MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
2 #1. Page 169; Exercise 2. Let A = −1 3
1 3 2 0 . −2 1
(a) Find adj A. (b) Compute det(A). (c) Verify Theorem 3.12; that is, show that A (adj A) = (adj A) A = det(A) I3 .
Solution: (a) First we compute the % % % 2 0 % 1+1 % % A11 = (−1) % −2 1 % = 2 % % % 1 3 % 2+1 % % A21 = (−1) % −2 1 % = −7 % % % % 3+1 % 1 3 % A31 = (−1) % 2 0 % = −6
Then the adjoint of A is the matrix
cofactors of A. We have % % % % 1+2 % −1 0 % A12 = (−1) % 3 1 %=1 % % % % 2+2 % 2 3 % A22 = (−1) % 3 1 % = −7 % % % 2 3 % 3+2 % % A32 = (−1) % −1 0 % = −3
A11 adj A = A12 A13
A21 A22 A23
A31 2 A32 = 1 A33 −4
A13 A23 A33
% % % −1 2 %% % = (−1) % 3 −2 % = −4 % % % 1 %% 2+3 % 2 = (−1) % 3 −2 % = 7 % % % 2 1 % 3+3 % % = (−1) % −1 2 % = 5 1+3
−7 −6 −7 −3 7 5
(b) We can compute the determinant by expanding along the first row of A:
det(A) = a11 A11 + a12 A12 + a13 A13 = (2) (2) + (1) (1) + (3) (−4) = −7 (c) We have the matrix products 2 1 3 2 2 0 1 A (adj A) = −1 3 −2 1 −4
2 (adj A) A = 1 −4
−7 −6 2 −7 −3 −1 7 5 3
−7 −6 −7 −7 −3 = 0 7 5 0
1 3 −7 2 0 = 0 −2 1 0
0 0 −7 0 = det(A) I3 . 0 −7 0 0 −7 0 = det(A) I3 . 0 −7
#2. Page 169; Exercise 4. Find the inverse of the matrix in Exercise 2 by the method given in Corollary 3.4. 1
2
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
Solution: The inverse of the matrix is A−1 =
− 27
1
4 7
−1
1 (adj A) = − 17 det(A)
1
6 7 3 7 − 57
#3. Page 172; Exercise 1. If possible, solve the following linear systems by Cramer’s rule: 2 x1 x1 2 x1
Solution: Denote 2 A= 1 2
+
4 x2
+
3 x2
+ + −
6 x3 2 x3 x3
the coefficient matrix of the system by % 4 6 % 0 2 0 2 =⇒ |A| = (2) %% 3 −1 3 −1
Then we have the values % % 2 4 6 % % 0 0 2 % % −5 3 −1 x1 = |A| % % 2 2 6 % % 1 0 2 % % 2 −5 −1 x2 = |A| % % % 2 4 2 %% % % 1 0 0 %% % % 2 3 −5 % x3 = |A|
% % % % % % % % % % % %
=
=
=
% % 0 (2) %% 3
% % 0 (2) %% −5
% % 0 (2) %% 3
= 2 = 0 = −5
% % % % % − (4) % 1 % % 2
% % % 0 2 %% 2 − (4) %% % −1 −5 −1 26 % % % 1 2 %% 2 − (2) %% 2 −1 −1 % 26
% % % 1 0 %% 0 − (4) %% % −5 2 −5 26
Hence the solution is x1 = −2, x2 = 0, and x3 = 1.
% % % 1 2 %% % + (6) % % 2 −1
% % % % % + (6) % 0 % −5 % % % % % % + (6) % 1 % % 2
% % % % % + (2) % 1 % % 2
% 0 %% 3 %
% 0 %% 3 %
% 0 %% −5 % =
% 0 %% = 26. 3 %
=
−52 = −2 26
=
0 =0 26
26 =1 26
#4. Page 172; Exercise 3. Solve the following linear system for x3 , by Cramer’s rule: 2 x1 3 x1 x1
+ + +
x2 2 x2 x2
+ x3 − 2 x3 + 2 x3
= 6 = −2 = −4
Solution: Denote the coefficient matrix of the system by % % % 2 1 1 % 2 −2 % % % − (1) % 3 A = 3 2 −2 =⇒ |A| = (2) %% % % 1 1 2 1 1 2
% % % 3 −2 %% % + (1) % 1 % 2
% 2 %% = 5. 1 %
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
Then we have the value % % 2 % % 3 % % 1 x3 =
1 6 2 −2 1 −4 |A|
% % % % % %
=
% % 2 (2) %% 1
3
% % % % % 3 −2 % % 3 −2 %% % % − (1) % + (6) %% −4 % 1 −4 % 1 5
#5. Page 187; Exercise 2. Determine the head of the vector
(
−2 5
)
% 2 %% 1 %
=
4 5
whose tail is (−3, 2). Make a sketch.
