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Chapter 1
Systems of Linear Equations 1.1
Soluti Solutions ons and Elemen Elementar tary y Opera Operatio tions ns
1. (a) 2(19t − 35) + 3(25 − 13 13tt) + t = 5 5(19t 5(19t − 35) + 7(25 − 13 13tt) − 4t = 0 (b) 2 ( 2s 2s + 12t 12 t + 13) + 5s + 9 (−s − 3t − 3) + 3t = −1 (2s (2s + 12t 12 t + 13) + 2s + 4 (−s − 3t − 3) = 1 2. (a) x = t = t,, y = 2 − 3t; x = x = 32 − 3t , y = t = t (b) x = t = t,, y = 31 (1 − 2t); x ); x = 21 (1 − 3s), y ), y = s = s (c) x = s = s,, y = 3s + 2t 2t − 5, z 5, z = t = t;; x = x = s s,, y = t = t,, z = 25 −
3s 2
+
t
2
(d) x = 1 + 2s 2 s − 5t, y = s = s,, z = t = t;; x = x = q q , y = p = p,, z = 51 (1 − p + p + 2q 2q ) 3. x = 5/2, y 2, y = t. = t. Master your semester with Scribd 4. x = 41 (2s (2s + 3), y 3), y = s = s,, z = t. = t. & The New York Times Special offer for students: $4.99/month. 5. (a) x (a)Only x = = t t (and b (and b = 0)
(b) x = b/a = b/a
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2
Chapter 1 — Systems of Linear
11. (a) x = 31 (2t + 5), y = t
(b) No solution.
12. The augmented matrix reduces to
1 2 −1 0 −3 5 0 0 0
a b − 2a c − (2b − 3a)
.
13. Zero – parallel lines, one – nonparallel lines, infinitely many – coincident lines.
14. x = 5, y = 1, so x = 23, y = −32. 15. The equations are: 2a + c = 1, −a + 2b = −1, −b + 2c = 3. Solution: a = − 19 , b = − 59 , c =
11 9 .
16. 100 mg of the first supplement and 160 mg of the second. 17. John gets $4.50 per hour, and Joe gets $5.20 per hour.
1.2
Gaussian Elimination
1. (a) No, Yes.
(b) No, No.
(c) No, Yes.
(d) No, Yes.
(e) No, Yes.
(f ) No, No.
2. (a)
0 0 0 0
1 −2 0 0 0 0 0 0
0 1 0 0
1 3 0 0
0 0 1 0
0 0 1 −3 0 0 0 0 You're Reading (b) a Preview 0 0 0 0 Unlock full access with a0free0trial.0 1
0 1 0 0
0 0 1 0
0 0 0 −1 0 0 1 1
3. (a) x1 = −1 − 2r − 3s − t, x 2 = r, x 3 = 2 + s − t, x 4 = s, x 5 = t, x6 = 3. Download With Free Trial Master your semester with Scribd For5on 30this Days Sign to vote title 1, x6 = t. (b) x1 = 2r − 2s − t + 1, x 2 = r, x 3 = −5s + 3t −Read 1, xup = s, x = −6t + 4Free & The New York Times Useful Not useful (c) x1 = −1 − 3s − 2t, x 2 = 1 + s − t, x 3 = s, x 4 = t, x 5 = t. Special offer for students: Only $4.99/month.
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(d) x1 = −4s − 5t − 4, x 2 = −2s + t − 2, x 3 = s, x 4 = 1, x 5 = t.
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Section 1.2 — Gaussian Elimination
(b) x1 = 0, x 2 = −t, x3 = 0, x 4 = t. (c) x1 = 4, x 2 = 3, x 3 = 0, x 4 = 0. (d) x1 = 1, x 2 = 1 − t, x 3 = 1 + t, x 4 = t. +10 8. (a) If b = −2a, unique solution x = bb+2 ,y = a a = 5 the solutions are x = 1 + 2t, y = t.
5−a b+2a
. If b = −2a : no solution if
(b) If ab = 2, unique solution x = 22 ab5b , y = 2a+5 . If ab = 2 : no solution if ab 2 a = −5 the solutions are x = −1 + 5 t, y = t. − − −
(c) If b = −a, unique solution x =
−
3b−a , y = a+b
4
a+b
. If b = −a there is no solution.
