⎥ , so ⎢ ⎥ ⎢ ⎥ ⎣ 1 0 1⎦ ⎣0 0 0 ⎦ rank(A) = 2 and nullity(A) = 1.

(b) V is closed under addition because addition in V is component-wise addition of real numbers and the real numbers are closed under addition. Similarly, V is closed under scalar multiplication because the real numbers are closed under multiplication.

⎡1 ⎢0 (b) A = ⎢ ⎢1 ⎣⎢ 0

(c) Axioms 1−5 hold in V because they hold in R3 .

(d) Axiom 7: k (u + v) = k (u1 + v1 , u2 + v2 , u3 + v3 ) = (k (u1 + v1 ), 0, 0) = (ku1, 0, 0) + (kv1 , 0, 0) = ku + kv Axiom 8: (k + m)u = ((k + m)u1 , 0, 0) = (ku1 , 0, 0) + (mu1 , 0, 0) = ku + mu Axiom 9: k (mu) = k (mu1 , 0, 0) = (kmu1 , 0, 0) = (km)u

0 1 0 1

1 0 1 0

0⎤ 1⎥ ⎥ reduces to 0⎥ 1⎦⎥

⎡ 1 0 1 0⎤ ⎢0 1 0 1⎥ ⎢ ⎥ , so rank(A) = 2 and ⎢0 0 0 0 ⎥ ⎣⎢0 0 0 0 ⎦⎥ nullity(A) = 2.

(c) The row vectors ri will satisfy ⎧r if i is odd . Thus, the reduced row ri = ⎨ 1 ⎩r2 if i is even echelon form will have 2 nonzero rows. Its rank will be 2 and its nullity will be n − 2. 11. (a) If p1 and p2 are in the set, then

(e) For any u = (u1 , u2 , u3 ) with u2 , u3 ≠ 0,

( p1 + p2 )(− x) = p1 (− x) + p2 (− x) = p1 ( x) + p2 ( x) = ( p1 + p2 )( x) so the set is closed under addition. Also (kp1 )(− x) = k ( p1 (− x)) = k ( p1 ( x)) = (kp1 )( x) so kp1 is in the set for any scalar k, and the set is a subspace of Pn , the subspace consisting of all even polynomials. A basis

1u = (1u1 , 0, 0) ≠ (u1, u2 , u3 ) = u. ⎡1 1 s⎤ 3. The coefficient matrix is A = ⎢ 1 s 1⎥ and ⎢ ⎥ ⎣ s 1 1⎦

det( A) = −( s − 1) 2 ( s + 2), so for s ≠ 1, −2, A is invertible and the solution space is the origin. ⎡1 1 1⎤ If s = 1, then A = ⎢1 1 1⎥ , which reduced to ⎢ ⎥ ⎣1 1 1⎦ ⎡ 1 1 1⎤ ⎢0 0 0 ⎥ so the solution space will require 2 ⎢ ⎥ ⎣0 0 0 ⎦ parameters, i.e., it is a plane through the origin. ⎡ 1 1 −2 ⎤ If s = −2, then A = ⎢ 1 −2 1⎥ , which ⎢ ⎥ 1 1⎦ ⎣ −2 ⎡ 1 0 −1⎤ reduces to ⎢0 1 −1⎥ , so the solution space ⎢ ⎥ ⎣0 0 0 ⎦ will require 1 parameter, i.e., it will be a line through the origin.

is {1, x 2 , x 4 , … , x 2m } where 2m = n if n is even and 2m = n − 1 if n is odd. (b) If p1 and p2 are in the set, then ( p1 + p2 )(0) = p1 (0) + p2 (0) = 0, so

p1 + p2 is in the set. Also, (kp1 )(0) = k ( p1 (0)) = k (0) = 0, so kp1 is in the set. Thus, the set is a subspace of Pn , the subspace consisting of polynomials with constant term 0. A basis is {x, x 2 , x3 , … , x n }.

7. A must be an invertible matrix for Av1 , Av 2 , … , Av n to be linearly independent. 118



13. (a) The entry aij , i > j, below the main diagonal is determined by the a ji entry above the main diagonal. A

⎧ ⎡ 1 0 0⎤ ⎡ 0 1 0 ⎤ ⎡ 0 0 1⎤ ⎡0 0 0 ⎤ ⎡0 0 0 ⎤ ⎪ basis is ⎨ ⎢ 0 0 0⎥ , ⎢ 1 0 0 ⎥ , ⎢ 0 0 0 ⎥ , ⎢0 1 0 ⎥ , ⎢0 0 1⎥ , ⎪⎩ ⎢⎣ 0 0 0⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 1 0 0 ⎥⎦ ⎢⎣0 0 0 ⎥⎦ ⎢⎣0 1 0 ⎥⎦

⎡0 0 0 ⎤ ⎫ ⎢0 0 0 ⎥ ⎪⎬ ⎢ ⎥ ⎣0 0 1⎦ ⎪⎭

⎧0 if i = j (b) In a skew-symmetric matrix, aij = ⎨ . Thus, the entries below the main diagonal are determined ⎩−a ji if i ≠ j ⎧ ⎡ 0 1 0 ⎤ ⎡ 0 0 1⎤ ⎡0 0 0 ⎤ ⎫ ⎪ ⎪ by the entries above the main diagonal. A basis is ⎨ ⎢ −1 0 0 ⎥ , ⎢ 0 0 0 ⎥ , ⎢0 0 1⎥ ⎬ . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎩⎪ ⎣ 0 0 0 ⎦ ⎣ −1 0 0 ⎦ ⎣0 −1 0 ⎦ ⎭⎪ 15. Every 3 × 3, 4 × 4, and 5 × 5 submatrix will have determinant 0, so the rank can be at most 2. For i < 5 and j < 6, ⎡ 0 ai 6 ⎤ det ⎢ ⎥ = −ai 6 a5 j so the rank is 2 if any of these products is nonzero, i.e., if the matrix has 2 nonzero ⎣ a5 j a56 ⎦

entries other than a56 . If exactly one entry in the last column is nonzero, the matrix will have rank 1, while if all entries are zero, the rank will be 0. Thus, the possible ranks are 0, 1, or 2.

119

Chapter 5 Eigenvalues and Eigenvectors 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡λ − 3 5. (a) ⎢ = − λ + 1⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣0 ⎥⎦ 8 ⎣


⎡ 0 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . λ = 3 gives ⎢ ⎢ ⎥= ⎣ −8 4 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦

⎡ 4 0 1⎤ ⎡1 ⎤ ⎡ 5⎤ 1. Ax = ⎢ 2 3 2 ⎥ ⎢ 2⎥ = ⎢10 ⎥ = 5x ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 1 0 4 ⎦ ⎣ 1 ⎦ ⎣ 5⎦ x corresponds to the eigenvalue λ = 5.

⎡ 0 0⎤ reduces to Since ⎢ ⎣ −8 4 ⎥⎦

⎡1 − 1⎤ 2 ⎥ , the ⎢ 0⎦ ⎣0

1 s, x2 = s, or 2 ⎡1⎤ ⎡1⎤ ⎡ x1 ⎤ ⎡ 12 s ⎤ 2 2 ⎢ x ⎥ = ⎢ ⎥ = s ⎢ ⎥ , so ⎢ ⎥ is a basis for ⎣ 2⎦ ⎣ s ⎦ ⎣1⎦ ⎣1 ⎦ the eigenspace corresponding to λ = 3. ⎡ −4 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . Since λ = −1 gives ⎢ ⎢ ⎥= ⎣ −8 0 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ general solution is x1 =

0 ⎤ ⎡λ − 3 = (λ − 3)(λ + 1) 3. (a) det ⎢ ⎣ −8 λ + 1⎥⎦ = λ 2 − 2λ − 3 The characteristic equation is λ 2 − 2λ − 3 = 0. 9 ⎤ ⎡ λ − 10 (b) det ⎢ = (λ − 10)(λ + 2) + 36 4 − λ + 2 ⎥⎦ ⎣ = λ 2 − 8λ + 16 The characteristic equation is

⎡1 0 ⎤ ⎡ −4 0 ⎤ ⎢⎣ −8 0 ⎥⎦ reduces to ⎢⎣0 0 ⎥⎦ , the general solution is x1 = 0, x2 = s, or

λ 2 − 8λ + 16 = 0.

⎡ x1 ⎤ ⎡0 ⎤ ⎡0 ⎤ ⎡0 ⎤ ⎢ ⎥ = ⎢ ⎥ = s ⎢1 ⎥ , so ⎢1 ⎥ is a basis for the ⎣ ⎦ ⎣ ⎦ ⎣ x2 ⎦ ⎣ s ⎦ eigenspace corresponding to λ = −1.

⎡ λ −3⎤ (c) det ⎢ = λ 2 − 12 ⎥ − 4 λ ⎣ ⎦ The characteristic equation is λ 2 − 12 = 0.

9 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡λ − 10 (b) ⎢ = − λ + 2 ⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣0 ⎥⎦ 4 ⎣

7 ⎤ ⎡λ + 2 (d) det ⎢ = (λ + 2)(λ − 2) + 7 ⎣ −1 λ − 2 ⎥⎦ = λ2 + 3

⎡ −6 9 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . λ = 4 gives ⎢ ⎢ ⎥= ⎣ −4 6 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ ⎡ −6 9 ⎤ reduces to Since ⎢ ⎣ −4 6 ⎥⎦

The characteristic equation is λ 2 + 3 = 0. ⎡λ 0 ⎤ (e) det ⎢ = λ2 ⎣ 0 λ ⎥⎦

⎡1 − 3⎤ 2 ⎥ , the ⎢ 0⎦ ⎣0

3 s, x2 = s, or 2 ⎡3⎤ ⎡3⎤ ⎡ x1 ⎤ ⎡ 32 s ⎤ 2 2 ⎢ x ⎥ = ⎢ ⎥ = s ⎢ ⎥ , so ⎢ ⎥ is a basis for ⎣ 2⎦ ⎣ s ⎦ ⎣1⎦ ⎣1 ⎦ the eigenspace corresponding to λ = 4.

general solution is x1 =

The characteristic equation is λ 2 = 0. ⎡λ − 1 0 ⎤ (f) det ⎢ = (λ − 1)2 = λ 2 − 2λ + 1 λ − 1⎥⎦ ⎣ 0 The characteristic equation is

⎡ λ −3⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ (c) ⎢ ⎢ ⎥= ⎣ −4 λ ⎥⎦ ⎣ x2 ⎦ ⎢⎣ 0 ⎥⎦

λ 2 − 2λ + 1 = 0.

⎡ 12 λ = 12 gives ⎢ ⎣ −4

120

−3 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . ⎥⎢ ⎥ = 12 ⎦ ⎣ x2 ⎦ ⎣⎢0 ⎦⎥


Section 5.1

⎡1 − 3 ⎤ −3 ⎤ 12 ⎥ , reduces to ⎢ ⎥ 12 ⎦ ⎢⎣0 0 ⎥⎦ 3 the general solution is x1 = s, x2 = s, 12

7. (a) The eigenvalues satisfy

⎡ 12 Since ⎢ ⎣ −4

λ3 − 6λ 2 + 11λ − 6 = 0 or (λ − 1)(λ − 2)(λ − 3) = 0, so the eigenvalues are λ = 1, 2, 3.

(b) The eigenvalues satisfy λ3 − 2λ = 0 or

3 s⎤ ⎡ 3 ⎤ ⎡ 3 ⎤ ⎡x ⎤ ⎡ or ⎢ 1 ⎥ = ⎢ 12 ⎥ = s ⎢ 12 ⎥ , so ⎢ 12 ⎥ is a ⎣ x2 ⎦ ⎢⎣ s ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ basis for the eigenspace corresponding to λ = 12.

(

)

are λ = 0, − 2 , 2 .

(c) The eigenvalues satisfy λ3 + 8λ 2 + λ + 8 = 0

λ = − 12 gives ⎡ − 12 −3 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ . ⎢ ⎥⎢ ⎥ = − 12 ⎦ ⎣ x2 ⎦ ⎢⎣ 0 ⎥⎦ ⎣ −4 ⎡ − 12 −3 ⎤ Since ⎢ ⎥ reduces to − 12 ⎦ ⎣ −4 ⎡1 ⎢ ⎢⎣0

)(

λ λ + 2 λ − 2 = 0, so the eigenvalues

or (λ + 8)(λ 2 + 1) = 0, so the only (real) eigenvalue is λ = −8. (d) The eigenvalues satisfy λ3 − λ 2 − λ − 2 = 0

or (λ − 2)(λ 2 + λ + 1) = 0, so the only (real eigenvalue is λ = 2.

3 ⎤ 12 ⎥ ,

the general solution is 0 ⎥⎦ 3 x1 = − s, x2 = s, or 12

(e) The eigenvalues satisfy

λ3 − 6λ 2 + 12λ − 8 = 0 or (λ − 2)3 = 0, so the only eigenvalue is λ = 2.

3 s⎤ ⎡− 3 ⎤ ⎡− 3 ⎤ ⎡ x1 ⎤ ⎡ − 12 ⎥ = s ⎢ 12 ⎥ , so ⎢ 12 ⎥ is = ⎢ ⎢x ⎥ ⎣ 2 ⎦ ⎣⎢ s ⎦⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ a basis for the eigenspace corresponding to λ = − 12.

(f) The eigenvalues satisfy

λ3 − 2λ 2 − 15λ + 36 = 0 or (λ + 4)(λ − 3)2 = 0, so the eigenvalues are λ = −4, 3.

(d) There are no eigenspaces since there are no real eigenvalues.

0 ⎤ −2 ⎡λ 0 ⎢ −1 λ −1 0 ⎥ 9. (a) det ⎢ ⎥ − λ + 0 1 2 0 ⎥ ⎢ λ − 1⎦⎥ 0 ⎣⎢ 0 0

⎡λ 0⎤ ⎡ x1 ⎤ ⎡ 0⎤ (e) ⎢ ⎢ ⎥= ⎣ 0 λ ⎥⎦ ⎣ x2 ⎦ ⎢⎣ 0⎥⎦

−2 ⎤ ⎡λ 0 = (λ − 1) det ⎢ −1 λ −1 ⎥ ⎢ ⎥ ⎣ 0 −1 λ + 2⎦ = (λ − 1){λ[λ(λ + 2) − 1] − 2(1 − 0)}

⎡0 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . λ = 0 gives ⎢ ⎢ ⎥= ⎣0 0 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ Clearly, every vector in R 2 is a solution, so ⎡1 ⎤ ⎡ 0 ⎤ ⎢⎣0 ⎥⎦ , ⎢⎣1 ⎥⎦ is a basis for the eigenspace corresponding to λ = 0.

= (λ − 1)(λ3 + 2λ 2 − λ − 2) = λ 4 + λ3 − 3λ 2 − λ + 2 The characteristic equation is λ 4 + λ3 − 3λ 2 − λ + 2 = 0.

⎡λ − 1 0 ⎤ ⎡ x1 ⎤ ⎡ 0⎤ (f) ⎢ = λ − 1⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣ 0⎥⎦ ⎣ 0 ⎡0 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . λ = 1 gives ⎢ ⎢ ⎥= ⎣0 0 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ Clearly, every vector in R 2 is a solution, so ⎡1 ⎤ ⎡ 0 ⎤ ⎢⎣0 ⎥⎦ , ⎢⎣1 ⎥⎦ is a basis for the eigenspace corresponding to λ = 1. 121

Chapter 5: Eigenvalues and Eigenvectors


9 0 0 ⎤ ⎡ λ − 10 0 0 ⎤ 0 0 ⎤ ⎡λ + 2 ⎡ −4 ⎢ −4 λ+2 0 0 ⎥ ⎢ ⎥ ⎢ = λ − − 10 9 (b) det ⎢ ( ) det det λ+2 0 7 0 λ+2 7 ⎥ ⎥ 0 7 ⎥ λ+2 ⎢ ⎥ ⎢ ⎥ ⎢ 0 −1 λ − 2 ⎦ −1 λ − 2 ⎦ ⎣ 0 ⎣0 0 −1 λ − 2 ⎥⎦ ⎢⎣ 0 = (λ − 10)(λ + 2)[(λ + 2)(λ − 2) + 7] − 9(−4)[(λ + 2)(λ − 2) + 7] = (λ 2 − 8λ − 20 + 36)(λ 2 − 4 + 7) = (λ 2 − 8λ + 16)(λ 2 + 3) = λ 4 − 8λ3 + 19λ 2 − 24λ + 48 The characteristic equation is λ 4 − 8λ3 + 19λ 2 − 24λ + 48 = 0. 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ −2 ⎡λ 0 ⎢ −1 λ 0 ⎥ ⎢ x2 ⎥ ⎢0 ⎥ −1 11. (a) ⎢ ⎥⎢ ⎥ = ⎢ ⎥ 0 ⎥ ⎢ x3 ⎥ ⎢0 ⎥ ⎢ 0 −1 λ + 2 0 λ − 1⎥⎦ ⎢⎣ x4 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 0 0 ⎡ 1 0 −2 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ −1 1 −1 0 ⎥ ⎢ x2 ⎥ ⎢0 ⎥ λ = 1 gives ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥. ⎢ 0 −1 3 0 ⎥ ⎢ x3 ⎥ ⎢0 ⎥ ⎣⎢ 0 0 0 0 ⎥⎦ ⎢⎣ x4 ⎥⎦ ⎣⎢0 ⎦⎥ ⎡ 1 0 −2 0 ⎤ ⎢ −1 1 −1 0 ⎥ Since ⎢ ⎥ reduces to ⎢ 0 −1 3 0 ⎥ ⎣⎢ 0 0 0 0 ⎥⎦

⎡1 ⎢0 ⎢ ⎢0 ⎣⎢0

0 −2 0 ⎤ 1 −3 0 ⎥ ⎥ , the general solution is x1 = 2 s, x2 = 3s, x3 = s, 0 0 0⎥ 0 0 0 ⎦⎥

⎡ x1 ⎤ ⎡ 2 s ⎤ ⎡ 2 ⎤ ⎡ 0⎤ ⎡2⎤ ⎢ x ⎥ ⎢ 3s ⎥ ⎢3⎥ ⎢ ⎥ ⎢ ⎥ 3 0 x4 = t or ⎢ 2 ⎥ = ⎢ ⎥ = s ⎢ ⎥ + t ⎢ ⎥ , so ⎢ ⎥ , x s 1 0 ⎢ 3⎥ ⎢ ⎥ ⎢1 ⎥ ⎢ ⎥ ⎢ ⎥ 0 1 ⎢⎣ x4 ⎥⎦ ⎣⎢ t ⎦⎥ ⎢⎣ 0 ⎥⎦ ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥

⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ is a basis for the eigenspace corresponding to λ = 1. ⎢0 ⎥ ⎣⎢1 ⎦⎥

⎡ −2 0 −2 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ −1 −2 −1 0 ⎥ ⎢ x2 ⎥ ⎢0 ⎥ λ = −2 gives ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥. ⎢ 0 −1 0 0 ⎥ ⎢ x3 ⎥ ⎢0 ⎥ ⎣⎢ 0 0 0 −3⎦⎥ ⎢⎣ x4 ⎥⎦ ⎣⎢0 ⎦⎥ ⎡ −2 0 −2 0 ⎤ ⎢ −1 −2 −1 0 ⎥ Since ⎢ ⎥ reduces to ⎢ 0 −1 0 0 ⎥ ⎣⎢ 0 0 0 −3⎦⎥

⎡1 ⎢0 ⎢ ⎢0 ⎣⎢0

0 1 0 0

1 0 0 0

0⎤ 0⎥ ⎥ , the general solution is x1 = − s, x2 = 0, x3 = s, 1⎥ 0 ⎦⎥

⎡ x1 ⎤ ⎡ − s ⎤ ⎡ −1⎤ ⎡ −1⎤ ⎢ x ⎥ ⎢ 0⎥ ⎢ ⎥ ⎢ 0⎥ 0 x4 = 0 or ⎢ 2 ⎥ = ⎢ ⎥ = s ⎢ ⎥ , so ⎢ ⎥ is a basis for the eigenspace corresponding to λ = −2. ⎢ x3 ⎥ ⎢ s ⎥ ⎢ 1⎥ ⎢ 1⎥ ⎢⎣ x4 ⎥⎦ ⎣⎢ 0 ⎦⎥ ⎣⎢ 0⎦⎥ ⎣⎢ 0 ⎦⎥ ⎡ −1 0 −2 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ −1 −1 −1 0 ⎥ ⎢ x2 ⎥ ⎢0 ⎥ λ = −1 gives ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥. ⎢ 0 −1 1 0 ⎥ ⎢ x3 ⎥ ⎢0 ⎥ ⎢⎣ 0 0 0 −2 ⎥⎦ ⎢⎣ x4 ⎥⎦ ⎢⎣0 ⎥⎦

122


Section 5.1

⎡ −1 0 −2 0 ⎤ ⎢ −1 −1 −1 0 ⎥ Since ⎢ ⎥ reduces to ⎢ 0 −1 1 0 ⎥ ⎢⎣ 0 0 0 −2 ⎥⎦

15. If a line is invariant under A, then it can be expressed in terms of an eigenvector of A. λ−4 1 −2 λ − 1 = (λ − 4)(λ − 1) + 2

(a) det(λI − A) =

⎡ 1 0 2 0⎤ ⎢ 0 1 −1 0 ⎥ ⎢ ⎥ , the general solution is ⎢0 0 0 1⎥ ⎢⎣0 0 0 0 ⎥⎦ x1 = −2s, x2 = s, x3 = s, x4 = 0 or

= λ 2 − 5λ + 6 = (λ − 3)(λ − 2) 1 ⎤ ⎡ x ⎤ ⎡0⎤ ⎡λ − 4 ⎢⎣ −2 λ − 1⎥⎦ ⎢⎣ y ⎥⎦ = ⎢⎣0 ⎥⎦ ⎡ −1 1⎤ ⎡ x ⎤ ⎡ 0⎤ λ = 3 gives ⎢ . = ⎣ −2 2 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ 0⎥⎦ ⎡ −1 1⎤ ⎡ 1 −1⎤ ⎢⎣ −2 2 ⎥⎦ reduces to ⎢⎣0 0⎥⎦ , so a general solution is x = y. The line y = x is invariant ⎡ −2 1⎤ ⎡ x ⎤ ⎡ 0 ⎤ = . under A. λ = 2 gives ⎢ ⎣ −2 1⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡ −2 1⎤ ⎡ 2 −1⎤ ⎢⎣ −2 1⎥⎦ can be reduced to ⎢⎣ 0 0 ⎥⎦ , so a general solution is 2x = y. The line y = 2x is invariant under A.

⎡ x1 ⎤ ⎡ −2s ⎤ ⎡ −2 ⎤ ⎡ −2 ⎤ ⎢x ⎥ ⎢ s⎥ ⎢ ⎥ ⎢ 1⎥ 1 ⎢ 2⎥ = ⎢ ⎥ = s ⎢ ⎥ , so ⎢ ⎥ is a basis ⎢ x3 ⎥ ⎢ s ⎥ ⎢ 1⎥ ⎢ 1⎥ ⎢⎣ x4 ⎥⎦ ⎣⎢ 0 ⎦⎥ ⎣⎢ 0 ⎦⎥ ⎣⎢ 0⎦⎥ for the eigenspace corresponding to λ = −1. 9 0 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡λ − 10 ⎢ −4 0 0 ⎥ ⎢ x2 ⎥ ⎢0 ⎥ λ+2 (b) ⎢ ⎥⎢ ⎥ = ⎢ ⎥ 0 7 ⎥ ⎢ x3 ⎥ ⎢0 ⎥ λ+2 ⎢ 0 0 −1 λ − 2 ⎦⎥ ⎢⎣ x4 ⎥⎦ ⎣⎢0 ⎦⎥ ⎣⎢ 0 ⎡ −6 ⎢ −4 λ = 4 gives ⎢ ⎢ 0 ⎣⎢ 0 ⎡ −6 ⎢ −4 Since ⎢ ⎢ 0 ⎣⎢ 0

⎡1 − 3 2 ⎢ 0 ⎢0 ⎢0 0 ⎢0 0 ⎣

9 0 0 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ 6 0 0 ⎥ ⎢ x2 ⎥ ⎢ 0 ⎥ ⎥ ⎢ ⎥ = ⎢ ⎥. 0 6 −7 ⎥ ⎢ x3 ⎥ ⎢ 0 ⎥ 0 −1 2 ⎦⎥ ⎣⎢ x4 ⎦⎥ ⎣⎢ 0 ⎦⎥

(b) det(λI − A) =

9 0 0⎤ 6 0 0⎥ ⎥ reduces to 0 6 −7 ⎥ 0 −1 2 ⎦⎥

Since the characteristic equation λ 2 + 1 = 0 has no real solutions, there are no lines that are invariant under A.

0 0⎤ ⎥ 1 0 ⎥ , the general solution is 0 1⎥ 0 0 ⎥⎦

(c) det(λI − A) =

λ − 2 −3 = (λ − 2)2 0 λ−2

⎡ λ − 2 −3 ⎤ ⎡ x ⎤ ⎡ 0 ⎤ = ⎢⎣ 0 λ − 2 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡0 −3⎤ ⎡ x ⎤ ⎡ 0 ⎤ λ = 2 gives ⎢ . = ⎣0 0 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡0 −3⎤ ⎡0 1⎤ Since ⎢ reduces to ⎢ , the ⎥ ⎣0 0 ⎥⎦ ⎣0 0⎦ general solution is y = 0, and the line y = 0 is invariant under A.

3 s, x2 = s, x3 = 0, x4 = 0 or 2 ⎡3⎤ ⎡3⎤ ⎡ x1 ⎤ ⎡ 3 s ⎤ ⎢2⎥ ⎢2⎥ ⎢x ⎥ ⎢ 2 ⎥ 2 s 1 ⎢ ⎥ ⎢ ⎥ , so ⎢ 1 ⎥ is a basis for ⎢ ⎥= =s ⎢0⎥ ⎢0⎥ ⎢ x3 ⎥ ⎢ 0 ⎥ ⎢0⎥ ⎢0⎥ ⎣⎢ x4 ⎦⎥ ⎢⎣ 0 ⎥⎦ ⎣ ⎦ ⎣ ⎦ x1 =

23. We have that Ax = λx.

the eigenspace corresponding to λ = 4.

A−1 ( Ax) = Ix = x, but also

13. Since A is upper triangular, the eigenvalues of A 1 are the diagonal entries: 1, , 0, 2. Thus the 2

A−1 ( Ax) = A−1 (λx) = λ( A−1x) so A−1x = 1 is an eigenvalue of A−1 , with λ corresponding eigenvector x.

9 1 ⎛1⎞ , eigenvalues of A are 1 = 1, ⎜ ⎟ = 512 ⎝2⎠ 9

λ −1 = λ2 + 1 1 λ

Thus

9

09 = 0, and 29 = 512. 123

1 x. λ



25. We have that Ax = λx. Since s is a scalar, (sA)x = s(Ax) = s(λx) = (sλ)x so sλ is an eigenvalue of sA with corresponding eigenvector x.

5. Since the geometric multiplicity of an eigenvalue is less than or equal to its algebraic multiplicity, the eigenspace for λ = 0 can have dimension 1 or 2, the eigenspace for λ = 1 has dimension 1, and the eigenspace for λ = 2 can have dimension 1, 2, or 3.

True/False 5.1 (a) False; the vector x must be nonzero to be an eigenvector⎯if x = 0, then Ax = λx for all λ.

7. Since the matrix is lower triangular, with 2s on the diagonal, the only eigenvalue is λ = 2. ⎡ 0 0⎤ For λ = 2, 2 I − A = ⎢ , which has rank 1, ⎣ −1 0 ⎥⎦ hence nullity 1. Thus, the matrix has only 1 distinct eigenvector and is not diagonalizable.

(b) False; if λ is an eigenvalue of A, then det(λI − A) = 0, so (λI − A)x = 0 has nontrivial solutions. (c) True; since p(0) = 02 + 1 ≠ 0, λ = 0 is not an eigenvalue of A and A is invertible by Theorem 5.1.5.

9. Since the matrix is lower triangular with diagonal entries of 3, 2, and 2, the eigenvalues are 3 and 2, where 2 has algebraic multiplicity 2. ⎡0 0 0 ⎤ For λ = 3, 3I − A = ⎢0 1 0 ⎥ which has rank ⎢ ⎥ ⎣0 −1 1⎦ 2, hence nullity 1, so the eigenspace corresponding to λ = 3 has dimension 1. ⎡ −1 0 0 ⎤ For λ = 2, 2 I − A = ⎢ 0 0 0 ⎥ which has rank ⎢ ⎥ ⎣ 0 −1 0 ⎦ 2, hence nullity 1, so the eigenspace corresponding to λ = 2 has dimension 1. The matrix is not diagonalizable, since it has only 2 distinct eigenvectors.

(d) False; the eigenspace corresponding to λ contains the vector x = 0, which is not an eigenvector of A. (e) True; if 0 is an eigenvalue of A, then 02 = 0 is an eigenvalue of A2 , so A2 is singular. (f) False; the reduced row echelon form of an invertible n × n matrix A is I n which has only one eigenvalue, λ = 1, whereas A can have eigenvalues that are not 1. For instance, the ⎡3 0 ⎤ matrix A = ⎢ has the eigenvalues λ = 3 ⎣8 −1⎥⎦ and λ = −1.

11. Since the matrix is upper triangular with diagonal entries 2, 2, 3, and 3, the eigenvalues are 2 and 3 each with algebraic multiplicity 2. ⎡0 1 0 −1⎤ ⎢0 0 −1 1⎥ For λ = 2, 2 I − A = ⎢ ⎥ which has ⎢ 0 0 −1 −2 ⎥ ⎣⎢0 0 0 −1⎦⎥ rank 3, hence nullity 1, so the eigenspace corresponding to λ = 2 has dimension 1. ⎡ 1 1 0 −1⎤ ⎢0 1 −1 1⎥ For λ = 3, 3I − A = ⎢ ⎥ which has ⎢ 0 0 0 −2 ⎥ ⎢⎣0 0 0 0 ⎥⎦ rank 3, hence nullity 1, so the eigenspace corresponding to λ = 3 has dimension 1. The matrix is not diagonalizable since it has only 2 distinct eigenvectors. Note that showing that the geometric multiplicity of either eigenvalue is less than its algebraic multiplicity is sufficient to show that the matrix is not diagonalizable.

(g) False; if 0 is an eigenvalue of A, then A is singular so the set of column vectors of A cannot be linearly independent. Section 5.2 Exercise Set 5.2 1. Since the determinant is a similarity invariant and det(A) = 2 − 3 = −1 while det(B) = −2 − 0 = −2, A and B are not similar matrices. ⎡ 1 0 0⎤ 3. A reduces to ⎢0 1 0 ⎥ while B reduces to ⎢ ⎥ ⎣0 0 1⎦ ⎡ 1 2 0⎤ ⎢0 0 1⎥ , so they have different ranks. Since ⎢ ⎥ ⎣0 0 0 ⎦ rank is a similarity invariant, A and B are not similar matrices. 124


Section 5.2

13. A has eigenvalues λ1 = 1 and λ 2 = −1 (the diagonal entries, since A is triangular). ⎡λ − 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . (λI − A)x = 0 is ⎢ ⎢ ⎥= ⎣ −6 λ + 1⎦⎥ ⎣ x2 ⎦ ⎣⎢0 ⎦⎥

⎡0 0 2 ⎤ ⎡0 1 0 ⎤ Since ⎢0 −1 0 ⎥ reduces to ⎢0 0 1⎥ , the ⎢ ⎥ ⎢ ⎥ ⎣0 0 −1⎦ ⎣0 0 0 ⎦ general solution is x1 = s, x2 = 0, x3 = 0 or ⎡ x1 ⎤ ⎡1 ⎤ ⎡1 ⎤ ⎢ x ⎥ = s ⎢0 ⎥ and p = ⎢0 ⎥ is a basis for the 1 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣0 ⎦ ⎣ x3 ⎦ eigenspace corresponding to λ1 = 2.

⎡ 0 0⎤ ⎡ x1 ⎤ ⎡ 0⎤ λ1 = 1 gives ⎢ . ⎢ ⎥= ⎣ −6 2⎥⎦ ⎣ x2 ⎦ ⎢⎣ 0⎥⎦ ⎡1 − 1⎤ 3 ⎥ , the general ⎢ 0 0 ⎣ ⎦ ⎡1 ⎤ x 1 ⎡ ⎤ solution is x1 = s, x2 = s, or ⎢ 1 ⎥ = s ⎢ 3 ⎥ , 3 ⎣ x2 ⎦ ⎣1 ⎦ ⎡ 0 0⎤ Since ⎢ reduces to ⎣ −6 2 ⎥⎦

⎡ 1 0 2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ 2 = 3 gives ⎢0 0 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ which has ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 0 ⎦ ⎣ x3 ⎦ ⎣0 ⎦ general solution x1 = −2s, x2 = t , x3 = s, or

⎡1⎤ and p1 = ⎢ 3 ⎥ is a basis for the eigenspace ⎣1⎦ corresponding to λ1 = 1.

⎡ x1 ⎤ ⎡0 ⎤ ⎡ −2 ⎤ ⎡ −2 ⎤ ⎡ 0 ⎤ ⎢ x ⎥ = s ⎢ 0 ⎥ + t ⎢1 ⎥ , and p = ⎢ 0 ⎥ , p = ⎢1 ⎥ 2 3 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1⎦ ⎣ 0⎦ ⎣0 ⎦ ⎣ 1⎦ ⎣ x3 ⎦ is a basis for the eigenspace corresponding to ⎡ 1 −2 0 ⎤ λ 2 = 3. Thus P = [p1 p 2 p3 ] = ⎢0 0 1⎥ ⎢ ⎥ 1 0⎦ ⎣0 diagonalizes A. ⎡ 1 0 2 ⎤ ⎡ 2 0 −2 ⎤ ⎡ 1 −2 0 ⎤ P −1 AP = ⎢0 0 1⎥ ⎢ 0 3 0⎥ ⎢ 0 0 1⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ 3⎦ ⎣ 0 1 0⎦ ⎣0 1 0 ⎦ ⎣ 0 0 0 ⎤ ⎡λ1 0 = ⎢0 λ 2 0 ⎥ ⎢ ⎥ λ2 ⎦ ⎣0 0 ⎡2 0 0⎤ = ⎢0 3 0⎥ ⎢ ⎥ ⎣ 0 0 3⎦

⎡ −2 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . λ 2 = −1 gives ⎢ ⎢ ⎥= ⎣ −6 0 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ ⎡ −2 0 ⎤ ⎡ 1 0⎤ Since ⎢ reduces to ⎢ , the general ⎥ − 6 0 ⎣0 0 ⎥⎦ ⎣ ⎦ ⎡x ⎤ ⎡0 ⎤ solution is x1 = 0, x2 = s or ⎢ 1 ⎥ = s ⎢ ⎥ , and ⎣1 ⎦ ⎣ x2 ⎦ ⎡0 ⎤ p 2 = ⎢ ⎥ is a basis for the eigenspace ⎣1 ⎦ corresponding to λ 2 = −1. ⎡ 1 0⎤ Thus, P = [p1 p 2 ] = ⎢ 3 ⎥ diagonalizes A. ⎣ 1 1⎦ ⎡ 3 0 ⎤ ⎡ 1 0 ⎤ ⎡ 13 0 ⎤ P −1 AP = ⎢ ⎢ ⎥ ⎣ −3 1⎥⎦ ⎢⎣ 6 −1⎥⎦ ⎣ 1 1⎦

0⎤ ⎡λ =⎢ 1 ⎥ 0 λ 2⎦ ⎣ ⎡ 1 0⎤ =⎢ ⎣0 −1⎥⎦

17. det(λI − A) =

λ + 1 −4 2 3 λ−4 0 3 −1 λ − 3

= x3 − 6 x 2 + 11x − 6 = ( x − 1)( x − 2)( x − 3) The eigenvalues are λ1 = 1, λ 2 = 2, and λ3 = 3, each with algebraic multiplicity 1. Since each eigenvalue must have nonzero geometric multiplicity, the geometric multiplicity of each eigenvalue is also 1 and A is diagonalizable. (λI − A)x = 0 is 2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡ λ + 1 −4 ⎢ 3 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . λ−4 ⎢ ⎥⎢ ⎥ ⎢ ⎥ −1 λ − 3⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎣ 3

15. A has eigenvalues λ1 = 2 and λ 2 = 3 (the diagonal entries, since A is triangular). (λI − A)x = 0 is 0 2 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ ⎡λ − 2 ⎢ 0 λ−3 0 ⎥ ⎢ x2 ⎥ = ⎢ 0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 λ − 3⎦ ⎣ x3 ⎦ ⎣ 0 ⎦ ⎣ 0 ⎡0 0 2⎤ ⎡ x1 ⎤ ⎡ 0⎤ λ1 = 2 gives ⎢0 −1 0⎥ ⎢ x2 ⎥ = ⎢ 0⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 0 −1⎦ ⎣ x3 ⎦ ⎣ 0⎦

125



⎡ 2 −4 2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ1 = 1 gives ⎢ 3 −3 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 3 −1 −2 ⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡ 2 −4 2 ⎤ ⎡ 1 0 −1⎤ Since ⎢ 3 −3 0 ⎥ reduces to ⎢0 1 −1⎥ , the ⎢ ⎥ ⎢ ⎥ ⎣ 3 −1 −2 ⎦ ⎣0 0 0 ⎦ general solution is x1 = s, x2 = s, x3 = s, or

diagonalizes A and 0 ⎤ ⎡ 1 0 0⎤ ⎡λ1 0 P −1 AP = ⎢0 λ 2 0 ⎥ = ⎢0 2 0 ⎥ . ⎢ ⎥ ⎢ ⎥ λ 3 ⎦ ⎣ 0 0 3⎦ ⎣0 0 19. Since A is triangular with diagonal entries of 0, 0, 1, the eigenvalues are λ1 = 0 (with algebraic

multiplicity 2) and λ 2 = 1. λ 2 = 1 must also have geometric multiplicity 1.

⎡ x1 ⎤ ⎡1⎤ ⎡1⎤ ⎢ x ⎥ = s ⎢1⎥ and p = ⎢1⎥ is a basis for the 2 1 ⎢ ⎥ ⎢⎥ ⎢⎥ ⎣1⎦ ⎣1⎦ ⎣ x3 ⎦ eigenspace corresponding to λ1 = 1.

0 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ ⎡λ 0 (λI − A)x = 0 is ⎢ 0 λ 0 ⎥ ⎢ x2 ⎥ = ⎢ 0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −3 0 λ − 1⎦ ⎣ x3 ⎦ ⎣ 0 ⎦ ⎡ 0 0 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ1 = 0 gives ⎢ 0 0 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −3 0 −1⎦ ⎣ x3 ⎦ ⎣0 ⎦

⎡3 −4 2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ 2 = 2 gives ⎢3 −2 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣3 −1 −1⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡ 3 −4 2 ⎤ Since ⎢3 −2 0⎥ reduces to ⎢ ⎥ ⎣3 −1 −1⎦

⎡1 0 − 2⎤ 3⎥ ⎢ , ⎢0 1 −1⎥ ⎢⎣0 0 0 ⎥⎦

⎡ 0 0 0⎤ Since ⎢ 0 0 0 ⎥ reduces to ⎢ ⎥ ⎣ −3 0 −1⎦

2 s, x2 = s, x3 = s 3 or x1 = 2t , x2 = 3t , x3 = 3t which is the general solution is x1 =

1 general solution is x1 = − r , x2 = t , x3 = r or 3 x1 = s, x2 = t , x3 = −3s which is

⎡ x1 ⎤ ⎡2⎤ ⎡ 2⎤ ⎢ x ⎥ = t ⎢ 3⎥ and p = ⎢ 3 ⎥ is a basis for the 2 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 3⎦ ⎣3⎦ ⎣ x3 ⎦ eigenspace corresponding to λ 2 = 2.

⎡ x1 ⎤ ⎡ 1⎤ ⎡0 ⎤ ⎢ x ⎥ = s ⎢ 0 ⎥ + t ⎢ 1⎥ . Thus λ = 0 has 1 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −3 ⎦ ⎣ 0 ⎦ ⎣ x3 ⎦ geometric multiplicity 2, and A is diagonalizable. ⎡ 1⎤ ⎡0 ⎤ p1 = ⎢ 0⎥ , p 2 = ⎢ 1⎥ are a basis for the ⎢ ⎥ ⎢ ⎥ ⎣ −3⎦ ⎣0 ⎦ eigenspace corresponding to λ1 = 0.

⎡ 4 −4 2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ3 = 3 gives ⎢ 3 −1 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 3 −1 0 ⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡1 0 − 1⎤ 4⎥ ⎢ ⎢0 1 − 34 ⎥ , a ⎢ ⎥ 0⎦ ⎣0 0 1 3 general solution is x1 = s, x2 = s, x3 = s or 4 4 ⎡ x1 ⎤ ⎡1 ⎤ x1 = t , x2 = 3t , x3 = 4t which is ⎢ x2 ⎥ = t ⎢ 3⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 4⎦ ⎣ x3 ⎦

⎡ 4 −4 2 ⎤ Since ⎢ 3 −1 0 ⎥ reduces to ⎢ ⎥ ⎣ 3 −1 0 ⎦

0 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ 1 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎥⎢ ⎥ ⎢ ⎥ 0 0 ⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡1 0 reduces to ⎢0 1 ⎢ ⎣0 0 general solution is x1 = 0, x2 = 0, x3 ⎡ 1 λ 2 = 1 gives ⎢ 0 ⎢ ⎣ −3 ⎡ 1 0 0⎤ Since ⎢ 0 1 0 ⎥ ⎢ ⎥ ⎣ −3 0 0 ⎦

0⎤ 0⎥ , a ⎥ 0⎦ = s or

⎡ x1 ⎤ ⎡0 ⎤ ⎡0 ⎤ ⎢ x ⎥ = s ⎢0 ⎥ and p = ⎢0 ⎥ is a basis for the 3 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1⎦ ⎣ 1⎦ ⎣ x3 ⎦ eigenspace corresponding to λ 2 = 1.