Solution: Say that the tail is P (−3, 2) and the head is Q(x, y). Then we have the directed line segment ( ) ( ) −−→ x − (−3) −2 PQ = = . y−2 5 Since x + 3 = −2 and y − 2 = 5, we see that x = −5 and y = 7. Hence the head is (−5, 7). A sketch can be found below. y# •!! (−5,7) !! !! !! !! !! !! !! !! !! !! !! ! •
7
− 5
3
|
|
−3
− −
1
|
− −
(−3,2)
−5
−
|
−
|
|
−1
|
1
|
|
3
#6. Page 187; Exercise 5. For what values of a and b are the vectors
(
a−b 2
|
"x
and
(
5
)
4 a+b
)
equal?
Solution: Upon equating both components, we find the system of equations a − b a + b
= =
4 2
Adding these two equations together gives 2 a = 6, so that a = 3. Similarly, subtracting the two equations gives −2 b = 2, so that b = −1. Hence the values are a = 3 and b = −1.
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MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
−−→ #7. Page 187; Exercise 7. In Exercises 7 and 8, determine the components of each vector P Q. (a) P (1, 2), Q(3, 5) (b) P (−2, 2, 3), Q(−3, 5, 2) Solution: (a) We form the directed line segment by subtracting: * ( ) ( ) P (1, 2) −−→ 3−1 2 =⇒ PQ = = . 5−2 3 Q(3, 5) Hence the components are 2 and 3. (b) Again, we form the directed line segment by subtracting: * (−3) − (−2) −1 P (−2, 2, 3) −−→ = 3 . 5−2 =⇒ PQ = Q(−3, 5, 2) 2−3 −1 Hence the components are −1, 3, and −1.
#8. Page 187; Exercise 12. Compute u + v, 2 u − v, 3 u − 2 v, and 0 − 3 v if 1 2 (a) u = 2 , v = 0 ; 3 1 2 1 (b) u = −1 , v = 2 ; 4 −3 1 −1 (c) u = 0 , v = 1 . 4 −1 + ,T + ,T Solution: (a) We perform operations component-wise on u = 1 2 3 and v = 2 0 1 : 3 0 −1 −6 u + v = 2 , 2u − v = 4 , 3u − 2v = 6 , 0 − 3v = 0 4 5 7 −3 + ,T + ,T (b) For u = 2 −1 4 and v = 1 2 −3 : 3 3 4 −3 u + v = 1 , 2 u − v = −4 , 3 u − 2 v = −7 , 0 − 3 v = −6 1 11 18 9 + ,T + ,T (c) For u = 1 0 −1 and v = −1 1 4 : 0 3 5 3 u + v = 1 , 2 u − v = −1 , 3 u − 2 v = −2 , 0 − 3 v = −3 3 −6 −11 −12
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
#9. Page 187; Exercise 14. Let ( ) ( ) 1 −3 x= , y= , 2 4
z=
Find r and s so that
(
r 4
)
,
and
u=
5
(
−2 s
)
.
(a) z = 2 x, (b) 32 u = y, (c) z + u = x. Solution: (a) We have the equation ( ) ( ) 2 r = z = 2x = 4 4 (b) We have the equation ( ) −3 = 3 2 s
3 2
u=y=
(
(c) We have the equation ( ) ( ) r−2 1 =z+u=x= 4+s 2
−3 4
)
=⇒
=⇒
=⇒
3 2
s=4
r−2 = 1 s+4 = 2
r=2
=⇒
=⇒
s=
8 3
r = 3 s = −2
#10. Page 187; Exercise 16. If possible, find scalars c1 and c2 so that ( ) ( ) ( ) 1 3 −5 c1 + c2 = . −2 −4 6 Solution: The left-hand side of the equation simplifies to give the equation ( ) ( ) c1 + 3 c2 −5 c1 + 3 c2 = =⇒ −2 c1 − 4 c2 6 −2 c1 − 4 c2
= −5 = 6
Upon adding twice the first equation to the second equation, we see that 2 c2 = −4, so that c2 = −2. Substituting this back into the first equation, we see that c1 − 6 = −5, so that c1 = 1. Hence the desired scalars are c1 = 1 and c2 = −2.