2 (d) If a = 2, unique solution x = a1 2b , y = ab . If a = 2 : There is no solution if a 2 1 b = 1 the solutions are x = 2 (1 − t), y = t. −
−
−
−
9. (a) If c = 2a − b, no solution. If c = 2a − b the solutions are x = 41 (a+b − t), y = 41 ( z = t. (b) Unique solution x = −2a + b + 5c, y = 3a − b − 6c, z = −2a + b + 4c for any
(c) If a = 6, unique solution x = 2a a b 6 12 , y = 2aa b6+12 , z = a b 6 . If a = 6 : no so b = 0; if b = 0, the solutions are x = 2 − t, y = −2 − t, z = t. − −
−
−
−
−
−
(d) If abc = −1, unique solution x = 0, y = 0, z = 0; if abc = −1 the solutions are y = −bt, z = t. (e) If b = 1, no solution. If b = 1 the solutions are x = 41 (5 − t), y = 41 (3 + 5t), z You're Reading a 1. Preview (f) If a = 1, solutions x = − t, y = t, z=− If a = 0, there is no solution. If a 1 a = 0, unique solution x = , y = 0, z = a1trial. . Unlockafull access with a free −
10. (a) 2
(b) 1
(c) 2
(d) 3
(e) 2
−
(f ) 1
Download With Free Trial Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title 11. (a) 3 (b) 2 (c) 2 (d) 3 & The New York Times Useful Not useful (e) 2 if a = 0 or 1; 3 otherwise. Special offer for students: Only $4.99/month.
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4
Chapter 1 — Systems of Linear
(h) True. The sequence of row operations carrying C to row-echelon form (with 3 1’s) will also carry A to row-echelon form with 3 leading 1’s, none of which is in 5. So an equation 0x1 + · · · + 0x5 = b = 0 does not occur. 13.
bi + ci ci + ai ai + bi
→
2(ai + bi + ci ) ci + ai ai + bi
→
14. (a) Since none of a, b and c are 0,
bi ci + ai ai + bi
p 0 a b 0 0 q c r
(b) Since one of b − a and c − a is nonzero,
1 a b + c 1 b c + a 1 c a + b
→
1 a b + c 0 b−a a−b 0 c−a a−c
bi ci + ai ai
→
→
→
1 0 0 0 c r 0 0 a
→
ai bi ci
1 0 0 0 1 0 0 0 1
→
1 a b + c 0 1 −1 0 0 0
→
bi ci ai
→
1 0 b 0 1 0 0
15. There can be at most two leading variables so there is at least one non-leading vari 16. (a) x 2 + y 2 − 2x + 6y − 15 = 0
(b) x 2 + y 2 − 2x + 6y − 6 = 0.
17. Nissans, $10 per day; Fords, $12 per day; Chevrolets, $13 per day. 18.
5 20
in A,
7 20
in B ,
8 20
in C .
You're Reading a Preview
19. Since p 1 , p 2 , p 3 are distinct, the augmented matrix for the equation giving a, b and Unlock full access with a free trial. as follows: 1 p1 p21 q 1 1 p1 p21 ∗ Download With Free Trial 2 2 2 1 p2 p2 q 2 0 pRead p2For → − p30 ∗ 2 − pFree Days Sign up1 to vote on12this title 2 2 1 p3 p3 q 3 0 p3 − p1 p3 − p1 ∗
Scribd Master your semester with & The New York Times Special offer for students: Only $4.99/month. 1 p1
→
0
1
p22 p + p
∗ ∗
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1 0 0 1
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∗ ∗
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∗ ∗
→
1 0 0 0 1 0
∗ ∗
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Section 1.3 — Homogeneous Equations
(b) With four players, the augmented matrix reduces as follows:
1 0 0 1
1 1 0 0
0 1 1 0
0 0 1 1
t1 t2 t3 t4
→
1 1 0 1 0 0 0 −1
0 1 1 0
0 0 1 1
∗ ∗ ∗ ∗
→
1 0 0 0
1 1 0 0
0 1 1 1
0 0 1 1
∗ ∗ ∗ ∗
→
1 0 0 0
1 1 0 0
Hence there is either no solution, or a parameter (so infinitely many solutions).
21. The equations are p + n + d = 17, p + 5n + 10d = 105. The solution is p = 5t − 5, n d = 4t. Since p ≥ 0 and n ≥ 0, we have t ≥ 1, t ≤ 2. In addition, t must be a int t = 1 or t = 2. Hence p = 0, n = 13, d = 4, or p = 5, n = 4, d = 4.
1.3
Homogeneous Equations
Master your semester with Scribd 1. (a) False. A =
1 0 1 0 1 1
0 0
(b) False. A =
1 0 1 0 1 1
0 0
(c) True. If a1 x1 + a 2 x2 + · · · + a n xn = b is any equation in the system, b = 0 x1 = x 2 = · · · = x n = 0 is a solution.
(d) False. A =
1 0 1 0 1 1
1 0
(e) False. The trivial solution always exists for a homogeneous system.