⎡1 ⎤ and p3 = ⎢ 3 ⎥ is a basis for the eigenspace ⎢ ⎥ ⎣4⎦ corresponding to λ3 = 3. Thus P = [p1 p 2

⎡1 0 1⎤ 3⎥ ⎢ , a ⎢0 0 0 ⎥ ⎣⎢0 0 0 ⎥⎦

⎡1 2 1⎤ p3 ] = ⎢1 3 3⎥ ⎢ ⎥ ⎣1 3 4 ⎦

Thus P = [p1 p 2

126

⎡ 1 0 0⎤ p3 ] = ⎢ 0 1 0 ⎥ ⎢ ⎥ ⎣ −3 0 1⎦


Section 5.2

diagonalizes A and ⎡λ1 0 0 ⎤ ⎡0 0 0 ⎤ P −1 AP = ⎢ 0 λ1 0 ⎥ = ⎢0 0 0 ⎥ . ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 λ 2 ⎦ ⎣0 0 1⎦

⎡ 1 0 0 0⎤ ⎢0 1 −1 1⎥ ⎢ ⎥ , a general solution is x1 = 0, ⎢0 0 0 0 ⎥ ⎢⎣0 0 0 0 ⎥⎦ x2 = s − t , x3 = s, x4 = t , or

21. Since A is triangular with entries −2, −2, 3, and 3 on the diagonal, the eigenvalues are λ1 = −2 and

⎡ x1 ⎤ ⎡0⎤ ⎡ 0⎤ ⎢x ⎥ ⎢1 ⎥ ⎢ −1⎥ 2 ⎢ ⎥ = s ⎢ ⎥ +t ⎢ ⎥. ⎢ x3 ⎥ ⎢1 ⎥ ⎢ 0 ⎥ ⎢⎣ x4 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1⎥⎦ Thus λ 2 = 3 has geometric multiplicity 2, and

λ 2 = 3, both with algebraic multiplicity 2. (λI − A)x = 0 is 0 0 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡λ + 2 ⎢ 0 λ + 2 −5 5 ⎥ ⎢ x2 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥. λ−3 0 0 ⎥ ⎢ x3 ⎥ ⎢0 ⎥ ⎢ 0 λ − 3⎥⎦ ⎢⎣ x4 ⎥⎦ ⎢⎣0 ⎥⎦ 0 0 ⎢⎣ 0 ⎡0 ⎢0 λ1 = −2 gives ⎢ ⎢0 ⎣⎢0 ⎡0 ⎢0 Since ⎢ ⎢0 ⎣⎢0 ⎡0 ⎢0 ⎢ ⎢0 ⎢⎣0

0 0 0 0

1 0 0 0

⎡0 ⎤ ⎡ 0⎤ ⎢1 ⎥ ⎢ −1⎥ p3 = ⎢ ⎥ , p 4 = ⎢ ⎥ are a basis for the ⎢1 ⎥ ⎢ 0⎥ ⎢⎣ 1⎥⎦ ⎣⎢0 ⎦⎥ eigenspace corresponding to λ 2 = 3. Since the geometric and algebraic multiplicities are equal for both eigenvalues, A is diagonalizable. ⎡ 1 0 0 0⎤ ⎢ 0 1 1 −1⎥ P = [p1 p 2 p3 p 4 ] = ⎢ ⎥ ⎢0 0 1 0⎥ ⎣⎢ 0 0 0 1⎦⎥ diagonalizes A, and 0⎤ ⎡λ1 0 0 ⎢0 λ 0 0⎥ −1 1 ⎥ P AP = ⎢ ⎢ 0 0 λ2 0 ⎥ ⎢⎣ 0 0 0 λ 2 ⎥⎦

0 0 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ 0 −5 5⎥ ⎢ x2 ⎥ ⎢0 ⎥ ⎥ ⎢ ⎥ = ⎢ ⎥. 0 −5 0 ⎥ ⎢ x3 ⎥ ⎢0 ⎥ 0 0 −5⎦⎥ ⎣⎢ x4 ⎦⎥ ⎣⎢0 ⎦⎥

0 0 0⎤ 0 −5 5⎥ ⎥ reduces to 0 −5 0 ⎥ 0 0 −5⎦⎥ 0⎤ 1⎥ ⎥ , a general solution is x1 = s, 0⎥ 0 ⎥⎦

⎡ x1 ⎤ ⎡1 ⎤ ⎡ 0 ⎤ ⎢x ⎥ ⎢0 ⎥ ⎢1 ⎥ x2 = t , x3 = 0, x4 = 0 or ⎢ 2 ⎥ = s ⎢ ⎥ + t ⎢ ⎥ . ⎢ x3 ⎥ ⎢0 ⎥ ⎢ 0⎥ ⎢⎣ x4 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 0⎥⎦ Thus λ1 = −2 has geometric multiplicity 2, and

⎡ −2 0 ⎢ 0 −2 =⎢ ⎢ 0 0 ⎢⎣ 0 0

⎡0 ⎤ ⎡1 ⎤ ⎢1 ⎥ ⎢0⎥ p1 = ⎢ ⎥ , p 2 = ⎢ ⎥ are a basis for the ⎢0 ⎥ ⎢0⎥ ⎢⎣0 ⎥⎦ ⎢⎣ 0 ⎥⎦ eigenspace corresponding to λ1 = −2. 0 0 5 −5 0 0 0 0

⎡5 ⎢0 Since ⎢ ⎢0 ⎣⎢0

0⎤ 5⎥ ⎥ reduces to 0⎥ 0 ⎦⎥

0⎤ 0⎥ ⎥. 0⎥ 3⎥⎦

1 λ + 1 −7 0 0 λ −1 0 −15 λ + 2

= λ3 + 2λ 2 − λ − 2 = (λ + 2)(λ + 1)(λ − 1) Thus, the eigenvalues of A are λ1 = −2, λ 2 = −1, and λ3 = 1. (λI − A)x = 0 is 1 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡ λ + 1 −7 ⎢ 0 λ −1 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ −15 λ + 2 ⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎣ 0 ⎡ −1 −7 1⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ1 = −2 gives ⎢ 0 −3 0⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 −15 0⎦ ⎣ x3 ⎦ ⎣0 ⎦

0 ⎤ ⎡ x1 ⎤ ⎡ 0⎤ 5⎥ ⎢ x2 ⎥ ⎢ 0⎥ ⎥ ⎢ ⎥ = ⎢ ⎥. 0 ⎥ ⎢ x3 ⎥ ⎢ 0⎥ 0 ⎦⎥ ⎢⎣ x4 ⎥⎦ ⎣⎢ 0⎦⎥

⎡5 ⎢0 λ 2 = 3 gives ⎢ ⎢0 ⎣⎢0 0 0 5 −5 0 0 0 0

23. det(λI − A) =

0 0 3 0

127



and D = P −1 AP ⎡ 0 −5 = ⎢1 4 ⎢ 1 ⎣0 ⎡ λ1 0 = ⎢ 0 λ2 ⎢ ⎣0 0 ⎡ −2 0 = ⎢ 0 −1 ⎢ ⎣ 0 0

⎡ −1 −7 1⎤ ⎡ 1 0 −1⎤ Since ⎢ 0 −3 0 ⎥ reduces to ⎢0 1 0 ⎥ , a ⎢ ⎥ ⎢ ⎥ ⎣ 0 −15 0 ⎦ ⎣0 0 0 ⎦ general solution is x1 = s, x2 = 0, x3 = s, or ⎡ x1 ⎤ ⎡1 ⎤ ⎡1 ⎤ ⎢ x ⎥ = s ⎢0 ⎥ , so p = ⎢0 ⎥ is a basis for the 1 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣1 ⎦ ⎣1 ⎦ ⎣ x3 ⎦ eigenspace corresponding to λ1 = −2. ⎡0 −7 1⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ λ 2 = −1 gives ⎢0 −2 0 ⎥ ⎢ x2 ⎥ = ⎢ 0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 −15 1⎦ ⎣ x3 ⎦ ⎣ 0 ⎦ ⎡0 −7 1⎤ ⎡0 1 0 ⎤ Since ⎢0 −2 0 ⎥ reduces to ⎢0 0 1 ⎥ , a ⎢ ⎥ ⎢ ⎥ ⎣0 −15 1⎦ ⎣0 0 0 ⎦ general solution is x1 = s, x2 = 0, x3 = 0, or

11

A

⎡ x1 ⎤ ⎡1 ⎤ ⎡1 ⎤ ⎢ x ⎥ = s ⎢0 ⎥ so p = ⎢0 ⎥ is a basis for the 2 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣0 ⎦ ⎣ x3 ⎦ eigenspace corresponding to λ 2 = −1. ⎡ 2 −7 1⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ3 = 1 gives ⎢ 0 0 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 −15 3⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡1 0 − 1⎤ ⎡ 2 −7 1⎤ 5⎥ ⎢ Since ⎢ 0 0 0⎥ reduces to ⎢0 1 − 1 ⎥ , a 5 ⎢ ⎥ ⎢ ⎥ ⎣ 0 −15 3⎦ 0⎦ ⎣0 0 1 1 general solution is x1 = s, x2 = s, x3 = s or 5 5 ⎡ x1 ⎤ ⎡1 ⎤ x1 = t , x2 = t , x3 = 5t which is ⎢ x2 ⎥ = t ⎢1⎥ so ⎢ ⎥ ⎢ ⎥ ⎣5 ⎦ ⎣ x3 ⎦

= PD P

⎡(−2)11 0 0⎤ ⎢ ⎥ −1 11 = P⎢ 0 (−1) 0 ⎥P ⎢ 0 0 111 ⎥⎦ ⎣ ⎡ −2048 0 0 ⎤ = P⎢ 0 −1 0 ⎥ P −1 ⎢ ⎥ 0 0 1⎦ ⎣ ⎡ −1 10, 237 −2047 ⎤ =⎢ 0 1 0⎥ ⎢ ⎥ ⎣ 0 10, 245 −2048⎦

λ −3 1 0 25. det(λI − A) = 1 λ−2 1 0 1 λ −3 = λ3 − 8λ 2 + 19λ − 12 = (λ − 1)(λ − 3)(λ − 4) The eigenvalues of A are λ1 = 1, λ 2 = 3, and λ3 = 4. (λI − A)x = 0 is 1 0 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ ⎡λ − 3 ⎢ 1 λ−2 1 ⎥ ⎢ x2 ⎥ = ⎢ 0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ λ − 3⎦ ⎣ x3 ⎦ ⎣ 0 ⎦ 1 ⎣ 0 ⎡ −2 1 0 ⎤ ⎡ x1 ⎤ ⎡ 0⎤ λ1 = 1 gives ⎢ 1 −1 1⎥ ⎢ x2 ⎥ = ⎢ 0⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 1 −2 ⎦ ⎣ x3 ⎦ ⎣ 0⎦ ⎡ −2 1 0 ⎤ ⎡ 1 0 −1⎤ Since ⎢ 1 −1 1⎥ reduces to ⎢0 1 −2⎥ , a ⎢ ⎥ ⎢ ⎥ ⎣ 0 1 −2 ⎦ ⎣0 0 0⎦ general solution is x1 = s, x2 = 2s, x3 = s or

⎡1 ⎤ p3 = ⎢1⎥ is a basis for the eigenspace ⎢ ⎥ ⎣5⎦ corresponding to λ3 = 1. P = [p1 p 2

11 −1

1⎤ ⎡ −1 7 −1⎤ ⎡ 1 1 1⎤ −1⎥ ⎢ 0 1 0 ⎥ ⎢0 0 1⎥ ⎥⎢ ⎥⎢ ⎥ 0 ⎦ ⎣ 0 15 −2 ⎦ ⎣ 1 0 5⎦ 0⎤ 0⎥ ⎥ λ3 ⎦ 0⎤ 0⎥ ⎥ 1⎦

⎡1 1 1 ⎤ p3 ] = ⎢0 0 1⎥ diagonalizes A, ⎢ ⎥ ⎣1 0 5⎦

⎡ x1 ⎤ ⎡1 ⎤ ⎡1 ⎤ ⎢ x ⎥ = s ⎢ 2⎥ and p = ⎢ 2 ⎥ is a basis for the 1 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣1 ⎦ ⎣1 ⎦ ⎣ x3 ⎦ eigenspace corresponding to λ1 = 1. ⎡0 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ 2 = 3 gives ⎢1 1 1 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 1 0 ⎦ ⎣ x3 ⎦ ⎣0 ⎦ 128


Section 5.2

⎡0 1 0 ⎤ ⎡1 0 Since ⎢1 1 1 ⎥ reduces to ⎢0 1 ⎢ ⎥ ⎢ ⎣0 1 0 ⎦ ⎣0 0 general solution is x1 = s, x2 = 0,

31. If A is diagonalizable, then P −1 AP = D. For the

1⎤ 0 ⎥ , the ⎥ 0⎦ x3 = − s or

same P, P −1 Ak P

−1 = ) A" ( PP −1 ) AP P −1 A( PP −1 ) A( PP

⎡ x1 ⎤ ⎡ 1⎤ ⎡ 1⎤ ⎢ x ⎥ = s ⎢ 0 ⎥ and p = ⎢ 0 ⎥ is a basis for the 2 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −1⎦ ⎣ −1⎦ ⎣ x3 ⎦ eigenspace corresponding to λ 2 = 3.

= ( P AP)( P AP)" ( P AP

) k factors of P −1 AP

=D

33. (a) The dimension of the eigenspace corresponding to an eigenvalue is the geometric multiplicity of the eigenvalue, which must be less than or equal to the algebraic multiplicity of the eigenvalue. Thus, for λ = 1, the dimension is 1; for λ = 3, the dimension is 1 or 2; for λ = 4, the dimension is 1, 2, or 3.

⎡ x1 ⎤ ⎡ 1⎤ ⎡ 1⎤ ⎢ x ⎥ = s ⎢ −1⎥ and p = ⎢ −1⎥ is a basis for the 3 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1⎦ ⎣ 1⎦ ⎣ x3 ⎦ eigenspace corresponding to λ3 = 4.

(b) If A is diagonalizable then each eigenvalue has the same algebraic and geometric multiplicity. Thus the dimensions of the eigenspaces are: λ = 1: dimension 1 λ = 3: dimension 2 λ = 4: dimension 3

⎡ 1 1 1⎤ p3 ] = ⎢ 2 0 −1⎥ ⎢ ⎥ ⎣ 1 −1 1⎦

diagonalizes A and ⎡ λ1 0 P −1 AP = D = ⎢ 0 λ 2 ⎢ ⎣0 0

0 ⎤ ⎡ 1 0 0⎤ 0 ⎥ = ⎢0 3 0⎥ . ⎥ ⎢ ⎥ λ3 ⎦ ⎣0 0 4⎦

(c) Since only λ = 4 can have an eigenspace with dimension 3, the eigenvalue must be 4.

1 1⎤ ⎡1 6 3 6⎥ ⎢ P −1 = ⎢ 12 0 − 12 ⎥ ⎢1 1 1⎥ ⎢⎣ 3 − 3 3 ⎥⎦

n

True/False 5.2 (a) True; use P = I. (b) True; since A is similar to B, B = P1−1 AP1 for

n −1

A = PD P

n ⎡ 1 1 1⎤ ⎡1 ⎢ = ⎢ 2 0 −1⎥ ⎢ 0 ⎢ ⎥ ⎣ 1 −1 1⎦ ⎢⎣ 0

0 3n 0

⎡ 0 ⎤ ⎢6 ⎥ 1 0 ⎥ ⎢2 ⎢ 4n ⎥⎦ ⎢ 13 ⎣ 1

1 3

0 − 13

some invertible matrix P1. Since B is similar to

1⎤ 6⎥ − 12 ⎥ 1⎥ 3 ⎥⎦

C, C = P2−1 BP2 for some invertible matrix P2 . Then C = P2−1 BP2

= P2−1 ( P1−1 AP1 ) P2

= ( P2−1P1−1 ) A( P1 P2 )

27. Using the result in Exercise 20 of Section 5.1, −b ⎤ ⎡ −b one possibility is P = ⎢ ⎥ where ⎣ a − λ1 a − λ 2 ⎦ 1 λ1 = ⎡ a + d + (a − d )2 + 4bc ⎤ and ⎦ 2⎣

λ2 =

k

Since D is a diagonal matrix, so is D k , so Ak is also diagonalizable.

⎡1 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ3 = 4 gives ⎢1 2 1 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣0 1 1 ⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡1 1 0 ⎤ ⎡ 1 0 −1⎤ Since ⎢1 2 1 ⎥ reduces to ⎢0 1 1⎥ , a ⎢ ⎥ ⎢ ⎥ ⎣0 1 1 ⎦ ⎣0 0 0 ⎦ general solution is x1 = s, x2 = − s, x3 = s, or

Thus P = [p1 p 2

with k factors of A −1 −1

−1

= ( P1P2 )−1 A( P1P2 ). Since the product of invertible matrices is invertible, P = P1 P2 is invertible and A is similar to C.

1⎡ a + d − (a − d )2 + 4bc ⎤⎦ . 2⎣

(c) True; since B = P −1 AP, then

B −1 = ( P −1 AP)−1 = P −1 A−1P so A−1 and B −1 are similar. 129



Im(4) ⎤ ⎡ −5 0⎤ ⎡ Im(−5i) Im( A) = ⎢ = ⎣ Im(2 − i) Im(1 + 5i) ⎥⎦ ⎢⎣ −1 5⎥⎦ det( A) = −5i(1 + 5i) − 4(2 − i) = −5i + 25 − 8 + 4i = 17 − i tr(A) = −5i + (1 + 5i) = 1

(d) False; a matrix P that diagonalizes A is not unique⎯it depends on the order of the eigenvalues in the diagonal matrix D and the choice of basis for each eigenspace. (e) True; if P −1 AP = D, then

P −1 A−1P = ( P −1 AP)−1 = D −1, so A−1 is also diagonalizable.

11. u ⋅ v = (i, 2i, 3) ⋅ (4, − 2i, 1 + i) = i( 4) + 2i(−2i) + 3(1 + i) = 4i + 2i(2i) + 3(1 − i) = 4i − 4 + 3 − 3i = −1 + i u ⋅ w = (i, 2i, 3) ⋅ (2 − i, 2i, 5 + 3i) = i(2 − i) + 2i(2i) + 3(5 + 3i) = i(2 + i) + 2i(−2i) + 3(5 − 3i) = 2i − 1 + 4 + 15 − 9i = 18 − 7i v ⋅ w = (4, − 2i, 1 + i) ⋅ (2 − i, 2i, 5 + 3i) = 4(2 − i) − 2i(2i) + (1 + i)(5 + 3i) = 4(2 + i) − 2i(−2i) + (1 + i)(5 − 3i) = 8 + 4i − 4 + 5 + 2i + 3 = 12 + 6i v ⋅ u = 4(−i) − 2i(−2i) + (1 + i)(3) = −4i − 4 + 3 + 3i = −1 − i = u⋅v u ⋅ ( v + w) = (i, 2i, 3) ⋅ (6 − i, 0, 6 + 4i) = i(6 + i) + 2i(0) + 3(6 − 4i) = 6i − 1 + 18 − 12i = 17 − 6i = (−1 + i) + (18 − 7i) = u⋅v + u⋅w k(u ⋅ v) = 2i(−1 + i) = −2i − 2 = −2 − 2i (ku) ⋅ v = (2i(i, 2i, 3)) ⋅ (4, − 2i, 1 + i) = (−2, − 4, 6i) ⋅ (4, − 2i, 1 + i) = −2(4) − 4(2i) + 6i(1 − i) = −8 − 8i + 6i + 6 = −2 − 2i = k (u ⋅ v )

(f) True; DT = ( P −1 AP)T = PT AT ( P −1 )T , so AT is also diagonalizable. (g) True; if A is n × n and has n linearly independent eigenvectors, then the geometric multiplicity of each eigenvalue must be equal to its algebraic multiplicity, hence A is diagonalizable. (h) True; if every eigenvalue of A has algebraic multiplicity 1, then every eigenvalue also has geometric multiplicity 1. Since the algebraic and geometric multiplicities are the same for each eigenvalue, A is diagonalizable. Section 5.3 Exercise Set 5.3 1. u = ( 2 − i, 4i, 1 + i ) = (2 + i, − 4i, 1 − i) Re(u) = (Re(2 − i), Re(4i), Re(1 + i)) = (2, 0, 1) Im(u) = (Im(2 − i), Im(4i), Im(1 + i)) = (−1, 4, 1)

u = =

2 2 2 2 − i + 4i + 1 + i

(

22 + 12 = 5 + 16 + 2 = 23

) +( 2

42

) +( 2

12 + 12

)

2

5. If ix − 3v = u, then ix = 3v + u, and 1 x = (3v + u) = −i(3v + u). i x = −i(3(1 + i, 2 − i, 4) + (3 − 4i, 2 + i, − 6i)) = −i((3 + 3i, 6 − 3i, 12) + (3 + 4i, 2 − i, 6i)) = −i(6 + 7i, 8 − 4i, 12 + 6i) = (7 − 6i, − 4 − 8i, 6 − 12i)

13. u ⋅ v = (i, 2i, 3) ⋅ (4, 2i, 1 − i) = i(4) + 2i(−2i) + 3(1 + i) = 4i + 4 + 3 + 3i = 7 + 7i

⎡ −5i 4 ⎤ ⎡ 5i 4 ⎤ 7. A = ⎢ ⎥ = ⎢ 2 + i 1 − 5i ⎥ 2 i 1 5 i − + ⎣ ⎦ ⎣ ⎦ Re(4) ⎤ ⎡ 0 4 ⎤ ⎡ Re(−5i) Re(A) = ⎢ = ⎣ Re(2 − i) Re(1 + 5i) ⎥⎦ ⎢⎣ 2 1 ⎥⎦

w ⋅ u = (u ⋅ w) = u ⋅ w = 18 − 7i

(u ⋅ v) − w ⋅ u = 7 + 7i − (18 − 7i) = −11 + 14i = −11 − 14i 130


Section 5.3

15. tr(A) = 4 + 0 = 4 det(A) = 4(0) − 1(−5) = 5 The characteristic equation of A is

21. Here a = 1, b = − 3. The angle from the positive x-axis to the ray that joins the origin to π ⎛ ⎞ the point (1, − 3 ) is φ = tan −1 ⎜ − 3 ⎟ = − . 3 ⎝ 1 ⎠

λ 2 − 4λ + 5 = 0, which has solutions λ = 2 ± i. ⎡λ − 4 5 ⎤ ⎡ x1 ⎤ ⎡ 0⎤ (λI − A) = 0 is ⎢ . ⎢ ⎥= ⎣ −1 λ ⎦⎥ ⎣ x2 ⎦ ⎣⎢ 0⎦⎥

λ = 1− i 3 = 1+ 3 = 2

5 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡ −2 − i λ1 = 2 − i gives ⎢ . ⎢ ⎥= ⎣ −1 2 − i ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ 5 ⎤ ⎡ −2 − i reduces to Since ⎢ 1 2 − − i ⎥⎦ ⎣ general solution is x1 = (2 − i) s,

23. tr(A) = −1 + 7 = 6 det(A) = −1(7) − 4(−5) = 13 The characteristic equation of A is

⎡ 1 −2 + i ⎤ , a ⎢⎣0 0 ⎥⎦ x2 = s or

λ 2 − 6λ + 13 = 0 which has solutions λ = 3 ± 2i. For λ = 3 − 2i, det(λI − A)x = 0 is 5 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ ⎡ 4 − 2i . Since = ⎢⎣ −4 −4 − 2i ⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣ 0 ⎥⎦

⎡ x1 ⎤ ⎡2 − i ⎤ ⎡2 − i ⎤ ⎢ x ⎥ = s ⎢ 1 ⎥ so x1 = ⎢ 1 ⎥ is a basis for the ⎣ ⎦ ⎣ ⎦ ⎣ 2⎦ eigenspace corresponding to λ1 = 2 − i.

⎡1 1 + 1 i ⎤ 5 ⎤ ⎡ 4 − 2i 2 ⎥, a reduces to ⎢ ⎢⎣ −4 −4 − 2i ⎥⎦ 0 ⎦ ⎣0 1 ⎞ ⎛ general solution is x1 = ⎜ −1 − i ⎟ s, x2 = s or 2 ⎠ ⎝ x1 = (−2 − i)t , x2 = 2t which is

⎡2 + i ⎤ x 2 = x1 = ⎢ is a basis for the eigenspace ⎣ 1 ⎥⎦ corresponding to λ 2 = λ1 = 2 + i.

⎡ x1 ⎤ ⎡ −2 − i ⎤ ⎡ −2 − i ⎤ ⎢ x ⎥ = t ⎢ 2 ⎥ so x = ⎢ 2 ⎥ is a basis for ⎣ ⎦ ⎣ ⎦ ⎣ 2⎦ the eigenspace corresponding to λ = 3 − 2i. ⎡ −2 ⎤ ⎡ −1⎤ Re(x) = ⎢ ⎥ and Im(x) = ⎢ ⎥ . ⎣ 2⎦ ⎣ 0⎦ ⎡ −2 −1⎤ Thus A = PCP −1 where P = ⎢ and ⎣ 2 0 ⎥⎦ ⎡ 3 −2 ⎤ C=⎢ . 3⎥⎦ ⎣2

17. tr(A) = 5 + 3 = 8 det(A) = 5(3) − 1(−2) = 17 The characteristic equation of A is λ 2 − 8λ + 17 = 0 which has solutions λ = 4 ± i. 2 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ ⎡λ − 5 = . (λI − A)x = 0 is ⎢ 1 − λ − 3⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎣ ⎡ −1 − i 2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ1 = 4 − i gives ⎢ . ⎢ ⎥= ⎣ −1 1 − i ⎦⎥ ⎣ x2 ⎦ ⎣⎢0 ⎦⎥ ⎡ −1 − i 2 ⎤ ⎡1 −1 + i ⎤ reduces to ⎢ , a Since ⎢ 0 ⎥⎦ ⎣0 ⎣ −1 1 − i ⎥⎦ general solution is x1 = (1 − i) s, x2 = s or

25. tr(A) = 8 + 2 = 10 det(A) = 8(2) − (−3)(6) = 34 The characteristic equation of A is

⎡ x1 ⎤ ⎡1 − i ⎤ ⎡1 − i ⎤ ⎢ x ⎥ = s ⎢ 1 ⎥ so x1 = ⎢ 1 ⎥ is a basis for the ⎣ ⎦ ⎣ ⎦ ⎣ 2⎦ eigenspace corresponding to λ1 = 4 − i.

λ 2 − 10λ + 34 = 0 which has solutions λ = 5 ± 3i. For λ = 5 − 3i, (λI − A)x = 0 is ⎡ −3 − 3i −6 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ . = ⎢⎣ 3 3 − 3i ⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣0 ⎥⎦

⎡1 + i ⎤ x 2 = x1 = ⎢ is a basis for the eigenspace ⎣ 1 ⎥⎦ corresponding to λ 2 = λ1 = 4 + i.

⎡ −3 − 3i −6 ⎤ ⎡1 1 − i ⎤ Since ⎢ reduces to ⎢ , a ⎥ 3 − 3i ⎦ ⎣ 3 ⎣0 0 ⎥⎦ general solution is x1 = (−1 + i) s, x2 = s or

19. Here a = b = 1. The angle from the positive xaxis to the ray that joins the origin to the point ⎛1⎞ π (1, 1) is φ = tan −1 ⎜ ⎟ = . ⎝1⎠ 4

⎡ x1 ⎤ ⎡1 − i ⎤ ⎡1 − i ⎤ ⎢ ⎥ = s ⎢ −1 ⎥ , so x = ⎢ −1 ⎥ is a basis for the ⎣ ⎦ ⎣ ⎦ ⎣ x2 ⎦ eigenspace corresponding to λ = 5 − 3i. ⎡ 1⎤ ⎡ −1⎤ Re(x) = ⎢ ⎥ and Im(x) = ⎢ ⎥ . − 1 ⎣ ⎦ ⎣ 0⎦

2 2 λ = 1 + 1i = 1 + 1 = 2

131



⎡ 1 −1⎤ Thus, A = PCP −1 where P = ⎢ and ⎣ −1 0 ⎥⎦ ⎡ 5 −3⎤ C=⎢ . ⎣3 5⎥⎦

Section 5.4 Exercise Set 5.4 1. (a) The coefficient matrix for the system is ⎡1 4 ⎤ A=⎢ . ⎣ 2 3 ⎥⎦

27. (a) u ⋅ v = (2i, i, 3i) ⋅ (i, 6i, k ) = 2i(−i) + i(−6i) + 3i(k ) = 2 + 6 + 3ik = 8 + 3ik

If u ⋅ v = 0, then 3ik = −8 or k = −

det (λI − A) = λ 2 − 4λ − 5 = (λ − 5)(λ + 1) The eigenvalues of A are λ1 = 5 and λ 2 = −1.

8 8 = i 3i 3

⎡λ − 1 −4 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ (λI − A)x = 0 is ⎢ . ⎢ ⎥= ⎣ −2 λ − 3⎥⎦ ⎣ x2 ⎦ ⎢⎣ 0 ⎥⎦

8 so k = i. 3

⎡ 4 −4 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ1 = 5 gives ⎢ . ⎢ ⎥= ⎣ −2 2 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ ⎡ 4 −4 ⎤ ⎡ 1 −1⎤ reduces to ⎢ a Since ⎢ ⎥ − 2 2 ⎣0 0⎥⎦ ⎣ ⎦ general solution is x1 = s, x2 = s, or

(b) u ⋅ v = (k , k , 1 + i) ⋅ (1, − 1, 1 − i) = k (1) + k (−1) + (1 + i)(1 + i) = (1 + i)2 = 2i Thus, there is no complex scalar k for which u ⋅ v = 0.

⎡ x1 ⎤ ⎡1⎤ ⎡1⎤ ⎢ x ⎥ = s ⎢1⎥ and p1 = ⎢1⎥ is a basis for the ⎣⎦ ⎣⎦ ⎣ 2⎦ eigenspace corresponding to λ1 = 5.

True/False 5.3

⎡ −2 −4 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ 2 = −1 gives ⎢ . ⎢ ⎥= ⎣ −2 −4 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ ⎡ −2 −4 ⎤ ⎡1 2 ⎤ reduces to ⎢ a Since ⎢ ⎥ − − 2 4 ⎣0 0 ⎥⎦ ⎣ ⎦ general solution is x1 = −2s, x2 = s or

(a) False; non-real complex eigenvalues occur in conjugate pairs, so there must be an even number. Since a real 5 × 5 matrix will have 5 eigenvalues, at least one must be real. (b) True; λ 2 − Tr( A)λ + det( A) = 0 is the characteristic equation of a 2 × 2 matrix A.

⎡ x1 ⎤ ⎡ −2 ⎤ ⎡ −2 ⎤ ⎢ x ⎥ = ⎢ 1⎥ and p 2 = ⎢ 1⎥ is a basis for ⎣ ⎦ ⎣ 2⎦ ⎣ ⎦ the eigenspace corresponding to λ 2 = −1.

(c) False; if A is a matrix with real entries and complex eigenvalues and k is a real scalar, k ≠ 0, 1, then kA has the same eigenvalues as A with the same algebraic multiplicities but tr(kA) = ktr(A) ≠ tr(A).

⎡1 −2 ⎤ diagonalizes Thus P = [p1 p 2 ] = ⎢ ⎣1 1⎥⎦ 0 ⎤ ⎡5 0⎤ ⎡λ A and P −1 AP = D = ⎢ 1 ⎥=⎢ ⎥. λ 0 2 ⎦ ⎣ 0 −1⎦ ⎣ The substitutions y = Pu and y ′ = Pu ′ give

(d) True; complex eigenvalues and eigenvectors occur in conjugate pairs.

⎡ 5 0⎤ u the “diagonal system” u′ = Du = ⎢ ⎣ 0 −1⎥⎦ u ′ = 5u1 or 1 which has the solution u2′ = −u2

(e) False; real symmetric matrices have real eigenvalues. (f) False; this is only true if the eigenvalues of A satisfy λ = 1.

⎡ c e5 x ⎤ u = ⎢ 1 − x ⎥ . Then y = Pu gives the ⎣⎢c2 e ⎦⎥ solution 5x 5x −x ⎡1 −2 ⎤ ⎡ c1e ⎤ ⎡c1e − 2c2 e ⎤ y=⎢ ⎢ − x ⎥ = ⎢ 5x ⎥ and ⎥ ⎣1 1⎦ ⎣⎢c2 e ⎦⎥ ⎣⎢ c1e + c2 e − x ⎦⎥ 132


Section 5.4

the solution of the system is

y1 = c1e5 x − 2c2 e− x y2 = c1e5 x + c2 e− x

1 x1 = − s, x2 = s, x3 = s or x1 = −t , 2 ⎡ x1 ⎤ ⎡ −1⎤ ⎢ x2 = 2t , x3 = 2t which is x2 ⎥ = t ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 2⎦ ⎣ x3 ⎦ ⎡ −1⎤ and p 2 = ⎢ 2 ⎥ is a basis for the eigenspace ⎢ ⎥ ⎣ 2⎦ corresponding to λ 2 = 2.

.

c − 2c2 = 0 (b) The initial conditions give 1 c1 + c2 = 0 which has the solution c1 = c2 = 0, so the solution satisfying the given initial y =0 . conditions is 1 y2 = 0

⎡ −1 0 −1⎤ ⎡ x1 ⎤ ⎡ 0⎤ λ3 = 3 gives ⎢ 2 2 0 ⎥ ⎢ x2 ⎥ = ⎢ 0⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 2 0 2 ⎦ ⎣ x3 ⎦ ⎣ 0⎦ ⎡ −1 0 −1⎤ Since ⎢ 2 2 0 ⎥ reduces to ⎢ ⎥ ⎣ 2 0 2⎦ ⎡ 1 0 1⎤ ⎢0 1 −1⎥ , the general solution is x1 = s, ⎢ ⎥ ⎣0 0 0 ⎦ ⎡ x1 ⎤ ⎡ 1⎤ x2 = − s, x3 = − s or ⎢ x2 ⎥ = s ⎢ −1⎥ and ⎢ ⎥ ⎢ ⎥ ⎣ −1⎦ ⎣ x3 ⎦ ⎡ 1⎤ p3 = ⎢ −1⎥ is a basis for the eigenspace ⎢ ⎥ ⎣ −1⎦ corresponding to λ3 = 3.

3. (a) The coefficient matrix for the system is ⎡ 4 0 1⎤ A = ⎢ −2 1 0 ⎥ . ⎢ ⎥ ⎣ −2 0 1⎦

det(λI − A) = λ3 − 6λ 2 + 11λ − 6 = (λ − 1)(λ − 2)(λ − 3) The eigenvalues of A are λ1 = 1, λ 2 = 2, and λ3 = 3. (λI − A)x = 0 is 0 −1 ⎤ ⎡ x1 ⎤ ⎡ 0⎤ ⎡λ − 4 ⎢ 2 λ − 1 0 ⎥ ⎢ x2 ⎥ = ⎢ 0⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 λ − 1⎦ ⎣ x3 ⎦ ⎣ 0⎦ ⎣ 2 ⎡ −3 0 −1⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ1 = 1 gives ⎢ 2 0 0 ⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 2 0 0 ⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡ −3 0 −1⎤ ⎡1 0 0 ⎤ Since ⎢ 2 0 0 ⎥ reduces to ⎢0 0 1 ⎥ , ⎢ ⎥ ⎢ ⎥ ⎣ 2 0 0⎦ ⎣0 0 0 ⎦ the general solution is x1 = 0, x2 = s,

Thus P = [p1 p 2

diagonalizes A and ⎡ λ1 0 −1 ⎢ P AP = D = 0 λ 2 ⎢ ⎣0 0

0 ⎤ ⎡1 0 0⎤ 0 ⎥ = ⎢0 2 0⎥ . ⎥ ⎢ ⎥ λ3 ⎦ ⎣0 0 3⎦ The substitutions y = Pu and y ′ = Pu ′ give the “diagonal system” u1′ = u1 ⎡1 0 0 ⎤ ⎢ ⎥ ′ u = Du = 0 2 0 u or u2′ = 2u2 which ⎢ ⎥ u3′ = 3u3 ⎣0 0 3⎦

⎡ x1 ⎤ ⎡0 ⎤ ⎡0 ⎤ x3 = 0 or ⎢ x2 ⎥ = s ⎢1 ⎥ and p1 = ⎢1 ⎥ is a ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣0 ⎦ ⎣ x3 ⎦ basis for the eigenspace corresponding to λ1 = 1. ⎡ −2 λ 2 = 2 gives ⎢ 2 ⎢ ⎣ 2 ⎡ −2 0 −1⎤ Since ⎢ 2 1 0 ⎥ ⎢ ⎥ ⎣ 2 0 1⎦

⎡ 0 −1 1⎤ p3 ] = ⎢ 1 2 −1⎥ ⎢ ⎥ ⎣ 0 2 −1⎦

0 −1⎤ ⎡ x1 ⎤ ⎡ 0⎤ 1 0 ⎥ ⎢ x1 ⎥ = ⎢ 0⎥ . ⎥⎢ ⎥ ⎢ ⎥ 0 1⎦ ⎣ x3 ⎦ ⎣ 0⎦

⎡ c ex ⎤ ⎢ 1 ⎥ has the solution u = ⎢c2 e2 x ⎥ . Then y = Pu ⎢ 3x ⎥ ⎣ c3e ⎦ gives the solution

reduces to

⎡1 0 1⎤ 2⎥ ⎢ , the general solution is ⎢0 1 −1⎥ ⎢⎣0 0 0 ⎥⎦ 133



x ⎤ ⎡ ⎡0 −1 1⎤ ⎢ c1e ⎥ y = ⎢ 1 2 −1⎥ ⎢ c2 e2 x ⎥ ⎢ ⎥ ⎣0 2 −1⎦ ⎢ c e3 x ⎥ ⎣ 3 ⎦ ⎡ − c e 2 x + c e3 x ⎤ 2 3 ⎢ ⎥ = ⎢c1e x + 2c2 e 2 x − c3e3 x ⎥ ⎢ ⎥ 2x 3x ⎣ 2c2 e − c3e ⎦ and the solution of the system is

⎡ 3 −1⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ1 = 3 gives ⎢ . ⎢ ⎥= ⎣ −6 2 ⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦ ⎡1 − 1⎤ ⎡ 3 −1⎤ 3 ⎥ , the reduces to Since ⎢ ⎢ ⎣ −6 2 ⎥⎦ 0⎦ ⎣0 1 general solution is x1 = s, x2 = s or x1 = t , 3 ⎡x ⎤ ⎡1 ⎤ ⎡1 ⎤ x2 = 3t which is ⎢ 1 ⎥ = t ⎢ ⎥ , so p1 = ⎢ ⎥ is a x 3 ⎣ ⎦ ⎣ 3⎦ ⎣ 2⎦ basis for the eigenspace corresponding to λ1 = 3.

y1 = −c2 e2 x + c3e3 x y2 = c1e x + 2c2 e2 x − c3e3 x . y3 = 2c2 e2 x − c3e3 x

⎡ −2 −1⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ λ 2 = −2 gives ⎢ . Since ⎢ ⎥= ⎣ −6 −3⎥⎦ ⎣ x2 ⎦ ⎢⎣ 0 ⎥⎦

(b) The initial conditions give −c2 + c3 = −1 c1 + 2c2 − c3 = 1 2c2 − c3 = 0

⎡ −2 −1⎤ ⎢⎣ −6 −3⎥⎦ reduces to

⎡1 1⎤ 2 ⎥ , the general ⎢ 0 0 ⎣ ⎦

1 solution is x1 = − s, x2 = s or x1 = t , 2 ⎡x ⎤ ⎡ 1⎤ ⎡ 1⎤ x2 = −2t which is ⎢ 1 ⎥ = t ⎢ ⎥ so p 2 = ⎢ ⎥ x 2 − ⎣ ⎦ ⎣ −2 ⎦ ⎣ 2⎦ is a basis for the eigenspace corresponding to ⎡ 1 1⎤ λ 2 = −2. Thus P = [p1 p 2 ] = ⎢ ⎣3 −2 ⎥⎦ diagonalizes A and 0 ⎤ ⎡ 3 0⎤ ⎡λ P −1 AP = D = ⎢ 1 ⎥=⎢ ⎥. ⎣ 0 λ 2 ⎦ ⎣0 −2 ⎦ The substitutions y = Pu and y ′ = Pu ′ give the

which has the solution c1 = 1, c2 = −1, c3 = −2, so the solution satisfying the given y1 = e2 x − 2e3 x initial conditions is y2 = e x − 2e2 x + 2e3 x . y3 = −2e 2 x + 2e3 x

5. Let y = f(x) be a solution of y ′ = ay. Then for g ( x) = f ( x)e − ax we have g ′( x) = f ′( x)e− ax + f ( x)(− ae −ax ) = af ( x)e −ax − af ( x)e− ax =0

⎡ 3 0⎤ “diagonal system” u′ = Du = ⎢ u or ⎣ 0 −2 ⎥⎦ u1′ = 3u1 u2′ = −2u2

Hence g ( x) = f ( x)e −ax = c where c is a constant or f ( x) = ceax .

⎡ c e3 x ⎤ which has the solution u = ⎢ 1 −2 x ⎥ . Then ⎢⎣c2 e ⎥⎦ y = Pu gives the solution 3x 3x −2 x ⎤ ⎡ 1 1⎤ ⎡ c1e ⎤ ⎡ c1e + c2 e y=⎢ = ⎢ ⎥ ⎢ ⎥. ⎣3 −2⎦⎥ ⎣⎢ c2 e−2 x ⎦⎥ ⎣⎢3c1e3 x − 2c2 e−2 x ⎦⎥

7. If y1 = y and y2 = y ′, then y2′ = y ′′ = y ′ + 6 y = y2 + 6 y1 which yields the system y1′ = y2 y2′ = 6 y1 + y2 The coefficient matrix of this system is ⎡0 1⎤ A=⎢ . ⎣6 1⎥⎦

Thus y1 = y = c1e3 x + c2 e−2 x is the solution of the original differential equation.

9. Use the substitutions y1 = y, y2 = y ′ = y1′ and y3 = y ′′ = y2′ . This gives the system

det(λI − A) = λ 2 − λ − 6 = (λ − 3)(λ + 2) The eigenvalues of A are λ1 = 3 and λ 2 = −2.

1 0⎤ ⎡0 A = ⎢0 0 1⎥ . ⎢ ⎥ ⎣6 −11 6 ⎦

−1 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡λ (λI − A)x = 0 is ⎢ . ⎢ ⎥= ⎣ −6 λ − 1⎥⎦ ⎣ x2 ⎦ ⎢⎣0 ⎥⎦

134


Section 5.4

⎡ x1 ⎤ ⎡1 ⎤ ⎡1 ⎤ ⎢ x ⎥ = t ⎢ 3⎥ so p = ⎢3⎥ is a basis for the 3 ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎣9 ⎦ ⎣9 ⎦ ⎣ x3 ⎦ eigenspace corresponding to λ3 = 3.

det(λI − A) = λ3 − 6λ 2 + 11λ − 6 = (λ − 1)(λ − 2)(λ − 3) The eigenvalues of A are λ1 = 1, λ 2 = 2, and λ3 = 3. 0 ⎤ ⎡ x1 ⎤ ⎡ 0⎤ ⎡ λ −1 ⎢ (λI − A)x = 0 is 0 λ −1 ⎥ ⎢ x2 ⎥ = ⎢ 0⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −6 11 λ − 6 ⎦ ⎣ x3 ⎦ ⎣0 ⎦

Thus P = [p1 p 2

⎡1 1 1 ⎤ p3 ] = ⎢1 2 3⎥ ⎢ ⎥ ⎣1 4 9 ⎦

diagonalizes A and ⎡ λ1 0 −1 P AP = D = ⎢ 0 λ 2 ⎢ ⎣0 0

⎡ 1 −1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ1 = 1 gives ⎢ 0 1 −1⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −6 11 −5⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡ 1 −1 0 ⎤ ⎡ 1 0 −1⎤ Since ⎢ 0 1 −1⎥ reduces to ⎢0 1 −1⎥ , ⎢ ⎥ ⎢ ⎥ ⎣ −6 11 −5⎦ ⎣0 0 0 ⎦ the general solution is x1 = s, x2 = s, x3 = s or

0 ⎤ ⎡1 0 0⎤ 0 ⎥ = ⎢0 2 0⎥ . ⎥ ⎢ ⎥ λ3 ⎦ ⎣0 0 3⎦ The substitutions y = Pu and y ′ = Pu ′ give the

⎡1 0 0 ⎤ “diagonal system” u′ = Du = ⎢0 2 0⎥ u or ⎢ ⎥ ⎣0 0 3⎦

⎡ x1 ⎤ ⎡1⎤ ⎡1⎤ ⎢ x ⎥ = s ⎢1⎥ so p = ⎢1⎥ is a basis for the 1 ⎢ 2⎥ ⎢⎥ ⎢⎥ ⎣1⎦ ⎣1⎦ ⎣ x3 ⎦ eigenspace corresponding to λ1 = 1.

⎡ c ex ⎤ u1′ = u1 ⎢ 1 ⎥ u2′ = 2u2 which has the solution u = ⎢c2 e2 x ⎥ . ⎢ 3x ⎥ u3′ = 3u3 ⎣ c3e ⎦ Then y = Pu gives the solution x ⎤ ⎡ ⎡1 1 1 ⎤ ⎢ c1e ⎥ y = ⎢1 2 3⎥ ⎢c2 e 2 x ⎥ ⎢ ⎥ ⎣1 4 9 ⎦ ⎢ c e3 x ⎥ ⎣ 3 ⎦ ⎡ c e x + c e 2 x + c e3 x ⎤ 2 3 ⎢ 1 ⎥ = ⎢ c1e x + 2c2 e2 x + 3c3e3 x ⎥ . ⎢ x 2x 3x ⎥ ⎣c1e + 4c2 e + 9c3e ⎦

⎡ 2 −1 0 ⎤ ⎡ x1 ⎤ ⎡ 0⎤ λ 2 = 2 gives ⎢ 0 2 −1⎥ ⎢ x2 ⎥ = ⎢ 0⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −6 11 −4 ⎦ ⎣ x3 ⎦ ⎣ 0⎦ ⎡1 0 − 1⎤ 4⎥ ⎢ ⎡ 2 −1 0 ⎤ 1⎥, reduces to Since ⎢ − 0 1 ⎢ 2 ⎣ 0 2 −1⎥⎦ ⎢ ⎥ 0⎦ ⎣0 0 1 1 s, x2 = s, 4 2 x3 = s or x1 = t , x2 = 2t , x3 = 4t which is

the general solution is x1 =

Thus y1 = y = c1e x + c2 e2 x + c3e3 x is the solution of the original differential equation.

⎡ x1 ⎤ ⎡1 ⎤ ⎡1 ⎤ ⎢ x ⎥ = t ⎢ 2⎥ so p = ⎢ 2 ⎥ is a basis for the 2 2 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 4⎦ ⎣4⎦ ⎣ x3 ⎦ eigenspace corresponding to λ 2 = 2.

True/False 5.4 (a) False; just as not every system Ax = b has a solution, not every system of differential equations has a solution.

⎡ 3 −1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ λ3 = 3 gives ⎢ 0 3 −1⎥ ⎢ x2 ⎥ = ⎢0 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −6 11 −3⎦ ⎣ x3 ⎦ ⎣0 ⎦ ⎡1 0 − 1⎤ ⎡ 3 −1 0 ⎤ 9⎥ ⎢ Since ⎢ 0 3 −1⎥ reduces to ⎢0 1 − 1 ⎥ , 3 ⎢ ⎥ ⎢ ⎥ ⎣ −6 11 −3⎦ 0⎦ ⎣0 0 1 1 the general solution is x1 = s, x2 = s, 9 3 x3 = s or x1 = t , x2 = 3t , x3 = 9t which is

(b) False; x and y may differ by any constant vector b. (c) True (d) True; if A is a square matrix with distinct real eigenvalues, it is diagonalizable. (e) False; consider the case where A is diagonalizable but not diagonal. Then A is similar to a diagonal matrix P which can be used to solve the system y ′ = Ay, but the systems y ′ = Ay and u′ = Pu do not have the same solutions. 135



⎡1 ⎤ ⎡1 ⎤ x 2 = ⎢ 1 ⎥ , x3 = ⎢ 2 ⎥ . ⎢ ⎥ ⎢ ⎥ ⎣0⎦ ⎣2⎦ ⎡1 1 1 ⎤ Thus P = ⎢ 0 1 2 ⎥ diagonalizes A, and ⎢ ⎥ ⎣0 0 2⎦

Chapter 5 Supplementary Exercises 1. (a) det(λI − A) =

λ − cos θ − sin θ

sin θ λ − cos θ

= λ 2 − (2 cos θ )λ + 1 Thus the characteristic equation of A is λ 2 − (2 cos θ )λ + 1 = 0. For this equation 2

2

⎡1 0 0 ⎤ P −1 AP = D = ⎢ 0 4 0 ⎥ . ⎢ ⎥ ⎣0 0 9 ⎦ ⎡1 0 0 ⎤ If S1 = ⎢ 0 2 0 ⎥ , then S12 = D. ⎢ ⎥ ⎣0 0 3⎦

2

b − 4ac = 4 cos θ − 4 = 4(cos θ − 1). Since 0 < θ < π, cos2 θ < 1 hence b2 − 4ac < 0, so the matrix has no real eigenvalues or eigenvectors.

S = PS1 P −1

(b) The matrix rotates vectors in R 2 through the angle θ. Since 0 < θ < π, no vector is transformed into a vector with the same or opposite direction. ⎡ d1 0 " ⎢0 d " 2 3. (a) If D = ⎢ # ⎢# ⎢⎣ 0 0 " ⎡ d1 0 " ⎢ 0 d2 " S=⎢ ⎢ # # ⎢ 0 0 " ⎣⎢

1 ⎡1 1 1 ⎤ ⎡1 0 0 ⎤ ⎡⎢ 1 −1 2 ⎤⎥ = ⎢0 1 2⎥ ⎢ 0 2 0 ⎥ ⎢ 0 1 −1⎥ ⎢ ⎥⎢ ⎥ ⎣0 0 2⎦ ⎣ 0 0 3 ⎦ ⎢⎣ 0 0 12 ⎥⎦ ⎡1 1 0 ⎤ = ⎢0 2 1 ⎥ ⎢ ⎥ ⎣0 0 3⎦

0⎤ 0⎥ ⎥ then # ⎥ d n ⎥⎦

S 2 = A, as required.

0 ⎤ ⎥ 0 ⎥ . # ⎥ ⎥ d n ⎦⎥

9. det(λI − A) =

λ − 3 −6 = λ 2 − 5λ −1 λ − 2

The characteristic equation of A is λ 2 − 5λ = 0, hence A2 − 5 A = 0 or A2 = 5 A. ⎡3 6 ⎤ ⎡15 30⎤ A2 = 5 A = 5 ⎢ = ⎣ 1 2 ⎥⎦ ⎢⎣ 5 10⎥⎦ ⎡15 30 ⎤ ⎡ 75 150 ⎤ A3 = 5 A2 = 5 ⎢ = ⎣ 5 10 ⎥⎦ ⎢⎣ 25 50 ⎥⎦

(b) If A is a diagonalizable matrix with nonnegative eigenvalues, there is a matrix P

such that P −1 AP = D and ⎡ λ1 0 " 0 ⎤ ⎢0 λ " 0 ⎥ 2 D=⎢ ⎥ is a diagonal # # ⎥ ⎢# ⎣⎢ 0 0 " λ n ⎦⎥ matrix with nonnegative entries on the main diagonal. By part (a) there is a matrix S1

⎡ 75 A4 = 5 A3 = 5 ⎢ ⎣ 25 ⎡375 A5 = 5 A4 = 5 ⎢ ⎣125

such that S12 = D. Let S = PS1P −1 , then

150 ⎤ ⎡375 750 ⎤ = 50 ⎥⎦ ⎢⎣125 250 ⎥⎦ 750 ⎤ ⎡1875 3750 ⎤ = 250 ⎥⎦ ⎢⎣ 625 1250 ⎥⎦

11. Suppose that Ax = λx for some nonzero ⎡ c1 x1 + c2 x2 + " + cn xn ⎤ ⎡ x1 ⎤ ⎢c x + c x + " + c x ⎥ ⎢x ⎥ n n⎥ x = ⎢ 2 ⎥ . Since Ax = ⎢ 1 1 2 2 # # ⎢ ⎥ ⎢ ⎥ ⎢⎣ c1 x1 + c2 x2 + " + cn xn ⎥⎦ ⎢⎣ xn ⎥⎦

S 2 = ( PS1P −1 )( PS1 P −1 ) = PS12 P −1 = PDP −1 = A.

then either x1 = x2 = " = xn ,

(c) A has eigenvalues λ1 = 1, λ 2 = 4, λ3 = 9

λ = c1 + c2 + " + cn = tr( A) or not all the xi are

⎡1 ⎤ with corresponding eigenvectors x1 = ⎢ 0 ⎥ , ⎢ ⎥ ⎣0⎦

equal and λ must be 0.

136



13. If A is a k × k nilpotent matrix where An = 0,

17. If λ is an eigenvalue of A, the λ3 will be an eigenvalue of A3 corresponding to the same eigenvector, i.e., if Ax1 = λx1, then

then det(λI − An ) = det(λI ) = λ k and An has characteristic polynomial λ k = 0, so the

A3 x1 = λ3 x1. Since A3 = A, the eigenvalues

eigenvalues of An are all 0. But if λ1 is an

must satisfy λ3 = λ and the only possibilities are −1, 0, and 1.

eigenvalue of A, then λ1n is an eigenvalue of An , so all the eigenvalues of A must also be 0.