#11. Page 187; Exercise 17. If possible, find scalars c1 , c2 , and c3 so that 1 −1 −1 2 c1 2 + c2 1 + c3 4 = −2 . −3 1 1 3
6
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
Solution: The left-hand side of the equation simplifies to give the equation c1 − c2 − c3 2 c1 − c2 2 c1 + c2 + 4 c3 = −2 2 c1 + c2 =⇒ −3 c1 + c2 − c3 3 −3 c1 + c2
− + −
= 2 = −2 = 3
c3 4 c3 c3
We list the augmented matrix for this system, and its reduced row echelon form: 1 0 1 0 2 1 −1 −1 0 1 2 0 2 1 4 −2 =⇒ −3 1 −1 3 0 0 0 1
Hence the system is inconsistent, so it is impossible; the scalars c1 , c2 , and c3 do not exist.
(
) a b #12. Page 196; Exercise 2. Let V be the set of all 2×2 matrices A = such that the product a b c d = c d 0. Let the operation ⊕ be standard addition of matrices and the operation & be standard multiplication of matrices. (a) (b) (c) (d) (e)
Is V closed under addition? Is V closed under scalar multiplication? What is the zero vector in the set V ? Does every matrix A in V have a negative that is in V ? Explain. Is V a vector space? Explain.
Solution: (a) No, it is not closed under addition. Consider the matrices ( ) ( ) ( 1 0 0 1 1 A= and B= =⇒ A⊕B = 1 0 0 1 1
1 1
)
.
Then A, B ∈ V yet A ⊕ B (∈ V . (b) Yes, it is closed under scalar multiplication. For any scalar r ∈ R, we have the matrix ( ) ( ) a b ra rb A= =⇒ rA = =⇒ (r a) (r b) (r c) (r d) = r4 (a b c d) = r4 · 0 = 0. c d rc rd ( ) 0 0 (c) The zero vector is O2 = . 0 0
(d) Yes, every matrix has a negative that is in V . For any A ∈ V , we can choose the negative −A = (−1) A. Since V is closed under scalar multiplication, this element is in V . (e) No, V is not a vector space. This is because V is not closed under ⊕.
#13. Page 196; Exercise 8. In Exercises 7 through 11, the given set together with the given operations is not a vector space. List the properties of Definition 4.4 that fail to hold: The set of all ordered pairs of real numbers with the operations (x, y) ⊕ (x" , y " ) = (x + x" , y + y " ) and
r & (x, y) = (x, r y) .
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
7
Solution: All properties hold except for Property (6). If we denote If we denote scalars c, d ∈ R and vectors u = (x, y), v = (x" , , y " ), and w = (x"" , y "" ), then we have (1) u ⊕ v = (x + x"-, y + y " ) = (x" + x, y " + y) = . v⊕ - u. . (2) u ⊕ (v ⊕ w) = x + (x" + x"" ), y + (y " + y "" ) = (x + x" ) + x"" , (y + y " ) + y "" = (u ⊕ v) ⊕ w. (3) 0 = (0, 0) satisfies u ⊕ 0 = 0 ⊕ u = u. (4) −u = (−x, −y) - satisfies u ⊕ −u.= −u ⊕ u = 0. (5) c & (u ⊕ v) = x + x" , c (y + y " ) = (x, c y) ⊕ (x" , c y " ) = c & u ⊕ c & v. (7) c & (d & u) = (x, c d y) = (c d) & u. (8) 1 & u = (x, 1 · y) = (x, y) = u. For (6), denote c = d = 1 and u = (1, 1). Then we have * / (c + d) & u = 2 & (1, 1) c & u ⊕ d & u = 1 & (1, 1) ⊕ 1 & (1, 1) = (1, 1) ⊕ (1, 1) yet = (1, 2) = (2, 2) .
#14. Page 197; Exercise 9. In Exercises 7 through 11, the given set together with the given operations is not a vector space. List the properties of Definition 4.4 that fail to hold: The set of all ordered triples of real numbers with the operations and
(x, y, z) ⊕ (x" , y " , z " ) = (x + x" , y + y " , z + z " ) r & (x, y, r) = (x, 1, z) .