(f) False. A =
1 0 0 1
0 You're Reading a Preview (g) False. A = 0
(h) False. A =
1 0 0 1 0 0
0 0 Download With Free Trial 0 Read Free Foron 30this Days Sign up to vote title
Unlock full access with a free trial.
& The New 2. York (a) a = Times 2, x = t, y = t, z = t Special offer for students: Only $4.99/month.
1 0 1 0 1 1
0 0
Not useful =Useful 9t, y (b) a 3, x = = −5t, z = t −Cancel anytime.
(c) a = −1, x = 2t, y = −t, z = t; or a = 0, x = −t, y = t, z = 0.
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6
Chapter 1 — Systems of Linear
(b) Three points lead to three equations ax + by + cz + d = 0 in the variables a,
7. Insisting that the graph of a(x2 + y 2 ) + bx + cy + d = 0 passes through the three poin three linear equations in a, b, c, d. This has a nontrivial solution by Theorem 1, an means the points all lie on the line bx + cy + d = 0, contrary to hypothesis. So required.
8. The number of parameters is n − r by Theorem 2 §1.2, so there are nontrivial solution only if n − r > 0.
1.4
An Application to Network Flow
1. (a) f 1 = f 3 + f 5 − 10 f 2 = −f 3 − f 5 + 60 f 3 = − f 5 + 50
(b) f 1 f 2 f 3 f 4
= 85 = 60 = −75 = 40
2. (a) f 1 = 55 − f 4 f 2 = 20 − f 4 + f 5 f 3 = 15 − f 5
(b)
3. (a) f 1 f 2 f 3 f 4
(b) The road CD.
= = = =
50 20 60 35
+ + + +
f 5 f 5 f 5 f 5
− f 4 − f 4
− f 7 − f 7
+ f 4 + f 6 − f 6 + f 7
f 5
= 15 25 ≤ f 4 ≤ 30
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial Master your semester withtoScribd 1.5 An Application Electrical Networks Read Free Foron 30this Days Sign up to vote title & The New York Times Useful Not useful 1. I 1 = 40 , I 2 = − 5 , I 3 = 45 . 11
11
11
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2. I
1
I
3
I
4
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Supplementary Exercises for Chapter 1
Supplementary Exercises for Chapter 1
1. (a) zero — parallel planes; one — no two planes parallel; infinitely many — co planes, or all three meeting in a line. (b) No. If the corresponding planes are parallel and distinct, there is no solution. they either coincide or have a whole common line of solutions. 2. (a) x1 = 7 − 27 s + 3t, x 2 = −4 + 25 s − 2t, x 3 = s, x 4 = t. (b) x1 =
1 1 10 (−6s − 6t + 16), x 2 = 10 (4s − t + 1), x 3 = s, x 4 = t.
a+11) +1) 1 3. (a) If a2 = 9, unique solution x = a7((4aa+3) , y = 5(2 7(a+3) , z = a+3 . If a = −3, no sol a = 3, the solutions are x = 71 (8 − 22t), y = 71 (10 + 25t), z = t.
(b) If a = 1 there is no solution. If a = 2, the solutions are x = 2 − 2t, y = −t a a = 1 and a = 2 the unique solution is x = 3(8a 5a1) , y = 3(a2 1) , z = a+2 3 . −
− −
−
4.
R1 R2
→
R1 + R2 R2
→
R1 + R2 −R1
→
R2 −R1
−
→
5. Either a or c is nonzero (as ad = bc), say a = 0. Then
R2 R1
a b c d
→
1 0
b a ad−bc a
1 0 . The case c = 0 is similar. 0 1 You're Reading a Preview 6. Substitute x = 3, y = −1, z = 2 to get the equations a − 2c = 3, 3b − c = 7, 3a Unlock full = access This has the unique solution a = 1, b 2, with c = a−free 1. trial. ad − bc = 0, this can be carried to
7. The Gaussian algorithm works even if some the coefficients and constants are not r Download WithofFree Trial Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title result is, x = 1, y = i, z = −2 − i. & The New York Times Useful Not useful 8. The (real) solution is x = 2, y = 3 − t, z = t where t is a parameter. The given Special offer for students: Only $4.99/month.
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solution occurs when t = 3 − i is complex. If the real system has a unique solu
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Chapter 1 — Systems of Linear
12. 2 adults, 8 youths, 140 children. 13. x2 = 4, xy = −2, y 2 = 1. Hence x = ± 2, y = ± 1, so xy = −2 gives x = 2 and y x = −2 and y = 1.
14. If there are m equations in n variables where n > m, and if rank A = r where augmented matrix, there are n − r parameters by Theorem 2 §1.2. So it suffices to pr n > r. But n > m, and m ≥ r because the row-echelon form of A has m rows and 1’s.
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