⎡ 0 15. The matrix P = ⎢ 1 ⎢ ⎣ −1 ⎡0 −1 and P AP = D = ⎢ 0 ⎢ ⎣0

1 0⎤ −1 1⎥ will diagonalize A ⎥ 1 1⎦ 0 0⎤ 1 0 ⎥ , thus ⎥ 0 −1⎦

A = PDP −1 1 1 ⎡ 0 1 0 ⎤ ⎡0 0 0 ⎤ ⎡⎢ 1 2 − 2 ⎤⎥ = ⎢ 1 −1 1⎥ ⎢0 1 0 ⎥ ⎢ 1 0 0⎥ ⎢ ⎥⎢ ⎥⎢ 1 1 ⎥ 1 1 1 0 0 1 − − ⎣ ⎦⎣ ⎦ ⎣0 2 2⎦ 0 0⎤ ⎡ 1 ⎢ −1 − 1 − 1 ⎥ =⎢ 2 2⎥ ⎢ 1 − 1 − 1⎥ 2 2⎦ ⎣

137

Chapter 6 Inner Product Spaces Section 6.1

(d)

Exercise Set 6.1 1. (a)

u, v = 1 ⋅ 3 + 1⋅ 2 = 5

(b)

kv, w = 3v, w = 9 ⋅ 0 + 6 ⋅ (−1) = −6

(c)

u + v, w = u, w + v, w = 1 ⋅ 0 + 1⋅ (−1) + 3 ⋅ 0 + 2 ⋅ (−1) = −3

(d)

v =

u, kv = (3, − 2), (−16, − 20) = 3(−16) + (−2)(−20) = −8 (e)

⎡ 2 1⎤ ⎡ 2 1⎤ = A so . Then AT = ⎢ 5. Let A = ⎢ ⎥ ⎣1 1⎥⎦ ⎣1 1⎦ ⎡5 3 ⎤ AT A = A2 = ⎢ . ⎣3 2 ⎥⎦

v, v = 32 + 22 = 13

= (−2)2 + (−1)2 = 5

(a)

u, v = vT AT Au ⎡ 5 3⎤ ⎡ 2 ⎤ = [ −1 1] ⎢ ⎣3 2 ⎥⎦ ⎢⎣1 ⎥⎦ ⎡2⎤ = [ −2 −1] ⎢ ⎥ ⎣1 ⎦ = −4 − 1 = −5

(b)

v, w = wT AT Av ⎡5 3⎤ ⎡ −1⎤ = [0 −1] ⎢ ⎣ 3 2 ⎥⎦ ⎢⎣ 1⎥⎦ ⎡ −1⎤ = [ −3 −2] ⎢ ⎥ ⎣ 1⎦ = 3−2 =1

(c)

u + v, w = wT AT A(u + v) ⎡ 5 3⎤ ⎡ 1 ⎤ = [0 −1] ⎢ ⎣3 2⎥⎦ ⎢⎣ 2 ⎥⎦ ⎡1 ⎤ = [ −3 −2] ⎢ ⎥ ⎣2⎦ = −3 − 4 = −7

u − kv = (1, 1) − 3(3, 2) = (−8, − 5) = (−8)2 + (−5)2 = 89

3. (a)

u, v = 3(4) + (−2)(5) = 2 v, u = 4(3) + 5(−2) = 2

(b)

(c)

0, v = (0, 0), (4, 5) = 0(4) + 0(5) = 0 v, 0 = (4, 5), (0, 0) = 4(0) + 5(0) = 0

(e) d (u, v) = u − v = (−2, − 1)

(f)

ku, v = (−12, 8), (4, 5) = −12(4) + 8(5) = −8 k u, v = −4(2) = −8

u + v, w = (7, 3), (−1, 6) = 7(−1) + 3(6) = 11 u, w + v, w = (3(−1) + (−2)(6)) + (4(−1) + 5(6)) = −15 + 26 = 11 u, v + w = (3, − 2), (3, 11) = 3(3) + (−2)(11) = −13 u, v + u, w = (3(4) + (−2)(5)) + (3(−1) + (−2)(6)) = 2 − 15 = −13 138


(d)

Section 6.1

v, v = vT AT Av ⎡5 3⎤ ⎡ −1⎤ = [ −1 1] ⎢ ⎣3 2 ⎥⎦ ⎢⎣ 1⎥⎦ ⎡ −1⎤ = [ −2 −1] ⎢ ⎥ ⎣ 1⎦ = 2 −1 =1

v =

⎡ 3 11. (a) The matrix is A = ⎢ ⎣ 0 ⎡ 3 0⎤ AT A = A2 = ⎢ . ⎣0 5⎥⎦

v, v = 1 = 1

⎡2 (b) The matrix is A = ⎢ ⎣0 ⎡4 0⎤ AT A = A2 = ⎢ . ⎣ 0 6 ⎥⎦

⎡ −1⎤ ⎡ 0 ⎤ ⎡ −1⎤ (e) v − w = ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ ⎣ 1⎦ ⎣ −1⎦ ⎣ 2⎦

v − w, v − w = ( v − w)T AT A( v − w ) ⎡5 3⎤ ⎡ −1⎤ = [ −1 2 ] ⎢ ⎣ 3 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎡ −1⎤ = [1 1] ⎢ ⎥ ⎣ 2⎦ = −1 + 2 =1 d ( v, w ) = v − w = v − w, v − w

v−w

2

A =

13. (a)

0⎤ ⎥ , since 5⎦

0⎤ , since 6 ⎥⎦

A, A

= (−2)2 + 52 + 32 + 62 = 74 A = 0 by inspection.

(b)

⎡6 −1⎤ 15. (a) A − B = ⎢ ⎣ 8 −2⎥⎦ d ( A, B) = A − B = A − B, A − B

= 1 =1 (f)

u, v = 9(−3)(1) + 4(2)(7) = −27 + 56 = 29

(b)

= 62 + (−1)2 + 82 + (−2)2

= (d ( v, w ))2 = 12 = 1

= 105

⎡ 3 4⎤ ⎡ −1 3⎤ ⎡ 1 13⎤ 7. (a) u v = ⎢ = ⎣ −2 8⎥⎦ ⎢⎣ 1 1⎥⎦ ⎢⎣10 2 ⎥⎦ T

⎡ 3 3⎤ (b) A − B = ⎢ ⎣ −5 −2 ⎥⎦ d ( A, B) = A − B = A − B, A − B

T

u, v = tr(u v) = 1 + 2 = 3

⎡ 1 −3⎤ ⎡ 4 6 ⎤ ⎡ 4 −18⎤ (b) uT v = ⎢ = ⎣ 2 5⎥⎦ ⎢⎣ 0 8⎥⎦ ⎢⎣ 8 52 ⎥⎦

= 32 + 32 + (−5)2 + (−2)2

u, v = tr(uT v) = 4 + 52 = 56

= 47

⎡ 3 0⎤ ⎡9 0 ⎤ 9. (a) AT = ⎢ = A, so AT A = A2 = ⎢ . ⎣0 2⎥⎦ ⎣0 4 ⎥⎦ ⎡v ⎤ ⎡u ⎤ For u = ⎢ 1 ⎥ and v = ⎢ 1 ⎥ , u ⎣v2 ⎦ ⎣ 2⎦

17.

u, v = vT AT Au

p, q = p(−1)q(−1) + p(0)q(0) + p(1)q(1) + p(2)q(2) = (−2)(2) + (0)(1) + (2)(2) + (10)(5) = −4 + 0 + 4 + 50 = 50 p =

⎡9 0 ⎤ ⎡ u1 ⎤ = [v1 v2 ] ⎢ ⎢ ⎥ ⎣0 4 ⎥⎦ ⎣u2 ⎦ ⎡u ⎤ = [9v1 4v2 ] ⎢ 1 ⎥ ⎣ u2 ⎦ = 9u1v1 + 4u2 v2

p, p

= [ p(−1)]2 + [ p(0)]2 + [ p(1)]2 + [ p(2)]2 = (−2)2 + 02 + 22 + 102 = 108 =6 3

139

Chapter 6: Inner Product Spaces


19. u − v = (−3, −3) (a) d (u, v) = u − v = u − v, u − v = (−3)2 + (−3)2 = 18 =3 2

(b) d (u, v) = u − v = u − v, u − v = 3(−3) 2 + 2(−3)2 = 45 =3 5

⎡ 1 −1⎤ ⎡ 1 2 ⎤ ⎡ 2 −1⎤ (c) AT A = ⎢ = ⎣ 2 3⎥⎦ ⎢⎣ −1 3⎥⎦ ⎢⎣ −1 13⎥⎦

u − v, u − v = (u − v)T AT A(u − v) ⎡ 2 −1⎤ ⎡ −3⎤ = [ −3 −3] ⎢ ⎣ −1 13⎥⎦ ⎢⎣ −3⎥⎦ ⎡ − 3⎤ = [ −3 −36] ⎢ ⎥ ⎣ − 3⎦ = 9 + 108 = 117 d (u, v) = u − v = u − v, u − v = 117 = 3 13

21. (a) For u = (x, y), u = u, u

1/ 2

=

1 2 1 2 x + y and the unit circle is 4 16

is the ellipse shown. y 4

2

–2

x

–4

140

1 2 1 2 x2 y 2 x + y = 1 or + = 1 which 4 16 4 16


(b) For u = (x, y), u = u, u

Section 6.1

1/ 2

= 2 x 2 + y 2 and the unit circle is

2 x 2 + y 2 = 1 or

ellipse shown. y 1

–

25.

u+v

2

1 2

1 –1 2

+ u−v

2

x

= u + v, u + v + u − v, u − v = u, u + v + v, u + v + u, u − v − v, u − v = u, u + u, v + v, u + v, v + u, u − u, v − v, u + v, v = 2 u, u + 2 v, v =2 u

2

+2 v

2

⎡ 0 1⎤ 27. Let V = ⎢ , then ⎣ −1 0 ⎥⎦ V , V = 0(0) + (1)(−1) + (−1)(1) + (0)(0) = −2 < 0 so Axiom 4 fails.

29. (a)

1

p, q = ∫ (1 − x + x 2 + 5 x3 )( x − 3 x 2 )dx −1 1

= ∫ ( x − 4 x 2 + 4 x3 + 2 x 4 − 15 x5 )dx −1

1

3 5 6⎞ ⎛ 2 = ⎜ x − 4 x + x4 + 2 x − 5x ⎟ 3 5 2 ⎠ −1 ⎝ 2 29 ⎛ 1 ⎞ = − −⎜− ⎟ 15 ⎝ 15 ⎠ 28 =− 15

(b)

1

p, q = ∫ ( x − 5 x3 )(2 + 8 x 2 )dx −1 1

= ∫ (2 x − 2 x3 − 40 x5 )dx −1

1

4 6⎞ ⎛ = ⎜ x 2 − x − 20 x ⎟ 2 3 ⎠ −1 ⎝ 37 ⎛ 37 ⎞ = − −⎜− ⎟ 6 ⎝ 6 ⎠ =0

True/False 6.1 (a) True; with w1 = w2 = 1, the weighted Euclidean inner product is the dot product on R 2 . 141

x2 1 2

+

y2 = 1 which is the 1



u, v = (−1)(2) + 5(4) + 2(−9) = 0

(b) False; for example, using the dot product on R 2 with u = (0, −1) and v = (0, 5), u, v = 0(0) − 1(5) = −5.

cos θ =

(c) True; first apply Axiom 1, then Axiom 2. (d)

(d) True; applying Axiom 3, Axiom 1, then Axiom 3 and Axiom 1 again gives ku, kv = k u, kv = k kv, u = k (k ) v, u =k

2

u, v . (e)

= v, v , then v = 0 by Axiom

(f)

Ax = 0 and x0 , x0 = Ax0 ⋅ Ax0 = 0 ⋅ 0 = 0, but x 0 ≠ 0.

Exercise Set 6.2 2 2 u = 1 + (−3) = 10

3. (a)

2

(b)

u v

−10 10 20

=−

10 200

=−

1

=

u v

−6 6 2

1

=−

2

u, v u v

=

8 4 55

2

=

55

A, A

B, B = 32 + 22 + 12 + 02 = 14

A, B = 2(3) + 6(2) + 1(1) + (−3)(0) = 19

cos θ =

2

u, v

A =

B =

2

v = 3 + 8 = 73 u, v = (−1)(3) + 0(8) = −3 cos θ =

u v

=

= 50 =5 2

2 2 u = (−1) + 0 = 1 2

u, v

= 22 + 62 + 12 + (−3) 2

v = 2 + 4 = 20 u, v = 1(2) + (−3)(4) = −10 =

9 10

2 2 2 2 v = 4 +0 +0 +0 = 4 u, v = 2(4) + 1(0) + 7(0) + (−1)(0) = 8

cos θ =

u, v

20

2 2 2 2 u = 2 + 1 + 7 + (−1) = 55

Section 6.2

cos θ =

u v

=−

2 2 2 2 u = 1 + 0 +1 + 0 = 2

cos θ =

2

u, v

2 2 2 2 v = (−3) + (−3) + (−3) + (−3) = 36 =6 u, v = 1(−3) + 0(−3) + 1(−3) + 0(−3) = −6

(g) False; the matrix A must be invertible, otherwise there is a nontrivial solution x0 to the system

1. (a)

=0

30 101

2 2 2 u = 4 + 1 + 8 = 81 = 9

cos θ =

orthogonal. 2

u v

0

=

2 2 2 v = 1 + 0 + (−3) = 10 u, v = 4(1) + 1(0) + 8(−3) = −20

(e) False; for example, using the dot product on R 2 with u = (1, 0) and v = (0, 1), u, v = 1(0) + 0(1) = 0, i.e., u and v are

(f) True; if 0 = v 4.

u, v

(b)

−3

A =

A, B A B

=

19 5 2 14

=

A, A

= 22 + 42 + (−1)2 + 32

73

= 30 (c)

2

2

2

u = (−1) + 5 + 2 = 30 2

2

B =

B, B

= (−3)2 + 12 + 42 + 22

2

v = 2 + 4 + (−9) = 101

= 30 142

19 10 7


Section 6.2

A, B = 2(−3) + 4(1) + (−1)(4) + 3(2) = 0

cos θ = 5.

A, B A B

=

0 =0 30

p, q = 1(0) + (−1)(2) + 2(1) = 0

7. 0 = u, w = 2(1) + k (2) + 6(3) = 20 + 2k Thus k = −10 and u = (2, −10, 6). 0 = v, w = l (1) + 5(2) + 3(3) = l + 19 Thus, l = −19 and v = (−19, 5, 3). u, v = 2(−19) − 10(5) + 6(3) = −70 ≠ 0 Thus, there are no scalars such that the vectors are mutually orthogonal. 9. (a)

u, v = 2(1) + 1(7) + 3k = 9 + 3k u and v are orthogonal for k = −3.

(b)

u, v = k (k ) + k (5) + 1(6) = k 2 + 5k + 6 k 2 + 5k + 6 = (k + 2)(k + 3) u and v are orthogonal for k = −2, −3.

11. (a)

u, v = 3(4) + 2(−1) = 10 = 10 2 2 2 2 u v = 3 + 2 4 + (−1) = 13 17 = 221 ≈ 14.9

(b)

u, v = −3(2) + 1(−1) + 0(3) = −7 = 7 2 2 2 2 2 2 u v = (−3) + 1 + 0 2 + (−1) + 3 = 10 14 = 140 ≈ 11.8

(c)

u, v = −4(8) + 2(−4) + 1(−2) = −42 = 42 2 2 2 2 2 2 u v = (−4) + 2 + 1 8 + (−4) + (−2) = 21 84 = 1764 = 42

143



u, v = 0(−1) + (−2)(−1) + 2(1) + 1(1) =5 =5

(d)

2 2 2 2 2 2 2 2 u v = 0 + (−2) + 2 + 1 (−1) + (−1) + 1 + 1 = 9 4 =6

13.

u, w1 = 0 by inspection. u, w 2 = −1(1) + 1(−1) + 0(3) + 2(0) = −2 Since u is not orthogonal to w 2 , it is not orthogonal to the subspace.

15. (a) W ⊥ will have dimension 1. A normal to the plane is u = (1, −2, −3), so W ⊥ will consist of all scalar multiples of u or tu = (t, −2t, −3t) so parametric equations for W ⊥ are x = t, y = −2t, z = −3t. (b) The line contains the vector u = (2, −5, 4), thus W ⊥ is the plane through the origin with normal u, or 2x − 5y + 4z = 0. (c) The intersection of the planes is the line x + z = 0 in the xz plane (y = 0) which can be parametrized as (x, y, z) = (t, 0, −t) or t(1, 0, −1). W ⊥ is the plane with normal (1, 0, −1) or x − z = 0. 17.

u−v

2

= u − v, u − v = u, u − v − v, u − v = u, u − u, v − v, u + v, v

= u 2 −0−0+ v 2 =2 Thus u − v = 2. 19. span{u1 , u 2 , … , u r } = k1u1 + k2 u 2 +

+ kr u r where k1, k2 , … , kr are arbitrary scalars. Let

v ∈ span{u1, u 2 , … , u r }. w, v = w, k1u1 + k2u 2 + + kr u r = k1 w, u1 + k2 w, u 2 + + kr w, u r = 0+0+ +0 =0 Thus if w is orthogonal to each vector u1 , u 2 , …, u r , then w must be orthogonal to every vector in span{u1 , u 2 , …, u r }. 21. Suppose that v is orthogonal to every basis vector. Then, as in exercise 19, v is orthogonal to the span of the set of

basis vectors, which is all of W, hence v is in W ⊥ . If v is not orthogonal to every basis vector, then v clearly cannot be in W ⊥ . Thus W ⊥ consists of all vectors orthogonal to every basis vector.

144


Section 6.3

→

Section 6.3

27. Using the figure in the text, AB = v + u and →

Exercise Set 6.3

BC = v − u. Use the Euclidean inner product on R 2 and note that u = v . →

1. (a)

→

AB, BC = v + u, v − u = v, v − v, u + u, v − u, u

(b)

= v2− u2 =0 →

→

Thus AB and BC are orthogonal, so the angle at B is a right angle.

(c)

31. (a) W ⊥ is the line through the origin which is perpendicular to y = x. Thus W ⊥ is the line y = −x.

(d)

(b) W ⊥ is the plane through the origin which is perpendicular to the y-axis. Thus, W ⊥ is the xz-plane.

3. (a)

(c) W ⊥ is the line through the origin which is perpendicular to the yz-plane. Thus W ⊥ is the x-axis.

True/False 6.2 (a) False; if u is orthogonal to every vector of a subspace W, then u is in W ⊥ .

(b) True; W ∩ W ⊥ = {0}. (c) True; for any vector w in W, u + v, w = u, w + v, w = 0, so u + v is in W ⊥. (d) True; for any vector w in W, ku, w = k u, w = k (0) = 0, so ku is in W ⊥ . (b)

(e) False; if u and v are orthogonal u, v = 0 = 0. (f) False; if u and v are orthogonal, u+ v

2

= u 2 + v 2 thus

u+v =

u

2

(0, 1), (2, 0) = 0 + 0 = 0 The set is orthogonal. 1 ⎞ ⎛ 1 1 ⎞ 1 1 ⎛ 1 , , ⎜− ⎟, ⎜ ⎟ =−2+2 =0 2 2⎠ ⎝ 2 2⎠ ⎝ The set is orthogonal.

1 ⎞ ⎛ 1 1 ⎞ 1 1 ⎛ 1 ,− , ⎜− ⎟, ⎜ ⎟ =−2−2 2 2⎠ ⎝ 2 2⎠ ⎝ = −1 ≠0 The set is not orthogonal. (0, 0), (0, 1) = 0 + 0 = 0 The set is orthogonal. 1 ⎞ 1 ⎞ ⎛ 1 1 ⎛ 1 , ,− , 0, ⎟ ⎜ ⎟, ⎜ 3⎠ 2⎠ ⎝ 3 3 ⎝ 2 1 1 = +0− 6 6 =0 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 , 0, , 0, ⎜ ⎟, ⎜ − ⎟ 2⎠ ⎝ 2 2⎠ ⎝ 2 1 1 = − +0+ 2 2 =0 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 , ,− , 0, ⎜ ⎟, ⎜ − ⎟ 3⎠ ⎝ 2 2⎠ ⎝ 3 3 1 1 =− +0− 6 6 2 =− ≠0 6 The set is not orthogonal.

2 ⎛2 ⎜ ,− , 3 ⎝3

4 2 2 1⎞ ⎛ 2 1 2⎞ ⎟, ⎜ , , − ⎟ = − − 9 9 9 3⎠ ⎝ 3 3 3 ⎠ =0 2 4 2 2 1⎞ ⎛1 2 2⎞ ⎛2 ⎜ , − , ⎟, ⎜ , , ⎟ = − + = 0 9 9 9 3 3⎠ ⎝3 3 3⎠ ⎝3

2 2 4 2⎞ ⎛1 2 2⎞ ⎛2 1 ⎜ , , − ⎟, ⎜ , , ⎟ = + − = 0 9 9 9 ⎝3 3 3⎠ ⎝3 3 3⎠ The set is orthogonal.

+ v2 ≠ u + v

145


(c)


2⎛1⎞ 1⎛ 2⎞ 2⎛ 2⎞ p2 ( x), p3 ( x) = ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ 3⎝ 3⎠ 3⎝ 3 ⎠ 3⎝ 3⎠ 2 2 4 = + − 9 9 9 =0 The set is an orthonormal set.

1 1 ⎞ ⎛ , (1, 0, 0), ⎜ 0, ⎟ = 0+0+0 = 0 2 2⎠ ⎝ (1, 0, 0), (0, 0, 1) = 0 + 0 + 0 = 0 1 1 ⎞ 1 ⎛ , ⎜ 0, ⎟ , (0, 0, 1) = 0 + 0 + 2 2 2 ⎝ ⎠ 1 = ≠0 2 The set is not orthogonal.

(d)

1 2 ⎞ ⎛ 1 1 ⎛ 1 , ,− ,− , ⎜ ⎟, ⎜ 6 6 6 2 2 ⎝ ⎠ ⎝ 1 1 = − +0 12 12 =0 The set is orthogonal. 2

5. (a)

⎛ 2⎞ ⎛ 2⎞ ⎛1⎞ p1 ( x) = ⎜ ⎟ + ⎜ − ⎟ + ⎜ ⎟ ⎝ 3⎠ ⎝ 3⎠ ⎝3⎠ 4 4 1 = + + 9 9 9 = 1 =1

2

2

2

2

2

⎛ 2⎞ ⎛1⎞ ⎛ 2⎞ p2 ( x) = ⎜ ⎟ + ⎜ ⎟ + ⎜ − ⎟ ⎝ 3⎠ ⎝3⎠ ⎝ 3⎠ 4 1 4 = + + 9 9 9 = 1 =1 2

2

(b)

p1 ( x) = 12 = 1 2

⎛ 1 ⎞ ⎛ 1 ⎞ p2 ( x) = ⎜ ⎟ +⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ 1 1 = + 2 2 = 1 =1

⎞ 0⎟ ⎠

2

p3 ( x) = 12 = 1 p1 ( x), p2 ( x) = 1(0) + 0(1) + 0(1) = 0 p1 ( x), p3 ( x) = 1(0) + 0(0) + 0(1) = 0 p2 ( x), p3 ( x) = 0(0) +

=

1 2

(0) +

1 2

(1)

1

≠0 2 The set is not an orthonormal set. 7. (a)

(−1, 2), (6, 3) = −6 + 6 = 0 (−1, 2) = (−1) 2 + 22 = 5 2 ⎞ (−1, 2) (−1, 2) ⎛ 1 , = = ⎜− ⎟ (−1, 2) 5 5⎠ 5 ⎝

2

(6, 3) = 62 + 32 = 45 = 3 5

⎛1⎞ ⎛ 2⎞ ⎛ 2⎞ p3 ( x) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ 1 4 4 = + + 9 9 9 = 1 =1 2⎛ 2⎞ 2⎛1⎞ 1⎛ 2⎞ p1 ( x), p2 ( x) = ⎜ ⎟ − ⎜ ⎟ + ⎜ − ⎟ 3⎝ 3⎠ 3⎝ 3⎠ 3⎝ 3⎠ 4 2 2 = − − 9 9 9 =0 2⎛1⎞ 2⎛ 2⎞ 1⎛ 2⎞ p1 ( x), p3 ( x) = ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ 3⎝ 3⎠ 3⎝ 3⎠ 3⎝ 3⎠ 2 4 2 = − + 9 9 9 =0

(6, 3) (6, 3) = (6, 3) 3 5 3 ⎞ ⎛ 6 , =⎜ ⎟ ⎝3 5 3 5⎠ ⎛ 2 1 ⎞ , =⎜ ⎟ 5⎠ ⎝ 5 2 ⎞ ⎛ 1 ⎛ 2 1 ⎞ , , The vectors ⎜ − ⎟ and ⎜ ⎟ 5 5⎠ 5⎠ ⎝ ⎝ 5 are an orthonormal set. (b)

(1, 0, − 1), (2, 0, 2) = 2 + 0 − 2 = 0 (1, 0, − 1), (0, 5, 0) = 0 + 0 + 0 = 0 (2, 0, 2), (0, 5, 0) = 0 + 0 + 0 = 0 (1, 0, − 1) = 12 + 02 + (−1)2 = 2

146


Section 6.3

1 ⎞ (1, 0, − 1) (1, 0, − 1) ⎛ 1 , 0, − = =⎜ ⎟ (1, 0, − 1) 2⎠ 2 ⎝ 2

2

(2, 0, 2) = 22 + 02 + 22 = 8 = 2 2 (2, 0, 2) (2, 0, 2) = (2, 0, 2) 2 2 2 ⎞ ⎛ 2 , 0, =⎜ ⎟ 2 2⎠ ⎝2 2 1 ⎞ ⎛ 1 , 0, =⎜ ⎟ 2⎠ ⎝ 2 2

2

( − 12 , 12 , 0) ( − 12 , 12 , 0)

2⎞ ⎛1 1 ⎛1⎞ ⎜ , ,− ⎟ = ⎜ ⎟ ⎝3 3 3⎠ ⎝ 3⎠ 6 = 9 6 = 3 1, 1, − 2 3 ⎛1 3 3 3 = ⎜ , 1, 1, − 2 6 ⎝3

1 ⎞ ⎛ 1 , 0, − The vectors ⎜ ⎟, 2⎠ ⎝ 2 1 ⎞ ⎛ 1 , 0, ⎜ ⎟ , and (0, 1, 0) are an 2 2⎠ ⎝ orthonormal set.

( (3

1 1 ⎛1 1 1⎞ ⎛ 1 1 ⎞ ⎜ , , ⎟, ⎜ − , , 0⎟ = − + + 0 10 10 ⎝5 5 5⎠ ⎝ 2 2 ⎠ =0 1 1 2 2⎞ ⎛1 1 1⎞ ⎛1 1 ⎜ , , ⎟, ⎜ , , − ⎟ = + − 15 15 15 ⎝5 5 5⎠ ⎝3 3 3⎠ =0 1 1 2⎞ ⎛ 1 1 ⎞ ⎛1 1 ⎜ − , , 0⎟, ⎜ , , − ⎟ = − + + 0 6 6 ⎝ 2 2 ⎠ ⎝3 3 3⎠ =0 2

2

( (5

5

) 5)

2⎛ 1 1 ⎞ ⎜ − , , 0⎟ 1 ⎝ 2 2 ⎠

2

(0, 5, 0) (0, 5, 0) = = (0, 1, 0) 5 (0, 5, 0)

⎛1 1 1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎜ , , ⎟ = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎝5 5 5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ 3 = 25 3 = 5 1, 1, 1 5 ⎛1 1 1⎞ 5 5 5 = ⎜ , , ⎟ 1, 1, 1 3 ⎝5 5 5⎠

=

1 ⎛ 1 ⎞ , , 0⎟ = ⎜− 2 2 ⎠ ⎝

2

(0, 5, 0) = 0 + 5 + 0 = 5

(c)

2

⎛ 1 1 ⎞ ⎛ 1⎞ ⎛1⎞ ⎜ − , , 0⎟ = ⎜ − ⎟ + ⎜ ⎟ + 0 ⎝ 2 2 ⎠ ⎝ 2⎠ ⎝ 2⎠ 1 = 2 1 = 2

) 3)

3

2

⎛1⎞ ⎛ 2⎞ +⎜ ⎟ +⎜− ⎟ ⎝3⎠ ⎝ 3⎠

2

1 2⎞ ,− ⎟ 3 3⎠

1 2 ⎞ ⎛ 1 , ,− =⎜ ⎟ 6 6⎠ ⎝ 6 ⎛ 1 1 1 ⎞ , , The vectors ⎜ ⎟, ⎝ 3 3 3⎠ 1 1 2 ⎞ ⎛ 1 ⎛ 1 ⎞ , , 0 ⎟ and ⎜ , ,− ⎟ are ⎜− 2 2 6 6 6⎠ ⎝ ⎠ ⎝ an orthonormal set.

2

2

9.

2

⎛ 3⎞ ⎛ 4⎞ v1 = ⎜ − ⎟ + ⎜ ⎟ + 02 ⎝ 5⎠ ⎝ 5⎠ 9 16 = + 25 25 = 1 =1 2

2

⎛ 4⎞ ⎛ 3⎞ v 2 = ⎜ ⎟ + ⎜ ⎟ + 02 ⎝ 5⎠ ⎝5⎠ 16 9 = + 25 25 = 1 =1

⎛ 1 1 1 ⎞ , , =⎜ ⎟ ⎝ 3 3 3⎠

v3 = 02 + 02 + 12 = 1 v1 , v 2 = − 147

12 12 + +0 = 0 25 25



v1, v3 = 0 + 0 + 0 = 0

linearly independent. Since S contains 4 linearly independent orthogonal vectors in

v 2 , v3 = 0 + 0 + 0 = 0

R 4 , it must be an orthogonal basis for R 4 .

(a) u = (1, −1, 2) 3 4 7 u, v1 = − − + 0 = − 5 5 5 4 3 1 u, v 2 = − + 0 = 5 5 5 u, v3 = 0 + 0 + 2 = 2

(b)

v2 v3 v4

= 12 + (−2)2 + 32 + (−4)2 = 30

2

= 22 + 12 + (−4) 2 + (−3)2 = 30

2

= (−3) 2 + 42 + 12 + (−2)2 = 30

2

= 42 + 32 + 22 + 12 = 30

u, v 2 = −2 + 2 − 12 − 21 = −33 u, v3 = 3 + 8 + 3 − 14 = 0

(b) u = (3, −7, 4) 9 28 37 u, v1 = − − + 0 = − 5 5 5 12 21 9 u, v 2 = − + 0 = − 5 5 5 u, v3 = 0 + 0 + 4 = 4

u, v 4 = −4 + 6 + 6 + 7 = 15 −24 −33 0 15 u= v1 + v 2 + v3 + v 4 30 30 30 30 4 11 1 = − v1 − v 2 + 0 v3 + v 4 5 10 2

37 9 v1 − v 2 + 4 v3 . 5 5

13. (a)

3 5⎞ ⎛1 (c) u = ⎜ , − , ⎟ 7 7⎠ ⎝7 3 12 15 3 u, v1 = − − + 0 = − = − 35 35 35 7 4 9 5 1 − +0= − = − u, v 2 = 35 35 35 7 5 5 u, v3 = 0 + 0 + = 7 7 3 1 5 3 5⎞ ⎛1 Thus, ⎜ , − , ⎟ = − v1 − v 2 + v3 . 7 7 7 7 7⎠ ⎝7 11. (a)

2

u, v1 = −1 − 4 + 9 − 28 = −24

7 1 Thus, (1, − 1, 2) = − v1 + v 2 + 2 v3 . 5 5

Thus, (3, − 7, 4) = −

v1

(b)

4 10 14 +0+ = 3 3 3 2 10 8 w, u 2 = + 0 − = − 3 3 3 4 5 1 w , u3 = + 0 − = − 3 3 3 14 8 1 Thus, w = u1 − u 2 − u3 . 3 3 3 w, u1 =

w, u 2 =

v1 , v 2 = 2 − 2 − 12 + 12 = 0 v1 , v3 = −3 − 8 + 3 + 8 = 0

3

w, u1 = −

w, u3 =

11 1 6 1

v 2 , v 3 = −6 + 4 − 4 + 6 = 0 v2 , v4 = 8 + 3 − 8 − 3 = 0 v3 , v 4 = −12 + 12 + 2 − 2 = 0 Thus, since S = {v1 , v 2 , v3 , v 4 } is a set of nonzero orthogonal vectors in R 4 , S is

148

1

2

+

+

66

−

Thus, w = 0u1 +

v1 , v 4 = 4 − 6 + 6 − 4 = 0

+

11 6 4

66 5 6

+

2

2

=

5

14

=

6 +

u2 +

=0

11

66 11

6

66

11 66

u3 .


15. (a)

v1 v2 v3

2

= 12 + 12 + 12 + 12 = 4

2

= 12 + 12 + (−1)2 + (−1) 2 = 4

2

= 12 + (−1)2 + 12 + (−1)2 = 4

Section 6.3

x, v1 = 1 + 2 + 0 − 2 = 1 x, v 2 = 1 + 2 + 0 + 2 = 5 x, v3 = 1 − 2 + 0 + 2 = 1 1 5 1 v1 + v 2 + v3 4 4 4 5 5⎞ ⎛1 1 1 1⎞ ⎛1 1 1 1⎞ ⎛5 5 = ⎜ , , , ⎟+⎜ , , − , − ⎟+⎜ , − , , − ⎟ 4 4⎠ ⎝4 4 4 4⎠ ⎝4 4 4 4⎠ ⎝4 4 3 5⎞ ⎛7 5 =⎜ , , − , − ⎟ 4 4⎠ ⎝4 4

projW x =

(b)

v1 v2 v3

2

= 02 + 12 + (−4)2 + (−1) 2 = 18

2

= 32 + 52 + 12 + 12 = 36

2

= 12 + 02 + 12 + (−4) 2 = 18

x, v1 = 0 + 2 + 0 + 2 = 4 x, v 2 = 3 + 10 + 0 − 2 = 11 x, v3 = 1 + 0 + 0 + 8 = 9 4 11 9 v1 + v 2 + v3 18 36 18 2 11 1 = v1 + v 2 + v3 9 36 2 2 ⎞ ⎛ 33 55 11 11 ⎞ ⎛ 1 1 ⎛ 2 8 ⎞ = ⎜ 0, , − , − ⎟ + ⎜ , , , ⎟ + ⎜ , 0, , − 2 ⎟ 9 9 ⎠ ⎝ 36 36 36 36 ⎠ ⎝ 2 2 ⎝ 9 ⎠ 1 23 ⎞ ⎛ 17 7 =⎜ , , − , − ⎟ ⎝ 12 4 12 12 ⎠

projW x =

17. (a)

1 3 +0+ = 18 18 18 1 10 1 x, v 2 = + + 0 − = 2 2 6 6 1 4 5 x, v3 = +0+0+ = 18 18 18 3 5 projW x = v1 + 2 v 2 + v3 18 18 12 3 ⎞ ⎛ 5 1 1⎞ ⎛ 5 5 20 ⎞ ⎛ 3 = ⎜ 0, , − , − ⎟ + ⎜1, , , ⎟ + ⎜ , 0, , − ⎟ 18 18 ⎠ ⎝ 18 18 18 ⎠ ⎝ 3 3 3 ⎠ ⎝ 18 1 17 ⎞ ⎛ 23 11 =⎜ , , − , − ⎟ 18 18 ⎠ ⎝ 18 6 x, v1 = 0 +

2

149


(b)


1 1 +1+ 0 − = 1 2 2 1 1 x, v 2 = + 1 + 0 + = 2 2 2 1 1 x, v3 = − 1 + 0 + = 0 2 2 projW x = 1v1 + 2 v 2 + 0 v3 ⎛1 1 1 1⎞ = ⎜ , , , ⎟ + (1, 1, − 1, − 1) ⎝2 2 2 2⎠ 1 1⎞ ⎛3 3 =⎜ , , − , − ⎟ ⎝2 2 2 2⎠ x, v1 =

q2 =

v2 v2

( 125 , 54 ) 4 10 5

⎛ 12 4 ⎞ ⎜ , ⎟ 4 10 ⎝ 5 5 ⎠ 1 ⎞ ⎛ 3 , =⎜ ⎟ ⎝ 10 10 ⎠ =

5

y 1 –1

(b) From Exercise 15(a), 3 5⎞ ⎛7 5 w1 = projW x = ⎜ , , − , − ⎟ , so ⎝4 4 4 4⎠ 3⎞ ⎛ 3 3 3 w 2 = x − projW x = ⎜ − , , , − ⎟ . ⎝ 4 4 4 4⎠

–1

q2 q1 1

x

(b) First, transform the given basis into an orthogonal basis {v1 , v 2 }. v1 = u1 = (1, 0) v1 = 12 + 02 = 1

21. (a) First, transform the given basis into an orthogonal basis {v1 , v 2 }.

v2 = u2 −

u 2 , v1

v1 2 v1 3+0 = (3, − 5) − (1, 0) 12 = (3, − 5) − (3, 0) = (0, − 5)

v1 = u1 = (1, − 3) v1 = 12 + (−3) 2 = 10

u 2 , v1

v1 2 v1 2−6 = (2, 2) − (1, − 3) 10 ⎛ 2 6⎞ = (2, 2) − ⎜ − , ⎟ ⎝ 5 5⎠ ⎛ 12 4 ⎞ =⎜ , ⎟ ⎝ 5 5⎠ 2

v1 3 ⎞ (1, − 3) ⎛ 1 ,− = =⎜ ⎟, v1 10 ⎠ 10 ⎝ 10

=

19. (a) From Exercise 14(a), ⎛3 3 ⎞ w1 = projW x = ⎜ , , − 1, − 1⎟ , so ⎝2 2 ⎠ ⎛ 1 1 ⎞ w 2 = x − projW x = ⎜ − , , 1, − 1⎟ . ⎝ 2 2 ⎠

v2 = u2 −

q1 =

v 2 = 02 + (−5)2 = 5

The orthonormal basis is v (1, 0) q1 = 1 = = (1, 0) v1 1 q2 =

2

v2 (0, − 5) = = (0, − 1). v2 5 y

⎛ 12 ⎞ ⎛ 4 ⎞ v2 = ⎜ ⎟ + ⎜ ⎟ ⎝ 5 ⎠ ⎝5⎠ 160 = 25 4 10 = 5 The orthonormal basis is

1 –1

150

–1

q1 q2 1

x


Section 6.3

23. First, transform the given basis into an orthogonal basis {v1 , v 2 , v3 , v 4 }. v1 = u1 = (0, 2, 1, 0) v1 = 02 + 22 + 12 + 02 = 5 v2 = u2 −

u 2 , v1 v1

2

v1

0−2+0+0 (0, 2, 1, 0) 5 ⎛ 4 2 ⎞ = (1, − 1, 0, 0) + ⎜ 0, , , 0 ⎟ ⎝ 5 5 ⎠ 1 2 ⎞ ⎛ = ⎜1, − , , 0 ⎟ 5 5 ⎠ ⎝ = (1, − 1, 0, 0) −

2 2 30 30 ⎛ 1⎞ ⎛2⎞ v 2 = 12 + ⎜ − ⎟ + ⎜ ⎟ + 02 = = 25 5 ⎝ 5⎠ ⎝5⎠

v 3 = u3 −

u3 , v1 v1

v1 −

2

u3 , v 2 v2

2

v2

1− 5 + 0 + 0 ⎛ 0+4+0+0 1 2 ⎞ (0, 2, 1, 0) − ⎜1, − , , 0 ⎟ 6 5 5 5 ⎠ ⎝ 5 2

= (1, 2, 0, − 1) −

1 1 ⎞ ⎛ 8 4 ⎞ ⎛1 = (1, 2, 0, − 1) − ⎜ 0, , , 0 ⎟ − ⎜ , − , , 0 ⎟ ⎝ 5 5 ⎠ ⎝ 2 10 5 ⎠ ⎛1 1 ⎞ = ⎜ , , − 1, − 1⎟ ⎝2 2 ⎠ 2

2

⎛1⎞ ⎛1⎞ v3 = ⎜ ⎟ + ⎜ ⎟ + (−1)2 + (−1) 2 ⎝2⎠ ⎝2⎠ 5 = 2 10 = 2 u ,v u ,v u ,v v 4 = u 4 − 4 1 v1 − 4 2 v 2 − 4 3 v3 2 2 2 v1 v2 v3 = (1, 0, 0, 1) −

1 0+0+0+0 1+ 0 + 0 + 0 ⎛ 1 2 ⎞ 2 + 0 + 0 −1⎛ 1 1 ⎞ v1 − − − , , , 1 0 ⎜ ⎟ ⎜ , , − 1, − 1⎟ 6 5 5 5 5 ⎠ ⎝ ⎠ 2 2 ⎝ 5 2

1 1 ⎞ ⎛ 1 1 1 1⎞ ⎛5 = (1, 0, 0, 1) − ⎜ , − , , 0 ⎟ − ⎜ − , − , , ⎟ 6 3 ⎠ ⎝ 10 10 5 5 ⎠ ⎝6 8 4⎞ ⎛ 4 4 =⎜ , , − , ⎟ ⎝ 15 15 15 5 ⎠ 2

2

2

⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 8 ⎞ ⎛4⎞ v4 = ⎜ ⎟ + ⎜ ⎟ + ⎜ − ⎟ + ⎜ ⎟ ⎝ 15 ⎠ ⎝ 15 ⎠ ⎝ 15 ⎠ ⎝ 5 ⎠ 240 = 225 4 = 15

2

151


The orthonormal basis is q1 =

(


2 1 v1 (0, 2, 1, 0) ⎛ ⎞ , , 0⎟, = = ⎜ 0, v1 5 5 ⎠ 5 ⎝

)

1, − 15 , 25 , 0 ⎛ 5 v2 1 ,− , q2 = = =⎜ 30 v2 30 ⎝ 30 q3 =

v3 = v3

q4 =

v4 = v4

(

5 1 , 1 , − 1, 2 2 10 2

⎞ , 0⎟, 30 ⎠

2

) =⎛

−1

1 1 2 2 ⎞ , ,− ,− ⎜ ⎟, 10 10 ⎠ ⎝ 10 10

( 154 , 154 , − 158 , 54 ) = ⎛ 4 15

1 1 2 3 ⎞ , ,− , ⎜ ⎟. 15 15 ⎠ ⎝ 15 15

25. First, transform the given basis into an orthogonal basis {v1 , v 2 , v3}. v1 = u1 = (1, 1, 1) v1 =

v1 , v1

= 12 + 2(1)2 + 3(1) 2 = 1+ 2 + 3 = 6 u ,v v 2 = u 2 − 2 1 v1 2 v1 1(1) + 2(1)(1) + 3(0)(1) (1, 1, 1) = (1, 1, 0) − 6 1 = (1, 1, 0) − (1, 1, 1) 2 1⎞ ⎛1 1 =⎜ , , − ⎟ 2⎠ ⎝2 2 v2 =

v2 , v2 2

2

2

⎛1⎞ ⎛1⎞ ⎛ 1⎞ = ⎜ ⎟ + 2 ⎜ ⎟ + 3⎜ − ⎟ ⎝2⎠ ⎝2⎠ ⎝ 2⎠ ⎛1⎞ = 6⎜ ⎟ ⎝4⎠ 6 = 2 u ,v u ,v v3 = u3 − 3 1 v1 − 3 2 v 2 2 2 v1 v2 1 +0+0 1+ 0 + 0 1⎞ ⎛1 1 = (1, 0, 0) − (1, 1, 1) − 2 ⎜ , ,− ⎟ 6 6 ⎝2 2 2⎠ 4

1⎞ ⎛1 1 1⎞ ⎛1 1 = (1, 0, 0) − ⎜ , , ⎟ − ⎜ , , − ⎟ 6⎠ ⎝6 6 6⎠ ⎝6 6 ⎛2 1 ⎞ = ⎜ , − , 0⎟ ⎝3 3 ⎠

152


v3 =

Section 6.3

w1 = projW w w, v1 w, v 2 v1 + v2 = 2 2 v1 v2

v3 , v3 2

2

⎛2⎞ ⎛ 1⎞ = ⎜ ⎟ + 2 ⎜ − ⎟ + 3(0) 2 ⎝3⎠ ⎝ 3⎠ 4 2 = + 9 9 6 = 3 The orthonormal basis is 1 1 ⎞ v (1, 1, 1) ⎛ 1 , , =⎜ q1 = 1 = ⎟, v1 6 6⎠ 6 ⎝ 6 q2 =

q3 =

(

=

9

9 ⎛5 1 4⎞ = 2(1, 1, 1) − ⎜ , − , − ⎟ 14 ⎝ 3 3 3 ⎠ 3 6⎞ ⎛ 15 = (2, 2, 2) − ⎜ , − , − ⎟ 7⎠ ⎝ 14 14 ⎛ 13 31 20 ⎞ =⎜ , , ⎟ ⎝ 14 14 7 ⎠ w 2 = w − w1 ⎛ 13 31 20 ⎞ = (1, 2, 3) − ⎜ , , ⎟ ⎝ 14 14 7 ⎠ 3 1⎞ ⎛1 =⎜ , − , ⎟ ⎝ 14 14 7 ⎠

)

1, 1, − 1 1 1 ⎞ v2 ⎛ 1 2 , ,− = 2 2 =⎜ ⎟, 6 v2 6 6⎠ ⎝ 6

v3 = v3

(

2

− 13 , 0

2, 3

6 3

) =⎛

2 1 ⎞ ,− , 0⎟. ⎜ 6 6 ⎝ ⎠

27. Let W = span{u1 , u 2 }. v1 = u1 = (1, 1, 1) v2 = u2 −

u 2 , v1 v1

= (2, 0, − 1) −

⎡ 1 −1⎤ 29. (a) A = ⎢ ⎣ 2 3⎥⎦ Since det(A) = 3 + 2 = 5 ≠ 0, A is invertible, so it has a QR-decomposition. Let u1 = (1, 2) and u 2 = (−1, 3). Apply the Gram-Schmidt process with normalization to u1 and u 2 .

v1

2

2 + 0 −1

(1, 1, 1) 1 + 12 + 12 ⎛1 1 1⎞ = (2, 0, − 1) − ⎜ , , ⎟ ⎝3 3 3⎠ 4⎞ ⎛5 1 =⎜ , − , − ⎟ ⎝3 3 3⎠ {v1 , v 2 } is an orthogonal basis for W. 2

v1 = u1 = (1, 2) v1 = 12 + 22 = 5 v2 = u2 −

2

2

u 2 , v1

v1 2 v1 −1 + 6 (1, 2) = (−1, 3) − 5 = (−1, 3) − (1, 2) = (−2, 1)

v1 = 12 + 12 + 12 = 3 ⎛5⎞ ⎛ 1⎞ ⎛ 4⎞ v2 = ⎜ ⎟ + ⎜ − ⎟ + ⎜ − ⎟ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ 42 = 9 42 = 3

5 − 2 − 12 1+ 2 + 3 v1 + 3 3 3 v 2 42 3

2

v 2 = (−2)2 + 12 = 5 q1 =

v1 2 ⎞ (1, 2) ⎛ 1 , = =⎜ ⎟ v1 5⎠ 5 ⎝ 5

q2 =

v2 (−2, 1) ⎛ 2 1 ⎞ , = = ⎜− ⎟ v2 5 5⎠ 5 ⎝

u1 , q1 = u 2 , q1 u2 , q 2 153

1

4

= 5 5 1 6 =− + = 5 5 5 2 3 = + = 5 5 5 5

+



⎡ 1 1⎤ (c) A = ⎢ −2 1⎥ ⎢ ⎥ ⎣ 2 1⎦ By inspection, the column vectors of A are linearly independent, so A has a QR-decomposition. Let u1 = (1, − 2, 2) and u 2 = (1, 1, 1). Apply the Gram-Schmidt process with normalization to u1 and u 2 .