Solution: All properties hold except for Properties (5), (6), and (8). If we denote scalars c, d ∈ R and vectors u = (x, y, z), v = (x" , y " , z " ), and w = (x"" , y "" , z "" ), then we have (1) u ⊕ v = (x + x-" , y + y " , z + z " ) = (x" + x, y " + y, z " .+ z)-= v ⊕ u. . (2) u⊕(v ⊕ w) = x+(x" +x"" ), y +(y " +y "" ), z +(z " +z "" ) = (x+x" )+x"" , (y +y " )+y "" , (z +z " )+z "" = (u ⊕ v) ⊕ w. (3) 0 = (0, 0, 0) satisfies u ⊕ 0 = 0 ⊕ u = u. (4) −u = (−x, −y, −z) satisfies u ⊕ −u = −u ⊕ u = 0. (7) c & (d & u) = c & (x, 1, z) = (x, 1, z) = (c d) & u. For (5), (6), and (8), we have the following counterexamples: (5) Let c = 1 and u = v = (1, 0, 0). Then c & (u ⊕ v) = (2, 1, 0), yet c & u ⊕ c & v = (2, 2, 0). (6) Let c = d = 1 and u = (1, 0, 0). Then (c + d) & (u ⊕ v) = (1, 1, 0), yet c & u ⊕ d & v = (2, 2, 0). (8) Let u = (1, 0, 0). Then 1 & u = (1, 1, 0) is different from u.
#15. Page 197; Exercise 10. In Exercises 7 through 11, the given set together with the given operations ( )is x not a vector space. List the properties of Definition 4.4 that fail to hold: The set of all 2 × 1 matrices , y 2 where x ≤ 0, with the usual operations in R . Solution: All properties hold except for Properties (4) and (b). The set V is closed under addition because ( ) ( " ) x x if we choose the vectors u = and v = for some x ≤ 0 and x" ≤ 0, then x + x" ≤ 0 so that y y" u ⊕ v ∈ V as well. Properties (1) and (2) hold because ⊕ is the usual operation for R2 . Property (3)
8
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
holds because the 2 × 1 zero(matrix ) has a nonpositive x-coordinate. For (4) and (b), denote the scalar −1 c = −1 and the vector u = . It is a vector in V because it has negative x-coordinate. However 0 ( ) 1 c & u = −u = is not in V because it has positive x-coordinate. Hence V does not contain negatives, 0 and it is not closed under scalar multiplication.
#16. Page 197; Exercise 12. Let V be the set of all positive real numbers; define ⊕ by u ⊕ v = u v (⊕ is ordinary multiplication) and define & by c & v = vc . Prove that V is a vector space. Solution: V is closed under ⊕ and & because both u v and vc are positive real numbers. In order to check that V is a vector space, we verify eight properties. For scalars c, d ∈ R and positive real numbers u, v, w ∈ V , we have (1) u ⊕ v = u v = v u = v ⊕ u. (2) u ⊕ (v ⊕ w) = u (v w) = (u v) w = (u ⊕ v) ⊕ w. (3) Denote 0 = 1. This is a positive real number, so 0 ∈ V . Moreover, u ⊕ 0 = 0 ⊕ u = u 1 = u. (4) Denote −u = (1/u). This is a positive real number, so −u ∈ V . Moreover, u ⊕ −u = −u ⊕ u = u (1/u) = 1 = 0. c (5) c & (u ⊕ v) = c & (u v) = (u v) = uc vc = c & u ⊕ c & v. c+d c d (6) (c + d) & u = u = u u = c & u ⊕ d & u. (7) c & (d & u) = c & ud = uc d = (c d) & u. (8) 1 & u = u1 = u. Hence, according to Definition 4.4, we see that V is a vector space.
#17. Page 205; Exercise 2. Let W be the set of all points in R3 that lie in the x y-plane. Is W a subspace of R3 ? Explain. Solution: Yes, W is a subspace of R3 . Let us write W in the form % x % W = x = y ∈ R3 %% z = 0 . z
We verify three properties according to Theorem 4.3. First, W is nonempty because the origin 0 ∈ W . If u, v ∈ W , then " x x x + x" u= y and v = y" =⇒ u + v = y + y " ∈ W. 0 0 0 Hence W is closed under addition. Similarly, if c is any real number and u ∈ W , then x cx u= y =⇒ c u = u = c y ∈ W. 0 0
Hence W is closed under scalar multiplication. This shows that W is indeed a subspace.