The QR-decomposition of A is ⎡ u1 , q1 u 2 , q1 ⎤ ⎡ 1 −1⎤ ⎥ ⎢⎣ 2 3⎥⎦ = [q1 q 2 ] ⎢ 0 u2 , q 2 ⎦ ⎣ ⎡1 − 2 ⎤ 5⎤ 5⎥⎡ 5 = ⎢ 25 ⎥. 1 ⎥⎢ 0 ⎢ 5⎦ ⎣ 5 5 ⎣ ⎦ ⎡ 1 2⎤ (b) A = ⎢0 1⎥ ⎢ ⎥ ⎣ 1 4⎦ By inspection, the column vectors of A are linearly independent, so A has a QR-decomposition. Let u1 = (1, 0, 1) and

v1 = u1 = (1, − 2, 2) v1 = 12 + (−2)2 + 22 = 3 v2 = u2 −

v1 2 v1 1− 2 + 2 = (1, 1, 1) − (1, − 2, 2) 9 2 2⎞ ⎛1 = (1, 1, 1) − ⎜ , − , ⎟ 9 9⎠ ⎝9 ⎛ 8 11 7 ⎞ =⎜ , , ⎟ ⎝9 9 9⎠

u 2 = (2, 1, 4). Apply the Gram-Schmidt process with normalization to u1 and u 2 . v1 = u1 = (1, 0, 1) v1 = 12 + 02 + 12 = 2 v2 = u2 −

u 2 , v1 v1

2

v1

2

2+0+4 (1, 0, 1) 2 = (2, 1, 4) − (3, 0, 3) = (−1, 1, 1) v 2 = (−1)2 + 12 + 12 = 3 q1 =

v1 1 ⎞ (1, 0, 1) ⎛ 1 , 0, = =⎜ ⎟ v1 2⎠ 2 ⎝ 2

q2 =

v2 (−1, 1, 1) ⎛ 1 1 1 ⎞ , , = = ⎜− ⎟ v2 3 3 3⎠ 3 ⎝

u1 , q1 = u 2 , q1 =

2 2 2

+0+ +0+

1 2 4 2

q2 =

=

= 2

2

v2 v2

( 89 , 119 , 97 ) 234 9

11 7 ⎞ ⎛ 8 , , =⎜ ⎟ 234 234 ⎠ ⎝ 234 1 4 4 u1 , q1 = + + = 3 3 3 3 1 2 2 1 u 2 , q1 = − + = 3 3 3 3 8 11 7 + + u2 , q 2 = 234 234 234 26 = 3 26 26 = 3

=3 2

1 4 + + = 3 3 3 3 The QR-decomposition of A is ⎡ 1 2⎤ ⎡u ,q u 2 , q1 ⎤ ⎢0 1⎥ = [q1 q 2 ] ⎢ 1 1 ⎥ 0 u2 , q 2 ⎦ ⎢ ⎥ ⎣ 1 4 ⎣ ⎦ ⎡ 1 − 1 ⎤ 3⎥ ⎢ 2 1 ⎥ ⎡ 2 3 2⎤ =⎢ 0 ⎥ 3⎥⎢ 0 ⎢ 3⎦ 1 ⎥⎣ ⎢ 1 ⎢⎣ 2 3 ⎥⎦ u2 , q 2 = −

2

⎛ 8 ⎞ ⎛ 11 ⎞ ⎛ 7 ⎞ v2 = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝9⎠ ⎝ 9 ⎠ ⎝9⎠ 234 = 81 234 = 9 v1 (1, − 2, 2) ⎛ 1 2 2⎞ = =⎜ , − , ⎟ q1 = 3 v1 ⎝3 3 3⎠

= (2, 1, 4) −

1

u 2 , v1

2

154


Section 6.3

The QR-decomposition of A is ⎡ 1 1⎤ ⎡ u ,q u 2 , q1 ⎤ ⎢ −2 1⎥ = [q1 q 2 ] ⎢ 1 1 ⎥ u 0 ⎢ ⎥ 2 , q2 ⎦ ⎣ ⎣ 1 1⎦ 8 ⎤ ⎡ 1 234 ⎥ ⎡ ⎢ 3 3 13 ⎤ 11 ⎥ ⎢ ⎥ = ⎢ − 23 234 ⎥ ⎢ 26 ⎥ ⎢ 0 3 ⎦ 7 ⎥⎣ ⎢ 2 ⎢⎣ 3 234 ⎥⎦

⎡1 0 2 ⎤ (d) A = ⎢0 1 1 ⎥ ⎢ ⎥ ⎣1 2 0 ⎦ Since det(A) = −4 ≠ 0, A is invertible, so it has a QR-decomposition. Let u1 = (1, 0, 1), u 2 = (0, 1, 2), and u3 = (2, 1, 0). Apply the Gram-Schmidt process with normalization to u1 , u 2 , and u3 . v1 = u1 = (1, 0, 1) v1 = 12 + 02 + 12 = 2 v2 = u2 −

u 2 , v1 v1

v1 = (0, 1, 2) −

2

0+0+2 (1, 0, 1) = (0, 1, 2) − (1, 0, 1) = (−1, 1, 1) 2

v 2 = (−1)2 + 12 + 12 = 3 v 3 = u3 −

u3 , v1 v1

v1 −

2

u3 , v 2 v2

2

v2

2+0+0 −2 + 1 + 0 (1, 0, 1) − (−1, 1, 1) 2 3 ⎛ 1 1 1⎞ = (2, 1, 0) − (1, 0, 1) + ⎜ − , , ⎟ ⎝ 3 3 3⎠ 2⎞ ⎛2 4 =⎜ , , − ⎟ 3⎠ ⎝3 3 = (2, 1, 0) −

2

2

2

⎛2⎞ ⎛4⎞ ⎛ 2⎞ v3 = ⎜ ⎟ + ⎜ ⎟ + ⎜ − ⎟ = ⎝3⎠ ⎝3⎠ ⎝ 3⎠

24 2 6 = 9 3

q1 =

v1 1 ⎞ (1, 0, 1) ⎛ 1 , 0, = =⎜ ⎟ v1 2 2⎠ 2 ⎝

q2 =

v2 (−1, 1, 1) ⎛ 1 1 1 ⎞ , , = = ⎜− ⎟ v2 3 3 3⎠ 3 ⎝

(

)

2, 4, − 2 v3 2 1 ⎞ ⎛ 1 3 , ,− = 3 3 =⎜ q3 = ⎟ 2 6 v3 6 6⎠ ⎝ 6 3

u1 , q1 =

1 2

1

+0+

u 2 , q1 = 0 + 0 +

2

2 2

= 2

= 2

155


u3 , q1 =

2

+0+0 = 2

2

1

u2 , q 2 = 0 +

u3 , q3 =

3

+

2

2

+

3

2

u3 , q 2 = −


3

1 3

= 3

+0 = −

2

1 3

4

+ +0 = 6 6 6 The QR-decomposition of A is ⎡ u1, q1 ⎡1 0 2 ⎤ ⎢0 1 1 ⎥ = [q1 q 2 q3 ] ⎢ 0 ⎢ ⎢ ⎥ ⎢⎣ 0 ⎣1 2 0 ⎦

⎡ 1 − ⎢ 2 =⎢ 0 ⎢ ⎢ 1 ⎣⎢ 2

1 3 1 3 1 3

−

1 ⎤ ⎡ 6⎥⎢ 2 ⎥⎢ 6⎥⎢ 1 ⎥⎢ 6 ⎦⎥ ⎣

2 0 0

u 2 , q1 u2 , q2 0 2 3 0

u3 , q1 u3 , q 2 u3 , q 3

⎤ ⎥ ⎥ ⎥⎦

2⎤ ⎥ − 1 ⎥. 3 4 ⎥ ⎥ 6 ⎦

⎡1 2 1⎤ (e) A = ⎢1 1 1⎥ ⎢ ⎥ ⎣ 0 3 1⎦ Since det(A) = −1 ≠ 0, A is invertible, so it has a QR-decomposition. Let u1 = (1, 1, 0), u 2 = (2, 1, 3), and u3 = (1, 1, 1). Apply the Gram-Schmidt process with normalization to u1 , u 2 , and u3 . v1 = u1 = (1, 1, 0) v1 = 12 + 12 + 02 = 2 v2 = u2 −

u 2 , v1 v1

2

v1

2 +1+ 0 (1, 1, 0) 2 ⎛3 3 ⎞ = (2, 1, 3) − ⎜ , , 0 ⎟ ⎝2 2 ⎠ 1 ⎞ ⎛1 = ⎜ , − , 3⎟ 2 ⎠ ⎝2 = (2, 1, 3) −

2 2 19 ⎛1⎞ ⎛ 1⎞ v 2 = ⎜ ⎟ + ⎜ − ⎟ + 32 = 2 ⎝2⎠ ⎝ 2⎠

156


v 3 = u3 −

u3 , v1 v1

= (1, 1, 1) −

v1 −

2

Section 6.3

u3 , v 2 v2 2 v2

1 − 1 +3 1+1+ 0 1 ⎞ ⎛1 (1, 1, 0) − 2 2 ⎜ , − , 3⎟ 19 2 2 ⎠ ⎝2 2

3 18 ⎞ ⎛ 3 = (1, 1, 1) − (1, 1, 0) − ⎜ , − , ⎟ ⎝ 19 19 19 ⎠ ⎛ 3 3 1⎞ = ⎜− , , ⎟ ⎝ 19 19 19 ⎠

2 2 2 1 1 ⎛ 3⎞ ⎛ 3⎞ ⎛1⎞ v3 = ⎜ − ⎟ + ⎜ ⎟ + ⎜ ⎟ = = 19 ⎝ 19 ⎠ ⎝ 19 ⎠ ⎝ 19 ⎠ 19

q1 =

v1 1 (1, 1, 0) ⎛ 1 ⎞ , , 0⎟ = =⎜ v1 2 ⎠ 2 ⎝ 2

q2 =

1, − 1, 3 ⎛ 2 v2 2 3 2⎞ 2 = 2 =⎜ ,− , ⎟ 19 v2 2 19 19 ⎠ ⎝ 2 19

q3 =

(

v3 = v3

u1 , q1 = u 2 , q1 = u3 , q1 = u2 , q 2 = u3 , q 2 =

(

)

2

3 , 3 , 1 − 19 19 19 1 19

1 2 2 2 1 2

+ + +

2 2 2 19 2

1

2 1 2

−

⎜ ⎝

3 19

,

3 19

,

1 ⎞ ⎟ 19 ⎠

+0 = 2

2 1

−

) = ⎛−

+0 =

3 2

+0 = 2 2

2 19 2

+

9 2

+

3 2

19

=

19

=

3 2

2

2 19 2 19 19 19 3 3 1 1 u3 , q3 = − + + = 19 19 19 19 The QR-decomposition of A is ⎡ u1, q1 u 2 , q1 ⎡1 2 1⎤ ⎢ ⎢1 1 1⎥ = [q1 q 2 q3 ] u2 , q2 ⎢ 0 ⎢ ⎥ 0 3 1 ⎢ 0 ⎣ ⎦ ⎣ 0

⎡ 1 ⎢ 2 ⎢ =⎢ 1 2 ⎢ ⎢0 ⎣

2 2 19 − 2 2 19 3 2 19

− 3 ⎤⎡ 2 19 ⎥ ⎢ 3 ⎥⎢ 0 19 ⎥ ⎢ ⎥ 1 ⎥⎢ 0 ⎢ 19 ⎦ ⎣

3 2 19 2

0

u3 , q1 u3 , q 2 u3 , q 3 2⎤ ⎥ 3 2⎥ 19 ⎥ 1 ⎥ 19 ⎥⎦

157

⎤ ⎥ ⎥ ⎥⎦



⎡ 1 0 1⎤ ⎢ −1 1 1⎥ (f) A = ⎢ ⎥ = [c1 c 2 c3 ] ⎢ 1 0 1⎥ ⎢⎣ −1 1 1⎥⎦ By inspection, c3 − c1 = 2c2 , so the column vectors of A are not linearly independent and A does not have a QR-decomposition.

1

p3 , v1 = ∫ x 2 ⋅1 dx = 0

1

1 ⎞ ⎛ 1 = ⎜ − x 3 + x3 ⎟ 4 ⎠0 ⎝ 6 1 = 12 p ,v p ,v v3 = p3 − 3 1 v1 − 3 2 v 2 2 2 v1 v2

normalization of a vector vi that is the result of applying the Gram-Schmidt process to {u1 , u 2 , ..., u n }. Thus, vi is ui minus a linear combination of the vectors v1, v 2 , ..., vi −1 , so ui = vi + k1v1 + k2 v 2 + + ki −1 vi −1 . Thus,

ui , q i = u i ,

1

and

12

vi vi

=

1 vi

1 1 =x − + −x 3 2 1 = − x + x2 6

vi , v i = v i .

2

v3

33. First transform the basis

2

2

1

1

= v1 , v1 = ∫ 12 dx = x 10 = 1 0

1 4 1 1 ⎞ ⎛ 1 = ⎜ x − x 2 + x3 − x 4 + x5 ⎟ 6 9 2 5 ⎠0 ⎝ 36 1 = 180

v1 = 1 = 1 1

p 2 , v1 = ∫ x ⋅1 dx = 0

v2 = p2 − v2

2

p 2 , v1 v1

2

1 2 x 2

1

= 0

= v3 , v3 1⎛ 1 ⎞ = ∫ ⎜ − x + x 2 ⎟ dx 0⎝ 6 ⎠ 1⎛ 1 1 4 ⎞ = ∫ ⎜ − x + x 2 − 2 x3 + x 4 ⎟ dx 0 ⎝ 36 3 3 ⎠

S = {p1, p 2 , p3} = {1, x, x 2} into an orthogonal basis {v1 , v 2 , v3}. v1 = p1 = 1 2

1

⎛ 1 ⎞ = x 2 − 3 (1) − 12 ⎜ − + x ⎟ 1 ⎝ 2 1 ⎠

Since each vector vi is nonzero, each diagonal entry of R is nonzero.

v1

1

1 1 3 x = 3 0 3

1 ⎛ 1 ⎞ p3 , v 2 = ∫ x 2 ⎜ − + x ⎟ dx 0 ⎝ 2 ⎠ 1⎛ 1 2 ⎞ = ∫ ⎜ − x + x3 ⎟ dx 0⎝ 2 ⎠

31. The diagonal entries of R are ui , qi for v i = 1, 2, ..., n, where qi = i is the vi

ui , v i = v i , v i

1 1 = 12 2 3

v2 =

1 2

1 1 = 180 6 5 The orthonormal basis is v 1 q1 = 1 = = 1, v1 1 v3 =

1 2

1 v1 = x − (1) = − + x 1 2

= v2 , v2 2

1⎛ 1 ⎞ = ∫ ⎜ − + x ⎟ dx 0⎝ 2 ⎠ 1⎛ 1 ⎞ = ∫ ⎜ − x + x 2 ⎟ dx 0⎝ 4 ⎠

q2 = = 1

1 1 ⎞ ⎛1 = ⎜ x − x 2 + x3 ⎟ 2 3 ⎠0 ⎝4 1 = 12

v2 v2 − 12 + x 1 2 3

⎛ 1 ⎞ = 2 3⎜− + x⎟ ⎝ 2 ⎠ = 3 (−1 + 2 x)

158


q3 = =

Section 6.4

⎡ 2 −1 0 ⎤ ⎢ 3 1 2⎥ (b) A = ⎢ ⎥ ⎢ −1 4 5⎥ ⎢⎣ 1 2 4 ⎥⎦

v3 v3 1 6

− x + x2 1 6 5

⎡ 2 −1 0⎤ ⎡ 2 3 −1 1⎤ ⎢ 3 1 2⎥ AT A = ⎢ −1 1 4 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −1 4 5⎥ 0 2 5 4 ⎣ ⎦ ⎢ 1 2 4⎥ ⎣ ⎦ − 15 1 5 ⎡ ⎤ = ⎢ −1 22 30⎥ ⎢ ⎥ ⎣ 5 30 45⎦ ⎡ −1⎤ ⎡ 2 3 −1 1⎤ ⎢ ⎥ ⎡ −1⎤ 0 T A b = ⎢ −1 1 4 2 ⎥ ⎢ ⎥ = ⎢ 9 ⎥ ⎢ ⎥ ⎢ 1⎥ ⎢ ⎥ ⎣ 0 2 5 4 ⎦ ⎢ 2 ⎥ ⎣ 13⎦ ⎣ ⎦ The associated normal system is ⎡15 −1 5⎤ ⎡ x1 ⎤ ⎡ −1⎤ ⎢ −1 22 30 ⎥ ⎢ x2 ⎥ = ⎢ 9⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 5 30 45⎦ ⎣ x3 ⎦ ⎣ 13⎦

⎛1 ⎞ = 6 5 ⎜ − x + x2 ⎟ ⎝6 ⎠ 2 = 5 (1 − 6 x + 6 x )

True/False 6.3 (a) False; for example, the vectors (1, 2) and (−1, 3) in R 2 are linearly independent but not orthogonal. (b) False; the vectors must be nonzero for this to be true. (c) True; a nontrivial subspace of R3 will have a basis, which can be transformed into an orthonormal basis with respect to the Euclidean inner product.

⎡ 1 1⎤ ⎡1 −1 −1⎤ ⎢ ⎡ 3 −2 ⎤ 3. (a) A A = ⎢ −1 1⎥ = ⎢ ⎥ 1 1 2 ⎢ ⎥ ⎣ ⎦ −1 2 ⎣ −2 6⎥⎦ ⎣ ⎦ ⎡ 7⎤ ⎡1 −1 −1⎤ ⎢ ⎥ ⎡14 ⎤ AT b = ⎢ 0 = ⎣1 1 2 ⎥⎦ ⎢ −7 ⎥ ⎢⎣ −7 ⎥⎦ ⎣ ⎦ The associated normal system is ⎡ 3 −2 ⎤ ⎡ x1 ⎤ ⎡14 ⎤ ⎢⎣ −2 6 ⎥⎦ ⎢ x ⎥ = ⎢⎣ −7 ⎥⎦ . ⎣ 2⎦ ⎡ 1 0 5⎤ ⎡ 3 −2 14 ⎤ reduces to ⎢0 1 1 ⎥ , so ⎢⎣ −2 6 −7 ⎥⎦ 2⎦ ⎣ T

(d) True; a nonzero finite-dimensional inner product space will have finite basis which can be transformed into an orthonormal basis with respect to the inner product via the GramSchmidt process with normalization. (e) False; projW x is a vector in W. (f) True; every invertible n × n matrix has a QR-decomposition. Section 6.4

1 the least squares solution is x1 = 5, x2 = . 2

Exercise Set 6.4 ⎡ 1 −1⎤ 1. (a) A = ⎢ 2 3⎥ ⎢ ⎥ ⎣ 4 5⎦

⎡1 ⎡ 1 2 1 1⎤ ⎢ 2 (b) AT A = ⎢ 0 1 1 1⎥ ⎢ ⎢ ⎥ 1 ⎣ −1 −2 0 −1⎦ ⎢⎢ 1 ⎣ ⎡ 7 4 −6 ⎤ = ⎢ 4 3 −3⎥ ⎢ ⎥ ⎣ −6 −3 6 ⎦

⎡ 1 −1⎤ ⎡ 1 2 4⎤ ⎢ ⎡ 21 25⎤ AT A = ⎢ 2 3⎥ = ⎢ ⎥ − 1 3 5 ⎢ ⎥ ⎣ ⎦ 4 5 ⎣ 25 35 ⎥⎦ ⎣ ⎦ ⎡ 2⎤ ⎡ 1 2 4 ⎤ ⎢ ⎥ ⎡ 20⎤ AT b = ⎢ −1 = ⎣ −1 3 5⎥⎦ ⎢ 5⎥ ⎢⎣ 20⎥⎦ ⎣ ⎦ The associated normal system is ⎡ 21 25⎤ ⎡ x1 ⎤ ⎡ 20 ⎤ ⎢⎣ 25 35 ⎥⎦ ⎢ x ⎥ = ⎢⎣ 20 ⎥⎦ ⎣ 2⎦

0 −1⎤ 1 −2 ⎥ ⎥ 1 0⎥ 1 −1⎦⎥

⎡ 6⎤ ⎡ 1 2 1 1⎤ ⎢ ⎥ ⎡ 18⎤ 0 A b=⎢ 0 1 1 1⎥ ⎢ ⎥ = ⎢ 12⎥ ⎢ ⎥ ⎢ 9⎥ ⎢ ⎥ ⎣ −1 −2 0 −1⎦ ⎢ 3⎥ ⎣ −9⎦ ⎣ ⎦ The associated normal system is T

159



⎡ 2 1⎤ ⎡ 2 4 −2 ⎤ ⎢ ⎡ 24 8⎤ 7. (a) AT A = ⎢ 4 2⎥ = ⎢ ⎥ 1 2 1 ⎣ ⎦ ⎢ −2 1⎥ ⎣ 8 6 ⎥⎦ ⎣ ⎦ ⎡ 3⎤ ⎡ 2 4 −2⎤ ⎢ ⎥ ⎡12 ⎤ AT b = ⎢ 2 = 1⎥⎦ ⎢ ⎥ ⎢⎣ 8⎥⎦ ⎣1 2 ⎣ 1⎦ The associated normal system is ⎡ 24 8⎤ ⎡ x1 ⎤ ⎡12 ⎤ ⎢⎣ 8 6⎥⎦ ⎢ x ⎥ = ⎢⎣ 8⎥⎦ . ⎣ 2⎦

−6 ⎤ ⎡ x1 ⎤ ⎡ 18⎤ −3⎥ ⎢ x2 ⎥ = ⎢ 12 ⎥ . ⎥⎢ ⎥ ⎢ ⎥ 6 ⎦ ⎣ x3 ⎦ ⎣ −9 ⎦ −6 18⎤ −3 12 ⎥ reduces to ⎥ 6 −9 ⎦ ⎡ 1 0 0 12 ⎤ ⎢0 1 0 −3⎥ so the least squares ⎢ ⎥ ⎣0 0 1 9 ⎦ solution is x1 = 12, x2 = −3, x3 = 9. ⎡ 7 4 ⎢ 4 3 ⎢ ⎣ −6 −3 ⎡ 7 4 ⎢ 4 3 ⎢ ⎣ −6 −3

⎡ 24 8 12⎤ ⎢⎣ 8 6 8⎥⎦ reduces to

5. (a) e = b − Ax ⎡ 7 ⎤ ⎡ 1 1⎤ ⎡ 5 ⎤ = ⎢ 0 ⎥ − ⎢ −1 1⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −7 ⎦ ⎣ −1 2⎦ ⎣ 2 ⎦ 11 ⎡ 7 ⎤ ⎡⎢ 2 ⎤⎥ = ⎢ 0⎥ − ⎢− 9 ⎥ 2 ⎢ ⎥ ⎣ −7 ⎦ ⎣⎢ −4 ⎦⎥ ⎡ 3⎤ ⎢ 2⎥ = ⎢ 9⎥ ⎢ 2⎥ ⎣ −3⎦ ⎡ 3⎤ ⎡1 −1 −1⎤ ⎢ 92 ⎥ ⎡ 0⎤ , so e is AT e = ⎢ ⎢ ⎥= ⎣1 1 2 ⎥⎦ ⎢ 2 ⎥ ⎢⎣ 0⎥⎦ ⎣ −3⎦ orthogonal to the column space of A. ⎡6 ⎤ ⎡ 1 ⎢0 ⎥ ⎢ 2 (b) e = b − Ax = ⎢ ⎥ − ⎢ ⎢9 ⎥ ⎢ 1 ⎣⎢ 3⎦⎥ ⎣⎢ 1

⎡1 0 ⎢ ⎢⎣0 1

the least squares solution is x1 =

1⎤ 10 ⎥ , 6⎥ 5⎦

so

1 , 10

⎡1⎤ 6 or x = ⎢ 10 ⎥ . 5 ⎢⎣ 65 ⎥⎦ The error vector is e = b − Ax ⎡ 3⎤ ⎡ 2 1 ⎤ ⎡ 1 ⎤ ⎥ = ⎢ 2 ⎥ − ⎢ 4 2 ⎥ ⎢ 10 ⎢ ⎥ ⎢ ⎥⎢ 6 ⎥ − 1 2 1 5 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 7 ⎡ ⎤ ⎡ 3⎤ ⎢ 5 ⎥ = ⎢ 2 ⎥ − ⎢ 14 ⎥ ⎢ ⎥ ⎢5⎥ ⎣ 1⎦ ⎣ 1 ⎦ ⎡ 8⎤ ⎢ 5⎥ = ⎢− 4 ⎥ ⎢ 5⎥ ⎣ 0⎦ so the least squares error is x2 =

0 −1⎤ ⎡ 12 ⎤ 1 −2 ⎥ ⎢ ⎥ ⎥ −3 1 0⎥ ⎢ ⎥ 9 1 −1⎦⎥ ⎣ ⎦

2 2 80 4 ⎛8⎞ ⎛ 4⎞ 2 = 5. e = ⎜ ⎟ +⎜− ⎟ +0 = 25 5 5 5 ⎝ ⎠ ⎝ ⎠

⎡ 6 ⎤ ⎡ 3⎤ ⎢ 0 ⎥ ⎢ 3⎥ = ⎢ ⎥−⎢ ⎥ ⎢ 9 ⎥ ⎢9 ⎥ ⎣⎢ 3⎦⎥ ⎣⎢0 ⎦⎥ ⎡ 3⎤ ⎢ −3⎥ =⎢ ⎥ ⎢ 0⎥ ⎣⎢ 3⎦⎥

⎡ 1 3⎤ ⎡ 1 −2 3⎤ ⎢ ⎡ 14 42 ⎤ (b) AT A = ⎢ −2 −6⎥ = ⎢ ⎥ ⎣3 −6 9 ⎦ ⎢ 3 9⎥ ⎣ 42 126 ⎥⎦ ⎣ ⎦ ⎡ 1⎤ ⎡ 1 −2 3⎤ ⎢ ⎥ ⎡ 4 ⎤ AT b = ⎢ 0 = ⎣3 −6 9 ⎥⎦ ⎢ 1⎥ ⎢⎣12 ⎥⎦ ⎣ ⎦ The associated normal system is ⎡ 14 42 ⎤ ⎡ x1 ⎤ ⎡ 4 ⎤ ⎢⎣ 42 126 ⎥⎦ ⎢ x ⎥ = ⎢⎣12 ⎥⎦ . ⎣ 2⎦

⎡ 3⎤ ⎡ 1 2 1 1⎤ ⎢ ⎥ ⎡0 ⎤ −3 A e=⎢ 0 1 1 1⎥ ⎢ ⎥ = ⎢0 ⎥ , so e is ⎢ ⎥ ⎢ 0⎥ ⎢ ⎥ ⎣ −1 −2 0 −1⎦ ⎢ 3⎥ ⎣0 ⎦ ⎣ ⎦ orthogonal to the column space of A. T

⎡ 14 42 4 ⎤ ⎢⎣ 42 126 12 ⎥⎦ reduces to

160

⎡1 3 2⎤ 7 ⎥ , so ⎢ ⎣0 0 0⎦


Section 6.4

the least squares solutions are x1 =

⎡− 7 ⎤ −1 ⎢ 6⎥ ⎡ ⎤ x3 = t or x = ⎢ 7 ⎥ + t ⎢ −1⎥ for t a real ⎢ 6 ⎥ ⎢⎣ 1⎥⎦ ⎣ 0⎦ number. The error vector is e = b − Ax ⎛ 7 ⎞ ⎡ 7 ⎤ ⎡ −1 3 2 ⎤ ⎜ ⎢⎡ − 6 ⎥⎤ ⎡ −1⎤ ⎟ = ⎢ 0 ⎥ − ⎢ 2 1 3⎥ ⎜ ⎢ 7 ⎥ + t ⎢ −1⎥ ⎟ 6 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −7 ⎦ ⎣ 0 1 1⎦ ⎜⎜ ⎣⎢ 0 ⎦⎥ ⎣ 1⎦ ⎟⎟ ⎝ ⎠ ⎛ ⎡ 14 ⎤ ⎞ ⎡ 7 ⎤ ⎜ ⎢ 3 ⎥ ⎡0 ⎤ ⎟ = ⎢ 0 ⎥ − ⎜ ⎢ − 76 ⎥ + t ⎢0 ⎥ ⎟ ⎢ ⎥ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎣ −7 ⎦ ⎜ ⎢ 7 ⎥ ⎣ 0 ⎦ ⎟ ⎝⎣ 6⎦ ⎠ 7 ⎡ ⎤ ⎢ 3⎥ = ⎢ 76 ⎥ . ⎢ 49 ⎥ ⎢⎣ − 6 ⎥⎦ The least squares error is

2 − 3t , 7

⎡ 2 ⎤ ⎡ −3⎤ x2 = t or x = ⎢ 7 ⎥ + t ⎢ ⎥ for t a real ⎣ 0 ⎦ ⎣ 1⎦ number. The error vector is ⎡ 1⎤ ⎡ 1 3⎤ ⎛ ⎡ 2 ⎤ ⎡ −3⎤ ⎞ e = b − Ax = ⎢ 0⎥ − ⎢ −2 −6 ⎥ ⎜ ⎢ 7 ⎥ + t ⎢ ⎥ ⎟ ⎜ ⎢ ⎥ ⎢ ⎥ 0 ⎣ 1⎦ ⎠⎟ ⎣ 1⎦ ⎣ 3 9 ⎦ ⎝ ⎣ ⎦ ⎛⎡ 2⎤ ⎞ ⎡ 1⎤ ⎜ ⎢ 7 ⎥ ⎡ 0⎤ ⎟ = ⎢ 0⎥ − ⎜ ⎢ − 74 ⎥ + t ⎢ 0⎥ ⎟ ⎢ ⎥ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎣ 1⎦ ⎜ ⎢ 6 ⎥ ⎣ 0⎦ ⎟ ⎝⎣ 7⎦ ⎠ ⎡5⎤ ⎢7⎥ = ⎢ 74 ⎥ . ⎢1⎥ ⎢⎣ 7 ⎥⎦ The least squares error is 2

2

⎛5⎞ ⎛4⎞ ⎛1⎞ e = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎝7⎠ ⎝7⎠ ⎝7⎠ 42 = 49 1 = 42 . 7

2

2

2

⎛ 7 ⎞ ⎛ 7 ⎞ ⎛ 49 ⎞ e = ⎜ ⎟ +⎜ ⎟ +⎜− ⎟ ⎝3⎠ ⎝6⎠ ⎝ 6 ⎠ 294 = 4 7 = 6. 2

⎡ −1 2 0 ⎤ ⎡ −1 3 2 ⎤ (c) AT A = ⎢ 3 1 1⎥ ⎢ 2 1 3⎥ ⎢ ⎥⎢ ⎥ ⎣ 2 3 1⎦ ⎣ 0 1 1⎦ ⎡ 5 −1 4 ⎤ = ⎢ −1 11 10 ⎥ ⎢ ⎥ ⎣ 4 10 14 ⎦ ⎡ −1 2 0 ⎤ ⎡ 7 ⎤ ⎡ −7 ⎤ AT b = ⎢ 3 1 1⎥ ⎢ 0 ⎥ = ⎢ 14 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 2 3 1⎦ ⎣ −7 ⎦ ⎣ 7 ⎦ The associated normal system is ⎡ 5 −1 4 ⎤ ⎡ x1 ⎤ ⎡ −7 ⎤ ⎢ −1 11 10 ⎥ ⎢ x2 ⎥ = ⎢ 14 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 4 10 14 ⎦ ⎣ x3 ⎦ ⎣ 7 ⎦

⎡2 ⎢1 9. (a) Let A = ⎢ ⎢1 ⎢⎣ 1

1 −2 ⎤ 0 −1⎥ ⎥. 1 0⎥ 1 −1⎥⎦

⎡2 ⎡ 2 1 1 1⎤ ⎢ 1 A A = ⎢ 1 0 1 1⎥ ⎢ ⎢ ⎥ ⎢1 ⎣ −2 −1 0 −1⎦ ⎢ 1 ⎣ ⎡ 7 4 −6 ⎤ = ⎢ 4 3 −3⎥ ⎢ ⎥ ⎣ −6 −3 6 ⎦ T

⎡ 5 −1 4 −7 ⎤ ⎢ −1 11 10 14⎥ reduces to ⎢ ⎥ ⎣ 4 10 14 7 ⎦ ⎡1 0 1 − 7⎤ 6⎥ ⎢ 7 ⎥ , so the least squares ⎢0 1 1 6 ⎢ ⎥ 0 0 0 0 ⎣ ⎦ 7 7 solutions are x1 = − − t , x2 = − t , 6 6

2

1 −2 ⎤ 0 −1⎥ ⎥ 1 0⎥ 1 −1⎥⎦

det( AT A) = 3 ≠ 0, so the column vectors of A are linearly independent and the column space of A, W, is the subspace of R 4 spanned by v1 , v 2 , and v3 . T

( A A)

161

−1

⎡ 3 −2 2 ⎤ = ⎢⎢ −2 2 −1⎥⎥ 5 ⎢⎣ 2 −1 3 ⎥⎦



projW u = A( AT A)−1 AT u ⎡ 2 1 −2 ⎤ ⎡ 3 −2 2⎤ ⎡6 ⎤ ⎡ 2 1 1 1⎤ ⎢ ⎥ ⎢ 1 0 −1⎥ ⎢ 3 ⎥ =⎢ ⎥ ⎢ −2 2 −1⎥ ⎢ 1 0 1 1⎥ ⎢ ⎥ 1 1 0 9 ⎢ ⎥ ⎢ ⎥ ⎢ 2 −1 5 ⎥ ⎣ −2 −1 0 −1⎦ ⎢ ⎥ 3⎦ ⎢⎣ 1 1 −1⎥⎦ ⎣ ⎢⎣6 ⎥⎦ ⎡7 ⎤ ⎢ 2⎥ =⎢ ⎥ ⎢ 9⎥ ⎢⎣ 5⎥⎦ projW u = (7, 2, 9, 5) ⎡ 1 −2 −3⎤ ⎢ 1 −1 −1⎥ (b) Let A = ⎢ ⎥. 1⎥ ⎢ 3 −2 1 3⎥⎦ ⎢⎣ 0 ⎡ 1 −2 −3⎤ ⎡ 1 1 3 0⎤ ⎢ 1 −1 −1⎥ AT A = ⎢ −2 −1 −2 1⎥ ⎢ ⎥ 3 1⎥ ⎢ ⎥ ⎢ −2 − − 3 1 1 3 ⎣ ⎦ ⎢0 1 3⎦⎥ ⎣ ⎡ 11 −9 −1⎤ = ⎢ −9 10 8⎥ ⎢ ⎥ ⎣ −1 8 20⎦

det( AT A) = 10 ≠ 0 so the column vectors of A are linearly independent and the column space of A, W, is the subspace of R 4 spanned by v1 , v 2 , and v3 . 86 − 31 ⎤ ⎡ 68 5 5⎥ ⎢ 5 219 − 79 ⎥ ( AT A)−1 = ⎢ 86 5 10 10 ⎢ 31 ⎥ 79 29 ⎢⎣ − 5 − 10 10 ⎥⎦ projW u = A( AT A)−1 AT u 86 ⎡ 1 −2 −3⎤ ⎡ 68 5 5 ⎢ 1 −1 −1⎥ ⎢ 86 219 ⎢ =⎢ ⎥ 10 1⎥ ⎢ 5 ⎢ 3 −2 31 79 1 3⎥⎦ ⎢⎣ − 5 − 10 ⎢⎣0 ⎡ − 12 ⎤ ⎢ 5⎥ ⎢ − 4⎥ = ⎢ 125 ⎥ ⎢ 5⎥ ⎢ 16 ⎥ ⎢⎣ 5 ⎥⎦ 4 12 16 ⎞ ⎛ 12 projW u = ⎜ − , − , , ⎟ 5 5 5⎠ ⎝ 5

⎤ ⎡ −2 ⎤ − 31 5 ⎥⎡ 1 1 3 0⎤ ⎢ ⎥ 0 79 ⎥ ⎢ − 10 −2 −1 −2 1⎥ ⎢ ⎥ 2 ⎢ ⎥ ⎥ 1 3⎦ ⎢ ⎥ 29 ⎣ −3 −1 ⎢⎣ 4 ⎥⎦ 10 ⎥⎦

162


⎡ −1 11. (a) A A = ⎢ 3 ⎢ ⎣ 2 ⎡ 5 = ⎢ −1 ⎢ ⎣ 4 T

2 1 3 −1 11 10

Section 6.4

0 ⎤ ⎡ −1 3 2 ⎤ 1⎥ ⎢ 2 1 3⎥ ⎥⎢ ⎥ 1⎦ ⎣ 0 1 1⎦ 4⎤ 10 ⎥ ⎥ 14 ⎦

(b) W = the yz-plane is two-dimensional and one basis for W is {w1, w 2 } where w1 = (0, 1, 0) and w 2 = (0, 0, 1). ⎡0 0 ⎤ ⎡ 0 1 0⎤ . Thus A = ⎢ 1 0 ⎥ and AT = ⎢ ⎢ ⎥ ⎣ 0 0 1⎥⎦ 0 1 ⎣ ⎦

det( AT A) = 0, so A does not have linearly independent column vectors.

⎡0 0⎤ ⎡ 0 1 0⎤ ⎢ ⎡1 0 ⎤ AT A = ⎢ 1 0⎥ = ⎢ ⎥ 0 0 1 ⎢ ⎥ ⎣ ⎦ 0 1 ⎣0 1 ⎥⎦ ⎣ ⎦

⎡ 2 −1 3 ⎤ ⎡ 2 0 −1 4 ⎤ ⎢ 0 1 1⎥ (b) AT A = ⎢ −1 1 0 −5⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −1 0 −2 ⎥ − 3 1 2 3 ⎣ ⎦ ⎢ 4 −5 3 ⎥ ⎣ ⎦ ⎡ 21 −22 20 ⎤ = ⎢ −22 27 −17 ⎥ ⎢ ⎥ 23⎦ ⎣ 20 −17

[ P] = A( AT A)−1 AT ⎡0 0⎤ ⎡1 0 ⎤ ⎡0 1 0 ⎤ = ⎢ 1 0⎥ ⎢ ⎢ ⎥ ⎣ 0 1 ⎥⎦ ⎢⎣0 0 1⎥⎦ ⎣0 1⎦ ⎡0 0⎤ ⎡0 1 0⎤ = ⎢ 1 0⎥ ⎢ ⎢ ⎥ ⎣ 0 0 1⎥⎦ ⎣0 1⎦ ⎡0 0 0 ⎤ = ⎢0 1 0 ⎥ ⎢ ⎥ ⎣0 0 1⎦

det( AT A) = 0, so A does not have linearly independent column vectors. 13. (a) W = the xz-plane is two-dimensional and one basis for W is {w1, w 2 } where

15. (a) If x = s and y = t, then a point on the plane is (s, t, −5s + 3t) = s(1, 0, −5) + t(0, 1, 3). w1 = (1, 0, − 5) and w 2 = (0, 1, 3) are a basis for W.

w1 = (1, 0, 0), w 2 = (0, 0, 1). ⎡ 1 0⎤ Thus A = ⎢0 0 ⎥ and ⎢ ⎥ ⎣0 1⎦ ⎡1 ⎡ 1 0 0⎤ ⎢ AT A = ⎢ 0 ⎣ 0 0 1⎥⎦ ⎢ 0 ⎣

⎡ 1 0 0⎤ AT = ⎢ . ⎣ 0 0 1⎥⎦

⎡ 1 0⎤ ⎡1 (b) A = ⎢ 0 1⎥ , AT = ⎢ ⎢ ⎥ ⎣0 ⎣ −5 3⎦ ⎡ 1 ⎡ 1 0 −5⎤ ⎢ AT A = ⎢ 0 ⎣ 0 1 3⎥⎦ ⎢ −5 ⎣

0⎤ ⎡1 0 ⎤ 0⎥ = ⎢ 0 1 ⎥⎦ ⎥ 1⎦ ⎣

[ P] = A( AT A)−1 AT ⎡ 1 0⎤ ⎡1 0 ⎤ ⎡ 1 0 0 ⎤ = ⎢0 0⎥ ⎢ ⎢ ⎥ ⎣ 0 1 ⎥⎦ ⎢⎣0 0 1⎥⎦ ⎣0 1⎦ ⎡ 1 0⎤ ⎡ 1 0 0⎤ = ⎢0 0⎥ ⎢ ⎢ ⎥ ⎣ 0 0 1⎥⎦ ⎣0 1⎦ ⎡1 0 0 ⎤ = ⎢0 0 0 ⎥ ⎢ ⎥ ⎣0 0 1 ⎦

0 −5 ⎤ , 1 3⎥⎦ 0⎤ ⎡ 26 −15⎤ 1⎥ = ⎢ ⎥ ⎣ −15 10 ⎥⎦ 3⎦

det( AT A) = 35, ( AT A)−1 =

1 ⎡10 15⎤ 35 ⎢⎣15 26 ⎥⎦

[ P] = A( AT A)−1 AT ⎡ 1 0⎤ ⎛ 1 ⎡10 15 ⎤ ⎞ ⎡ 1 0 −5⎤ = ⎢ 0 1⎥ ⎜ ⎢ ⎢ ⎥ ⎝ 35 ⎣15 26 ⎥⎦ ⎟⎠ ⎢⎣0 1 3⎥⎦ ⎣ −5 3⎦ ⎡ 10 15⎤ 1 ⎢ ⎡ 1 0 −5 ⎤ = 15 26 ⎥ 35 ⎢ −5 3⎥ ⎢⎣0 1 3⎥⎦ ⎣ ⎦ ⎡ 10 15 −5⎤ 1 ⎢ = 15 26 3⎥ 35 ⎢ −5 3 34 ⎥ ⎣ ⎦

163



⎡ x0 ⎤ ⎡ 10 15 −5⎤ ⎡ x0 ⎤ 1 ⎢ (c) [ P] ⎢ y0 ⎥ = 15 26 3⎥ ⎢ y0 ⎥ ⎢ ⎥ 35 ⎢ ⎥⎢ ⎥ ⎣ −5 3 34 ⎦ ⎣⎢ z0 ⎦⎥ ⎣⎢ z0 ⎦⎥ ⎡ 10 x0 + 15 y0 − 5 z0 ⎤ 1 ⎢ 15 x0 + 26 y0 + 3z0 ⎥ = ⎥ 35 ⎢ ⎣⎢ −5 x0 + 3 y0 + 34 z0 ⎦⎥ ⎡ 10 x0 +15 y0 −5 z0 ⎤ 35 ⎢ ⎥ = ⎢ 15 x0 + 26 y0 +3 z0 ⎥ 35 ⎢ ⎥ ⎢ −5 x0 +3 y0 +34 z0 ⎥ 35 ⎣⎢ ⎦⎥ ⎛ 2 x − 3 y0 − z0 15 x0 + 26 y0 + 3 z0 −5 x0 + 3 y0 + 34 z0 ⎞ The orthogonal projection is ⎜ 0 , , ⎟. 7 35 35 ⎝ ⎠

(d) The orthogonal projection of P0 (1, − 2, 4) on W is

5 25 ⎞ ⎛ 2(1) − 3(−2) − 4 15(1) + 26(−2) + 3(4) −5(1) + 3(−2) + 34(4) ⎞ ⎛ 8 , , ⎜ ⎟ = ⎜− , − , ⎟. 7 35 35 7 7 ⎠ ⎝ ⎠ ⎝ 7 2

2

2

3 35 ⎛ ⎛ 8 ⎞⎞ ⎛ 25 ⎞ ⎛ 5 ⎞⎞ ⎛ The distance between P0 (1, − 2, 4) and W is d = ⎜1 − ⎜ − ⎟ ⎟ + ⎜ −2 − ⎜ − ⎟ ⎟ + ⎜ 4 − ⎟ = . 7 7 ⎠ ⎝ 7 ⎠⎠ ⎝ ⎝ ⎝ 7 ⎠⎠ ⎝ Using Theorem 3.3.4, the distance is

5(1) − 3(−2) + 4 52 + (−3)2 + 12

=

15 35

=

3 35 . 7

17. By inspection, when t = 1, the point (t, t, t) = (1, 1, 1) is on line l. When s = 1, the point (s, 2s − 1, 1) = (1, 1, 1) is on line m. Thus since P − Q ≥ 0, these are the values of s and t that minimize the distance between the lines. 19. If A has linearly independent column vectors, then AT A is invertible and the least squares solution of Ax = b is

the solution of AT Ax = AT b, but since b is orthogonal to the column space of A, AT b = 0, so x is a solution of AT Ax = 0. Thus, x = 0 since AT A is invertible. 21. AT will have linearly independent column vectors, and the column space of AT is the row space of A. Thus, the

standard matrix for the orthogonal projection of R n onto the row space of A is [ P] = AT [( AT )T AT ]−1 ( AT )T = AT ( AAT )−1 A. True/False 6.4 (a) True; AT A is an n × n matrix. (b) False; only square matrices have inverses, but AT A can be invertible when A is not a square matrix. (c) True; if A is invertible, so is AT , so the product AT A is also invertible. (d) True (e) False; the system AT Ax = AT b may be consistent. (f) True 164


Section 6.5

⎡ a0 ⎤ ⎡ 2⎤ v* = ⎢ a1 ⎥ = ( M T M )−1 M T y = ⎢ 5⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −3⎦ ⎣ a2 ⎦

(g) False; the least squares solution may involve a parameter. (h) True; if A has linearly independent column

The desired quadratic is y = 2 + 5 x − 3x 2 .

vectors, then AT A is invertible, so AT Ax = AT b has a unique solution.

5. The two column vectors of M are linearly independent if and only neither is a nonzero multiple of the other. Since all the entries in the first column are equal, the columns are linearly independent if and only if the second column has at least two different entries, i.e., if and only if at least two of the numbers x1 , x2 , ..., xn are distinct.

Section 6.5 Exercise Set 6.5 ⎡1 0⎤ ⎡ 1 1 1⎤ 1. M = ⎢1 1⎥ , M T = ⎢ , ⎢ ⎥ ⎣ 0 1 2 ⎥⎦ 1 2 ⎣ ⎦ ⎡1 0 ⎤ ⎡ 1 1 1⎤ ⎢ ⎡3 3⎤ MT M = ⎢ , 1 1⎥ = ⎣0 1 2 ⎥⎦ ⎢1 2 ⎥ ⎢⎣3 5⎥⎦ ⎣ ⎦

( M T M ) −1 =

1 11. With the substitution X = , the problem x becomes to find a line of the form y = a + b ⋅ X ⎛1 ⎞ that best fits the data points (1, 7), ⎜ , 3 ⎟ , ⎝3 ⎠ ⎛1 ⎞ ⎜ , 1⎟ . ⎝6 ⎠ ⎡1 1⎤ ⎡7 ⎤ ⎢ 1⎥ M = ⎢1 3 ⎥ , y = ⎢ 3⎥ , ⎢ ⎥ ⎢1 1 ⎥ ⎣ 1⎦ ⎣ 6⎦ ⎡1 1⎤ ⎡1 1 1⎤ ⎢ 1 ⎥ ⎡ 3 32 ⎤ T ⎥, M M = ⎢ 1 1 ⎥ ⎢1 3 ⎥ = ⎢ 3 41 1 ⎣ 3 6 ⎦ ⎢1 1 ⎥ ⎢⎣ 2 36 ⎥⎦ ⎣ 6⎦ 1 −54 ⎤ 41 ⎡ ( M T M )−1 = ⎢ 42 ⎣ −54 108⎥⎦

1 ⎡ 5 − 3 ⎤ 1 ⎡ 5 − 3⎤ = 15 − 9 ⎢⎣ −3 3⎥⎦ 6 ⎢⎣ −3 3⎥⎦

v* = ( M T M )−1 M T y ⎡0⎤ 1 ⎡ 5 −3⎤ ⎡ 1 1 1⎤ ⎢ ⎥ 2 6 ⎣⎢ −3 3⎦⎥ ⎣⎢0 1 2 ⎦⎥ ⎢7 ⎥ ⎣ ⎦ ⎡− 1 ⎤ = ⎢ 72 ⎥ ⎢⎣ 2 ⎥⎦ 1 7 The desired line is y = − + x. 2 2

=

⎡1 x 1 ⎢ ⎢1 x2 3. M = ⎢ ⎢1 x3 ⎢1 x 4 ⎣

x12 ⎤ ⎡1 ⎥ x22 ⎥ ⎢1 ⎥=⎢ x32 ⎥ ⎢1 ⎢1 x42 ⎥⎦ ⎣

2 4⎤ 3 9⎥ ⎥ 5 25⎥ 6 36 ⎥⎦

⎡5⎤ ⎡a ⎤ v* = ⎢ ⎥ = ( M T M )−1 M T y = ⎢ 21 ⎥ . ⎣b⎦ ⎢⎣ 48 ⎥ 7 ⎦

⎡ 0⎤ ⎢ −10 ⎥ y=⎢ ⎥ ⎢ −48⎥ ⎣⎢ −76 ⎦⎥

5 48 + X , so the 21 7 5 48 required curve is y = + . 21 7 x The line in terms of X is y =

⎡1 2 4 ⎤ ⎡ 1 1 1 1⎤ ⎢ 1 3 9⎥ M T M = ⎢2 3 5 6⎥ ⎢ ⎥ ⎢ ⎥ 1 5 25 ⎣ 4 9 25 36 ⎦ ⎢⎢1 6 36 ⎥⎥ ⎣ ⎦ 74 ⎤ ⎡ 4 16 = ⎢16 74 376 ⎥ ⎢ ⎥ ⎣74 376 2018⎦ ⎡ 1989 −1116 135⎤ 1 ⎢ ( M T M )−1 = −1116 649 −80 ⎥ ⎥ 90 ⎢ 135 −80 10 ⎦ ⎣

y 10

x 10

165



3. (a) The space W of continuous functions of the

True/False 6.5

form a + be x over [0, 1] is spanned by

(a) False; there is only a unique least squares straight line fit if the data points do not lie on a vertical line.

v1 = 1 and v 2 = e x . Let g1 = v1 = 1.

(b) True; if the points are not collinear, there is no solution to the system.

2

g1

1 = ∫ 12 dx = 1 0

1

d12

+ d 22

+ not the individual terms di .

(c) False; the sum

+ d n2

v 2 , g1 = ∫ e x dx = e x 0

is minimized,


ak = =

1 2π

π ∫0

1 2π

π

∫0

1 2π

π ∫0

1

= ∫ (1 − e + e x ) 2 dx

2π

(1 + x)dx =

0 1

1⎛ x2 ⎞ = 2 + 2π ⎜x+ ⎟ π⎝ 2 ⎠0

= ∫ (1 − 2e + e2 + 2e x − 2ee x + e 2 x )dx 0

1

1 ⎛ ⎞ = ⎜ x − 2ex + e2 x + 2e x − 2e x +1 + e2 x ⎟ 2 ⎝ ⎠0 3 1 2 = − + 2e − e 2 2 1 = − (1 − e)(3 − e) 2 g1 and g 2 are an orthogonal basis for W. The least squares approximation of x is the orthogonal projection of x on W, x, g1 x, g 2 g + g projW x = 2 1 2 2 g1 g2

(1 + x) cos kx dx cos kx dx + 2π

1 2π

π

∫0

x cos kx dx

1 sin kx + 0 kπ 0 =0 1 2π bk = ∫ (1 + x)sin kx dx =

π

0

1 2π

π ∫0

2

g2

1. With f(x) = 1 + x: a0 =

sin kx dx +

= e −1

g1

2

g1 e −1 (1) = ex − 1 = 1− e + ex

(d) True

=

v 2 , g1

g2 = v2 −

1 0

1 2π

π

∫0

x sin kx dx

2π 2 1 cos kx − k kπ 0 2 =− k

=−

1

x, g1 = ∫ x dx = 0

1

x, g 2 = ∫ x(1 − e + e x )dx 0

3 1 − e 2 2 1 = (3 − e) 2

(a) a0 = 2 + 2π , a1 = a2 = 0, b1 = −2, b2 = −1

=

1 (2 + 2π ) − 2 sin x − sin 2 x 2 = (1 + π ) − 2 sin x − sin 2 x.

Thus f ( x) ≈

1 1 (3 − e) 2 projW x = 2 (1) + (1 − e + e x ) 1 − 1 (1 − e)(3 − e)

(b) a1 = a2 = an = 0

2

2 ⎛ sin kx ⎞ bk = − so bk sin kx = −2 ⎜ ⎟. k ⎝ k ⎠ sin 2 x ⎛ f ( x) ≈ (1 + π ) − 2 ⎜ sin x + + 2 ⎝

1

1 1 2 x = 2 0 2

+

1 1 = − (1 − e + e x ) 2 1− e 1 1− e 1 x = − − e 2 1− e 1− e 1 1 x =− + e 2 e −1

sin nx ⎞ ⎟ n ⎠

166


Section 6.5

(b) Let f1 = x and f2 = projW f1, then the mean square error is f1 − f2

2

−1

. Recall that

f1 − f 2

= = = =

v3 , g 2 = ∫ x3 dx = −1

f1 − f 2 , f1 − f 2 f1, f1 − f2 − f2 , f1 − f2 f1, f1 − f2 f1, f1 − f1, f2 2

= f1

2

= f1

2

−

= f1

2

f1

2

g1

f1 , g1

−

g1

2

−

2

1 2 x dx 0

=∫

f1 , g 2

f1 , g1 −

2

g2

f1 , g 2

3 =x 3

0

f1 , g 2

2

f1 − f 2

3

1

( 12 (3 − e) )

2

−

2

f1 − f 2

2

2

−

= f1

g2

2

f1

1

g1

= ∫ x 2 dx = −1

π

1

2

−1

π

2

f1 , g1 g1

−1

=0

π

2

2

−

f1 , g 2 g2

2

2

−

f1 , g3 g3

1

= ∫ sin 2 π x dx = 1 −1

f1 − f 2

1

−1

1

cos π x

Thus, the mean square error is

1

2

−1

g3

(b) Let f2 = projW f1, then in a clear extension of the process in Exercise 3(b),

= ∫ 12 dx = x 1−1 = 2 −1

g2 = v2 −

1

2 3

spanned by v1 = 1, v 2 = x, and v3 = x . Let g1 = v1 = 1.

v 2 , g1

1

2

g3

2

2

v 2 , g1

2

f1 , g3

3 projW f1 = 0g1 + π x + 0g3 = x

form a0 + a1 x + a2 x 2 over [−1, 1] is

1 = ∫ x dx = x 2 −1 2

g2

g2 +

⎛ 1 ⎞ f1, g3 = ∫ ⎜ − + x 2 ⎟ sin π x dx = 0 −1 ⎝ 3 ⎠

− 12 (1 − e)(3 − e)

1

g2

1

5. (a) The space W of continuous functions of the

2

2

f1 , g 2 = ∫ x sin π x dx =

1 1 3−e = − + 3 4 2(1 − e) 1 2 − 2e + 1 + e = + 12 2(1 − e) 13 1+ e = + 12 2(1 − e)

g1

g1

f1 , g 2

g1 +

f1 , g1 = ∫ sin π x dx = −

1 = 3

Thus, 1 2 1 (2) = −

f1 , g1

projW f1 =

2

g2 1

2

2

g2

2

− f1 , f 2

f1 , g1

g1

=0

v3 , g 2

g1 −

2

−1

= x 2 − 3 (1) − 0 2 1 2 =− +x 3 {g1 , g 2 , g3} is an orthogonal basis for W. The least squares approximation of f1 = sin π x is the orthogonal projection of f1 on W,

Now decompose f2 in terms of g1 and g 2 from part (a). f1 − f 2

v3 , g1

g3 = v3 −

1

1 4 x 4

1

f1 − projW f1 = f1 − f2 and f2 are orthogonal. 2

2 3

1

v3 , g1 = ∫ x 2 dx =

=0

2

(2) = 1− 0 − π 2 3

2

− 0 = 1−

⎧1, 0 < x < π 9. Let f ( x) = ⎨ . ⎩0, π ≤ x ≤ 2π 1 2π 1 π a0 = ∫ f ( x) dx = ∫ dx = 1

g1 = v 2 = x 1

2 1 3 = x 3 −1 3

π

167

0

π

0

6

π2

2

2


2π

1

ak =

π ∫0

bk =

1

=

π

2π

∫0

f ( x) cos kx dx =


1 π

π

x = x22 + 4 x22 = 5 x22 = x2

∫0 cos kx dx = 0

5

, so

1 2 ⎛ ⎞ , , 0⎟. x = ± ⎜ 0, 5 5 ⎠ ⎝

1 π

sin kx dx π ∫0

1 (1 − (−1)k ) kπ So the Fourier series is 1 ∞ 1 (1 − (−1)k ) sin kx. +∑ 2 k =1 kπ =

⎡ u u2 ⎤ ⎡ v1 3. Recall that if U = ⎢ 1 ⎥ and V = ⎢v u u 4⎦ ⎣ 3 ⎣ 3 then U , V = u1v1 + u2 v2 + u3v3 + u4 v4 .

v2 ⎤ , v4 ⎥⎦

(a) If U is a diagonal matrix, then u2 = u3 = 0

True/False 6.6

and U , V = u1v1 + u4 v4 . For V to be in the orthogonal complement of the subspace of all diagonal matrices, then it must be the case that v1 = v4 = 0 and V must have zeros on the main diagonal.

(a) False; the area between the graphs is the error, not the mean square error. (b) True (c) True

(b) If U is a symmetric matrix, then u2 = u3 and U , V = u1v1 + u2 (v2 + v3 ) + u4 v4 .

2π

(d) False; 1 = 1, 1 = ∫ 12 dx = 2π ≠ 1 0

Since u1 and u4 can take on any values, for V to be in the orthogonal complement of the subspace of all symmetric matrices, it must be the case that v1 = v4 = 0 and v2 = −v3 , thus V must be skew-symmetric.

(e) True Chapter 6 Supplementary Exercises 1. (a) Let v = (v1, v2 , v3 , v4 ). v, u1 = v1,

1

If x = 1, then x2 = ±

f ( x)sin kx dx

5.

v, u 2 = v2 ,

5. Let u =

v, u3 = v3 ,

v, u 4 = v4 If v, u1 = v, u 4 = 0, then v1 = v4 = 0 and v = (0, v2 , v3 , 0). Since the angle θ

between u and v satisfies cos θ =

(

a1 , ..., an

⎛ 1 , ..., v=⎜ ⎝ a1

) and

1 ⎞ ⎟ . By the Cauchy-Schwarz an ⎠

Inequality, u ⋅ v

2

= (1 +

u, v

+ 1)2 ≤ u

2

v

2

or

n terms

, u v then v making equal angles with u 2 and u3

n 2 ≤ (a1 +

means that v2 = v3 . In order for the angle

⎛1 + a1 ) ⎜ + ⎝ a1

+

1 an

⎞ ⎟. ⎠

7. Let x = ( x1, x2 , x3 ).

between v and u3 to be defined v ≠ 0. Thus, v = (0, a, a, 0) with a ≠ 0.

x, u1 = x1 + x2 − x3 x, u 2 = −2 x1 − x2 + 2 x3

(b) As in part (a), since x, u1 = x, u 4 = 0,

x, u3 = − x1 + x3

x1 = x4 = 0.

x1, u3 = 0 ⇒ − x1 + x3 = 0, so x1 = x3 . Then x, u1 = x2 and x, u 2 = − x2 , so x2 = 0 and

Since u 2 = u3 = 1 and we want x = 1, then the cosine of the angle between x and u 2 is cos θ 2 = x, u 2 = x2 and, similarly,

x = ( x1, 0, x1 ). Then x = x12 + x12 = 2 x12 = x1

cos θ3 = x, u3 = x3 , so we want x2 = 2 x3 , and x = 0, x2 , 2 x2 , 0 . 168

2.


If x = 1 then x1 = ±

1

Section 6.6

⎛ u, vi ⎞ cos αi = ⎜ ⎜ u v ⎟⎟ i ⎠ ⎝

and the vectors are

2

1 ⎞ ⎛ 1 , 0, ±⎜ ⎟. 2⎠ ⎝ 2

⎛a ⎞ =⎜ i ⎟ ⎝ u ⎠

9. For u = (u1 , u2 ), v = (v1 , v2 ) in R 2 , let

=

u, v = au1v1 + bu2 v2 be a weighted inner

v

2

= a(1)2 + b(2)2 = a + 4b = 1,

2

= a(3)2 + b(−1)2 = 9a + b = 1, and

cos2 α1 +

k (k

(b) As n approaches ∞,

approaches

π 2

1 n

.

+ cos2 α n =

a12 + a22 +

+ an2

a12 + a22 +

+ an2

= 1.

To show that (W ⊥ ) ⊥ ⊆ W , let v be in (W ⊥ ) ⊥ . Since v is in V, we have, by the Projection Theorem, that v = w1 + w 2 where w1 is in W and w 2 is in W ⊥ . By definition, v, w 2 = w1 , w 2 = 0. But v, w 2 = w1 + w 2 , w 2 = w1 , w 2 + w 2 , w 2 = w2, w2

so that w 2 , w 2 = 0. Hence w 2 = 0 and therefore v = w1, so that v is in W. Thus

ui , u = k 2 , so ui u

+ an2

Thus W ⊆ (W ⊥ )⊥ .

be the edges of the ‘cube’ in R n and u = (k, k, k, ..., k) be the diagonal. Then ui = k , u = k n , and k2

a12 + a22 +

to every vector in W ⊥ , so that w is in (W ⊥ ) ⊥ .

11. (a) Let u1 = (k , 0, 0, ..., 0), u 2 = (0, k , 0, ..., 0), ..., u 2 = (0, 0, 0, ..., k )

=

ai2

= 1)

W ⊆ (W ⊥ )⊥ . If w is in W, then w is orthogonal

⎡ 1 4⎤ ⎡1 ⎤ ⎡a ⎤ This leads to the system ⎢9 1⎥ ⎢ ⎥ = ⎢1 ⎥ . ⎢ ⎥ ⎣b ⎦ ⎢ ⎥ ⎣ 3 −2 ⎦ ⎣0 ⎦ ⎡ 1 4 1⎤ ⎡ 1 0 0⎤ Since ⎢9 1 1⎥ reduces to ⎢0 1 0 ⎥ , the ⎢ ⎥ ⎢ ⎥ ⎣0 0 1⎦ ⎣ 3 −2 0 ⎦ system is inconsistent and there is no such weighted inner product.

ui , u

( vi

15. To show that (W ⊥ )⊥ = W , we first show that

u, v = a(1)(3) + b(2)(−1) = 3a − 2b = 0.

cos θ =

2

Therefore

product. If u = (1, 2) and v = (3, −1) form an orthonormal set, then u

2

2

(W ⊥ ) ⊥ ⊆ W . n)

=

1 n

⎡ 1 −1⎤ ⎡ 1 2 4⎤ 17. A = ⎢ 2 3⎥ , AT = ⎢ , ⎢ ⎥ ⎣ −1 3 5⎥⎦ 4 5 ⎣ ⎦ ⎡ 21 25⎤ , AT A = ⎢ ⎣ 25 35 ⎥⎦ ⎡1 ⎤ ⎡ 1 2 4 ⎤ ⎢ ⎥ ⎡ 4 s + 3⎤ AT b = ⎢ 1 = ⎣ −1 3 5⎥⎦ ⎢ s ⎥ ⎢⎣5s + 2 ⎥⎦ ⎣ ⎦ The associated normal system is ⎡ 21 25⎤ ⎡ x1 ⎤ ⎡ 4s + 3⎤ ⎢⎣ 25 35 ⎥⎦ ⎢ x ⎥ = ⎢⎣5s + 2 ⎥⎦ . ⎣ 2⎦ If the least squares solution is x1 = 1 and x2 = 2,

.

approaches 0, so θ

.

13. Recall that u can be expressed as the linear combination u = a1 v1 + + an v n where ai = u, vi for i = 1, ..., n. Thus

⎡ 21 25⎤ ⎡1 ⎤ ⎡ 71⎤ ⎡ 4 s + 3⎤ then ⎢ . = = ⎣ 25 35 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣95⎥⎦ ⎢⎣5s + 2⎥⎦ The resulting equations have solutions s = 17 and s = 18.6, respectively, so no such value of s exists. 169

Chapter 7 Diagonalization and Quadratic Forms A is orthogonal with inverse ⎡− 1 0 1 ⎤ 2⎥ ⎢ 2 1 ⎥ . AT = ⎢ 1 − 2 6 6⎥ ⎢ 6 1 1 ⎥ ⎢ 1 ⎢⎣ 3 3 3 ⎥⎦

Section 7.1 Exercise Set 7.1 1. (b) Since A is orthogonal, ⎡ 4 −9 25 ⎢ 5 −1 T 4 A =A =⎢ 0 5 ⎢ 3 12 ⎢⎣ − 5 − 25

12 ⎤ 25 ⎥ 3⎥. 5 ⎥ 16 25 ⎥⎦

(e) For the given matrix A,

AT A 1 1 1⎤ ⎡1 1 ⎡1 2 2 2⎥ ⎢2 2 ⎢2 5 5 1 1 1 1 −6 −6 ⎢ ⎥⎢ 6 6 2 = ⎢2 ⎥⎢ 1 1 1 −5 1 1 ⎢2 6 6 6⎥ ⎢2 6 ⎢1 1 −5 1⎥ ⎢1 1 ⎢⎣ 2 6 6 6 ⎥⎦ ⎢⎣ 2 6 ⎡ 1 0 0 0⎤ ⎢0 1 0 0⎥ =⎢ ⎥ ⎢0 0 1 0⎥ ⎣⎢ 0 0 0 1⎦⎥ A is orthogonal with inverse 1 1 1⎤ ⎡1 2 2 2⎥ ⎢2 1 −5 1 1⎥ ⎢ 6 6 6 . AT = ⎢ 12 ⎥ 1 1 −5 ⎢2 6 6 6⎥ ⎢1 1 −5 1⎥ ⎢⎣ 2 6 6 6 ⎥⎦

3. (a) For the given matrix A, ⎡ 1 0 ⎤ ⎡1 0 ⎤ ⎡ 1 0 ⎤ AT A = ⎢ = . ⎣ 0 1 ⎥⎦ ⎢⎣0 1 ⎥⎦ ⎢⎣ 0 1 ⎥⎦

⎡1 0 ⎤ A is orthogonal with inverse AT = ⎢ . ⎣ 0 1 ⎥⎦ (b) For the given matrix A, 1 ⎤⎡ 1 ⎡ 1 − 1 ⎤ 2⎥ AT A = ⎢ 12 12 ⎥ ⎢ 12 1 ⎥ ⎢− ⎥⎢ 2⎦⎣ 2 2⎦ ⎣ 2 ⎡ 1 0⎤ =⎢ . ⎣ 0 1⎥⎦ A is orthogonal with inverse 1 ⎤ ⎡ 1 AT = ⎢ 12 12 ⎥ . ⎢− ⎥ 2⎦ ⎣ 2 Note that the given matrix is the standard matrix for a rotation of 45°. 2

(c)

⎛ 1 ⎞ r1 = 02 + 12 + ⎜ ⎟ = ⎝ 2⎠ matrix is not orthogonal.

2

(f)

3 ≠ 1 so the 2

1 ⎤ ⎡− 2⎥⎢ 1 ⎥⎢ 6⎥⎢ 1 ⎥⎢ 3 ⎥⎦ ⎢⎣

1 2

0 − 1 2

1 6 2 6 1 6

2 7 ⎛ 1 ⎞ ⎛ 1⎞ ≠1 r2 = ⎜ ⎟ +⎜− ⎟ = 12 ⎝ 3⎠ ⎝ 2⎠ The matrix is not orthogonal.

1 π 3 = , sin = 2 3 2 3⎤ ⎡ 1 ⎡ x′ ⎤ ⎢ 2 2 ⎥ ⎡ −2⎤ ⎡ −1 + 3 3 ⎤ = =⎢ ⎥ ⎢⎣ y ′⎥⎦ ⎢ 3 1 ⎥ ⎢⎣ 6 ⎥⎦ ⎣ 3 + 3 ⎦ − ⎣ 2 2⎦

7. (a) cos

(d) For the given matrix A,

AT A ⎡− 1 0 ⎢ 2 =⎢ 1 − 2 6 ⎢ 6 1 ⎢ 1 ⎢⎣ 3 3 ⎡1 0 0⎤ = ⎢ 0 1 0⎥ ⎢ ⎥ ⎣0 0 1⎦

1 2 1 6 1 6 − 56

1 ⎤ 3⎥ 1 ⎥ 3⎥ 1 ⎥ 3 ⎥⎦

π

3

Thus, ( x′, y ′) = ( −1 + 3 3, 3 + 3 ) .

170

1⎤ 2⎥ 1⎥ 6 ⎥ − 56 ⎥ 1⎥ 6 ⎥⎦


Section 7.1

⎡ sin θ ⎤ [u3′ ]B = ⎢ 0 ⎥ . Thus the transition matrix ⎢ ⎥ ⎣cos θ ⎦ ⎡ cos θ 0 sin θ ⎤ from B′ to B is P = ⎢ 0 1 0 ⎥, ⎢ ⎥ θ θ⎦ sin 0 cos − ⎣ ⎡x⎤ ⎡ x′ ⎤ i.e., ⎢ y ⎥ = P ⎢ y ′⎥ . Then ⎢ ⎥ ⎢ ′⎥ ⎣z⎦ ⎣z ⎦ ⎡cos θ 0 − sin θ ⎤ −1 ⎢ A=P = 0 1 0 ⎥. ⎢ ⎥ ⎣ sin θ 0 cos θ ⎦

(b) Since a rotation matrix is orthogonal, Equation (2) gives 3⎤ ⎡ 1 ⎡5 ⎤ ⎡ x ⎤ ⎢ 2 − 2 ⎥ ⎡5⎤ ⎢ 2 − 3 ⎥ . = = ⎢⎣ y ⎥⎦ ⎢ 3 1 ⎥ ⎢⎣ 2 ⎥⎦ ⎢ 5 3 + 1⎥ 2 ⎣ ⎦ 2⎦ ⎣2 5 ⎛5 ⎞ Thus, ( x, y) = ⎜ − 3, 3 + 1⎟ . ⎝2 ⎠ 2

9. If B = {u1 , u 2 , u3} is the standard basis for R3 , and B ′ = {u1′ , u′2 , u3′ } is the rotated basis, then

the transition matrix from B′ to B is ⎡ cos π 0 sin π ⎤ ⎡ 1 0 3 3⎥ ⎢ 2 ⎢ P=⎢ 0 1 0 ⎥=⎢ 0 1 ⎢ − sin π 0 cos π ⎥ ⎢ 3 0 3 3 ⎦ ⎢⎣ − 2 ⎣

3⎤ 2 ⎥

0⎥ 1 ⎥ 2 ⎥⎦

⎡1 ⎤ (b) With the same notation, [u1′ ]B = ⎢ 0 ⎥ , ⎢ ⎥ ⎣0⎦ ⎡ 0 ⎤ ⎡ 0 ⎤ ⎢ ⎥ ′ ′ [u 2 ]B = cos θ , and [u3 ]B = ⎢ − sin θ ⎥ , so ⎢ ⎥ ⎢ ⎥ ⎣ sin θ ⎦ ⎣ cos θ ⎦ the transition matrix from B′ to B is 0 0 ⎤ ⎡1 P = ⎢ 0 cos θ − sin θ ⎥ and ⎢ ⎥ ⎣ 0 sin θ cos θ ⎦ 0 0 ⎤ ⎡1 A = ⎢ 0 cos θ sin θ ⎥ . ⎢ ⎥ ⎣ 0 − sin θ cos θ ⎦

⎡ x′ ⎤ ⎡ x⎤ (a) ⎢ y ′⎥ = P −1 ⎢ y ⎥ ⎢ ′⎥ ⎢ ⎥ ⎣z ⎦ ⎣z⎦ ⎡ 1 0 − 3⎤ 2 ⎥ ⎡ −1⎤ ⎢ 2 =⎢ 0 1 0 ⎥ ⎢ 2⎥ ⎢ 3 ⎥ ⎢ 5⎥ 1 ⎣ ⎦ ⎢⎣ 2 0 2 ⎥⎦ ⎡− 1 − 5 3 ⎤ ⎢ 2 2 ⎥ =⎢ 2 ⎥ ⎢− 1 3 + 5 ⎥ 2⎦ ⎣ 2 Thus

⎡a + b b − a ⎤ 13. Let A = ⎢ . Then ⎣ a − b b + a ⎥⎦ ⎡ 2(a 2 + b 2 ) ⎤ 0 AT A = ⎢ ⎥ , so a and b 0 2(a 2 + b2 ) ⎦⎥ ⎣⎢

1 5⎞ ⎛ 1 5 ( x′, y ′, z ′) = ⎜ − − 3, 2, − 3 + ⎟. ⎝ 2 2 2 2⎠ ⎡ 1 ⎡x⎤ ⎡ x′ ⎤ ⎢ 2 (b) ⎢ y ⎥ = P ⎢ y ′⎥ = ⎢ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎣z⎦ ⎣ z′ ⎦ ⎢− 3 ⎣ 2

0 1 0

3⎤ 2 ⎥⎡

1⎤ 0 ⎥ ⎢ 6⎥ ⎢ ⎥ 1 ⎥ ⎣ − 3⎦ 2 ⎥⎦

1 must satisfy a 2 + b 2 = . 2

⎡ 1−3 3 ⎤ ⎢ 2 2 ⎥ =⎢ 6 ⎥ ⎢− 1 3 − 3 ⎥ ⎣ 2 2⎦ 1 3⎞ ⎛1 3 Thus ( x, y, z ) = ⎜ − 3, 6, − 3 − ⎟. ⎝2 2 2 2⎠

17. The row vectors of an orthogonal matrix are an orthonormal set. 2

2

⎛ 1 ⎞ ⎛ 1 ⎞ 2 = a2 + ⎜ ⎟ +⎜− ⎟ = a +1 2 2 ⎝ ⎠ ⎝ ⎠ Thus a = 0. r1

2

2

2

2

2

⎛ 1 ⎞ ⎛ 1 ⎞ 2 2 r2 = b + ⎜ ⎟ +⎜ ⎟ =b +6 ⎝ 6⎠ ⎝ 6⎠ 2 . Thus b = ± 6 2

11. (a) If B = {u1 , u 2 , u3} is the standard basis for R3 and B ′ = {u1′ , u′2 , u′3}, then

⎡ cos θ ⎤ ⎡0 ⎤ [u1′ ]B = ⎢ 0 ⎥ , [u′2 ]B = ⎢1 ⎥ , and ⎢ ⎥ ⎢ ⎥ ⎣ − sin θ ⎦ ⎣0 ⎦

r3 171

2

2

⎛ 1 ⎞ ⎛ 1 ⎞ 2 2 = c2 + ⎜ ⎟ +⎜ ⎟ =c +3 3 3 ⎝ ⎠ ⎝ ⎠

Chapter 7: Diagonalization and Quadratic Forms

Thus c = ±

1


Section 7.2

.

3 The column vectors of an orthogonal matrix are an orthonormal set, from which it is clear that b and c must have opposite signs. Thus the only 2 1 possibilities are a = 0, b = , c=− or 6 3 2 1 , c= . a = 0, b = − 2 3

Exercise Set 7.2 1. (a)

The characteristic equation is λ 2 − 5λ = 0 and the eigenvalues are λ = 0 and λ = 5. Both eigenspaces are one-dimensional.

21. (a) Rotations about the origin, reflections about any line through the origin, and any combination of these are rigid operators.

(b)

λ3 − 27λ − 54 = 0 and the eigenvalues are λ = 6 and λ = −3. The eigenspace for λ = 6 is one-dimensional; the eigenspace for λ = −3 is two-dimensional.

(c) All rigid operators on R 2 are angle preserving. Dilations and contractions are angle preserving operators that are not rigid.

(c)

True/False 7.1

λ − 1 −1 −1 −1 λ − 1 −1 = λ3 − 3λ 2 = λ 2 (λ − 3) −1 −1 λ − 1 The characteristic equation is λ3 − 3λ 2 = 0 and the eigenvalues are λ = 3 and λ = 0. The eigenspace for λ = 3 is one-dimensional; the eigenspace for λ = 0 is two-dimensional.

(a) False; only square matrices can be orthogonal. (b) False; the row and column vectors are not unit vectors. (c) False; only square matrices can be orthogonal.

(d)

(d) False; the column vectors must form an orthonormal set.

λ − 4 −2 −2 −2 λ − 4 −2 = λ3 − 12λ 2 + 36λ − 32 −2 −2 λ − 4 = (λ − 8)(λ − 2) 2 The characteristic equation is

(e) True; since AT A = I for an orthogonal matrix A, A must be invertible.

λ3 − 12λ 2 + 36λ − 32 = 0 and the eigenvalues are λ = 8 and λ = 2. The eigenspace for λ = 8 is one-dimensional; the eigenspace for λ = 2 is two-dimensional.

(f) True; a product of orthogonal matrices is orthogonal, so A2 is orthogonal, hence det( A2 ) = (det A)2 = (±1)2 = 1. (g) True; since Ax = x for an orthogonal matrix.

vector, so A

λ −1 4 −2 4 2 = λ3 − 27λ − 54 λ −1 2 −2 λ+2 = (λ − 6)(λ + 3)2 The characteristic equation is

(b) Rotations about the origin, dilations, contractions, reflections about lines through the origin, and combinations of these are angle preserving.

(h) True; for any nonzero vector x,

λ − 1 −2 = λ 2 − 5λ −2 λ − 4

(e)

x is a unit x

λ − 4 −4 −4 λ − 4 0 0 0 0

0 0 λ 0

0 0 = λ 4 − 8λ3 = λ3 (λ − 8) 0 λ

The characteristic equation is λ 4 − 8λ3 = 0 and the eigenvalues are λ = 0 and λ = 8. The eigenspace for λ = 0 is three-dimensional; the eigenspace for λ = 8 is one-dimensional.

x = 1. It follows that x

1 Ax = 1 and Ax = x , so A is orthogonal x by Theorem 7.1.3. 172


(f)

Section 7.2

⎡3⎤ ⎡3⎤ 4 ⎢5⎥ x x3 = ⎢ 0 ⎥ . v3 = 3 = ⎢ 0 ⎥ is an orthonormal ⎢ ⎥ x3 ⎢4⎥ ⎣⎢ 1 ⎦⎥ ⎣5⎦ 4 ⎡− 0 35 ⎤ ⎢ 5 ⎥ basis. Thus P = ⎢ 0 1 0 ⎥ orthogonally ⎢ 3 0 4⎥ 5⎦ ⎣ 5 diagonalizes A and 0 ⎤ ⎡ 25 0 ⎡λ1 0 0⎤ P −1 AP = ⎢ 0 λ 2 0 ⎥ = ⎢ 0 −3 0⎥ . ⎢ ⎥ ⎢ ⎥ − 0 0 50 ⎦ ⎣ 0 0 λ3 ⎦ ⎣

λ−2 1 0 0 1 λ−2 0 0 0 0 λ−2 1 0 0 1 λ−2 = λ 4 − 8λ3 + 22λ 2 − 24λ + 9 = (λ − 1)2 (λ − 3)2 The characteristic equation is λ 4 − 8λ3 + 22λ 2 − 24λ + 9 = 0 and the eigenvalues are λ = 1 and λ = 3. Both eigenspaces are two-dimensional.

3. The characteristic equation of A is λ 2 − 13λ + 30 = (λ − 3)(λ − 10) = 0. λ1 = 3: A basis for the eigenspace is

7. The characteristic equation of A is λ3 − 6λ 2 + 9λ = λ(λ − 3)2 = 0.

⎡− 2 ⎤ ⎡− 2 ⎤ x1 ⎢ 7 ⎥ x1 = ⎢ 3 ⎥ . v1 = = is an x1 ⎢ 3 ⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 7 ⎥⎦ orthonormal basis. λ 2 = 10: A basis for the eigenspace is ⎡ ⎡ 3⎤ x x2 = ⎢ 2 ⎥ . v 2 = 2 = ⎢ ⎢ x2 ⎣⎢ 1 ⎦⎥ ⎢⎣

3⎤ 7⎥ 2 ⎥ 7 ⎥⎦

⎡1⎤ λ1 = 0: A basis for the eigenspace is x1 = ⎢1⎥ . ⎢⎥ ⎣1⎦ 1 ⎡ ⎤ ⎢ 3⎥ x v1 = 1 = ⎢ 1 ⎥ is an orthonormal basis. x1 ⎢ 3 ⎥ ⎢1 ⎥ ⎢⎣ 3 ⎥⎦ ⎡ −1⎤ λ 2 = 3: A basis for the eigenspace is x 2 = ⎢ 1⎥ , ⎢ ⎥ ⎣ 0⎦

is an

⎡− ⎢ orthonormal basis. Thus P = ⎢ ⎢⎣ orthogonally diagonalizes A and 0 ⎤ ⎡ 3 0⎤ ⎡λ P −1 AP = ⎢ 1 ⎥=⎢ ⎥. λ 0 2 ⎦ ⎣ 0 10 ⎦ ⎣

2 7 3 7

3⎤ 7⎥

2 ⎥ ⎥ 7⎦

⎡ −1⎤ x3 = ⎢ 0⎥ . The Gram-Schmidt process gives the ⎢ ⎥ ⎣ 1⎦ ⎡− 1 ⎤ ⎡− 1 ⎤ ⎢ 6⎥ ⎢ 2⎥ 1 1 orthonormal basis v 2 = ⎢ ⎥ , v3 = ⎢− ⎥ . 6⎥ ⎢ ⎢ 2⎥ 2 ⎥ ⎢ ⎣⎢ 0 ⎦⎥ ⎣⎢ 6 ⎦⎥

5. The characteristic equation of A is λ3 + 28λ 2 − 1175λ − 3750 = (λ − 25)(λ + 3)(λ + 50) = 0. λ1 = 25: A basis for the eigenspace is

⎡1 − ⎢ 3 Thus, P = ⎢ 1 ⎢ 3 ⎢1 ⎢⎣ 3 diagonalizes A and ⎡λ1 0 P −1 AP = ⎢ 0 λ 2 ⎢ ⎣0 0

⎡− 4 ⎤ ⎡− 4 ⎤ 3⎥ x1 ⎢ 5 ⎥ ⎢ = ⎢ 0 ⎥ is an orthonormal x1 = 0 . v1 = ⎢ ⎥ x1 ⎢ 3 ⎥ ⎢⎣ 1⎥⎦ ⎣ 5⎦ basis. λ 2 = −3: A basis for the eigenspace is ⎡ 0⎤ x 2 = ⎢1 ⎥ = v 2 , since x 2 = 1. ⎢ ⎥ ⎣ 0⎦

− 1 ⎤ 6⎥ − 1 ⎥ orthogonally 6⎥ 2 ⎥ 0 6 ⎥⎦

1 2 1 2

0 ⎤ ⎡0 0 0 ⎤ 0 ⎥ = ⎢0 3 0 ⎥ . ⎥ ⎢ ⎥ λ 2 ⎦ ⎣0 0 3 ⎦

9. The characteristic equation of A is

λ 4 − 1250λ 2 + 390, 625 = (λ + 25)2 (λ − 25)2 = 0. λ1 = −25: A basis for the eigenspace is

λ3 = −50: A basis for the eigenspace is

173



⎡− 4 ⎤ ⎡ 0⎤ ⎢ 3⎥ ⎢ 0⎥ x1 = ⎢ 1⎥ , x 2 = ⎢ 4 ⎥ . The Gram-Schmidt ⎢ 0⎥ ⎢− 3 ⎥ ⎢ 0⎥ ⎢ 1⎥ ⎣ ⎦ ⎣ ⎦ ⎡− 4 ⎤ ⎢ 5⎥ 3 process gives the orthonormal basis v1 = ⎢ 5 ⎥ , ⎢ ⎥ ⎢ 0⎥ ⎢⎣ 0 ⎥⎦ ⎡ 0⎤ ⎢ 0⎥ v 2 = ⎢− 4 ⎥ . ⎢ 5⎥ ⎢ 3⎥ ⎢⎣ 5 ⎥⎦ λ 2 = 25: A basis for the eigenspace is

15. No, a non-symmetric matrix A can have eigenvalues that are real numbers. For instance, ⎡3 0 ⎤ the eigenvalues of the matrix ⎢ are 3 and ⎣8 −1⎥⎦ −1. 17. If A is an orthogonal matrix then its eigenvalues have absolute value 1, but may be complex. Since the eigenvalues of a symmetric matrix must be real numbers, the only possible eigenvalues for an orthogonal symmetric matrix are 1 and −1. 19. Yes; to diagonalize A, follow the steps on page 399. The Gram-Schmidt process will ensure that columns of P corresponding to the same eigenvalue are an orthonormal set. Since eigenvectors from distinct eigenvalues are orthogonal, this means that P will be an orthogonal matrix. Then since A is orthogonally diagonalizable, it must be symmetric.

⎡3⎤ ⎡0⎤ ⎢4⎥ ⎢0⎥ x3 = ⎢ 1 ⎥ , x 4 = ⎢ 3 ⎥ . The Gram-Schmidt ⎢0⎥ ⎢4⎥ ⎢0⎥ ⎢1 ⎥ ⎣ ⎦ ⎣ ⎦

True/False 7.2

⎡3⎤ ⎢5⎥ 4 process gives the orthonormal basis v3 = ⎢ 5 ⎥ , ⎢ ⎥ ⎢0⎥ ⎢⎣ 0 ⎥⎦ ⎡0⎤ ⎢0⎥ v 4 = ⎢ 3 ⎥ . Thus, ⎢5⎥ ⎢4⎥ ⎢⎣ 5 ⎥⎦ ⎡− 4 3 0 0⎤ ⎢ 5 5 ⎥ ⎢ 53 54 0 0⎥ P = [ v1 v 2 v3 v 4 ] = ⎢ ⎥ 4 3 ⎢ 0 0 − 5 5⎥ ⎢ 3 4⎥ 0 0 5 5 ⎦⎥ ⎣⎢ orthogonally diagonalizes A and ⎡λ1 0 0 0 ⎤ ⎢0 λ 0 0⎥ 2 P −1 AP = ⎢ ⎥ ⎢ 0 0 λ1 0 ⎥ ⎢⎣ 0 0 0 λ 2 ⎥⎦

(a) True; for any square matrix A, both AAT and AT A are symmetric, hence orthogonally diagonalizable.

(b) True; since v1 and v 2 are from distinct eigenspaces of a symmetric matrix, they are orthogonal, so

v1 + v 2

2

= v1 + v 2 , v1 + v 2 = v1, v1 + 2 v1, v 2 + v 2 , v 2 = v1

2

+ 0 + v2

2

.

(c) False; an orthogonal matrix is not necessarily symmetric. (d) True; ( P −1 AP)−1 = P −1 A−1P since P −1 and A−1 both exist and since P −1 AP is diagonal, so

is ( P −1 AP)−1 . (e) True; since Ax = x for an orthogonal matrix.

0 0⎤ ⎡ −25 0 ⎢ 0 25 0 0⎥ =⎢ ⎥. ⎢ 0 0 −25 0 ⎥ 0 25⎥⎦ ⎢⎣ 0 0

(f) True; if A is an n × n orthogonally diagonalizable matrix, then A has an orthonormal set of n eigenvectors, which form a basis for R n .

174


Section 7.3

⎡− 2 ⎤ ⎢ 3⎥ bases for the eigenspaces are λ = 1: ⎢ 23 ⎥ ; ⎢ 1⎥ ⎣⎢ 3 ⎦⎥

(g) True; if A is orthogonally diagonalizable, then A must be symmetric, so the eigenvalues of A will all be real numbers. Section 7.3

⎡2⎤ ⎢3⎥ λ = 4: ⎢ 13 ⎥ ; λ = 7: ⎢2⎥ ⎢⎣ 3 ⎥⎦

Exercise Set 7.3 1. (a) 3x12 + 7 x22 = [ x1

⎡ 3 0 ⎤ ⎡ x1 ⎤ x2 ] ⎢ ⎢ ⎥ ⎣ 0 7 ⎥⎦ ⎣ x2 ⎦

(b) 4 x12 − 9 x22 − 6 x1 x2 = [ x1

⎡− 2 ⎢ 3 For P = ⎢ 23 ⎢ 1 ⎢⎣ 3

⎡ 4 −3⎤ ⎡ x1 ⎤ x2 ] ⎢ ⎢ ⎥ ⎣ −3 −9 ⎦⎥ ⎣ x2 ⎦

(c) 9 x12 − x22 + 4 x32 + 6 x1 x2 − 8 x1 x3 + x2 x3 ⎡ 9 3 −4 ⎤ ⎡ x ⎤ 1 ⎢ 1⎥ = [ x1 x2 x3 ] ⎢ 3 −1 2 ⎥ ⎢ x2 ⎥ ⎢ ⎥ ⎢ −4 1 4 ⎥⎦ ⎣ x3 ⎦ ⎣ 2 3. [ x

1⎤ 3⎥ 2⎥, 3 ⎥ 2 − 3 ⎥⎦

let x = Py, then

xT Ax = yT ( PT AP)y

= [ y1

y2

⎡1 0 0 ⎤ ⎡ y1 ⎤ and y3 ] ⎢ 0 4 0 ⎥ ⎢ y2 ⎥ ⎢ ⎥⎢ ⎥ ⎣ 0 0 7 ⎦ ⎣ y3 ⎦

Q = y12 + 4 y22 + 7 y32 .

⎡ 2 −3⎤ ⎡ x ⎤ = 2 x 2 + 5 y 2 − 6 xy y] ⎢ ⎥ ⎢ ⎥ − 3 5 y ⎣ ⎦⎣ ⎦

9. (a) 2 x 2 + xy + x − 6 y + 2 = 0 can be written as

2 x 2 + 0 y 2 + xy + ( x − 6 y) + 2 = 0 or

⎡ 2 −1⎤ ⎡ x1 ⎤ 5. Q = x Ax = [ x1 x2 ] ⎢ ⎢ ⎥ ⎣ −1 2 ⎥⎦ ⎣ x2 ⎦ The characteristic equation of A is T

[x

λ 2 − 4λ + 3 = (λ − 3)(λ − 1) = 0, so the eigenvalues are λ = 3, 1. Orthonormal bases for

⎡2 y] ⎢ ⎢⎣ 12

1⎤ 2 ⎥ ⎡ x ⎤ + [1

0⎥⎦ ⎢⎣ y ⎥⎦

⎡ x⎤ −6] ⎢ ⎥ + 2 = 0. ⎣ y⎦

(b) y 2 + 7 x − 8 y − 5 = 0 can be written as

⎡ 1 ⎤ the eigenspaces are λ = 3: ⎢ 12 ⎥ ; λ = 1: ⎢− ⎥ ⎣ 2⎦ ⎡ 1 ⎤ ⎢ 2 ⎥. ⎢ 1 ⎥ ⎣ 2⎦ 1 ⎤ ⎡ 1 2 2⎥ ⎢ For P = , let x = Py, then 1 ⎥ ⎢− 1 2⎦ ⎣ 2 ⎡ 3 0 ⎤ ⎡ y1 ⎤ xT Ax = yT ( PT AP)y = [ y1 y2 ] ⎢ ⎢ ⎥ ⎣ 0 1 ⎥⎦ ⎣ y2 ⎦

0 x 2 + y 2 + 0 xy + (7 x − 8 y) − 5 = 0 or [x

⎡0 0 ⎤ ⎡ x ⎤ ⎡ x⎤ + [7 −8] ⎢ ⎥ − 5 = 0. y] ⎢ ⎥ ⎢ ⎥ ⎣0 1 ⎦ ⎣ y ⎦ ⎣ y⎦

x2 y 2 + = 1 which is the 10 4 equation of an ellipse.

11. (a) 2 x 2 + 5 y 2 = 20 is

(b) x 2 − y 2 − 8 = 0 is x 2 − y 2 = 8 or

x2 y 2 − = 1 which is the equation of a 8 8 hyperbola.

and Q = 3 y12 + y22 .

7 2 y which is the 2 equation of a parabola.

⎡ 3 2 0 ⎤ ⎡ x1 ⎤ 7. Q = x Ax = [ x1 x2 x3 ] ⎢ 2 4 −2 ⎥ ⎢ x2 ⎥ ⎢ ⎥⎢ ⎥ 5⎦ ⎣ x3 ⎦ ⎣ 0 −2 The characteristic equation of A is

(c) 7 y 2 − 2 x = 0 is x =

T

3

2 3 1 3 2 3

⎡ 1⎤ ⎢ 3⎥ ⎢ 23 ⎥ . ⎢ 2⎥ ⎢⎣ − 3 ⎥⎦

(d) x 2 + y 2 − 25 = 0 is x 2 + y 2 = 25 which is the equation of a circle.

2

λ − 12λ + 39λ − 28 = (λ − 1)(λ − 4)(λ − 7) = 0, so the eigenvalues are λ = 1, 4, 7. Orthonormal 175



⎡ −1 0 ⎤ (b) The eigenvalues of ⎢ are λ = −1 ⎣ 0 −2 ⎥⎦ and λ = −2, so the matrix is negative definite.

13. The equation can be written in matrix form as ⎡ 2 −2 ⎤ xT Ax = −8 where A = ⎢ . The ⎣ −2 −1⎥⎦ eigenvalues of A are λ1 = 3 and λ 2 = −2, with

⎡ 2⎤ corresponding eigenvectors v1 = ⎢ ⎥ and ⎣ −1⎦ ⎡1 ⎤ v 2 = ⎢ ⎥ respectively. Thus the matrix ⎣2⎦ 1 ⎤ ⎡ 2 P = ⎢ 15 25 ⎥ orthogonally diagonalizes A. ⎢− ⎥ 5⎦ ⎣ 5 Note that det(P) = 1, so P is a rotation matrix. The equation of the conic in the rotated x ′y ′-coordinate system is [ x′

⎡ −1 0 ⎤ are λ = −1 and (c) The eigenvalues of ⎢ ⎣ 0 2 ⎥⎦ λ = 2, so the matrix is indefinite. ⎡1 0 ⎤ (d) The eigenvalues of ⎢ are λ = 1 and ⎣0 0 ⎥⎦ λ = 0, so the matrix is positive semidefinite. ⎡0 0 ⎤ (e) The eigenvalues of ⎢ are λ = 0 and ⎣0 −2 ⎥⎦ λ = −2, so the matrix is negative semidefinite.

⎡ 3 0 ⎤ ⎡ x′ ⎤ = −8 which can be written y ′] ⎢ ⎣ 0 −2 ⎦⎥ ⎣⎢ y ′⎦⎥

as 2 y ′2 − 3x′2 = 8; thus the conic is a hyperbola. The angle through which the axes have been ⎛ 1⎞ rotated is θ = tan −1 ⎜ − ⎟ ≈ −26.6°. ⎝ 2⎠

19. We have Q = x12 + x22 > 0 for ( x1 , x2 ) ≠ (0, 0); thus Q is positive definite. 21. We have Q = ( x1 − x2 )2 > 0 for x1 ≠ x2 and Q = 0 for x1 = x2 ; thus Q is positive semidefinite.