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
9
#18. Page 206; Exercise (4. Consider the unit square shown in the accompanying figure. Let W be the set ) x of all vectors of the form , where 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. That is, W is the set of all vectors whose tail y is at the origin and whose head is a point inside or on the square. Is W a subspace of R2 ? Explain. Solution: No, W is not a subspace of R2 . W is not closed under either addition or scalar multiplication, because if we choose ( ) ( ) 1 2 (∈ W. u=v= ∈W =⇒ 2u = u + v = 2 1
#19. Page 206; Exercise 6. In Exercises 5 and 6, which of the given subsets of R3 are subspaces? The set of vectors the form of a (a) b 0 a (b) b , where a > 0. c a (c) a c a (b) b , where 2 a − b + c = 1. c Solution: (a) This is a subspace. This follows from Exercise 2 (Problem #17 above). (b) This is not a subspace. We define a W = x = b ∈ R3 c
% % %a>0∈R . %
If W were a subspace, then it would be closed under scalar multiplication. In particular 0 u = 0 would be an element. However, a 0 b = 0 0 = a ≤ 0. c 0 This means 0 (∈ W . Hence W cannot be a subspace. (c) This is a subspace. We define % a % W = x = a ∈ R3 %% a, c ∈ R . c We verify three properties according to Theorem 4.3. First, W is can be seen by choosing a = c = 0. If u, v ∈ W , then a1 a2 u = a1 and v = a2 =⇒ u+v= c1 c2
nonempty because the origin 0 ∈ W ; this a a c
in terms of
6
a = a1 + a2 c = c1 + c2
10
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
Similarly, if r is any real number and u ∈ W , then a a1 =⇒ ru+v = a u = a1 c1 c (d) This is not a subspace. We define a W = x = b ∈ R3 c
in terms of
6
a = r a1 c = r c1
% % % 2a − b + c = 1 ∈ R . %
If W were a subspace, then it would be closed under scalar multiplication. In particular 0 u = 0 would be an element. However, a 0 b = 0 0 = a − 2 b + c (= 1. c 0 This means 0 (∈ W . Hence W cannot be a subspace.
#20. Page 206; Exercise 10. In Exercises 9 and 10, which of the given subsets of the vector space, M23 , or 2 × 3 matrices are subspaces? The set of all matrices of the form ( ) a b c (a) , where a = 2 c + 1. d e f ( ) 0 1 a (b) b c 0 ( ) a b c (c) , where a + c = 0 and b + d + f = 0 d e f Solution: (a) This is not a subspace. If W were a subspace, then it would be closed under scalar multiplication. In particular 0 u = 0 would be an element. However, ( ) ( ) a b c 0 0 0 = =⇒ 0 = a (= 2 c + 1 = 1. d e f 0 0 0 This means 0 (∈ W . Hence W cannot be a subspace. (b) This is not a subspace. The 2 × 3 zero matrix is not in the space because ( ) ( ) 0 0 0 0 1 a (= 0 0 0 b c 0 for any real numbers a, b, and c. (c) This is a subspace. We define 6( a b W = d e
c f
)
% 7 % ∈ M23 %% a + c = 0, b + d + f = 0 .
We verify three properties according to Theorem 4.3. First, W is nonempty because the origin 0 ∈ W ; this can be seen by choosing a = b = c = d = e = f = 0. If u, v ∈ W , then ( ) ( ) ( ) a1 b1 c1 a2 b2 c2 a1 + a2 b1 + b2 c1 + c2 u= and v = =⇒ u+v= . d1 e1 f1 d2 e2 f2 d1 + d2 e1 + e2 f1 + f2 We verify that this is in W :
(a1 + a2 ) + (c1 + c2 ) = (a1 + c1 ) + (a2 + c2 ) = 0 + 0 = 0 (b1 + b2 ) + (d1 + d2 ) + (f1 + f2 ) = (b1 + d1 + f1 ) + (b2 + d2 + f2 ) = 0 + 0 = 0
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
Similarly, if r is any real number and u ∈ W , then ( ) a1 b1 c1 u= =⇒ d1 e1 f1 We verify that this is in W :
ru =
(
r a1 r d1
r b1 r e1
11
r c1 r f1
(r a1 ) + (r c1 ) = r (a1 + c1 ) = r 0 = 0 (r b1 ) + (r d1 ) + (r f1 ) = r (b1 + d1 + f1 ) = r 0 = 0
)
.