15. The equation can be written in matrix form as ⎡11 12 ⎤ xT Ax = 15 where A = ⎢ . The ⎣12 4 ⎥⎦ eigenvalues of A are λ1 = 20 and λ 2 = −5, with

23. We have x12 − x22 > 0 for x1 ≠ 0, x2 = 0 and Q < 0 for x1 = 0, x2 ≠ 0; thus Q is indefinite.

⎡4⎤ corresponding eigenvectors v1 = ⎢ ⎥ and ⎣3⎦ ⎡ −3⎤ v 2 = ⎢ ⎥ respectively. Thus the matrix ⎣ 4⎦ ⎡4 − 3⎤ 5 ⎥ orthogonally diagonalizes A. Note P = ⎢5 4⎥ ⎢ 35 5⎦ ⎣ that det(P) = 1, so P is a rotation matrix. The equation of the conic in the rotated x ′y ′-coordinate system is [ x′

25. (a) The eigenvalues of the matrix ⎡ 5 −2 ⎤ are λ = 3 and λ = 7; thus A is A=⎢ 5⎥⎦ ⎣ −2 positive definite. Since 5 = 5 and 5 −2 = 21 are positive, we reach the −2 5 same conclusion using Theorem 7.3.4.

⎡ 2 −1 0 ⎤ (b) The eigenvalues of A = ⎢ −1 2 0 ⎥ are ⎢ ⎥ ⎣ 0 0 5⎦ λ = 1, λ = 3, and λ = 5; thus A is positive definite. The determinants of the principal 2 −1 submatrices are 2 = 2, = 3, and −1 2

⎡ 20 0⎤ ⎡ x ′ ⎤ = 15 which we can write y ′] ⎢ ⎣ 0 −5⎦⎥ ⎣⎢ y ′⎦⎥

as 4 x ′2 − y ′2 = 3; thus the conic is a hyperbola. The angle through which the axes have been ⎛3⎞ rotated is θ = tan −1 ⎜ ⎟ ≈ 36.9°. ⎝4⎠

2 −1 0 −1 2 0 = 15; thus we reach the same 0 0 5 conclusion using Theorem 7.3.4.

⎡1 0 ⎤ 17. (a) The eigenvalues of ⎢ are λ = 1 and ⎣0 2 ⎥⎦ λ = 2, so the matrix is positive definite. 176


Section 7.3

27. The quadratic form Q = 5 x12 + x22 + kx32 + 4 x1 x2 − 2 x1 x3 − 2 x2 x3 can be expressed in matrix notation as Q = xT Ax

⎡ 5 2 where A = ⎢ 2 1 ⎢ ⎣ −1 −1 5 2 −1 2 1 −1 = k − 2. −1 −1 k

−1⎤ 5 2 = 1, and −1⎥ . The determinants of the principal submatrices of A are 5 = 5, 2 1 ⎥ k⎦ Thus Q is positive definite if and only if k > 2.

31. (a) For each i = 1, ..., n we have

( xi − x )2 = xi2 − 2 xi x + x 2 2

n ⎞ 1 n 1 ⎛ ⎜ ∑ xj ⎟ = xi2 − 2 xi ∑ x j + n j =1 n2 ⎜⎝ j =1 ⎟⎠ n n n ⎞ 2 n 1 ⎛ 2 = xi2 − ∑ xi x j + ⎜ ∑ x j + 2 ∑ ∑ x j xk ⎟ . 2 ⎟ n j =1 n ⎜⎝ j =1 j =1 k = j +1 ⎠ 1 Thus in the quadratic form s x2 = [( x1 − x )2 + ( x2 − x )2 + " + ( xn − x )2 ] the coefficient of xi2 is n −1 1 ⎡ 2 1 ⎤ 1 1 ⎡ 2 2 2 ⎤ 2 1− + n = , and the coefficient of xi x j for i ≠ j is n =− − − + . It follows n − 1 ⎢⎣ n n 2 ⎥⎦ n n − 1 ⎢⎣ n n n 2 ⎥⎦ n(n − 1)

⎡ 1 ⎢ n ⎢− 1 that s x2 = xT Ax where A = ⎢ n( n −1) ⎢ # ⎢− 1 ⎣⎢ n( n −1)

− n( n1−1) " − n( n1−1) ⎤ ⎥ 1 1 ⎥ " − n n( n −1) ⎥ . # % # ⎥ 1 ⎥ − n( n1−1) " ⎥⎦ n

1 [( x1 − x ) 2 + ( x2 − x )2 + " + ( xn − x )2 ] ≥ 0, and s x2 = 0 if and only if x1 = x , (b) We have s x2 = n −1 x2 = x , ..., xn = x , i.e., if and only if x1 = x2 = " = xn . Thus s x2 is a positive semidefinite form. 33. The eigenvalues of A must be positive and equal to each other. That is, A must have a positive eigenvalue of multiplicity 2. True/False 7.3 (a) True (b) False; because of the term 4 x1 x2 x3 . (c) True; ( x1 − 3x2 )2 = x12 − 6 x1 x2 + 9 x22 . (d) True; none of the eigenvalues will be 0. (e) False; a symmetric matrix can be positive semidefinite or negative semidefinite. (f) True (g) True; x ⋅ x = x12 + x22 + " + xn2 177



⎡0 ⎤ ⎡0 ⎤ v 2 = ⎢1 ⎥ , v3 = ⎢0 ⎥ . Thus the constrained ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣1 ⎦ maximum is 9 occurring at (x, y, z) = (±1, 0, 0), and the constrained minimum is 3 occurring at (x, y, z) = (0, 0, ±1).

(h) True (i) False; this is only true for symmetric matrices. (j) True (k) False; there will be no cross product terms if aij = −a ji for all i ≠ j.

2

2 ⎛x⎞ ⎛ y ⎞ 7. Rewrite the constraint as ⎜ ⎟ + ⎜ ⎟ = 1, ⎝2⎠ ⎝ 2 ⎠

(l) False; if c < 0, xT Ax = c has no graph.

then let x = 2 x1 and y = 2 y1 . The problem is

Section 7.4

now to maximize z = xy = 2 2 x1 y1 subject to

Exercise Set 7.4

x12 + y12 = 1. Write

1. The quadratic form 5 x 2 − y 2 can be written in

z = xT Ax = [ x1

⎡5 0⎤ . matrix notation as xT Ax where A = ⎢ ⎣ 0 −1⎥⎦ The eigenvalues of A are λ1 = 5 and λ 2 = −1,

Since

⎡1 ⎤ with corresponding eigenvectors v1 = ⎢ ⎥ and ⎣0 ⎦ ⎡0 ⎤ v 2 = ⎢ ⎥ . Thus the constrained maximum is 5 ⎣1 ⎦ occurring at ( x, y) = (±1, 0), and the constrained

⎡ 0 y1 ] ⎢ ⎣ 2

2 ⎤ ⎡ x1 ⎤ ⎥ ⎢ ⎥. 0 ⎦ ⎣ y1 ⎦

λ − 2 = λ 2 − 2 = ( λ + 2 )( λ − 2 ) , − 2 λ

the largest eigenvalue of A is

2, with

⎡ corresponding positive unit eigenvector ⎢ ⎢ ⎣ Thus the maximum value is ⎛ 1 ⎞⎛ 1 ⎞ z = 2 2⎜ ⎟⎜ ⎟ = 2 occurs when ⎝ 2 ⎠⎝ 2 ⎠

minimum is −1 occurring at ( x, y) = (0, ± 1). 3. The quadratic form 3x 2 + 7 y 2 can be written in

1 ⎤ 2⎥ . 1 ⎥ 2⎦

x = 2 x1 = 2, y = 2 y1 = 1 or x = − 2, y = −1. The smallest eigenvalue of A is − 2,

⎡3 0⎤ . matrix notation as xT Ax where A = ⎢ ⎣0 7 ⎥⎦ The eigenvalues of A are λ1 = 7 and λ 2 = 3,

⎡− with corresponding unit eigenvectors ± ⎢ ⎢ ⎣

⎡0 ⎤ with corresponding unit eigenvectors v1 = ⎢ ⎥ ⎣1 ⎦ ⎡1 ⎤ and v 2 = ⎢ ⎥ . Thus the constrained maximum is ⎣0 ⎦ 7 occurring at (x, y) = (0, ±1), and the constrained minimum is 3 occurring at (x, y) = (±1, 0).

1 ⎤ 2⎥ . 1 ⎥ 2⎦

Thus the minimum value is z = − 2 which occurs at ( x, y) = ( − 2 , 1) and

9.

(0, 1)

5 x 2 − y 2 = −1

(–1, 0)

⎡9 0 0 ⎤ expressed as x Ax where A = ⎢ 0 4 0 ⎥ . The ⎢ ⎥ ⎣0 0 3⎦ eigenvalues of A are λ1 = 9, λ 2 = 4, λ3 = 3,

(0, –1)

T

⎡1 ⎤ with corresponding eigenvectors v1 = ⎢0 ⎥ , ⎢ ⎥ ⎣0 ⎦

2 , − 1) .

5x2 − y2 = 5 y 5 x2 + y 2 = 1

5. The quadratic form 9 x 2 + 4 y 2 + 3z 2 can be

(

x (1, 0)

5

13. The first partial derivatives of f are f x ( x, y) = 3 x 2 − 3 y and f y ( x, y) = −3x − 3 y 2 . To find the critical points we set f x and f y 178


Section 7.5

17. The problem is to maximize z = 4xy subject to

equal to zero. This yields the equations y = x 2

2

2

⎛ x⎞ ⎛ y⎞ x 2 + 25 y 2 = 25, or ⎜ ⎟ + ⎜ ⎟ = 1. Let ⎝5⎠ ⎝1⎠ x = 5 x1 and y = y1, so that the problem is to

and x = − y 2 . From this we conclude that y = y 4 and so y = 0 or y = 1. The corresponding values of x are x = 0 and x = −1 respectively. Thus there are two critical points: (0, 0) and (−1, 1). The Hessian matrix is ⎡ f xx ( x, y) f xy ( x, y) ⎤ ⎡6 x −3 ⎤ H ( x, y) = ⎢ ⎥=⎢ ⎥. ⎣ f yx ( x, y) f yy ( x, y) ⎦ ⎣ −3 −6 y ⎦

maximize z = 20 x1 y1 subject to ( x1 , y1 ) = 1. ⎡ 0 10⎤ ⎡ x1 ⎤ Write z = xT Ax = [ x1 y1 ] ⎢ ⎢ ⎥. ⎣10 0⎥⎦ ⎣ y1 ⎦ λ −10 = λ 2 − 100 = (λ + 10)(λ − 10). −10 λ

⎡ 0 −3⎤ are The eigenvalues of H (0, 0) = ⎢ ⎣ −3 0 ⎥⎦ λ = ±3; this matrix is indefinite and so f has a saddle point at (0, 0). The eigenvalues of ⎡ −6 −3⎤ are λ = −3 and H (−1, 1) = ⎢ ⎣ −3 −6 ⎥⎦ λ = −9; this matrix is negative definite and so f has a relative maximum at (−1, 1).

The largest eigenvalue of A is λ = 10 which has ⎡ 1 ⎤ positive unit eigenvector ⎢ 12 ⎥ . Thus the ⎢ ⎥ ⎣ 2⎦ ⎛ 1 ⎞⎛ 1 ⎞ maximum value of z = 20 ⎜ ⎟⎜ ⎟ = 10 ⎝ 2 ⎠⎝ 2 ⎠ 5 which occurs when x = 5 x1 = and 2 1 y = y1 = , which are the corner points of the 2 rectangle.

15. The first partial derivatives of f are

f x ( x, y) = 2 x − 2 xy and f y ( x, y) = 4 y − x 2 . To find the critical points we set f x and f y equal to zero. This yields the equations 2x(1 − y) = 0 1 and y = x 2 . From the first, we conclude that 4 x = 0 or y = 1. Thus there are three critical points: (0, 0), (2, 1), and (−2, 1). The Hessian matrix is ⎡ f xx ( x, y) f xy ( x, y) ⎤ H ( x, y ) = ⎢ ⎥ ⎣ f yx ( x, y) f yy ( x, y) ⎦

21. If x is a unit eigenvector corresponding to λ,

then q(x) = xT Ax = xT (λx) = λ(xT x) = λ(1) = λ. True/False 7.4 (a) False; if the only critical point of the quadratic form is a saddle point, then it will have neither a maximum nor a minimum value.

⎡ 2 − 2 y −2 x ⎤ =⎢ . 4 ⎥⎦ ⎣ −2 x

(b) True (c) True

⎡2 0⎤ The eigenvalues of the matrix H (0, 0) = ⎢ ⎣ 0 4⎥⎦ are λ = 2 and λ = 4; this matrix is positive definite and so f has a relative minimum at (0, 0). ⎡ 0 −4 ⎤ The eigenvalues of H (2, 1) = ⎢ are ⎣ −4 4 ⎥⎦

(d) False; the Hessian form of the second derivative test is inconclusive. (e) True; if det(A) < 0, then A will have negative eigenvalues.

λ = 2 ± 2 5. One of these is positive and one is negative; thus this matrix is indefinite and f has a saddle point at (2, 1). Similarly, the eigenvalues ⎡ 0 4⎤ of H (−2, 1) = ⎢ are λ = 2 ± 2 5; thus f ⎣ 4 4⎥⎦ has a saddle point at (−2, 1).


4 5 − i⎤ ⎡ −2i 1. A* = ⎢ + −i 1 i 3 0 ⎥⎦ ⎣

179



i 2 − 3i ⎤ ⎡ 1 3. A = ⎢ −i −3 1 ⎥ ⎢ ⎥ 2 ⎦ ⎣ 2 + 3i 1 5. (a) A is not Hermitian since a31 ≠ a13 . (b) A is not Hermitian since a22 is not real. 9. The following computations show that the row vectors of A are orthonormal: 2

r1 =

3 4 + i 5 5

r2 =

−

2

2

4 3 + i 5 5

9 16 + =1 25 25

= 2

16 9 + =1 25 25

=

⎛ 3 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞⎛ 3 ⎞ r1 ⋅ r2 = ⎜ ⎟ ⎜ − ⎟ + ⎜ i ⎟⎜ − i ⎟ ⎝ 5 ⎠ ⎝ 5 ⎠ ⎝ 5 ⎠⎝ 5 ⎠ 12 12 =− + 5 5 =0 ⎡ 3 − 54 ⎤ ⎥. Thus A is unitary, and A−1 = A* = ⎢ 5 ⎢ − 54 i − 53 i ⎥ ⎣ ⎦

11. The following computations show that the column vectors of A are orthonormal: c1 =

1

c2 =

1

2 2 4 4 = + 8 8 =1

= =1

c1 ⋅ c2 =

(

2 2 4 4 + 8 8 1 2 2

2

3 + i) +

(1 − i 3 )

(

3 + i)

1 2 2

2

+

1 2 2

(1 + i 3 )

1 2 2

(i −

(1 + i 3 ) +

3)

2

2

1 2 2

(1 + i 3 )

⎡ 1 ( 3 − i) Thus A is unitary, and A−1 = A* = ⎢ 21 2 ⎢ (1 + i 3 ) ⎣2 2

1 2 2

1 2 2 1 2 2

( −i −

3) = 0

(1 − i 3 ) ⎤ ⎥. ( −i − 3 ) ⎥⎦

13. The characteristic polynomial of A is λ 2 − 9λ + 18 = (λ − 3)(λ − 6); thus the eigenvalues of A are λ = 3 and λ = 6. ⎡ −1 −1 + i 0 ⎤ ⎡1 1 − i 0 ⎤ The augmented matrix of the system (3I − A)x = 0 is ⎢ , which reduces to ⎢ . Thus −2 0 ⎥⎦ ⎣0 0 0 ⎥⎦ ⎣ −1 − i ⎡ −1+i ⎤ ⎡ −1 + i ⎤ ⎢ 3 ⎥ v1 = ⎢ is a basis for the eigenspace corresponding to λ = 3, and p = 1 ⎢ 1 ⎥ is a unit eigenvector. A ⎣ 1 ⎥⎦ ⎣⎢ 3 ⎦⎥ 180


Section 7.5

⎡ 1−i ⎤ similar computation shows that p 2 = ⎢ 6 ⎥ is a unit eigenvector corresponding to λ = 6. The vectors p1 and p 2 ⎢ 2 ⎥ ⎣⎢ 6 ⎦⎥ are orthogonal, and the unitary matrix P = [p1 p 2 ] diagonalizes the matrix A: ⎡ −1−i P*AP = ⎢ 3 ⎢ 1+i ⎢⎣ 6 ⎡3 0⎤ =⎢ ⎣ 0 6 ⎥⎦

1 ⎤ 3⎥⎡ 4 2 ⎥ ⎢⎣1 + i 6 ⎥⎦

⎡ −1+i 1− i⎤ ⎢ 3 5 ⎥⎦ ⎢ 1 ⎣⎢ 3

1−i ⎤ 6⎥ 2 ⎥ 6 ⎦⎥

15. The characteristic polynomial of A is λ 2 − 10λ + 16 = (λ − 2)(λ − 8); thus the eigenvalues of A are λ = 2 and λ = 8. −2 − 2i 0 ⎤ ⎡ −4 ⎡2 1 + i 0⎤ The augmented matrix of the system (2I − A)x = 0 is ⎢ , which reduces to ⎢ . ⎥ −2 0⎦ ⎣ 0 0 0 ⎥⎦ ⎣ −2 + 2i ⎡ −1−i ⎤ ⎡ −1 − i ⎤ Thus v1 = ⎢ is a basis for the eigenspace corresponding to λ = 2, and p1 = ⎢ 6 ⎥ is a unit eigenvector. A ⎢ 2 ⎥ ⎣ 2 ⎥⎦ ⎣⎢ 6 ⎦⎥ ⎡ 1+i ⎤ similar computation shows that p 2 = ⎢ 3 ⎥ is a unit eigenvector corresponding to λ = 8. The vectors p1 and p 2 ⎢1 ⎥ ⎢⎣ 3 ⎥⎦ are orthogonal, and the unitary matrix P = [p1 p 2 ] diagonalizes the matrix A: ⎡ −1+i 6 ⎢ P*AP = ⎢ 1−i ⎣⎢ 3 ⎡ 2 0⎤ =⎢ ⎣ 0 8 ⎥⎦

2 ⎤ 6⎥⎡ 6 1 ⎥ ⎢⎣ 2 − 2i 3 ⎦⎥

⎡ −1−i 2 + 2i ⎤ ⎢ 6 4 ⎥⎦ ⎢ 2 ⎣⎢ 6

1+i ⎤ 3⎥ 1 ⎥ 3 ⎦⎥

17. The characteristic polynomial of A is (λ − 5)(λ 2 + λ − 2) = (λ + 2)(λ − 1)(λ − 5); thus the eigenvalues of A are 0 0⎤ ⎡ −7 0 λ1 = −2, λ 2 = 1, and λ3 = 5. The augmented matrix of the system (−2I − A)x = 0 is ⎢ 0 −1 1 − i 0 ⎥ , ⎢ ⎥ ⎣ 0 1 + i −2 0 ⎦ 0 0⎤ ⎡1 0 ⎡ 0 ⎤ which can be reduced to ⎢0 1 −1 + i 0 ⎥ . Thus v1 = ⎢1 − i ⎥ is a basis for the eigenspace corresponding to ⎢ ⎥ ⎢ ⎥ 0 0⎦ ⎣ 1 ⎦ ⎣0 0 ⎡0⎤ ⎡ 0 ⎤ ⎢ 1−i ⎥ ⎢ −1+i ⎥ λ1 = −2, and p1 = ⎢ 3 ⎥ is a unit eigenvector. Similar computations show that p 2 = ⎢ 6 ⎥ is a unit eigenvector ⎢ 1 ⎥ ⎢ 2 ⎥ ⎣⎢ 3 ⎦⎥ ⎣⎢ 6 ⎦⎥ ⎡1 ⎤ corresponding to λ 2 = 1, and p3 = ⎢0 ⎥ is a unit eigenvector corresponding to λ3 = 5. The vectors {p1, p 2 , p3} ⎢ ⎥ ⎣0 ⎦ form an orthogonal set, and the unitary matrix P = [p1 p 2 p3 ] diagonalizes the matrix A:

181



1 ⎤ 0 1⎤ ⎡ 0 1+i ⎡0 0 0 ⎤ ⎢ 1−i −1+i 3 3 ⎥ ⎡5 ⎢ 0⎥⎥ P*AP = ⎢ 0 −1−i 2 ⎥ ⎢0 −1 −1 + i ⎥ ⎢ 3 6 ⎥ 6 6 ⎥ ⎢ 0 −1 − i ⎢ ⎥ 2 0 ⎦⎢ 1 0⎥ ⎣ 0 0 ⎦⎥ 6 ⎣⎢ 3 ⎦ ⎣⎢1 ⎡ −2 0 0 ⎤ = ⎢ 0 1 0⎥ ⎢ ⎥ ⎣ 0 0 5⎦ Note: The matrix P is not unique. It depends on the particular choice of the unit eigenvectors. This is just one possibility.

i 2 − 3i ⎤ ⎡ 0 19. A = ⎢ i 0 1 ⎥ ⎢ ⎥ 4i ⎦ ⎣ −2 − 3i −1

21. (a) a21 = −i ≠ i = −a12 and a31 ≠ −a13 (b) a11 ≠ −a11 , a31 ≠ −a13 , and a32 ≠ −a23 ⎡ 0 −1 + i ⎤ 23. The characteristic polynomial of A = ⎢ is λ 2 − iλ + 2 = (λ − 2i)(λ + i); thus the eigenvalues of A are ⎥ i i + 1 ⎣ ⎦ λ = 2i and λ = −i.

27. We have A*A =

1 ⎡ e−iθ ⎢ 2 ⎣⎢ eiθ

−ie−iθ ⎤ 1 ⎡ eiθ ⎥ ⎢ ieiθ ⎦⎥ 2 ⎢⎣ieiθ

e −iθ ⎤ 1 ⎡ 1 + 1 ⎥= ⎢ −ie−iθ ⎦⎥ 2 ⎢⎣ e2iθ − e2iθ

e−2iθ − e −2iθ ⎤ ⎡1 0 ⎤ −1 ⎥=⎢ ⎥⎦ ; thus A* = A 0 1 ⎣ 1+1 ⎦⎥

and A is unitary. 1 1 29. (a) If B = ( A + A*), then B* = ( A + A*)* 2 2 1 = ( A * + A **) 2 1 = ( A * + A) 2 = B. Similarly, C* = C.

(b) We have B + iC =

1 1 1 1 ( A + A*) + ( A − A*) = A and B − iC = ( A + A*) − ( A − A*) = A * . 2 2 2 2

(c) AA* = ( B + iC )( B − iC ) = B 2 − iBC + iCB + C 2 and A*A = B 2 + iBC − iCB + C 2 . Thus AA* = A*A if and only if −iBC + iCB = iBC − iCB, or 2iCB = 2iBC. Thus A is normal if and only if B and C commute i.e., CB = BC.

31. If A is unitary, then A−1 = A * and so ( A*) −1 = ( A−1 )* = ( A*)*; thus A* is also unitary. 33. A unitary matrix A has the property that Ax = x for all x in C n . Thus if A is unitary and Ax = λx where x ≠ 0, we must have λ x = Ax = x and so λ = 1.

182



35. If H = I − 2uu*, then H* = (I − 2uu*)* = I* − 2u**u*= I − 2uu* = H; thus H is Hermitian. HH * = ( I − 2uu*)( I − 2uu*) = I − 2uu * −2uu * +4uu * uu *

Chapter 7 Supplementary Exercises ⎡3 1. (a) For A = ⎢ 5 ⎢ 45 ⎣

− 45 ⎤ ⎥, 3⎥ 5⎦ ⎡ 3 −1 T A =A =⎢ 5 ⎢ − 45 ⎣

2

= I − 4uu * +4u u u * =I and so H is unitary. ⎡ 37. A = ⎢ ⎢ ⎣

1 2 i 2

− i ⎤ 2⎥ is both Hermitian and unitary. 1 − ⎥ 2⎦

4⎤ 5⎥. 3⎥ 5⎦

⎡ 4 0 − 3⎤ 5⎥ ⎢ 5 9 4 − 12 ⎥ , (b) For A = ⎢ − 25 25 ⎢ 12 53 ⎥ 16 ⎢⎣ 25 5 25 ⎥⎦ ⎡1 0 0 ⎤ AT A = ⎢ 0 1 0 ⎥ , so ⎢ ⎥ ⎣0 0 1⎦ ⎡ 4 − 9 12 ⎤ 25 25 ⎥ ⎢ 5 −1 T 3⎥. 4 A =A =⎢ 0 5 5 ⎢ 3 ⎥ 12 16 ⎢⎣ − 5 − 25 25 ⎥⎦

39. If P = uu*, then

Px = (uu*)x = (uuT )x = u(uT x) = (x ⋅ u)u. Thus 2

multiplication of x by P corresponds to u times the orthogonal projection of x onto W = span{u}. If u = 1, then multiplication of x by H = I − 2uu* corresponds to reflection of x about the hyperplane u ⊥ .

True/False 7.5

5. The characteristic equation of A is λ3 − 3λ 2 + 2λ = λ(λ − 2)(λ − 1), so the eigenvalues are λ = 0, 2, 1. Orthogonal bases for ⎡− 1 ⎤ ⎢ 2⎥ the eigenspaces are λ = 0: ⎢ 0 ⎥ ; λ = 2: ⎢ 1 ⎥ ⎢⎣ 2 ⎥⎦ ⎡ 1 ⎤ ⎡0 ⎤ ⎢ 2⎥ ⎢ 0 ⎥ ; λ = 1: ⎢1 ⎥ . ⎢ ⎥ ⎢ 1 ⎥ ⎣0 ⎦ ⎢⎣ 2 ⎥⎦ 1 ⎡− 1 0⎤ 2 ⎢ 2 ⎥ Thus P = ⎢ 0 0 1⎥ orthogonally ⎢ 1 ⎥ 1 0⎥ 2 ⎣⎢ 2 ⎦ ⎡0 0 0 ⎤ diagonalizes A, and PT AP = ⎢0 2 0 ⎥ . ⎢ ⎥ ⎣0 0 1 ⎦

(a) False; i ≠ i . (b) False; for r1 = ⎡ − i ⎣ 2 i ⎤ r2 = ⎡0 − i , 6 3⎦ ⎣

r1 ⋅ r2 = −

⎡1 0 ⎤ , so AT A = ⎢ ⎣ 0 1 ⎥⎦

i 2

( 0) +

i ⎤ 3⎦

i 6

and

i ⎛ i ⎞ i ⎛ i ⎞ ⎜− ⎟+ ⎜ ⎟ 6⎝ 3⎝ 3⎠ 6⎠

2

2

⎛ i ⎞ ⎛ i ⎞ = 0+⎜ ⎟ −⎜ ⎟ ⎝ 6⎠ ⎝ 3⎠ 1 1 =− + 6 3 1 = 6 Thus the row vectors do not form an orthonormal set and the matrix is not unitary. (c) True; if A is unitary, so A−1 = A*, then ( A*) −1 = A = ( A*) *.

7. In matrix form, the quadratic form is ⎡ 1 − 3⎤ ⎡x ⎤ 2 ⎥ 1 . The xT Ax = [ x1 x2 ] ⎢ ⎢ ⎥ 4 ⎥⎦ ⎣ x2 ⎦ ⎢⎣ − 32

(d) False; normal matrices that are not Hermitian are also unitarily diagonalizable.

characteristic equation of A is λ 2 − 5λ +

(e) False; if A is skew-Hermitian, then ( A2 )* = ( A*)( A*) = (− A)(− A) = A2 ≠ − A2 .

183

7 =0 4



5±3 2 or 2 λ ≈ 4.62, 0.38. Since both eigenvalues of A are positive, the quadratic form is positive definite. which has solutions λ =

9. (a) y − x 2 = 0 or y = x 2 represents a parabola. (b) 3x − 11y 2 = 0 or x =

11 2 y represents a 3

parabola.

184

Chapter 8 Linear Transformations Section 8.1 Exercise Set 8.1 1. T (−u) = −u = u = T (u) ≠ −T (u), so the function is not a linear transformation.

3. T(kA) = (kA)B = k(AB) = kT(A) T ( A1 + A2 ) = ( A1 + A) B = A1 B + A2 B = T ( A1 ) + T ( A2 ) Thus T is a linear transformation. 5. F (kA) = (kA)T = kAT = kF ( A)

F ( A + B) = ( A + B)T = AT + BT = F ( A) + F ( B) Thus F is a linear transformation. 7. Let p( x) = a0 + a1 x + a2 x 2 and q( x) = b0 + b1 x + b2 x 2 . (a) T (kp( x)) = ka0 + ka1 ( x + 1) + ka2 ( x + 1)2 = kT ( p( x))

T ( p( x) + q( x)) = a0 + b0 + (a1 + b1 )( x + 1) + (a2 + b2 )( x + 1)2 = a0 + a1 ( x + 1) + a2 ( x + 1)2 + b0 + b1 ( x + 1) + b2 ( x + 1)2 = T ( p( x)) + T (q( x)) Thus T is a linear transformation. (b) T (kp( x)) = T (ka0 + ka1 x + ka2 x 2 )

= (ka0 + 1) + (ka1 + 1) x + (ka2 + 1) x 2 ≠ kT ( p( x)) T is not a linear transformation. 9. For x = ( x1, x2 ) = c1v1 + c2 v 2 , we have ( x1 , x2 ) = c1 (1, 1) + c2 (1, 0) = (c1 + c2 , c1 ) or

c1 + c2 = x1 c1 = x2 which has the solution c1 = x2 , c2 = x1 − x2 .

( x1 , x2 ) = x2 (1, 1) + ( x1 − x2 )(1, 0) = x2 v1 + ( x1 − x2 ) v 2 and T ( x1, x2 ) = x2T ( v1 ) + ( x1 − x2 )T ( v 2 ) = x2 (1, − 2) + ( x1 − x2 )(−4, 1) = (−4 x1 + 5 x2 , x1 − 3x2 ) T(5, −3) = (−20 − 15, 5 + 9) = (−35, 14).

185

Chapter 8: Linear Transformations


11. For x = ( x1, x2 , x3 ) = c1 v1 + c2 v 2 + c3 v3 , we have ( x1 , x2 , x3 ) = c1 (1, 1, 1) + c2 (1, 1, 0) + c3 (1, 0, 0) = (c1 + c2 + c3 , c1 + c2 , c1 )

(b), (c) Neither 1 + x nor 3 − x 2 can be expressed as xp(x) with p(x) in P2 , so neither are in R(T). 21. (a) Since −8x + 4y = −4(2x − y) and 2x − y can assume any real value, R(T) is the set of all vectors (x, −4x) or the line y = −4x. The vector (1, −4) is a basis for this space.

c1 + c2 + c3 = x1 or c1 + c2 = x2 which has the solution c1 = x3 c1 = x3 , c2 = x2 − x3 , c3 = x1 − ( x2 − x3 ) − x3 = x1 − x2 . ( x1 , x2 , x3 ) = x3 v1 + ( x2 − x3 ) v 2 + ( x1 − x2 ) v3

(b) T can be expressed as a matrix operator from R 4 to R3 with the matrix ⎡ 4 1 −2 −3⎤ A = ⎢ 2 1 1 −4 ⎥ . A basis for R(T) is a ⎢ ⎥ ⎣ 6 0 −9 9 ⎦ basis for the column space of A. A row ⎡ 1 1 1 −2 ⎤ 2 2 echelon form of A is B = ⎢0 1 4 −5⎥ . ⎢ ⎥ ⎢⎣0 0 0 1⎥⎦

T ( x1, x2 , x3 ) = x3T ( v1 ) + ( x2 − x3 )T ( v 2 ) + ( x1 − x2 )T ( v3 ) = x3 (2, − 1, 4) + ( x2 − x3 )(3, 0, 1) + ( x1 − x2 )(−1, 5, 1) = (− x1 + 4 x2 − x3 , 5 x1 − 5 x2 − x3 , x1 + 3 x3 ) T (2, 4, − 1) = (−2 + 16 + 1, 10 − 20 + 1, 2 − 3) = (15, − 9, − 1)

The columns of A corresponding to the columns of B containing leading 1s are ⎡ 4 ⎤ ⎡ 1⎤ ⎡ −3⎤ ⎢ 2 ⎥ , ⎢ 1⎥ , and ⎢ −4 ⎥ . Thus, (4, 2, 6), ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 6 ⎦ ⎣0 ⎦ ⎣ 9⎦

13. T (2 v1 − 3v 2 + 4 v3 ) = 2T ( v1 ) − 3T ( v 2 ) + 4T ( v3 ) = (2, − 2, 4) − (0, 9, 6) + (−12, 4, 8) = (−10, − 7, 6)

(1, 1, 0), and (−3, −4, 9) form a basis for R(T).

15. (a) T(5, 10) = (10 − 10, −40 + 40) = (0, 0) so (5, 10) is in ker(T).

(c) R(T) is the set of all polynomials in P3 with

(b) T(3, 2) = (6 − 2, −24 + 8) = (4, −16) so (3, 2) is not in ker(T).

constant term 0. Thus, {x, x 2 , x3} is a basis for R(T).

(c) T(1, 1) = (2 − 1, −8 + 4) = (1, −4) so (1, 1) is not in ker(T).

3⎤ ⎡ 1 −1 ⎢ 19 23. (a) A row echelon form of A is 0 1 − 11 ⎥ , ⎢ ⎥ 0 ⎦⎥ ⎣⎢0 0

17. (a) T(3, −8, 2, 0) = (12 − 8 − 4, 6 − 8 + 2, 18 − 18) = (0, 0, 0) so (3, −8, 2, 0) is in ker(T).

⎡1 ⎤ ⎡ −1⎤ thus columns 1 and 2 of A, ⎢ 5 ⎥ and ⎢ 6⎥ ⎢ ⎥ ⎢ ⎥ ⎣7 ⎦ ⎣ 4⎦ form a basis for the column space of A, which is the range of T.

(b) T(0, 0, 0, 1) = (0 + 0 + 0 − 3, 0 + 0 + 0 − 4, 0 − 0 + 9) = (−3, −4, 9) so (0, 0, 0, 1) is not in ker(T).

14 ⎤ ⎡1 0 11 ⎥ ⎢ (b) A reduces to ⎢0 1 − 19 ⎥ , so a general 11 ⎢ ⎥ 0 0 0⎦ ⎣ 14 19 solution of Ax = 0 is x1 = − s, x2 = s, 11 11 x3 = s or x1 = −14t , x2 = 19t , x3 = 11t , so

(c) T (0, − 4, 1, 0) = (−4 − 2, − 4 + 1, − 9) = (−6, − 3, − 9) so (0, −4, 1, 0) is not in ker(T). 19. (a) Since x + x 2 = x(1 + x), x + x 2 is R(T).

186


Section 8.1 (b) dim( P4 ) = 5, so nullity(T) = 5 − rank(T) = 4

⎡ −14 ⎤ ⎢ 19 ⎥ is a basis for the null space of A, ⎢ ⎥ ⎣ 11⎦ which is ker(T).

(c) Since R(T ) = R3 , T has rank 3. nullity(T) = 6 − rank(T) = 3

(c) R(T) is two-dimensional, so rank(T) = 2. ker(T) is one-dimensional, so nullity (T) = 1.

(d) nullity(T) = 4 − rank(T) = 1 31. (a) nullity(A) = 7 − rank(A) = 3, so the solution space of Ax = 0 has dimension 3.

(d) The column space of A is two-dimensional, so rank(A) = 2. The null space of A is onedimensional, so nullity(A) = 1.

(b) No, since rank(A) = 4 while R5 has dimension 5.

25. (a) A row echelon form of A is 0⎤ ⎡1 2 3 ⎢0 1 1 − 2 ⎥ , thus columns 1 and 2 of 7⎦ ⎣ ⎡4⎤ ⎡1 ⎤ A, ⎢ ⎥ and ⎢ ⎥ form a basis for the column ⎣1 ⎦ ⎣2⎦ space of A, which is the range of T.

33. R(T) must be a subspace of R3 , thus the possibilities are a line through the origin, a plane through the origin, the origin only, or all of R3 .

35. (b) No; F (kx, ky) = (a1k 2 x 2 + b1k 2 y 2 , a2 k 2 x 2 + b2 k 2 y 2 )

4⎤ ⎡1 0 1 7 ⎥ , so a general (b) A reduces to ⎢ ⎢⎣0 1 1 − 72 ⎥⎦ 4 solution of Ax = 0 is x1 = − s − t , 7 2 x2 = − s + t , x3 = s, x4 = t or 7 x1 = − s − 4r , x2 = − s + 2r , x3 = s,

= k 2 F ( x, y) ≠ kF ( x, y)

37. Let w = cv1 + c2 v 2 + " + cn v n be any vector in V. Then T (w ) = c1T ( v1 ) + c2T ( v 2 ) + " + cnT ( v n ) = c1 v1 + c2 v 2 + " + cn v n =w Since w was an arbitrary vector in V, T must be the identity operator.

⎡ −1⎤ ⎡ −4 ⎤ ⎢ −1⎥ ⎢ 2⎥ x4 = 7r , so ⎢ ⎥ and ⎢ ⎥ form a basis ⎢ 0⎥ ⎢ 1⎥ ⎢⎣ 7 ⎥⎦ ⎢⎣ 0⎥⎦ for the null space of A, which is ker(T).

41. If p′( x) = 0, then p(x) is a constant, so ker(D) consists of all constant polynomials. 43. (a) The fourth derivative of any polynomial of

(c) Both R(T) and ker(T) are two-dimensional, so rank(T) = nullity(T) = 2.

degree 3 or less is 0, so T ( f ( x)) = f ( 4) ( x) has ker(T ) = P3 .

(d) Both the column space and the null space of A are two-dimensional, so rank(A) = nullity(A) = 2.

(b) By similar reasoning, T ( f ( x)) = f ( n +1) ( x) has ker(T ) = Pn .

27. (a) The kernel is the y-axis, the range is the entire xz-plane.

True/False 8.1

(b) The kernel is the x-axis, the range is the entire yz-plane.

(a) True; c1 = k , c2 = 0 gives the homogeneity property and c1 = c2 = 1 gives the additivity property.

(c) The kernel is the line through the origin perpendicular to the plane y = x, the range is the entire plane y = x.

(b) False; every linear transformation will have T(−v) = −T(v).

29. (a) nullity(T) = 5 − rank(T) = 2 187



⎡ 1 0⎤ (c) A reduces to ⎢0 1⎥ , so nullity(A) = 0 or ⎢ ⎥ ⎣0 0 ⎦ Ax = 0 has only the trivial solution, so multiplication by A is one-to-one.

(c) True; only the zero transformation has this property. (d) False; T (0) = v 0 + 0 = v 0 ≠ 0, so T is not a linear transformation. (e) True

5. (a) All points on the line y = −x are mapped to 0, so ker(T) = {k(−1, 1)}.

(f) True

(b) Since ker(T) ≠ {0}, T is not one-to-one.

(g) False; T does not necessarily have rank 4.

7. (a) Since nullity(T) = 0, T is one-to-one.

(h) False; det(A + B) ≠ det(A) + det(B) in general.

(b) nullity(T) = n − (n − 1) = 1, so T is not oneto-one.

(i) False; nullity(T) = rank(T) = 2 Section 8.2

(c) Since n < m, T is not one-to-one.

Exercise Set 8.2

(d) Since R(T ) = R n , rank(T) = n, so nullity (T) = 0. Thus T is one-to-one.

1. (a) By inspection, ker(T) = {0}, so T is one-toone.

9. If there were such a transformation T, then it would have nullity 0 (because it would be oneto-one), and have rank no greater than the dimension of W (because its range is a subspace W). Then by the dimension Theorem, dim(V) = rank(T) + nullity(T) ≤ dim(W) which contradicts that dim(W) < dim(V). Thus there is no such transformation.

3 (b) T(x, y) = 0 if 2x + 3y = 0 or x = − y so 2 ⎧ ⎛ 3 ⎞⎫ ker(T ) = ⎨k ⎜ − , 1⎟ ⎬ and T is not one-to⎩ ⎝ 2 ⎠⎭ one.

(c) (x, y) is in ker(T) only if x + y = 0 and x − y = 0, so ker(T) = {0} and T is one-toone.

⎛ ⎡a b 11. (a) T ⎜ ⎢ b d ⎜⎜ ⎢ ⎝ ⎣c e

(d) By inspection, ker(T) = {0}, so T is one-toone. (e) (x, y) is in ker(T) if x − y = 0 or x = y, so ker(T) = {k(1, 1)} and T is not one-to-one.

⎡ a⎤ ⎢ b⎥ c ⎤⎞ ⎢ ⎥ c e ⎥⎟ = ⎢ ⎥ ⎟ d ⎥ ⎢ ⎥ f ⎦ ⎟⎠ ⎢ ⎥ e ⎢f⎥ ⎣ ⎦

⎡a ⎤ ⎛ ⎡a b ⎤ ⎞ ⎢ b ⎥ (b) T1 ⎜ ⎢ ⎥ ⎟ = ⎢ ⎥ and ⎝ ⎣c d ⎦ ⎠ ⎢ c ⎥ ⎣⎢ d ⎦⎥

(f) (x, y, z) is in ker(T) if both x + y + z = 0 and x − y − z = 0, which is x = 0 and y + z = 0. Thus, ker(T) = {k(0, 1, −1)} and T is not one-to-one.

⎡a⎤ ⎛ ⎡a b ⎤ ⎞ ⎢ c ⎥ T2 ⎜ ⎢ ⎥⎟ = ⎢ ⎥. ⎝ ⎣c d ⎦ ⎠ ⎢b ⎥ ⎣⎢ d ⎦⎥

⎡ 1 −2 ⎤ 3. (a) By inspection, A reduces to ⎢0 0⎥ , so ⎢ ⎥ ⎣0 0⎦ nullity(A) = 1 and multiplication by A is not one-to-one.

(c) If p(x) is a polynomial and p(0) = 0, then p(x) has constant term 0. ⎡a⎤ T (ax3 + bx 2 + cx) = ⎢ b ⎥ ⎢ ⎥ ⎣c ⎦

(b) A can be viewed as a mapping from R 4 to R3 , thus multiplication by A is not one-toone. 188


Section 8.3

⎡a⎤ (d) T (a + b sin( x) + c cos( x)) = ⎢ b ⎥ ⎢ ⎥ ⎣c ⎦

Section 8.3 Exercise Set 8.3 1. (a) (T2 D T1 )( x, y) = T2 (2 x, 3 y) = ( 2 x − 3 y, 2 x + 3 y )

13. T (kf ) = kf (0) + 2kf ′(0) + 3kf ′(1) = kT (f ) T (f + g) = f (0) + g (0) + 2( f ′(0) + g ′(0)) + 3( f ′(1) + g ′(1)) = T (f ) + T ( g ) Thus T is a linear transformation.

(b) (T2 D T1 )( x, y) = T2 ( x − 3 y, 0) = (4( x − 3 y) − 5(0), 3( x − 3 y) − 6(0)) = (4 x − 12 y, 3x − 9 y)

Let f = f ( x) = x 2 ( x − 1)2 , then f ′( x) = 2 x( x − 1)(2 x − 1) so f(0) = 0, f ′(0) = 0, and f ′(1) = 0. T (f ) = f (0) + 2 f ′(0) + 3 f ′(1) = 0, so ker(T) ≠ {0} and T is not one-to-one.

(c) (T2 D T1 )( x, y) = T2 (2 x, − 3 y, x + y) = ( 2 x + 3 y, − 3 y + x + y ) = ( 2 x + 3 y, x − 2 y ) (d) (T2 D T1 )( x, y) = T2 ( x − y, y, x) = (0, ( x − y) + y + x) = (0, 2 x)

15. Yes; if T(a, b, c) = 0, then a = b = c = 0, so ker(T) = {0}. 17. No; T is not one-to-one because ker(T) ≠ {0} as T(a) = a × a = 0.

⎛ ⎡a c ⎤ ⎞ 3. (a) (T1 D T2 )( A) = T1 ( AT ) = tr ⎜ ⎢ ⎥⎟ = a + d ⎝ ⎣b d ⎦ ⎠

19. Yes, since T (u), T ( v) = u, v , so

T (u) =

T (u), T (u) =

u, u = u = 1 and T(u) is orthogonal to T(v) if u is orthogonal to v.

(b) (T2 D T1 )( A) does not exist because T1 ( A) is not a 2 × 2 matrix.

True/False 8.2 1 v, then 4 ⎛1 ⎞ ⎛1 ⎞ (T1 D T2 )( v) = T1 ⎜ v ⎟ = 4 ⎜ v ⎟ = v and ⎝4 ⎠ ⎝4 ⎠ 1 (T2 D T1 )( v) = T2 (4 v) = (4 v) = v. 4

5. T2 ( v) =

2

(a) False; dim( R ) = 2 while dim( P2 ) = 3. (b) True; if ker(T) = {0} is one-to-one rank(T) = 4 so T is one-to-one and onto. (c) False; dim( M 33 ) = 9 while dim( P9 ) = 10.

9. By inspection, T(x, y, z) = (x, y, 0). Then T(T(x, y, z)) = T(x, y, 0) = (x, y, 0) = T(x, y, z) or T DT = T.

(d) True; for instance, if V consists of all matrices of ⎡a b 0⎤ the form ⎢ , then ⎣ c d 0 ⎥⎦ ⎡a⎤ ⎛ ⎡ a b 0⎤ ⎞ ⎢ b ⎥ T ⎜⎢ ⎥ ⎟ = ⎢ ⎥ is an isomorphism T: ⎝ ⎣ c d 0⎦ ⎠ ⎢ c ⎥ ⎣⎢ d ⎦⎥

11. (a) Since det(A) = 0, T does not have an inverse.

V → R4 . (e) False; T(I) = P − P = 0 so ker(T) ≠ 0. (f) False; if there were such a transformation T, then dim(ker(T)) = dim(R(T)), but since dim(ker(T )) + dim( R(T )) = dim( P4 ) = 5, there can be no such transformation. 189



(b) det(A) = −8 ≠ 0, so T has an inverse. ⎡ 1 1 − 3⎤ 4⎥ ⎢ 8 8 1 ⎥ so A−1 = ⎢ 18 18 4 ⎢ 3 5 ⎥ 1 − 4 ⎦⎥ ⎣⎢ 8 8

15. (a) Since T1 ( p( x)) = xp( x),

1 p( x). x Since T2 ( p( x)) = p( x + 1), T1−1 ( p( x)) =

T2−1 ( p( x)) = p( x − 1).

⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 ⎤ −1 ⎜ ⎢ ⎥ ⎟ −1 ⎢ ⎥ x x =A T ⎜⎢ 2⎥⎟ ⎢ 2⎥ ⎜ x ⎟ 3 ⎣ x3 ⎦ ⎝⎣ ⎦⎠ 1 ⎡ x1 + 1 x2 − 3 x3 ⎤ 8 4 ⎢ 8 ⎥ 1 1 = ⎢ 8 x1 + 8 x2 + 14 x3 ⎥ . ⎢ 3 ⎥ 5 1 ⎢⎣ − 8 x1 + 8 x2 + 4 x3 ⎥⎦

(T2 D T1 ) −1 ( p( x)) = (T1−1 D T2−1 )( p( x)) = T1−1 ( p( x − 1)) 1 = p( x − 1) x

17. (a) T(1 − 2x) = (1 − 2(0), 1 − 2(1)) = (1, −1) (c) Let p( x) = a0 + a1 x, then T ( p( x)) = (a0 , a0 + a1 ) so if T(p(x)) = (0, 0), then a0 = a1 = 0 and p is the zero polynomial, so ker(T) = {0}.

(c) det(A) = −2 ≠ 0, so T has an inverse. 1⎤ ⎡ 1 −1 2 2 2⎥ ⎢ 1 1 ⎥ so A−1 = ⎢ − 12 2 2 ⎢ 1 1 − 1⎥ ⎢⎣ 2 2 2 ⎥⎦ ⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 ⎤ −1 ⎜ ⎢ ⎥ ⎟ −1 ⎢ ⎥ T x x =A ⎜⎢ 2⎥⎟ ⎢ 2⎥ ⎜ x ⎟ ⎣ x3 ⎦ ⎝⎣ 3⎦⎠

(d) Since T ( p( x)) = (a0 , a0 + a1 ), then

T −1 (2, 3) has a0 = 2 and a0 + a1 = 3 or a1 = 1. Thus, T −1 (2, 3) = 2 + x. P(x)

⎡ 1 x1 − 1 x2 + 1 x3 ⎤ 2 2 ⎢ 2 ⎥ 1 1 = ⎢ − 2 x1 + 2 x2 + 12 x3 ⎥ . ⎢ 1 ⎥ 1 1 ⎢⎣ 2 x1 + 2 x2 − 2 x3 ⎥⎦

x

(d) det(A) = 1 ≠ 0, so T has an inverse. ⎡ 3 3 −1⎤ A−1 = ⎢ −2 −2 1⎥ so ⎢ ⎥ ⎣ −4 −5 2 ⎦

21. (a) T1 ( x, y) = ( x, − y) and T2 ( x, y) = (− x, y) (T1 D T2 )( x, y) = T1 (− x, y) = (− x, − y) (T2 D T1 )( x, y) = T2 ( x, − y) = (− x, − y) T1 D T2 = T2 D T1

⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 ⎤ T −1 ⎜ ⎢ x2 ⎥ ⎟ = A−1 ⎢ x2 ⎥ ⎜⎢ ⎥⎟ ⎢ ⎥ ⎜ x ⎟ ⎣ x3 ⎦ ⎝⎣ 3⎦⎠ ⎡ 3x1 + 3x2 − x3 ⎤ = ⎢ −2 x1 − 2 x2 + x3 ⎥ . ⎢ ⎥ ⎢⎣ −4 x1 − 5 x2 + 2 x3 ⎥⎦

(b) Consider (x, y) = (0, 1): (T2 D T1 )(0, 1) = T2 (0, 0) = (0, 0) but (T1 D T2 )(0, 1) = T1 (− sin θ , cosθ ) = (− sin θ , 0) Thus T1 D T2 ≠ T2 D T1 .

13. (a) For T to have an inverse, all the ai , i = 1, 2, ..., n must be nonzero. (b) T −1 ( x1 , x2 , ..., xn ) 1 1 ⎛1 ⎞ x2 , ..., xn ⎟ = ⎜ x1 , a a a 2 n ⎝ 1 ⎠

190


Section 8.4

(c) T1 ( x, y, z) = (kx, ky, kz ) and T2 ( x, y, z) = ( x cosθ − y sin θ , x sin θ + y cos θ , z ) (T1 D T2 )( x, y, z) = (k ( x cosθ − y sin θ ), k ( x sin θ + y sin θ ), kz) (T2 D T1 )( x, y, z) = (kx cos θ − ky sin θ , kx sin θ + ky cos θ , kz) T1 D T2 = T2 D T True/False 8.3 (a) True (b) False; the linear operators in Exercise 21(b) are an example. (c) False; a linear transformation need not have an inverse. (d) True; for T to have an inverse, it must be one-to-one. (e) False; T −1 does not exist. (f) True; if T1 is not one-to-one then there is some nonzero vector v1 with T1 ( v1 ) = 0, then (T2 D T1 )( v1 ) = T2 (0) = 0 and ker(T2 D T1 ) ≠ {0}. Section 8.4 Exercise Set 8.4 1. (a) T (u1 ) = T (1) = x = v 2

T (u 2 ) = T ( x) = x 2 = v3 T (u 3 ) = T ( x 2 ) = x 3 = v 4 ⎡0 ⎢1 Thus the matrix for T relative to B and B′ is ⎢ ⎢0 ⎢⎣0 ⎡0 ⎢1 (b) [T ]B′, B [ x]B = ⎢ ⎢0 ⎣⎢ 0

0 0 1 0

0 0 1 0

⎡0⎤ 0⎤ ⎡c ⎤ 0 ⎥ ⎢ 0 ⎥ ⎢ c0 ⎥ ⎥ c =⎢ ⎥ 0 ⎥ ⎢ 1 ⎥ ⎢ c1 ⎥ c 0 ⎦⎥ ⎣ 2 ⎦ ⎣⎢ c2 ⎦⎥

⎡0⎤ ⎢c ⎥ [T (x)]B′ = [c0 x + c1 x + c2 x ]B′ = ⎢ 0 ⎥ ⎢ c1 ⎥ ⎢⎣ c2 ⎥⎦ 2

3

3. (a) T(1) = 1 T(x) = x − 1 = −1 + x

T ( x 2 ) = ( x − 1)2 = 1 − 2 x + x 2 ⎡ 1 −1 1⎤ Thus the matrix for T relative to B is ⎢0 1 −2 ⎥ . ⎢ ⎥ 1⎦ ⎣0 0

191

0⎤ 0⎥ ⎥. 0⎥ 1⎥⎦



⎡ 1 −1 1⎤ ⎡ a0 ⎤ (b) [T ]B [ x]B = ⎢ 0 1 −2 ⎥ ⎢ a1 ⎥ ⎢ ⎥⎢ ⎥ 1⎦ ⎣ a2 ⎦ ⎣0 0 ⎡ a0 − a1 + a2 ⎤ = ⎢ a1 − 2a2 ⎥ ⎢ ⎥ a2 ⎣ ⎦

9. (a) Since A is the matrix for T relative to B, A = [[T ( v1 ) B [T ( v 2 )]B ] . That is, ⎡ 1⎤ ⎡3⎤ [T ( v1 )]B = ⎢ ⎥ and [T ( v 2 )]B = ⎢ ⎥ . − 2 ⎣5⎦ ⎣ ⎦ ⎡ 1⎤ (b) Since [T ( v1 )]B = ⎢ ⎥ , ⎣ −2 ⎦ ⎡ 1⎤ ⎡ −2⎤ ⎡ 3⎤ T ( v1 ) = 1v1 − 2 v 2 = ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ . ⎣ 3⎦ ⎣ 8 ⎦ ⎣ −5 ⎦ Similarly, ⎡ 3⎤ ⎡ −5 ⎤ ⎡ −2 ⎤ T ( v 2 ) = 3v1 + 5v 2 = ⎢ ⎥ + ⎢ ⎥ = ⎢ ⎥ . ⎣9 ⎦ ⎣ 20⎦ ⎣ 29⎦

2

For x = a0 + a1 x + a2 x , T (x) = a0 + a1 ( x − 1) + a2 ( x − 1) 2 = a0 − a1 + a2 + (a1 − 2a2 ) x + a2 x 2 , ⎡ a0 − a1 + a2 ⎤ so [T (x)]B = ⎢ a1 − 2a2 ⎥ . ⎢ ⎥ a2 ⎣ ⎦

⎡x ⎤ ⎡ 1⎤ ⎡ −1⎤ (c) If ⎢ 1 ⎥ = c1v1 + c2 v 2 = c1 ⎢ ⎥ + c2 ⎢ ⎥ , x 3 ⎣ ⎦ ⎣ 4⎦ ⎣ 2⎦ x = c −c then 1 1 2 . Solving for c1 and c2 x2 = 3c1 + 4c2

⎡ 7⎤ 1 8 ⎛ ⎡1⎤ ⎞ 5. (a) T (u1 ) = T ⎜ ⎢ ⎥ ⎟ = ⎢ −1⎥ = 0 v1 − v 2 + v 2 3 ⎢ ⎥ 2 3 ⎣ ⎦ ⎝ ⎠ ⎣ 0⎦ ⎡6 ⎤ 4 ⎛ ⎡ −2 ⎤ ⎞ ⎢ ⎥ T (u 2 ) = T ⎜ ⎢ ⎥ ⎟ = 2 = 0 v1 + v 2 + v3 4 ⎢ ⎥ 3 ⎝⎣ ⎦⎠ 0 ⎣ ⎦ ⎡ 0 0⎤ ⎢ 1 ⎥ [T ]B′, B = ⎢ − 2 1⎥ ⎢ 8 4⎥ ⎣ 3 3⎦

gives c1 =

4 1 3 1 x1 + x2 , c2 = − x1 + x2 , 7 7 7 7

so ⎛⎡ x ⎤⎞ T ⎜ ⎢ 1⎥ ⎟ ⎝ ⎣ x2 ⎦ ⎠ = c1T ( v1 ) + c2T ( v 2 ) 1 ⎞ ⎡ 3⎤ ⎛ 3 1 ⎞ ⎡ −2 ⎤ ⎛4 = ⎜ x1 + x2 ⎟ ⎢ ⎥ + ⎜ − x1 + x2 ⎟ ⎢ ⎥ − 5 7 ⎠⎣ ⎦ ⎝ 7 7 ⎠ ⎣ 29 ⎦ ⎝7 18 1 ⎡ x1 + 7 x2 ⎤ ⎥ =⎢ 7 24 x ⎥ ⎢ − 107 + x 1 2 7 ⎣ 7 ⎦ 1 ⎤ ⎡ 18 7 ⎥ ⎡ x1 ⎤ =⎢ 7 ⎢ ⎥ 107 24 ⎢− 7 ⎥ ⎣ x2 ⎦ 7 ⎦ ⎣

7. (a) T(1) = 1 T(x) = 2x + 1 = 1 + 2x T ( x)2 = (2 x + 1)2 = 1 + 4 x + 4 x 2 ⎡1 1 1 ⎤ [T ]B = ⎢0 2 4 ⎥ ⎢ ⎥ ⎣0 0 4 ⎦ ⎡ 2⎤ (b) The matrix for 2 − 3 x + 4 x 2 is ⎢ −3⎥ . ⎢ ⎥ ⎣ 4⎦

18 ⎛ ⎡1⎤ ⎞ ⎡ (d) T ⎜ ⎢ ⎥ ⎟ = ⎢ 7 ⎝ ⎣1⎦ ⎠ ⎢⎣ − 107 7

⎡ 2 ⎤ ⎡1 1 1 ⎤ ⎡ 2⎤ ⎡ 3⎤ [T ]B ⎢ −3⎥ = ⎢ 0 2 4 ⎥ ⎢ −3⎥ = ⎢10 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 4 ⎦ ⎣ 0 0 4 ⎦ ⎣ 4⎦ ⎣16 ⎦

1 ⎤ 7 ⎥ ⎡1⎤ 24 ⎥ ⎢⎣1⎥⎦ 7 ⎦

⎡ 19 ⎤ =⎢ 7⎥ ⎢ − 83 ⎥ ⎣ 7⎦

11. (a) Since A is the matrix for T relative to B, the columns of A are [T ( v1 )]B , [T ( v 2 )]B , and

[T ( v3 )]B , respectively. That is,

Thus T (2 − 3 x + 4 x 2 ) = 3 + 10 x + 16 x 2 .

⎡ 1⎤ ⎡ 3⎤ [T ( v1 )]B = ⎢ 2 ⎥ , [T ( v 2 )]B = ⎢ 0 ⎥ , and ⎢ ⎥ ⎢ ⎥ ⎣ −2 ⎦ ⎣6⎦ ⎡ −1⎤ [T ( v3 )]B = ⎢ 5⎥ . ⎢ ⎥ ⎣ 4⎦

(c) T (2 − 3 x + 4 x 2 ) = 2 − 3(2 x + 1) + 4(2 x + 1)2 = 2 − 6 x − 3 + 16 x 2 + 16 x + 4 = 3 + 10 x + 16 x 2

192


Section 8.4

⎡ 1⎤ (b) Since [T ( v1 )]B = ⎢ 2 ⎥ , ⎢ ⎥ ⎣6⎦

(d) In 1 + x 2 , a0 = 1, a1 = 0, a2 = 1.

T (1 + x 2 ) 239 + 289 201 + 247 61 + 107 2 = + x+ x 24 8 12 = 22 + 56 x + 14 x 2

T ( v1 ) = v1 + 2 v 2 + 6 v3 = 3 x + 3 x 2 − 2 + 6 x + 4 x 2 + 18 + 42 x + 12 x 2

13. (a) (T2 D T1 )(1) = T2 (2) = 6 x

= 16 + 51x + 19 x 2 . Similarly, T ( v 2 ) = 3v1 − 2 v3 2

= 9 x + 9 x − 6 − 14 x − 4 x

(T2 D T1 )( x) = T2 (−3x) = −9 x 2

2

⎡0 0⎤ ⎢6 0⎥ Thus, [T2 D T1 ]B′, B = ⎢ ⎥. ⎢ 0 −9 ⎥ ⎣⎢0 0⎦⎥

= −6 − 5 x + 5 x 2 , and T ( v3 ) = − v1 + 5v 2 + 4 v3 = −3 x − 3x 2 − 5 + 15 x + 10 x 2 + 12

T1 (1) = 2 T1 ( x) = −3x

+ 28 x + 8 x 2 2

= 7 + 40 x + 15 x .

⎡2 0⎤ Thus [T1 ]B′′, B = ⎢ 0 −3⎥ . ⎢ ⎥ ⎣0 0⎦ T2 (1) = 3x

(c) If a0 + a1 x + a2 x 2 = c1 v1 + c2 v 2 + c3 v3

= c1 (3x + 3x 2 ) + c2 (−1 + 3x + 2 x 2 )

T2 ( x) = 3 x 2

+ c3 (3 + 7 x + 2 x 2 ), a0 = −c2 + 3c3 Then a1 = 3c1 + 3c2 + 7c3 . a2 = 3c1 + 2c2 + 2c3

T2 ( x 2 ) = 3x3 ⎡0 ⎢3 Thus [T2 ]B′, B′′ = ⎢ ⎢0 ⎣⎢0

Solving for c1 , c2 , and c3 gives 1 c1 = (a0 − a1 + 2a2 ), 3 1 c2 = (−5a0 + 3a1 − 3a2 ), 8 1 c3 = (a0 + a1 − a2 ), so 8

0 0 3 0

0⎤ 0⎥ ⎥. 0⎥ 3⎦⎥

(b) [T2 D T1 ]B′, B = [T2 ]B′, B′′ [T1 ]B′′, B ⎡0 ⎢3 (c) ⎢ ⎢0 ⎣⎢0

T (a0 + a, x + a2 x 2 ) = c1T ( v1 ) + c2T ( v 2 ) + c3T ( v3 ) 1 = (a0 − a1 + 2a2 )(16 + 51x + 19 x 2 ) 3 1 + (−5a0 + 3a1 − 3a2 )(−6 − 5 x + 5 x 2 ) 8 1 + (a0 + a1 − a2 )(7 + 40 x + 15 x 2 ) 8 239a0 − 161a1 + 289a2 = 24 201a0 − 111a1 + 247 a2 + x 8 61a0 − 31a1 + 107a2 2 + x 12

0 0 3 0

0⎤ ⎡0 0 ⎤ ⎡2 0⎤ ⎢ ⎥ 0 ⎢ 6 0⎥ ⎥ 0 −3⎥ = ⎢ ⎥ 0⎥ ⎢ ⎥ ⎢ 0 −9 ⎥ 0 0 ⎦ ⎢0 0 ⎥ 3⎦⎥ ⎣ ⎣ ⎦

15. If T is a contraction or dilation of V, then T maps any basis B = {v1 , v 2 , ..., v n } of V to

{kv1, kv 2 , ..., kv n } where k is a nonzero constant. Thus the matrix for T relative to B is ⎡k 0 0 " 0 ⎤ ⎢0 k 0 " 0⎥ ⎢ ⎥ ⎢0 0 k " 0⎥ . #⎥ ⎢# # # ⎢⎣ 0 0 0 " k ⎥⎦

193


SSM: Elementary Linear Algebra 21. (a) [T2 D T1 ]B′, B = [T2 ]B′, B′′ [T1 ]B′′, B

17. The matrix for T relative to B is the matrix whose columns are the transforms of the basis vectors in B in terms of the standard basis. Since

(b) [T3 D T2 D T1 ]B′, B = [T3 ]B′, B′′′ [T2 ]B′′′, B′′ [T1 ]B′′, B

B is the standard basis for R n , this matrix is the standard matrix for T. Also, since B′ is the standard basis for R m , the resulting transformation will give vector components relative to the standard basis.

True/False 8.4 (a) False; the conclusion would only be true if T: V → V were a linear operator, i.e., if V = W.

19. (a) D(f1 ) = D(1) = 0

(b) False; the conclusion would only be true if T: V → V were a linear operator, i.e., if V = W.

D(f 2 ) = D(sin x) = cos x D(f3 ) = D(cos x) = − sin x The matrix for D relative to this basis is ⎡0 0 0 ⎤ ⎢0 0 −1⎥ . ⎢ ⎥ ⎣0 1 0 ⎦

(c) True; since the matrix for T is invertible, ker(T) = {0}. (d) False; the matrix of S D T relative to B is [ S ]B [T ]B .

(b) D(f1 ) = D(1) = 0

(e) True

D (f 2 ) = D ( e x ) = e x

Section 8.5

D(f3 ) = D(e2 x ) = 2e2 x The matrix for D relative to this basis is ⎡0 0 0 ⎤ ⎢0 1 0 ⎥ . ⎢ ⎥ ⎣0 0 2 ⎦

Exercise Set 8.5 ⎡1 ⎤ ⎡ −2 ⎤ 1. T (u1 ) = ⎢ ⎥ and T (u 2 ) = ⎢ ⎥ , so 0 ⎣ ⎦ ⎣ −1⎦ ⎡ 1 −2 ⎤ [T ]B = ⎢ . ⎣0 −1⎥⎦ By inspection, v1 = 2u1 + u 2 and

(c) D(f1 ) = D(e2 x ) = 2e2 x

D(f2 ) = D( xe 2 x ) = e 2 x + 2 xe2 x

v 2 = −3u1 + 4u 2 , so the transition matrix from

D(f3 ) = D( x 2 e2 x ) = 2 xe2 x + 2 x 2 e2 x The matrix for D relative to this basis is ⎡2 1 0⎤ ⎢ 0 2 2⎥ . ⎢ ⎥ ⎣ 0 0 2⎦

⎡ 2 −3⎤ B′ to B is P = ⎢ . Thus ⎣ 1 4⎥⎦ ⎡ 4 3⎤ PB→ B′ = P −1 = ⎢ 11 11 ⎥ . 1 2⎥ ⎢⎣ − 11 11 ⎦ [T ]B′ = PB → B′ [T ]B PB′→ B ⎡ 4 3 ⎤ ⎡1 −2 ⎤ ⎡ 2 −3⎤ = ⎢ 11 11 ⎥ ⎢ 1 2 ⎥ ⎣ 0 −1⎥⎦ ⎢⎣ 1 4 ⎥⎦ ⎢⎣ − 11 11 ⎦ ⎡ − 3 − 56 ⎤ 11 ⎥ = ⎢ 11 3⎥ 2 ⎢⎣ − 11 11 ⎦

(d) D(4e2 x + 6 xe2 x − 10 x 2 e 2 x ) ⎡ 2 1 0⎤ ⎡ 4⎤ = ⎢ 0 2 2⎥ ⎢ 6⎥ ⎢ ⎥⎢ ⎥ ⎣ 0 0 2 ⎦ ⎣ −10 ⎦ ⎡ 14 ⎤ = ⎢ −8 ⎥ ⎢ ⎥ ⎣ −20 ⎦

Thus, D(4e2 x + 6 xe2 x − 10 x 2 e 2 x ) = 14e 2 x − 8 xe2 x − 20 x 2 e2 x .

194


Section 8.5

7⎤ ⎡− 2 ⎡3 −1 ⎢ 4 9 ⎥ . Thus P = = P=⎢ 9 P B → B′ ⎢⎣ 13 − 61 ⎥⎦ ⎢⎣ 32 and [T ]B′ = PB→ B′ [T ]B PB′→ B 7⎤ ⎡ 3 7 ⎤ ⎡ 2 − 2 ⎤ ⎡− 2 9⎥ ⎢ 9 9⎥ = ⎢4 2⎥ ⎢3 4⎥ ⎢ 1 − 1⎥ ⎢⎣ 32 1⎥⎦ ⎢⎣ 12 3⎦ ⎣ 3 6⎦ ⎡1 1⎤ =⎢ . ⎣0 1⎥⎦

3. Since B is the standard basis for R 2 , ⎡ 1 − 1 ⎤ ⎡cos 45° − sin 45° ⎤ ⎢ 2 2⎥ = [T ]B = ⎢ . The 1 ⎥ ⎣ sin 45° cos 45° ⎥⎦ ⎢ 1 2⎦ ⎣ 2 matrices for PB→ B′ and PB′↔ B are the same as in Exercise 1, so [T ]B′ = PB → B′ [T ]B PB′→ B ⎡ 4 3 ⎤ ⎡ 1 − 1 ⎤ ⎡ 2 −3⎤ 2⎥ = ⎢ 11 11 ⎥ ⎢ 2 1 2 ⎥⎢ 1 1 ⎥ ⎢⎣ 1 4 ⎥⎦ ⎢⎣ − 11 11 ⎦ ⎣ 2 2⎦ ⎡ 13 − 25 ⎤ 11 2 ⎥ = ⎢ 115 2 . 9 ⎥ ⎢ 11 2 ⎦ ⎣ 11 2

1 ⎥⎦

11. (a) The matrix for T relative to the standard ⎡ 1 −1⎤ . The eigenvalues basis B is [T ]B = ⎢ ⎣ 2 4 ⎥⎦ of [T ]B are λ = 2 and λ = 3 with

⎡ 1⎤ corresponding eigenvectors ⎢ ⎥ and ⎣ −1⎦ ⎡ 1⎤ ⎢⎣ −2 ⎥⎦ . ⎡ 1 1⎤ , we have Then for P = ⎢ ⎣ −1 −2 ⎥⎦ ⎡ 2 1⎤ ⎡2 0⎤ and P −1[T ]B P = ⎢ P −1 = ⎢ . ⎥ − − 1 1 ⎣ ⎦ ⎣ 0 3⎥⎦ Since P represents the transition matrix from the basis B′ to the standard basis B, then ⎧ ⎡ 1⎤ ⎡ 1⎤ ⎫ B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ is a basis for which ⎩ ⎣ −1⎦ ⎣ −2 ⎦ ⎭ [T ]B′ is diagonal.

⎡1 ⎤ ⎡0 ⎤ ⎡0⎤ 5. T (u1 ) = ⎢ 0 ⎥ , T (u 2 ) = ⎢1 ⎥ and T (u3 ) = ⎢ 0 ⎥ , so ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0⎦ ⎣0 ⎦ ⎣0⎦ ⎡1 0 0⎤ [T ]B = ⎢ 0 1 0⎥ . ⎢ ⎥ ⎣ 0 0 0⎦ By inspection, v1 = u1 , v 2 = u1 + u 2 , and

v3 = u1 + u 2 + u3 , so the transition matrix from ⎡ 1 1 1⎤ B′ to B is P = ⎢ 0 1 1⎥ . ⎢ ⎥ ⎣ 0 0 1⎦ ⎡ 1 −1 0 ⎤ Thus PB→ B′ = P −1 = ⎢ 0 1 −1⎥ ⎢ ⎥ ⎣ 0 0 1⎦ [T ]B′ = PB→ B′ [T ]B PB′→ B ⎡ 1 −1 0 ⎤ ⎡ 1 0 0⎤ ⎡ 1 = ⎢0 1 −1⎥ ⎢ 0 1 0⎥ ⎢ 0 ⎢ ⎥⎢ ⎥⎢ ⎣0 0 1⎦ ⎣ 0 0 0⎦ ⎣ 0 ⎡ 1 0 0⎤ = ⎢0 1 1⎥ . ⎢ ⎥ ⎣0 0 0 ⎦

7⎤ 2⎥

and

(b) The matrix for T relative to the standard ⎡ 4 −1⎤ . basis B is [T ]B = ⎢ ⎣ −3 1⎥⎦

1 1⎤ 1 1⎥ ⎥ 0 1⎦

The eigenvalues of [T ]B are λ =

5 + 21 2

5 − 21 with corresponding 2 ⎡ −3− 21 ⎤ ⎡ −3+ 21 ⎤ eigenvectors ⎢ 6 ⎥ and ⎢ 6 ⎥ . ⎣⎢ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ and λ =

2 1 7. T (p1 ) = 6 + 3( x + 1) = 9 + 3x = p1 + p 2 , and 3 2 2 4 T (p 2 ) = 10 + 2( x + 1) = 12 + 2 x = − p1 + p 2 , 9 3 ⎡2 − 2⎤ 9⎥. so [T ]B = ⎢ 3 4⎥ ⎢⎣ 12 3⎦ 2 1 7 1 q1 = − p1 + p 2 and q 2 = p1 − p 2 , so the 9 3 9 6 transition matrix from B′ to B is

⎡ −3− 21 −3+ 21 ⎤ 6 Then for P = ⎢ 6 ⎥ , we have ⎢⎣ 1 1 ⎥⎦ −3+ 21 ⎤ ⎡− 3 −1 ⎢ 21 2 21 ⎥ P =⎢ and 3+ 21 ⎥ 3 ⎢⎣ 21 2 21 ⎥⎦

195



⎡ 5+ 21 P [ I ]B P = ⎢ 2 ⎢ 0 ⎣

0 ⎤ ⎥ . Since P 5− 21 ⎥ 2 ⎦ represents the transition matrix from the basis B′ to the standard basis B, then ⎧⎪ ⎡ −3− 21 ⎤ ⎡ −3+ 21 ⎤ ⎫⎪ B ′ = ⎨ ⎢ 6 ⎥ , ⎢ 6 ⎥ ⎬ is a basis for ⎩⎪ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎭⎪ which [T ]B′ is diagonal.

True/False 8.5

−1

(a) False; every matrix is similar to itself since

A = I −1 AI . (b) True; if A = P −1BP and B = Q −1CQ, then

A = P −1 (Q −1CQ) P = (QP)−1 C (QP). (c) True; invertibility is a similarity invariant. (d) True; if A = P −1 BP, then

13. (a) T (1) = 5 + x 2 T(x) = 6 − x 2

A−1 = ( P −1BP)−1 = P −1 B −1P. 2

T ( x ) = 2 − 8x − 2x Thus the matrix for T relative to the standard ⎡5 6 2⎤ basis B is [T ]B = ⎢ 0 −1 −8⎥ . The ⎢ ⎥ ⎣ 1 0 −2 ⎦

(e) True (f) False; for example, let T be the zero operator. (g) True

characteristic equation for [T ]B is

(h) False; if B and B′ are different, let [T ]B be given by the matrix PB→ B′ . Then

λ3 − 2λ 2 − 15λ + 36 = (λ + 4)(λ − 3)2 = 0, so the eigenvalues of T are λ = −4 and λ = 3.

[T ]B′, B = PB → B′ [T ]B = PB → B′ PB′→ B = I n .

(b) A basis for the eigenspace corresponding to ⎡ −2 ⎤ λ = −4 is ⎢⎢ 83 ⎥⎥ , so the polynomial ⎢⎣ 1⎥⎦

Chapter 8 Supplementary Exercises 1. No; T (x1 + x 2 ) = A(x1 + x 2 ) + B ≠ ( Ax1 + B) + ( Ax 2 + B) = T (x1 ) + T (x2 ), and if c ≠ 1, then T (cx) = cAx + B ≠ c( Ax + B) = cT (x).

8 −2 + x + x 2 is a basis in P 2 . A basis for 3 the eigenspace corresponding to λ = 3 is ⎡ 5⎤ ⎢ −2 ⎥ , so the polynomial so the polynomial ⎢ ⎥ ⎣ 1⎦

3. T (kv) = kv kv = k k v v ≠ kT ( v) if k ≠ 1. 5. (a) The matrix for T relative to the standard ⎡ 1 0 1 1⎤ basis is A = ⎢ 2 1 3 1⎥ . A basis for the ⎢ ⎥ ⎣ 1 0 0 1⎦ range of T is a basis for the column space of ⎡ 1 0 0 1⎤ A. A reduces to ⎢0 1 0 −1⎥ . ⎢ ⎥ ⎣0 0 1 0 ⎦ Since row operations don’t change the dependency relations among columns, the reduced form of A indicates that T (e3 ) and

5 − 2x + x 2 is a basis in P 2 . 15. If v is an eigenvector of T corresponding to λ, then v is a nonzero vector which satisfies the equation T(v) = λv or (λI − T)(v) = 0. Thus v is in the kernel of λI − T. 17. Since C[ x]B = D[ x]B for all x in V, then we can

let x = vi for each of the basis vectors v1, ..., v n of V. Since [ vi ]B = ei for each i

where {e1 , ..., e n } is the standard basis for R n ,

any two of T (e1 ), T (e2 ), T (e4 ) form a basis for the range. The reduced form of A shows that the general solution of Ax = 0 is x = −s, x2 = s,

this yields Cei = Dei for i = 1, 2, ..., n. But Cei and Dei are the ith columns of C and D, respectively. Since corresponding columns of C and D are all equal, C = D. 196



x3 = 0, x4 = s so a basis for the null space

one-dimensional. Hence, nullity(T) = 1 and rank(T ) = dim( M 22 ) − nullity(T ) = 3.

⎡ −1⎤ ⎢ 1⎥ of A, which is the kernel of T is ⎢ ⎥ . ⎢ 0⎥ ⎢⎣ 1⎥⎦

⎡1 0 ⎤ , 13. The standard basis for M 22 is u1 = ⎢ ⎣0 0 ⎥⎦ ⎡0 1 ⎤ ⎡0 0 ⎤ ⎡0 0 ⎤ , u3 = ⎢ , u4 = ⎢ . Since u2 = ⎢ ⎥ ⎥ 0 0 1 0 ⎣ ⎦ ⎣0 1 ⎥⎦ ⎣ ⎦ L(u1 ) = u1, L(u 2 ) = u3 , L(u3 ) = u 2 , and

(b) Since R(T) is three-dimensional and ker(T) is one-dimensional, rank(T) = 3 and nullity(T) = 1.

L(u 4 ) = u 4 , the matrix of L relative to the

7. (a) The matrix for T relative to B is ⎡ 1 1 2 −2 ⎤ ⎢ 1 −1 −4 6 ⎥ [T ]B = ⎢ ⎥. ⎢ 1 2 5 −6 ⎥ 3 −2 ⎥⎦ ⎢⎣3 2

⎡1 ⎢0 standard basis is ⎢ ⎢0 ⎢⎣0

0 0 1 0

0 1 0 0

0⎤ 0⎥ ⎥. 0⎥ 1⎥⎦

15. The transition matrix from B′ to B is ⎡ 1 1 1⎤ P = ⎢ 0 1 1⎥ , thus ⎢ ⎥ ⎣ 0 0 1⎦

⎡ 1 0 −1 2 ⎤ ⎢ 0 1 3 −4 ⎥ [T ]B reduces to ⎢ ⎥ which ⎢0 0 0 0 ⎥ ⎣⎢0 0 0 0 ⎦⎥ has rank 2 and nullity 2. Thus, rank(T) = 2 and nullity(T) = 2.

⎡ −4 0 9 ⎤ [T ]B′ = P −1[T ]B P = ⎢ 1 0 −2 ⎥ . ⎢ ⎥ ⎣ 0 1 1⎦

(b) Since rank(T) < dim(V), T is not one-to-one.

⎛ ⎡ 1⎤ ⎞ ⎡ 1⎤ ⎛ ⎡0 ⎤ ⎞ ⎡ −1⎤ 17. T ⎜ ⎢0 ⎥ ⎟ = ⎢0 ⎥ , T ⎜ ⎢ 1⎥ ⎟ = ⎢ 1⎥ , and ⎜⎜ ⎢ ⎥ ⎟⎟ ⎢ ⎥ ⎜⎜ ⎢ ⎥ ⎟⎟ ⎢ ⎥ ⎝ ⎣0 ⎦ ⎠ ⎣ 0⎦ ⎝ ⎣0 ⎦ ⎠ ⎣ 1⎦ ⎛ ⎡0 ⎤ ⎞ ⎡ 1⎤ ⎡ 1 −1 1⎤ T ⎜ ⎢0 ⎥ ⎟ = ⎢ 0⎥ , thus [T }B = ⎢0 1 0⎥ . ⎜⎜ ⎢ ⎥ ⎟⎟ ⎢ ⎥ ⎢ ⎥ ⎣ 1 0 −1⎦ ⎝ ⎣ 1⎦ ⎠ ⎣ −1⎦ Note that this can also be read directly from [T (x)]B .

9. (a) Since A = P −1 BP, then

AT = ( P −1BP)T = PT BT ( P −1 )T

= (( PT )−1 )−1 BT ( P −1 )T

= (( P −1 )T )−1 BT ( P −1 )T . Thus, AT and BT are similar. (b) A−1 = ( P −1 BP)−1 = P −1 B −1 P

Thus A−1 and B −1 are similar.

19. (a) D(f + g) = ( f ( x) + g ( x))′′ = f ′′( x) + g ′′( x) and D(kf ) = (kf ( x))′′ = kf ′′( x).

⎡a b ⎤ 11. For X = ⎢ , ⎣ c d ⎥⎦ ⎡a + c b + d ⎤ ⎡ b b ⎤ T(X ) = ⎢ + 0 ⎥⎦ ⎢⎣ d d ⎥⎦ ⎣ 0 ⎡ a + b + c 2b + d ⎤ . =⎢ d ⎥⎦ ⎣ d T(X) = 0 gives the equations a+b+c = 0 2b + d = 0 d = 0. Thus c = −a and X is in ker(T) if it has the form ⎧ ⎡ 1 0⎤ ⎫ ⎡ k 0⎤ ⎢⎣ − k 0 ⎥⎦ , so ker(T ) = ⎨k ⎣⎢ −1 0⎦⎥ ⎬ which is ⎩ ⎭

(b) If f is in ker(D), then f has the form f = f ( x) = a0 + a1 x, so a basis for ker(D) is f(x) = 1, g(x) = x. (c) D(f) = f if and only if f = f ( x) = ae x + be− x for arbitrary constants a and b. Thus,

f ( x) = e x and g ( x) = e − x form a basis for this subspace.

197



21. (c) Note that a1 P1 ( x) + a2 P2 ( x) + a3 P3 ( x) evaluated at x1 , x2 , and x3 gives the values

25.

J ( x) =

a1 , a2 , and a3 , respectively, since

J ( x2 ) =

Pi ( xi ) = 1 and Pi ( x j ) = 0 for i ≠ j.

#

Thus ⎡ a1 ⎤ T (a1 P1 ( x) + a2 P2 ( x) + a3 P3 ( x)) = ⎢ a2 ⎥ , or ⎢ ⎥ ⎣ a3 ⎦ ⎛ ⎡ a1 ⎤ ⎞ T −1 ⎜ ⎢ a2 ⎥ ⎟ = a1 P1 ( x) + a2 P2 ( x) + a3 P3 ( x). ⎜⎢ ⎥⎟ ⎜ a ⎟ ⎝⎣ 3⎦⎠

D(1) = 0 D( x) = 1 D( x 2 ) = #

1 2 x 2 1 3 x 3

1 n +1 x n +1 This gives the (n + 2) × (n + 1) matrix ⎡0 0 0 " 0 ⎤ ⎢1 0 0 " 0 ⎥ ⎢ ⎥ 1 ⎢0 2 0 " 0 ⎥ ⎢ ⎥. 1 ⎢0 0 3 " 0 ⎥ ⎢# # # # ⎥ ⎢ ⎥ 1 ⎢⎣0 0 0 " n+1 ⎥⎦ J ( xn ) =

(d) From the computations in part (c), the points lie on the graph. 23.

J (1) = x

2x

D( x n ) = nx n −1 This gives the matrix shown.

198

Chapter 9 Numerical Methods ⎡ 1 −1 −1⎤ ⎡ 1 −1 −1⎤ → ⎢ 0 −2 2 ⎥ → ⎢ 0 1 −1⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 4 1⎦ ⎣ 0 4 1⎦ ⎡ 1 −1 −1⎤ ⎡ 1 −1 −1⎤ → ⎢ 0 1 −1⎥ → ⎢0 1 −1⎥ = U ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 5⎦ ⎣0 0 1⎦ 1 1 The multipliers used were , 1, − , −4, and 2 2 ⎡ 2 0 0⎤ 1 , which leads to L = ⎢ 0 −2 0 ⎥ , so the ⎢ ⎥ 5 ⎣ −1 4 5⎦ system is ⎡ 2 0 0 ⎤ ⎡ 1 −1 −1⎤ ⎡ x1 ⎤ ⎡ −4 ⎤ ⎢ 0 −2 0 ⎥ ⎢0 1 −1⎥ ⎢ x2 ⎥ = ⎢ −2 ⎥ . ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −1 4 5⎦ ⎣0 0 1⎦ ⎣ x3 ⎦ ⎣ 6 ⎦ ⎡ 2 0 0 ⎤ ⎡ y1 ⎤ ⎡ −4 ⎤ ⎢ 0 −2 0 ⎥ ⎢ y2 ⎥ = ⎢ −2 ⎥ is ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −1 4 5⎦ ⎣ y3 ⎦ ⎣ 6 ⎦ = −4 2 y1 −2 y 2 = −2 which has the solution − y1 + 4 y2 + 5 y3 = 6 y1 = −2, y2 = 1, y3 = 0.

Section 9.1 Exercise Set 9.1 1. In matrix form, the system is ⎡ 3 −6 ⎤ ⎡ x1 ⎤ ⎡ 3 0⎤ ⎡ 1 −2 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ = = ⎢⎣ −2 5⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣ −2 1⎥⎦ ⎢⎣ 0 1⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣1 ⎥⎦

⎡ 1 −2 ⎤ ⎡ x1 ⎤ ⎡ y1 ⎤ = which reduces to ⎢ and 1⎥⎦ ⎢⎣ x2 ⎥⎦ ⎢⎣ y2 ⎥⎦ ⎣0 ⎡ 3 0 ⎤ ⎡ y1 ⎤ ⎡0 ⎤ ⎢⎣ −2 1⎥⎦ ⎢ y ⎥ = ⎢⎣1 ⎥⎦ . ⎣ 2⎦ The second system is

3 y1 = 0 , which has −2 y1 + y2 = 1

solution y1 = 0, y2 = 1, so the first system x1 − 2 x2 = 0 ⎡ 1 −2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ becomes ⎢ , ⎢ x ⎥ = ⎢1 ⎥ or ⎥ 0 1 x2 = 1 ⎣ ⎦⎣ 2⎦ ⎣ ⎦ which gives the solution to the original system: x1 = 2, x2 = 1. 3. Reduce the coefficient matrix. ⎡ 2 8⎤ ⎡ 1 4 ⎤ ⎡ 1 4 ⎤ ⎡ 1 4 ⎤ ⎢⎣ −1 −1⎥⎦ → ⎢⎣ −1 −1⎥⎦ → ⎢⎣ 0 3⎥⎦ → ⎢⎣0 1⎥⎦ =U 1 1 The multipliers used were , 1, and , which 2 3 ⎡ 2 0⎤ leads to L = ⎢ , so the system is ⎣ −1 3⎥⎦ ⎡ 2 0 ⎤ ⎡ 1 4 ⎤ ⎡ x1 ⎤ ⎡ −2 ⎤ ⎢⎣ −1 3⎥⎦ ⎢⎣0 1⎥⎦ ⎢ x ⎥ = ⎢⎣ −2 ⎥⎦ . ⎣ 2⎦

x1 − x2 − x3 = −2 ⎡ 1 −1 −1⎤ ⎡ x1 ⎤ ⎡ −2 ⎤ ⎢0 1 −1⎥ ⎢ x2 ⎥ = ⎢ 1⎥ is x2 − x3 = 1 ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 1 0 x x3 = 0 ⎣ ⎦⎣ 3⎦ ⎣ ⎦ which gives the solution to the original system: x1 = −1, x2 = 1, x3 = 0. 7. Reduce the coefficient matrix. 5 10 ⎤ ⎡ 1 1 2⎤ ⎡ 5 ⎢ −8 −7 −9 ⎥ → ⎢ −8 −7 −9⎥ ⎢ ⎥ ⎢ ⎥ 4 26⎦ ⎣ 0 4 26 ⎦ ⎣ 0 ⎡ 1 1 2 ⎤ ⎡ 1 1 2⎤ → ⎢0 1 7 ⎥ → ⎢0 1 7 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 4 26 ⎦ ⎣ 0 0 −2⎦ ⎡ 1 1 2⎤ → ⎢0 1 7 ⎥ = U ⎢ ⎥ ⎣ 0 0 1⎦

2 y1 = −2 ⎡ 2 0 ⎤ ⎡ y1 ⎤ ⎡ −2 ⎤ ⎢⎣ −1 3⎦⎥ ⎢ y ⎥ = ⎣⎢ −2 ⎦⎥ is − y + 3 y = −2 which ⎣ 2⎦ 1 2 has the solution y1 = −1, y2 = −1.

x1 + 4 x2 = −1 ⎡ 1 4 ⎤ ⎡ x1 ⎤ ⎡ −1⎤ which ⎢⎣0 1⎦⎥ ⎢ x ⎥ = ⎣⎢ −1⎦⎥ is x2 = −1 ⎣ 2⎦ gives the solution to the original system: x1 = 3, x2 = −1. 5. Reduce the coefficient matrix. ⎡ 2 −2 −2 ⎤ ⎡ 1 −1 −1⎤ ⎢ 0 −2 2 ⎥ → ⎢ 0 −2 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −1 5 2 ⎦ ⎣ −1 5 2 ⎦

The multipliers used were

199

1 1 , 8, −4, and − 2 5

Chapter 9: Numerical Methods


⎡ 5 0 0⎤ which leads to L = ⎢ −8 1 0 ⎥ , so the system ⎢ ⎥ ⎣ 0 4 −2 ⎦ ⎡ 5 0 0 ⎤ ⎡ 1 1 2 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ is ⎢ −8 1 0 ⎥ ⎢0 1 7 ⎥ ⎢ x2 ⎥ = ⎢ 1 ⎥ . ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 4 −2 ⎦ ⎣0 0 1⎦ ⎣ x3 ⎦ ⎣ 4 ⎦ ⎡ 5 0 0⎤ ⎡ y1 ⎤ ⎡ 0 ⎤ ⎢ −8 1 0⎥ ⎢ y2 ⎥ = ⎢ 1 ⎥ is ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 4 −2⎦ ⎣ y3 ⎦ ⎣ 4 ⎦ =0 5 y1 −8 y1 + y2 = 1 which has the solution 4 y2 − 2 y3 = 4 y1 = 0, y2 = 1, y3 = 0.

⎡ 1 0 −1 ⎢0 3 0 →⎢ ⎢ 0 −1 2 ⎣⎢ 0 0 1

0 ⎤ ⎡ y1 ⎤ ⎡ 5⎤ 0 ⎥ ⎢ y2 ⎥ ⎢ −1⎥ ⎥ ⎢ ⎥ = ⎢ ⎥ is 0 ⎥ ⎢ y3 ⎥ ⎢ 3⎥ 4 ⎦⎥ ⎢⎣ y4 ⎥⎦ ⎣⎢ 7 ⎦⎥ =5 = −1 , which has the =3 =7

0 −1 0 ⎤ ⎡ x1 ⎤ ⎡ −5⎤ 1 0 2 ⎥ ⎢ x2 ⎥ ⎢ 3⎥ ⎥ ⎢ ⎥ = ⎢ ⎥ is 0 1 1⎥ ⎢ x3 ⎥ ⎢ 3⎥ 0 0 1⎦⎥ ⎢⎣ x4 ⎥⎦ ⎣⎢ 1⎦⎥ x1 − x3 = −5 x2 + 2 x4 = 3 , which gives the solution x3 + x4 = 3 x4 = 1

⎡1 ⎢0 ⎢ ⎢0 ⎣⎢0

0⎤ 2⎥ ⎥ 0⎥ 5⎦⎥

11. (a) Reduce A to upper triangular form. ⎡ 1 1 − 1⎤ ⎡ 2 1 −1⎤ 2 2⎥ ⎢ ⎢ −2 −1 2⎥ → −2 −1 2 ⎢ ⎥ ⎢ ⎥ ⎢⎣ 2 1 ⎣ 2 1 0⎦ 0 ⎥⎦

0⎤ ⎡ 1 2⎥ ⎢0 ⎥→⎢ 2⎥ ⎢0 5⎥⎦ ⎢⎣ 0

0 −1 0 ⎤ 1 0 2⎥ ⎥ 0 1 1⎥ 0 1 5⎥⎦

⎡1 ⎢0 →⎢ ⎢0 ⎣⎢ 0

0 −1 1 0 0 1 0 0

0⎤ ⎡ 1 2⎥ ⎢0 ⎥→⎢ 1⎥ ⎢ 0 4 ⎦⎥ ⎣⎢ 0

0 −1 0 ⎤ 1 0 2⎥ ⎥ =U 0 1 1⎥ 0 0 1⎦⎥

⎡ −1 ⎢ 2 1 and , which leads to L = ⎢ 4 ⎢ 0 ⎣⎢ 0 the system is

0 0 2 1

x4 = 1.

0 −1 1 0 0 2 0 1

1 , 3 0 3 −1 0

⎡ −1 0 ⎢ 2 3 ⎢ ⎢ 0 −1 ⎣⎢ 0 0

to the original system: x1 = −3, x2 = 1, x3 = 2,

⎡1 ⎢0 →⎢ ⎢0 ⎢⎣ 0

The multipliers used were −1, −2,

0⎤ ⎡ 1 0 ⎥ ⎢0 ⎥⎢ 0 ⎥ ⎢0 4 ⎦⎥ ⎣⎢0

solution y1 = −5, y2 = 3, y3 = 3, y4 = 1.

0⎤ 6⎥ ⎥ 0⎥ 5⎦⎥

0⎤ ⎡ 1 0 −1 ⎢0 1 0 6⎥ ⎥→⎢ 0⎥ ⎢ 0 −1 2 5⎦⎥ ⎣⎢ 0 0 1

0 0 2 1

− y1 2 y1 + 3 y2 − y2 + 2 y3 y3 + 4 y4

x1 + x2 + 2 x3 = 0 ⎡ 1 1 2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢0 1 7 ⎥ ⎢ x2 ⎥ = ⎢1 ⎥ is x2 + 7 x3 = 1 ⎢ ⎥⎢ ⎥ ⎢ ⎥ x3 = 0 ⎣0 0 1⎦ ⎣ x3 ⎦ ⎣0 ⎦ which gives the solution to the original system: x1 = −1, x2 = 1, x3 = 0. 9. Reduce the coefficient matrix. 1 0⎤ ⎡ −1 0 ⎡ 1 0 −1 ⎢ 2 3 −2 6 ⎥ ⎢ 2 3 −2 ⎢ ⎥→⎢ ⎢ 0 −1 2 0 ⎥ ⎢ 0 −1 2 1 5⎦⎥ 1 ⎣⎢ 0 0 ⎣⎢ 0 0

0 −1 0 ⎤ ⎡ x1 ⎤ ⎡ 5⎤ 1 0 2 ⎥ ⎢ x2 ⎥ ⎢ −1⎥ ⎥ ⎢ ⎥ = ⎢ ⎥. 0 1 1⎥ ⎢ x3 ⎥ ⎢ 3⎥ 0 0 1⎦⎥ ⎣⎢ x4 ⎦⎥ ⎣⎢ 7 ⎦⎥

⎡ −1 0 ⎢ 2 3 ⎢ ⎢ 0 −1 ⎣⎢ 0 0

⎡1 1 − 1⎤ ⎡1 2 2⎥ ⎢ → ⎢0 0 1⎥ → ⎢0 ⎢ 0 ⎥⎦ ⎢⎣0 ⎣⎢ 2 1

1, 0 0 2 1

− 12 ⎤ ⎥ 0 1⎥ = U 0 1⎦⎥ 1 2

1 , 2, and −2, 2 ⎡ 2 0 0⎤ which leads to L = ⎢ −2 1 0⎥ where the ⎢ ⎥ ⎣ 2 0 1⎦ 1’s on the diagonal reflect that no multiplication was required on the 2nd and 3rd diagonal entries.

The multipliers used were

1 , −1, 2 0⎤ 0⎥ ⎥ , so 0⎥ 4 ⎦⎥

(b) To change the 2 on the diagonal of L to a 1, the first column of L is divided by 2 and the diagonal matrix has a 2 as the 1, 1 entry.

200


Section 9.1

A = L1 DU1

x1 + 2 x2 + 2 x3 = 1 ⎡ 1 2 2⎤ ⎡ x1 ⎤ ⎡ 1⎤ ⎢0 1 4⎥ ⎢ x2 ⎥ = ⎢ 2 ⎥ is x2 + 4 x3 = 2 ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 17 12 x 17 x3 = 12 ⎣ ⎦⎣ 3⎦ ⎣ ⎦ which gives the solution of the original system: 21 14 12 x1 = , x2 = − , x3 = . 17 17 17

⎡ 1 0 0 ⎤ ⎡ 2 0 0 ⎤ ⎡ 1 12 − 12 ⎤ ⎥ = ⎢ −1 1 0 ⎥ ⎢ 0 1 0 ⎥ ⎢0 0 1⎥ ⎢ ⎥⎢ ⎥⎢ ⎣ 1 0 1⎦ ⎣ 0 0 1⎦ ⎢⎣0 0 1⎥⎦ is the LDU-decomposition of A. (c) Let U 2 = DU , and L2 = L1, then

17. If rows 2 and 3 of A are interchanged, then the resulting matrix has an LU-decomposition. For ⎡ 1 0 0⎤ ⎡ 3 −1 0 ⎤ P = ⎢0 0 1⎥ , PA = ⎢ 0 2 1⎥ . ⎢ ⎥ ⎢ ⎥ ⎣0 1 0 ⎦ ⎣ 3 −1 1⎦ Reduce PA to upper triangular form. ⎡ 1 − 1 0⎤ ⎡ 3 −1 0 ⎤ 3 ⎥ ⎢0 2 1⎥ → ⎢0 2 1⎥ ⎢ ⎢ ⎥ ⎢⎣ 3 −1 1⎥⎦ ⎣ 3 −1 1⎦

⎡ 1 0 0 ⎤ ⎡ 2 1 −1⎤ A = L2U 2 = ⎢ −1 1 0 ⎥ ⎢ 0 0 1⎥ . ⎢ ⎥⎢ ⎥ ⎣ 1 0 1⎦ ⎣ 0 0 1⎦ 13. Reduce A to upper triangular form. ⎡ 3 −12 6 ⎤ ⎡ 1 −4 2 ⎤ ⎢0 2 0⎥ → ⎢0 2 0⎥ ⎢ ⎥ ⎢ ⎥ 6 28 13 6 28 13⎦ − − ⎣ ⎦ ⎣

⎡ 1 −4 2 ⎤ ⎡ 1 −4 2 ⎤ → ⎢ 0 2 0⎥ → ⎢0 1 0⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 −4 1⎦ ⎣0 −4 1⎦ ⎡ 1 −4 2 ⎤ → ⎢0 1 0⎥ = U ⎢ ⎥ ⎣ 0 0 1⎦ 1 1 The multipliers used were , −6, , and 4, 2 3 ⎡ 3 0 0⎤ which leads to L1 = ⎢0 2 0 ⎥ . Since ⎢ ⎥ ⎣6 −4 1⎦ ⎡ 1 0 0⎤ ⎡ 3 0 0⎤ L1 = ⎢ 0 1 0 ⎥ ⎢ 0 2 0 ⎥ , then ⎢ ⎥⎢ ⎥ 2 2 1⎦ ⎣ 0 0 1⎦ − ⎣ ⎡ 1 0 0⎤ ⎡ 3 0 0 ⎤ ⎡ 1 −4 2 ⎤ A = ⎢0 1 0⎥ ⎢ 0 2 0 ⎥ ⎢0 1 0 ⎥ is the ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ 2 −2 1⎦ ⎣ 0 0 1⎦ ⎣0 0 1⎦ LDU-decomposition of A.

⎡ 1 − 1 0⎤ ⎡ 1 − 1 0⎤ 3 3 ⎢ ⎥ ⎢ ⎥ → 0 → ⎢0 1 12 ⎥ = U 2 1 ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 1⎥⎦ 0 1⎦ ⎣0 1 1 The multipliers used were , −3, and , so 2 3 ⎡ 3 0 0⎤ L = ⎢0 2 0⎥ . Since P = P −1 , we have ⎢ ⎥ ⎣ 3 0 1⎦ 1 ⎡ 1 0 0⎤ ⎡ 3 0 0 ⎤ ⎢⎡ 1 − 3 0 ⎥⎤ A = PLU = ⎢ 0 0 1⎥ ⎢ 0 2 0 ⎥ ⎢0 1 12 ⎥ . ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ 0 1 0⎦ ⎣ 3 0 1⎦ ⎣0 0 1⎦ ⎡ −2 ⎤ Since Pb = ⎢ 4⎥ , the system to solve is ⎢ ⎥ ⎣ 1⎦ 1 ⎡ 3 0 0 ⎤ ⎢⎡ 1 − 3 0 ⎥⎤ ⎡ x1 ⎤ ⎡ −2 ⎤ ⎢ ⎥ ⎢0 2 0 ⎥ ⎢0 1 12 ⎥ ⎢ x2 ⎥ = ⎢⎢ 4 ⎥⎥ . ⎢ ⎥⎢ ⎥ ⎣ 3 0 1⎦ ⎣0 0 1⎦ ⎣ x3 ⎦ ⎣ 1⎦ 3 y1 = −2 ⎡ 3 0 0 ⎤ ⎡ y1 ⎤ ⎡ −2 ⎤ ⎢0 2 0 ⎥ ⎢ y2 ⎥ = ⎢ 4 ⎥ is 2 y2 =4 ⎢ ⎥⎢ ⎥ ⎢ ⎥ 3 0 1 1 3 y1 + y3 = 1 ⎣ ⎦ ⎣ y3 ⎦ ⎣ ⎦

⎡0 1 0⎤ ⎡1 ⎤ 15. P −1 = ⎢ 1 0 0 ⎥ and P −1b = ⎢ 2 ⎥ , so the ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 1⎦ ⎣5⎦ system P −1 Ax = P −1b is ⎡ 1 0 0⎤ ⎡ 1 2 2 ⎤ ⎡ x1 ⎤ ⎡1 ⎤ ⎢0 1 0⎥ ⎢ 0 1 4 ⎥ ⎢ x2 ⎥ = ⎢ 2⎥ . ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 3 −5 1⎦ ⎣ 0 0 17 ⎦ ⎣ x3 ⎦ ⎣ 5 ⎦

2 which has the solution y1 = − , y2 = 2, 3 y3 = 3.

y1 =1 ⎡ 1 0 0 ⎤ ⎡ y1 ⎤ ⎡ 1 ⎤ ⎢0 y2 =2 1 0 ⎥ ⎢ y2 ⎥ = ⎢ 2 ⎥ is ⎢ ⎥⎢ ⎥ ⎢ ⎥ − 3 5 1 5 3 y1 − 5 y2 + y3 = 5 ⎣ ⎦ ⎣ y3 ⎦ ⎣ ⎦ which has the solution y1 = 1, y2 = 2, y3 = 12.

⎡ 1 − 1 0⎤ ⎡ x ⎤ ⎡− 2 ⎤ 3 ⎢ ⎥ ⎢ 1 ⎥ ⎢ 3⎥ 1 12 ⎥ ⎢ x2 ⎥ = ⎢ 2 ⎥ is ⎢0 ⎢ ⎥ 0 1⎦ ⎣ x3 ⎦ ⎢⎣ 3⎥⎦ ⎣0

201


1 x1 − x2 3 x2 +

=− 1 x3 = 2 2 x3 = 3


(b) There is no dominant eigenvalue since λ1 = λ 4 = 5.

2 3 which gives the solution

3. The characteristic polynomial of A is

λ 2 − 4λ − 6 = 0, so the eigenvalues of A are

1 1 to the original system: x1 = − , x2 = , 2 2 x3 = 3.

λ = 2 ± 10. λ = 2 + 10 ≈ 5.16228 is the dominant eigenvalue and the dominant ⎡ 1 ⎤ ⎡ 1 ⎤ eigenvector is ⎢ ≈⎢ . ⎥ ⎥ 0 16228 − . ⎦ ⎣3 − 10 ⎦ ⎣

19. (a) If A has such an LU-decomposition, it can be written as y ⎤ ⎡a b ⎤ ⎡ 1 0⎤ ⎡ x y ⎤ ⎡ x ⎢⎣ c d ⎥⎦ = ⎢⎣ w 1 ⎥⎦ ⎢⎣ 0 z ⎥⎦ = ⎢⎣ wx wy + z ⎥⎦ which leads to the equations x=a y=b wx = c wy + z = d Since a ≠ 0, the system has the unique c solution x = a, y = b, w = , and a bc ad − bc z=d− = . a a Because the solution is unique, the LU-decomposition is also unique.

⎡ 5 −1⎤ ⎡1 ⎤ ⎡ 5⎤ Ax 0 = ⎢ = ⎣ −1 −1⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ −1⎥⎦ Ax 0 1 ⎡ 5⎤ ⎡ 0.98058⎤ x1 = = ≈ Ax 0 26 ⎢⎣ −1⎥⎦ ⎢⎣ −0.19612 ⎥⎦ ⎡ 5 −1⎤ ⎡ 0.98050⎤ ⎡ 5.09902 ⎤ Ax1 ≈ ⎢ ≈ ⎣ −1 −1⎥⎦ ⎢⎣ −0.19612⎥⎦ ⎢⎣ −0.78446 ⎥⎦ Ax1 ⎡ 0.98837 ⎤ x2 = ≈ Ax1 ⎢⎣ −0.15206⎥⎦ ⎡ 5 −1⎤ ⎡ 0.98837 ⎤ ⎡ 5.09391⎤ Ax 2 ≈ ⎢ ≈ ⎣ −1 −1⎥⎦ ⎢⎣ −0.15206 ⎥⎦ ⎢⎣ −0.83631⎥⎦ Ax 2 ⎡ 0.98679 ⎤ x3 = ≈⎢ Ax 2 ⎣ −0.16201⎥⎦ ⎡ 5 −1⎤ ⎡ 0.98679 ⎤ ⎡ 5.09596 ⎤ Ax 3 ≈ ⎢ ≈ ⎣ −1 −1⎥⎦ ⎢⎣ −0.16201⎥⎦ ⎢⎣ −0.82478⎥⎦ Ax 3 ⎡ 0.98715⎤ x4 = ≈⎢ Ax 3 ⎣ −0.15977 ⎥⎦

(b) From part (a) the LU-decomposition is b ⎤ ⎡a b ⎤ ⎡ 1 0⎤ ⎡ a ⎢⎣ c d ⎥⎦ = ⎢ c 1 ⎥ ⎢ 0 ad −bc ⎥ . a ⎦⎥ ⎣⎢ a ⎦⎥ ⎣⎢

⎡ 5 −1⎤ ⎡ 0.98715⎤ ⎡ 5.09552 ⎤ Ax 4 = ⎢ ≈ ⎣ −1 −1⎥⎦ ⎢⎣ −0.15977 ⎥⎦ ⎢⎣ −0.82738⎥⎦ ⎡ 0.98709 ⎤ The dominant unit eigenvector is ≈ ⎢ , ⎣ −0.16018⎥⎦ which x 4 approximates to three decimal place accuracy.

True/False 9.1 (a) False; if the matrix cannot be reduced to row echelon form without interchanging rows, then it does not have an LU-decomposition.

λ(1) = ( Ax1 )T x1 ≈ 5.24297

(b) False; if the row equivalence of A and U requires interchanging rows of A, then A does not have an LU-decomposition.

λ( 2) = ( Ax2 )T x2 ≈ 5.16138 λ(3) = ( Ax3 )T x3 ≈ 5.16226

(c) True

λ( 4) = ( Ax4 )T x4 ≈ 5.16223

(d) True

λ( 4) approximates λ to four decimal place accuracy.

(e) True

5. The characteristic equation of A is

Section 9.2

λ 2 − 6λ − 4 = 0, so the eigenvalues of A are

Exercise Set 9.2

λ = 3 ± 13. λ = 3 + 13 ≈ 6.60555 is the dominant eigenvalue with corresponding scaled

1. (a) λ3 = −8 is the dominant eigenvalue. 202


Section 9.2

⎡ 2 − 13 ⎤ ⎡ −0.53518⎤ eigenvector ⎢ 3 ⎥ ≈ ⎢ ⎥⎦ . 1 ⎣⎢ 1 ⎦⎥ ⎣

(b) λ(1) =

⎡ 1 −3⎤ ⎡1⎤ ⎡ −2⎤ Ax 0 = ⎢ = ⎣ −3 5⎥⎦ ⎢⎣1⎥⎦ ⎢⎣ 2 ⎥⎦ Ax 0 ⎡ −1⎤ x1 = =⎢ ⎥ max( Ax0 ) ⎣ 1⎦ λ(1) =

Ax1 ⋅ x1 =6 x1 ⋅ x1

Ax 2 ⋅ x 2 = 6.6 x 2 ⋅ x2

λ(3) =

Ax 3 ⋅ x 3 ≈ 2.997 x3 ⋅ x 3

(d) The percentage error is

3 − 2.997 = 0.001 3

or 0.1%.

9. By Formula (10),

A5 x0

⎡0.99180 ⎤ ≈⎢ ⎥. max( A x0 ) ⎣ 1 ⎦ Ax 5 ⋅ x 5 Thus λ(5) = ≈ 2.99993. x5 ⋅ x 5

x5 =

Ax 3 ⋅ x 3 ≈ 6.60550 x 3 ⋅ x3

⎡ 1 −3⎤ ⎡ −0.53846 ⎤ ⎡ −3.53846 ⎤ Ax 3 ≈ ⎢ ⎥⎦ ≈ ⎢⎣ 6.61538⎥⎦ 1 ⎣ −3 5⎥⎦ ⎢⎣ Ax 3 ⎡ −0.53488⎤ x4 = ≈ ⎥⎦ 1 max( Ax3 ) ⎢⎣ λ ( 4) =

Ax 2 ⋅ x 2 ≈ 2.976 x 2 ⋅ x2

λ 2 − 4λ + 3 = (λ − 3)(λ − 1) = 0 so the dominant eigenvalue of A is λ = 3, with ⎡ 1⎤ dominant eigenvector ⎢ ⎥ . ⎣ −1⎦

⎡ 1 −3⎤ ⎡ −0.5⎤ ⎡ −3.5⎤ Ax 2 = ⎢ = ⎣ −3 5⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 6.5⎥⎦ Ax 2 ⎡ −0.53846 ⎤ x3 = ≈ ⎥⎦ 1 max( Ax2 ) ⎢⎣ λ(3) =

λ ( 2) =

(c) The characteristic polynomial of A is

⎡ 1 −3⎤ ⎡ −1⎤ ⎡ −4 ⎤ Ax1 = ⎢ = ⎣ −3 5⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣ 8⎥⎦ Ax1 ⎡ −0.5⎤ x2 = = max( Ax1 ) ⎢⎣ 1 ⎥⎦ λ ( 2) =

Ax1 ⋅ x1 = 2.8 x1 ⋅ x1

5

⎡1 ⎤ 13. (a) Starting with x0 = ⎢ 0⎥ , it takes 8 iterations. ⎢ ⎥ ⎣ 0⎦ ⎡0.229 ⎤ x1 ≈ ⎢0.668 ⎥ , λ(1) ≈ 7.632 ⎢ ⎥ ⎣0.668 ⎦ ⎡0.507 ⎤ x 2 ≈ ⎢ 0.320 ⎥ , λ( 2) ≈ 9.968 ⎢ ⎥ ⎣ 0.800 ⎦ ⎡ 0.380 ⎤ x3 ≈ ⎢0.197 ⎥ , λ(3) ≈ 10.622 ⎢ ⎥ ⎣ 0.904 ⎦ ⎡ 0.344⎤ x 4 ≈ ⎢ 0.096⎥ , λ( 4) ≈ 10.827 ⎢ ⎥ ⎣ 0.934⎦ ⎡0.317 ⎤ x5 ≈ ⎢ 0.044 ⎥ , λ(5) ≈ 10.886 ⎢ ⎥ ⎣ 0.948 ⎦ ⎡0.302 ⎤ x6 ≈ ⎢0.016 ⎥ , λ(6) ≈ 10.903 ⎢ ⎥ ⎣ 0.953⎦

Ax 4 ⋅ x 4 ≈ 6.60555 x 4 ⋅ x4

⎡ 2 −1⎤ ⎡1 ⎤ ⎡ 2⎤ 7. (a) Ax0 = ⎢ = ⎣ −1 2 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ −1⎥⎦ Ax0 ⎡ 1 ⎤ x1 = = max( Ax0 ) ⎢⎣ −0.5⎥⎦ ⎡ 2 −1⎤ ⎡ 1 ⎤ ⎡ 2.5⎤ Ax1 = ⎢ = ⎣ −1 2 ⎥⎦ ⎢⎣ −0.5⎥⎦ ⎢⎣ 2 ⎥⎦ Ax1 ⎡ 1 ⎤ x2 = = max( Ax1 ) ⎢⎣ −0.8⎥⎦ ⎡ 2 −1⎤ ⎡ 1 ⎤ ⎡ 2.8⎤ Ax 2 = ⎢ = ⎣ −1 2 ⎥⎦ ⎢⎣ −0.8⎥⎦ ⎢⎣ −2.6⎥⎦ Ax 2 ⎡ 1 ⎤ x3 = ≈⎢ max( Ax2 ) ⎣ −0.929 ⎥⎦

203



⎡ 0.294⎤ x7 ≈ ⎢ 0.002⎥ , λ(7) ≈ 10.908 ⎢ ⎥ ⎣ 0.956⎦


⎡ 0.290 ⎤ x8 ≈ ⎢ −0.006 ⎥ , λ(8) ≈ 10.909 ⎢ ⎥ ⎣ 0.957 ⎦

⎡1 ⎤ 1. The row sums of A are h0 = ⎢ 2 ⎥ . The row sums ⎢ ⎥ ⎣2⎦ ⎡2⎤ of AT are a0 = ⎢ 0 ⎥ . ⎢ ⎥ ⎣3⎦

⎡1 ⎤ ⎢ 0⎥ (b) Starting with x0 = ⎢ ⎥ , it takes 8 iterations. ⎢ 0⎥ ⎢⎣ 0⎥⎦

⎡ 0 0 1⎤ ⎡ 2 ⎤ ⎡3⎤ 3. Aa0 = ⎢ 1 0 1⎥ ⎢ 0 ⎥ = ⎢5⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 1 0 1⎦ ⎣ 3 ⎦ ⎣5⎦ ⎡0.39057 ⎤ Aa 0 ≈ ⎢ 0.65094 ⎥ h1 = ⎢ ⎥ Aa 0 ⎣ 0.65094 ⎦ ⎡1.30189 ⎤ AT h1 ≈ ⎢ 0 ⎥ ⎢ ⎥ ⎣1.69246 ⎦

⎡0.577 ⎤ ⎢ 0 ⎥ (1) x1 ≈ ⎢ ⎥ , λ ≈ 6.333 ⎢0.577 ⎥ ⎢⎣0.577 ⎥⎦ ⎡0.249 ⎤ ⎢ 0 ⎥ ( 2) x2 ≈ ⎢ ⎥ , λ ≈ 8.062 ⎢ 0.498⎥ ⎣⎢0.830 ⎦⎥ ⎡ 0.193⎤ ⎢ 0.041⎥ (3) x3 = ⎢ ⎥ , λ ≈ 8.382 ⎢0.376 ⎥ ⎣⎢ 0.905⎦⎥

a1 =

⎡ 0.60971⎤ ≈⎢ 0 ⎥ AT h1 ⎢⎣ 0.79262 ⎥⎦

AT h1

⎡2 ⎢1 5. AT A = ⎢ ⎢0 ⎣⎢ 0

⎡ 0.175⎤ ⎢ 0.073⎥ ( 4) x4 ≈ ⎢ ⎥ , λ ≈ 8.476 ⎢ 0.305⎥ ⎢⎣ 0.933⎥⎦

1 2 0 0

0 0 1 0

0⎤ 0⎥ ⎥ 0⎥ 0 ⎦⎥

⎡0.167 ⎤ ⎢ 0.091⎥ (5) x5 ≈ ⎢ ⎥ , λ ≈ 8.503 ⎢ 0.266 ⎥ ⎢⎣ 0.945⎥⎦

⎡2⎤ ⎢2⎥ The column sums of A are ⎢ ⎥ , so we start with ⎢1 ⎥ ⎢⎣ 0 ⎥⎦

⎡ 0.162 ⎤ ⎢ 0.101⎥ (6) x6 ≈ ⎢ ⎥ , λ ≈ 8.511 ⎢ 0.245⎥ ⎣⎢ 0.951⎦⎥

⎡2⎤ ⎢3⎥ ⎢2⎥ a0 = ⎢ 3 ⎥ . 1 ⎢3⎥ ⎢0⎥ ⎣ ⎦

⎡ 0.159 ⎤ ⎢ 0.107 ⎥ ( 7) x7 ≈ ⎢ ⎥ , λ ≈ 8.513 ⎢ 0.234 ⎥ ⎣⎢ 0.953⎦⎥

⎡ 0.70225⎤ ⎢ 0.70225⎥ a1 = ≈⎢ ⎥ AT Aa0 ⎢0.11704 ⎥ ⎢⎣ 0 ⎥⎦

AT Aa0

⎡ 0.158⎤ ⎢0.110 ⎥ (8) x8 ≈ ⎢ ⎥ , λ ≈ 8.513 ⎢ 0.228⎥ ⎢⎣0.954 ⎥⎦

⎡0.70656 ⎤ ⎢0.70656 ⎥ ≈⎢ a2 = ⎥ AT Aa1 ⎢ 0.03925⎥ ⎢⎣ 0 ⎥⎦

AT Aa1

204


Section 9.4

⎡0.04370 ⎤ ⎢ 0.73455⎥ A Aa 2 ⎢ ⎥ ≈ ⎢0.58889 ⎥ a3 = T A Aa 2 ⎢0.32670 ⎥ ⎢⎣ 0.07075⎥⎦ ⎡ 0.02273⎤ ⎢ 0.73630⎥ T A Aa3 ⎢ ⎥ a4 = ≈ ⎢ 0.59045⎥ T A Aa3 ⎢ 0.32766⎥ ⎢⎣0.03677 ⎥⎦ ⎡0.01179 ⎤ ⎢0.73679 ⎥ AT Aa 4 ⎢ ⎥ ≈ ⎢0.59086 ⎥ a5 = T A Aa 4 ⎢0.32790 ⎥ ⎢⎣ 0.01908⎥⎦ ⎡ 0.00612⎤ ⎢ 0.73693⎥ T A Aa5 ⎢ ⎥ a6 = ≈ ⎢0.59097 ⎥ T A Aa5 ⎢ 0.32796⎥ ⎢⎣ 0.00990⎥⎦ From a6 , it is apparent that site 2 has the most authority, followed by sites 3 and 4, while sites 1 and 5 are irrelevant.

⎡ 0.70705⎤ ⎢ 0.70705⎥ ≈⎢ a3 = ⎥ AT Aa3 ⎢ 0.01309 ⎥ ⎢⎣ 0 ⎥⎦

AT Aa 2

T

⎡0.70710 ⎤ ⎢0.70710 ⎥ ≈⎢ a4 = ⎥ AT Aa3 ⎢0.00436 ⎥ ⎢⎣ 0 ⎥⎦

AT Aa3

⎡ 0.70711⎤ ⎢ 0.70711⎥ ≈⎢ a5 = ⎥ T A Aa 4 ⎢0.00145⎥ ⎣⎢ 0 ⎦⎥

AT Aa 4

⎡ 0.70711⎤ ⎢ 0.70711⎥ =⎢ a6 = ⎥ T A Aa5 ⎢0.00048⎥ ⎣⎢ 0 ⎦⎥

AT Aa5

From a6 , it is apparent that sites 1 and 2 are tied for the most authority, while sites 3 and 4 are irrelevant.

⎡1 ⎢0 ⎢ T 7. A A = ⎢ 0 ⎢0 ⎢⎣ 1

0 3 2 1 0

0 2 2 1 0

0 1 1 1 0

1⎤ 0⎥ ⎥ 0⎥ 0⎥ 2⎥⎦


⎡ 1⎤ ⎢ 3⎥ ⎢ ⎥ The column sums of A are ⎢ 2 ⎥ , so we start with ⎢ 1⎥ ⎢⎣ 2 ⎥⎦ ⎡ 1⎤ ⎡ 0.22942 ⎤ ⎢ 3⎥ ⎢ 0.68825⎥ 1 ⎢ ⎥ ⎢ ⎥ a0 = 2 ⎥ ≈ ⎢ 0.45883⎥ . ⎢ 19 ⎢ 1⎥ ⎢ 0.22942 ⎥ ⎢⎣ 2 ⎥⎦ ⎢⎣ 0.45883⎥⎦ ⎡0.15250 ⎤ ⎢0.71166 ⎥ T A Aa 0 ⎢ ⎥ a1 = ≈ ⎢0.55916 ⎥ T A Aa 0 ⎢0.30500 ⎥ ⎢⎣0.25416 ⎥⎦ ⎡0.08327 ⎤ ⎢ 0.72861⎥ T A Aa1 ⎢ ⎥ ≈ ⎢ 0.58289⎥ a2 = T A Aa1 ⎢0.32267 ⎥ ⎢⎣ 0.13531⎥⎦

1. (a) For n = 1000 = 103 , the flops for both phases is 2 3 3 3 3 2 7 3 (10 ) + (10 ) − (10 ) = 668,165, 500, 3 2 6 which is 0.6681655 gigaflops, so it will take 0.6681655 × 10−1 ≈ 0.067 second.

(b) n = 10, 000 = 104 : 2 4 3 3 4 2 7 4 (10 ) + (10 ) − (10 ) 3 2 6 = 666, 816, 655, 000 flops or 666.816655 gigaflops The time is about 66.68 seconds. (c) n = 100, 000 = 105 2 5 3 3 5 2 7 5 (10 ) + (10 ) − (10 ) 3 2 6 ≈ 666, 682 × 109 flops or 666,682 gigaflops The time is about 66,668 seconds which is about 18.5 hours. 205


SSM: Elementary Linear Algebra in AB requires 2n − 1 flops, for a total of

3. n = 10, 000 = 104 (a)

n 2 (2n − 1) = 2n3 − n2 flops. Note that adding two numbers requires 1 flop, adding three numbers requires 2 flops, and in general, n − 1 flops are required to add n numbers.

2 3 2 12 n ≈ (10 ) ≈ 666.67 × 109 3 3 666.67 gigaflops are required, which will 666.67 ≈ 9.52 seconds. take 70

Section 9.5

(b) n 2 ≈ 108 = 0.1× 109 0.1 gigaflop is required, which will take about 0.0014 second.

Exercise Set 9.5 1. The characteristic polynomial of ⎡1 ⎤ ⎡1 2 0⎤ AT A = ⎢ 2 ⎥ [1 2 0] = ⎢ 2 4 0⎥ is λ 2 (λ − 5); ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣ 0 0 0⎦

(c) This is the same as part (a); about 9.52 seconds.

thus the eigenvalues of AT A are λ1 = 5 and

(d) 2n3 ≈ 2 ×1012 = 2000 × 109 2000 gigaflops are required, which will take about 28.57 seconds.

λ 2 = 0, and σ1 = 5 and σ 2 = 0 are singular values of A.

5. (a) n = 100, 000 = 105 2 3 2 n ≈ × 1015 3 3 ≈ 0.667 × 1015

3. The eigenvalues of ⎡ 1 2 ⎤ ⎡ 1 −2 ⎤ ⎡ 5 0 ⎤ AT A = ⎢ = are λ1 = 5 1⎥⎦ ⎢⎣0 5⎥⎦ ⎣ −2 1⎥⎦ ⎢⎣ 2 and λ 2 = 5 (i.e., λ = 5 is an eigenvalue of multiplicity 2); thus the singular value of A is σ1 = 5.

= 6.67 × 105 ×109 Thus, the forward phase would require about 6.67 ×105 seconds.

5. The only eigenvalue of ⎡ 1 1⎤ ⎡1 −1⎤ ⎡ 2 0⎤ AT A = ⎢ = is λ = 2 ⎣ −1 1⎥⎦ ⎢⎣1 1⎥⎦ ⎢⎣ 0 2⎥⎦ ⎡1 ⎤ (multiplicity 2), and the vectors v1 = ⎢ ⎥ and ⎣0 ⎦ ⎡0 ⎤ v 2 = ⎢ ⎥ form an orthonormal basis for the ⎣1 ⎦

n 2 = 1010 = 10 × 109 The backward phase would require about 10 seconds. (b) n = 10, 000 = 104 2 3 2 n ≈ × 1012 3 3 ≈ 0.667 ×1012

eigenspace (which is all of R 2 ). The singular

≈ 6.67 × 102 × 109 About 667 gigaflops are required, so the computer would have to execute 2(667) = 1334 gigaflops per second.

values of A are σ1 = 2 and σ 2 = 2. We have ⎡ 1 ⎡1 −1⎤ ⎡1 ⎤ ⎢ Av1 = u1 = = σ1 2 ⎢⎣1 1⎥⎦ ⎢⎣0 ⎥⎦ ⎢ ⎣

1

7. Multiplying each of the n 2 entries of A by c

1 ⎤ 2⎥ , 1 ⎥ 2⎦

and

⎡ 1 ⎤ 1 ⎡1 −1⎤ ⎡ 0⎤ ⎢ − 2 ⎥ Av = . This u2 = = σ2 2 2 ⎢⎣1 1⎥⎦ ⎢⎣1 ⎥⎦ ⎢ 1 ⎥ 2 ⎣ ⎦ results in the following singular value decomposition of A: ⎡ 1 − 1 ⎤⎡ 2 0 ⎤ ⎡1 0 ⎤ T 2⎥ ⎢ A = U ΣV = 12 ⎢ ⎥ 1 ⎢ ⎥⎣ 0 2 ⎦ ⎢⎣0 1 ⎥⎦ 2⎦ ⎣ 2

requires n 2 flops.

1

9. Let C = [cij ] = AB. Computing cij requires first multiplying each of the n entries aik by the

corresponding entry bkj , which requires n flops. Then the n terms aik bkj must be summed, which requires n − 1 flops. Thus, each of the n 2 entries 206


Section 9.5

2⎤ ⎡ 6 ⎥ ⎢ u3 = ⎢ − 2 2 ⎥ . This results in the following ⎢ 3 ⎥ ⎢ ⎥ − 62 ⎥ ⎣⎢ ⎦ singular value decomposition:

7. The eigenvalues of ⎡ 4 0⎤ ⎡ 4 6⎤ ⎡ 16 24 ⎤ AT A = ⎢ = are λ1 = 64 ⎣ 6 4⎥⎦ ⎢⎣ 0 4⎥⎦ ⎢⎣ 24 52 ⎥⎦ and λ 2 = 4, with corresponding unit

⎡1 ⎤ ⎡− 2 ⎤ 5⎥ ⎢ eigenvectors v1 = 2 and v 2 = ⎢ 15 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 5⎦ ⎣ 5⎦ respectively. The singular values of A are σ1 = 8

A = U ΣV T 2⎤ ⎡ 2 1 6 ⎥⎡ ⎢ 3 1 ⎤ 2 3 2 0⎤ ⎡− 1 ⎢ 2 2⎥ 2 2⎥⎢ ⎥ ⎢ . =⎢ 1 0 − 3 ⎥ 0 0⎥ 1 ⎥ 3 ⎢ ⎢ 1 ⎢ 2 1 ⎥⎢ 0 0 ⎦⎥ ⎣ 2 2⎦ − 62 ⎥ ⎣ ⎢− 3 2 ⎣ ⎦ Note: The singular value decomposition is not unique. It depends on the choice of the

and σ 2 = 2. We have ⎡ 1 ⎡4 6⎤ ⎢ Av1 = ⎢ u1 = 8 ⎣ 0 4 ⎦⎥ ⎢ σ1 ⎣

⎡ 2 ⎤ = ⎢ 15 ⎥ , and ⎢ ⎥ ⎣ 5⎦ 2 ⎡ ⎤ ⎡ 1 ⎤ 1 1 ⎡4 6⎤ ⎢ − 5 ⎥ ⎢ − 5 ⎥ Av = . u2 = = σ 2 2 2 ⎢⎣ 0 4 ⎥⎦ ⎢ 1 ⎥ ⎢ 2 ⎥ ⎣ 5⎦ ⎣ 5⎦ This results in the following singular value decomposition: 2 ⎤ ⎡ 2 − 1 ⎤ ⎡ 1 T 5 5 ⎥ ⎡8 0 ⎤ ⎢ 5 15 ⎥ ⎢ A = U ΣV = 1 2 ⎥ ⎢⎣ 0 2 ⎥⎦ ⎢ − 2 1 ⎥ ⎢ 5⎦ 5⎦ ⎣ 5 ⎣ 5

1

1 ⎤ 5⎥ 2 ⎥ 5⎦

(extended) orthonormal basis for R3 . This is just one possibility. 11. The eigenvalues of

⎡ 1 0⎤ ⎡ 1 1 −1⎤ ⎢ ⎡3 0 ⎤ AT A = ⎢ are λ1 = 3 1 1⎥ = ⎢ ⎥ 0 1 1 ⎢ ⎥ ⎣ ⎦ −1 1 ⎣0 2 ⎥⎦ ⎣ ⎦ and λ 2 = 2, with corresponding unit ⎡1 ⎤ ⎡0 ⎤ eigenvectors v1 = ⎢ ⎥ and v 2 = ⎢ ⎥ ⎣0 ⎦ ⎣1 ⎦ respectively. The singular values of A are σ1 = 3 and σ 2 = 2. We have

9. The eigenvalues of

⎡ −2 2 ⎤ ⎡ −2 −1 2 ⎤ ⎢ ⎡ 9 −9 ⎤ AT A = ⎢ are −1 1⎥ = ⎣ 2 1 −2 ⎦⎥ ⎢ 2 −2 ⎥ ⎣⎢ −9 9⎦⎥ ⎣ ⎦ λ1 = 18 and λ 2 = 0, with corresponding unit

⎡ ⎢ ⎡ 1 0⎤ 1 1 ⎢ ⎡1 ⎤ ⎢ ⎥ = Av = u1 = 1 1 σ1 1 3 ⎢ −1 1⎥ ⎢⎣ 0 ⎥⎦ ⎢ ⎣ ⎦ ⎢− ⎣⎢

⎡− 1 ⎤ ⎡ 1 ⎤ eigenvectors v1 = ⎢ 12 ⎥ and v 2 = ⎢ 12 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 2⎦ ⎣ 2⎦ respectively. The only nonzero singular value of A is σ1 = 18 = 3 2, and we have ⎡ −2 2 ⎤ ⎡ − 1 ⎢ Av1 = u1 = −1 1⎥ ⎢ σ1 3 2 ⎢ 2 −2 ⎥ ⎢ ⎣ ⎦⎣

1

1 ⎤ 2⎥ 1 ⎥ 2⎦

and

⎡0⎤ ⎡ 1 0⎤ ⎢ 1 ⎥ 1 ⎢ 0 ⎡ ⎤ u2 = 1 1⎥ ⎢ ⎥ = ⎢ 2 ⎥ . We choose 2 ⎢ −1 1⎥ ⎣1 ⎦ ⎢ 1 ⎥ ⎣ ⎦ ⎢⎣ 2 ⎥⎦ ⎡ 2 ⎤ ⎢ 6⎥ u3 = ⎢ − 1 ⎥ so that {u1, u 2 , u3} is an ⎢ 6⎥ ⎢ 1 ⎥ ⎢⎣ 6 ⎥⎦

⎡ 2⎤ ⎢ 3⎥ = ⎢ 13 ⎥ . ⎢ 2⎥ ⎣⎢ − 3 ⎦⎥

We must choose the vectors u 2 and u3 so that {u1, u 2 , u3} is an orthonormal basis R3 . ⎡ ⎢ A possible choice is u 2 = ⎢ ⎢ ⎢⎣

1 ⎤ 3⎥ 1 ⎥ 3⎥ 1 ⎥ 3 ⎦⎥

1 ⎤ 2⎥

orthonormal basis for R3 . This results in the following singular value decomposition:

0 ⎥ and 1 ⎥ 2 ⎥⎦

A = U ΣV T ⎡ 1 ⎢ 3 =⎢ 1 ⎢ 3 ⎢− 1 ⎢⎣ 3 207

0 1 2 1 2

−

2 ⎤ 6⎥⎡ 1 ⎥⎢ 6⎥⎢ 1 ⎥ ⎢⎣ 6 ⎥⎦

3 0 0

0 ⎤ ⎥ ⎡1 0 ⎤ 2 ⎥ ⎢0 1 ⎥ ⎣ ⎦ 0 ⎥⎦



5. The reduced singular value expansion of A is ⎡ 2⎤ ⎢ 3⎥ 1 ⎤ 3 2 ⎢ 13 ⎥ ⎡ − 1 . 2 2⎦ ⎣ ⎢ 2⎥ ⎢⎣ − 3 ⎥⎦

True/False 9.5 (a) False; if A is an m × n matrix, then AT is an

n × m matrix, and AT A is an n × n matrix. (b) True

7. The reduced singular value decomposition of A ⎡ 1 ⎤ ⎡0⎤ ⎢ 3⎥ ⎢ 1 ⎥ is 3 ⎢ 1 ⎥ [1 0] + 2 ⎢ 2 ⎥ [0 1]. 3 ⎢ ⎥ ⎢ 1 ⎥ ⎢− 1 ⎥ ⎢⎣ 2 ⎥⎦ ⎣⎢ 3 ⎦⎥

(c) False; AT A may have eigenvalues that are 0. (d) False; A would have to be symmetric to be orthogonally diagonalizable. (e) True; since AT A is a symmetric n × n matrix.

9. A rank 100 approximation of A requires storage space for 100(200 + 500 + 1) = 70,100 numbers, while A has 200(500) = 100,000 has entries.

(f) False; the eigenvalues of AT A are the squares of the singular values of A. (g) True

True/False 9.6

Section 9.6

(a) True

Exercise Set 9.6

(b) True

1. From Exercise 9 in Section 9.5, A has the singular value decomposition 2 ⎤ ⎡ 2 1 6 ⎥⎡ 1 ⎤ ⎢ 3 2 3 2 0⎤ ⎡− 1 ⎢ 2 2⎥ ⎢ 2 2⎥⎢ ⎥ A=⎢ 1 0 − 3 ⎥ 0 0⎥ ⎢ 1 1 ⎥ 3 ⎢ ⎢ 2 1 ⎥⎢ 0 0 ⎦⎥ ⎣⎢ 2 2 ⎦⎥ − 62 ⎥ ⎣ ⎢− 3 2 ⎣ ⎦ where the lines show the relevant blocks. Thus the reduced singular value decomposition of A is ⎡ 2⎤ ⎢ 3⎥ 1 ⎤ A = U1Σ1V1T = ⎢ 13 ⎥ [3 2 ] ⎡ − 1 . 2 2⎦ ⎣ ⎢ 2⎥ ⎢⎣ − 3 ⎥⎦

(c) False; V1 has size n × k so that V1T has size k × n. Chapter 9 Supplementary Exercises 1. Reduce A to upper triangular form. ⎡ −6 2 ⎤ ⎡ −3 1⎤ ⎡ −3 1⎤ ⎢⎣ 6 0 ⎥⎦ → ⎢⎣ 6 0⎥⎦ → ⎢⎣ 0 2 ⎥⎦ = U 1 The multipliers used were and 2, so 2 ⎡ 2 0⎤ ⎡ 2 0 ⎤ ⎡ −3 1⎤ and A = ⎢ . L=⎢ ⎣ −2 1⎥⎦ ⎣ −2 1⎥⎦ ⎢⎣ 0 2 ⎥⎦ 3. Reduce A to upper triangular form. ⎡2 4 6⎤ ⎡1 2 3⎤ ⎡ 1 2 3⎤ ⎢ 1 4 7 ⎥ → ⎢1 4 7 ⎥ → ⎢0 2 4⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1 3 7⎦ ⎣1 3 7 ⎦ ⎣ 1 3 7⎦ ⎡ 1 2 3⎤ ⎡ 1 2 3⎤ ⎡ 1 2 → ⎢ 0 2 4 ⎥ → ⎢0 1 2 ⎥ → ⎢ 0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎣0 1 4⎦ ⎣0 1 4 ⎦ ⎣ 0 0

3. From Exercise 11 in Section 9.5, A has the singular value decomposition 2 ⎤ ⎡ 1 0 6⎥⎡ 3 ⎢ 3 0 ⎤ ⎥ ⎡1 0 ⎤ 1 1 1 ⎥⎢ ⎢ − A= 2 ⎥ ⎢0 1 ⎥ 2 6⎥⎢ 0 ⎢ 3 ⎣ ⎦ 0 ⎥⎦ 1 1 ⎥ ⎢⎣ 0 ⎢− 1 2 6 ⎦⎥ ⎣⎢ 3 where the lines indicate the relevant blocks. Thus the reduced singular value decomposition of A is ⎡ 1 0⎤ ⎢ 3 ⎥ 0 ⎤ ⎡1 0 ⎤ 1 1 ⎥⎡ 3 T ⎢ . A = U1Σ1V1 = ⎥ 2⎥⎢ 0 ⎢ 3 2 ⎦ ⎢⎣ 0 1 ⎥⎦ ⎣ 1 1 ⎢− ⎥ 2 ⎦⎥ ⎣⎢ 3

3⎤ 2⎥ ⎥ 2⎦

⎡ 1 2 3⎤ → ⎢0 1 2⎥ = U ⎢ ⎥ ⎣ 0 0 1⎦ The multipliers used were

208

1 1 , −1, −1, , −1, 2 2


⎡2 1 so L = ⎢ 1 ⎢ 2 ⎣1 ⎡2 0 0⎤ ⎡ 1 A = ⎢ 1 2 0 ⎥ ⎢0 ⎢ ⎥⎢ ⎣ 1 1 2 ⎦ ⎣0

and


0 0⎤ 2 0 ⎥ and ⎥ 1 2⎦ 2 3⎤ 1 2⎥ . ⎥ 0 1⎦

x5 =

⎡ 1 ⎤ x5 ≈ ⎢ as compared to the exact ⎣0.9918⎥⎦ ⎡1⎤ eigenvector v = ⎢ ⎥ . ⎣1⎦

5. (a) The characteristic equation of A is

7. The Rayleigh quotients will converge to the dominant eigenvalue λ 4 = −8.1. However, since

λ 2 − 4λ + 3 = (λ − 3)(λ − 1) = 0 so the dominant eigenvalue of A is λ1 = 3, with corresponding positive unit eigenvector ⎡ 1 ⎤ ⎡ 0.7071⎤ . v = ⎢ 12 ⎥ ≈ ⎢ ⎢ ⎥ ⎣ 0.7071⎥⎦ ⎣ 2⎦

the ratio

Ax1 ⎡ 0.7809 ⎤ ≈ Ax1 ⎢⎣0.6247 ⎥⎦

x3 =

Ax 2 ⎡0.7328⎤ ≈⎢ Ax 2 ⎣0.6805⎥⎦

λ1

=

8.1 = 1.0125 is very close to 1, 8

⎡2 2⎤ are λ1 = 4 9. The eigenvalues of AT A = ⎢ ⎣ 2 2 ⎥⎦ and λ 2 = 0 with corresponding unit ⎡− 1 ⎤ ⎡− 1 ⎤ 2⎥ ⎢ eigenvectors v1 = and v 2 = ⎢ 12 ⎥ , ⎢ ⎥ ⎢− 1 ⎥ ⎣ 2⎦ ⎣ 2⎦ respectively. The only nonzero singular value A is σ1 = 4 = 2, and we have ⎡− ⎡ 1 1⎤ ⎡ − 1 ⎤ ⎢ 1⎢ u1 = Av = 0 0 ⎥ ⎢ 2 ⎥ = ⎢ σ1 1 2 ⎢ 1 1⎥ ⎢ − 1 ⎥ ⎢ ⎣ ⎦ ⎣ 2 ⎦ ⎢− ⎣ We must choose the vectors u 2 and 1

Ax 3 ⎡0.7158⎤ x4 = ≈⎢ Ax 3 ⎣ 0.6983⎥⎦ x5 =

λ4

the rate of convergence is likely to be quite slow.

⎡ 2 1 ⎤ ⎡1 ⎤ ⎡ 2 ⎤ = (b) Ax0 = ⎢ ⎣ 1 2 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣1 ⎥⎦ Ax 0 1 ⎡ 2⎤ ⎡ 0.8944 ⎤ = ≈ x1 = Ax 0 5 ⎢⎣1 ⎥⎦ ⎢⎣ 0.4472 ⎥⎦ x2 =

Ax 4 ⎡ 1 ⎤ ≈ max( Ax4 ) ⎢⎣ 0.9918⎥⎦

Ax 4 ⎡0.7100 ⎤ ≈⎢ Ax 4 ⎣0.7042 ⎥⎦

1 ⎤ 2⎥

0⎥ .

1 ⎥ 2 ⎦⎥

u3 so that

{u1, u 2 , u3} is an orthonormal basis for R3 . A ⎡ 1 ⎤ ⎡0 ⎤ ⎢ 2⎥ possible choice is u 2 = ⎢1 ⎥ and u3 = ⎢ 0 ⎥ . ⎢ ⎥ ⎢ 1 ⎥ − ⎣0 ⎦ ⎣⎢ 2 ⎦⎥ This results in the following singular value decomposition:

⎡0.7100⎤ x5 ≈ ⎢ as compared to ⎣0.7042⎥⎦ ⎡0.7071⎤ . v≈⎢ ⎣0.7071⎥⎦ ⎡2⎤ (c) Ax0 = ⎢ ⎥ ⎣1 ⎦ Ax 0 ⎡1 ⎤ = x1 = max( Ax0 ) ⎢⎣ 0.5⎥⎦ Ax1 ⎡1 ⎤ =⎢ ⎥ x2 = max( Ax1 ) ⎣ 0.8⎦

A = U ΣV T 1 ⎤ ⎡− 1 0 2 ⎥ ⎡ 2 0⎤ ⎡ − ⎢ 2 =⎢ 0 1 0⎥ ⎢ 0 0 ⎥ ⎢ ⎢ ⎢ 1 1 ⎥ ⎢ 0 0 ⎦⎥ − ⎣ ⎢⎣ − 2 0 − 2 ⎥⎦ ⎣

Ax 2 ⎡ 1 ⎤ ≈ max( Ax2 ) ⎢⎣ 0.9286 ⎥⎦ Ax 3 ⎡ 1 ⎤ ≈⎢ x4 = max( Ax3 ) ⎣ 0.9756 ⎥⎦ x3 =

209

1 2 1 2

−

1 ⎤ 2⎥ 1 ⎥ 2⎦



11. A has rank 2, thus U1 = [u1 u 2 ] and ⎡ vT ⎤ V1T = ⎢ 1T ⎥ and the reduced singular value ⎣⎢ v 2 ⎦⎥ decomposition of A is

A = U1Σ1V1T 1⎤ ⎡1 2⎥ ⎢2 ⎢ 12 − 12 ⎥ ⎡ 24 0 ⎤ ⎡ 23 = ⎢1 ⎥ ⎢2 1⎥ ⎢ ⎢ 2 − 2 ⎥ ⎣ 0 12 ⎦ ⎢⎣ 3 ⎢1 1⎥ ⎣⎢ 2 2 ⎦⎥

− 13 2 3

2⎤ 3⎥ − 13 ⎥ ⎦

210

Anton - Elementary Linear Algebra With Applications 10e Student Solutions Manual

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