EXERCISE 4-3 4-21
70.
131122
152220
262244
1310 13
~
131122
023302
0 0000 0
000131
R1 + R2 → R2, (-2)R1 + R3 → R3 , R1 + R4 → R4 (1/2)R2 →R2
131122
33
01 01
11 11
10 2 5
22
33
01 0 1
11 11
10 2 5
22
33
01 0 1
~
29
11 21
10 0
222
395
01 0
222
72. y = ax2 + bx + c
For (–1, –5): –5 = a(–1)2 + b(–1) + c = a – b + c (1)
For (2, 7): 7 = a(2)2 + b(2) + c = 4a + 2b + c (2)
4-22 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
74. Let x1 = Number of one-person boats
x2 = Number of two-person boats
x3 = Number of four-person boats
(A) The mathematical model is:
0.5x1 + x2 + 1.5x3 = 350
The corresponding augmented matrix is:
0.5 1 1.5 350
0.6 0.9 1.2 330
~ 1 2 3 700
0.6 0.9 1.2 330
~ 12 3700
00.30.690
EXERCISE 4-3 4-23
The corresponding augmented matrix is:
0.5 1 350
0.6 0.9 330
0.2 0.3 115
~
12700
0.6 0.9 330
0.2 0.3 115
~
1 2 700
00.390
00.125
~
12700
01300
00.125
76. Let x1 = number of 10-passenger planes
x2 = number of 15-passenger planes
x3 = number of 20-passenger planes
Then, the mathematical model is:
Divide the second equation by 5 to get the following system:
The augmented matrix corresponding to this system is:
78. Let C = 8,000x1 + 14,000x2 + 16,000x3. Then
for x1 = 0, x2 = 4, x3 = 8, C = $184,000;
EXERCISE 4-3 4-25
–0.11x1 – 0.09x2 + x3 – 0.08x4 = 3.8
–0.06x1 – 0.02x2 – 0.14x3 + x4 = 4.4
The augmented coefficient matrix is:
1 0.08 0.03 0.07 3.2
0.12 1 0.11 0.13 2.6
84. Let x1 = number of ounces of food A
x2 = number of ounces of food B
x3 = number of ounces of food C
(A) The mathematical model is: 30x1 + 10x2 + 20x3 = 400
10x1 + 10x2 + 20x3 = 160
10 R1 → R1 (–3)R1 + R2 → R2 1
2
1
10 R2 → R2 (–1)R1 + R3 → R3
1
4-26 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
(B) The mathematical model is:
30x1 + 10x2 = 400
10x1 + 10x2 = 160
10 30 240
10 30 240
1324
028
22
1
1
10
1
10
RR
13 3
(1)
RR R
1
1116
02 8
1012
00 0
21 1
23 3
(1)
(2)
RR R
RR R
Thus, x1 = 12, x2 = 4, i.e. 12 ounces of food A and 4 ounces of food B.
(C) The mathematical model is:
30x1 + 10x2 + 20x3 = 400
86. Let x1 = number of packets of brand A
EXERCISE 4-3 4-27
The mathematical model is
x1 + x2 + x3 + x4 = 5
5x1 + 10x2 + 15x3 + 20x4 = 80
88. From Problem 86 we have:
Solution: x1 = 0, x2 = 2, x3 = 0, x4 = 3 cost: 6 + 6.75 = $12.75
$12.00, which is the minimum amount.
90. Let y = ax2 + bx + c be the quadratic equation for which we need to find a, b, c based on the information
provided.
For (0, 30): 30 = a(0)2 + b(0) + c or c = 30
4-28 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
For 2030, x = 2030 – 1990 = 40 and
y = –0.005(40)2 + 0.45(40) + 30 = 40 million.
This estimate is plausible.
92. Let y = ax2 + x + c
For (0, 70.7): 70.7 = a(0)2 + b(0) + c or c = 70.7
a = 0.006
Use this to find b from (1): 25(0.006) + 5b = 0.4
= 0.05.
94.
96. Let x1 = number of hours for company A, and
x2 = number of hours for company B
The mathematical model is:
30x1 + 20x2 = 650
10x1 + 20x2 = 350
30 20 650
~ 10 20 350
~ 1235
~ 1235
~
EXERCISE 4-4 4-29
Copyright © 2019 Pearson Education, Inc.
1235
0110
~ 1015
0110
1
2
Thus, 15
10
x
x
(–2)R2 + R1 → R1
Solution: company A: 15 hours; company B: 10 hour
98. (A) 5th Street and Washington Avenue: x1 + x4 = 1,500
6th Street and Washington Avenue: x1 + x2 = 800
(B) The augmented matrix is:
1 0 0 1 1500
1100 800
100 1 1500
0 1 0 1 700
100 1 1500
0 1 0 1 700
EXERCISE 4-4
2. 37 4
4. 92 90 02
6. 261 8 244
485 3 3220 12
10. 11
4
= 6
4-30 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
Copyright © 2019 Pearson Education, Inc.
12. 32
25
= 49
14. 25
32
= 26 9
16. 39
18.
39
20. 00
22.
40
1
1
113
32. CA is not defined; the number of columns of C (3) does not equal the number of rows of A (2).
213
36. C2 = CC =
102
431
102
431
=
368
18 12 10
38. C + DA =
102
431
+
32
01
213
102
235
61113
271
51115
0104
40. (0.2)CD = (0.2)
102
431
32
01
= (0.2)
16
13 3
=
0.2 1.2
2.6 0.6
42. (2)DB + (5)CD = (2)
32
01
31
+ (5)
16
13 3
13 7
16
222
555
377
44. (–1)AC + (3)DB is not defined; AC is 3 × 2 and DB is 2 × 3.
EXERCISE 4-4 4-31
46. From problem 38, DA =
61113
042
. Thus
102
235
61113
271
22515
24525
48. BAD = B(AD) = 31
32
213
01
25
93
28
834
50. A = ab
, B = aa
aa
22
aabaab
aa
ab
22
a a ab ab
00
52. A =
2
1
ab b ab
aab
2
ab b ab
2
ab b ab
2 2 22 22
12
abaab abab abab
1
aab
54. Bn → 0.75 0.25
, ABn → [0.75 0.25]
56. 42
+ wx
= 42
wx
= 23
. Thus
4-32 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
58. 13
14
ab
cd
= 65
77
acbd
0022 2200
62. False: For [a][b] = [0], we must have [ab] = [0], or ab = 0. This is impossible if a and b are two nonzero
64. Let A = 11
1
0
ab
d
and B = 22
2
0
ab
d
(A) Always true:
12
0
aa bb
dd
(B) Always true:
1
0
ab
d
22
2
0
ab
d
12
0
aa ab bd
dd
(C) False: Let A = 11
, B = 11
. Then
01
01
EXERCISE 4-4 4-33
66. (1.2)A = $56.4 $46.8
$108 $150
2(1.2A + B) = 1
2
$108 $150 $84 $115
$96.00 $132.50 Labor
68. The dealer is increasing the retail price by 15%. Thus, the new retail price matrix M‘ (to the nearest dollar)
is given by:
M ‘ = M + 0.15M = 1.15M = 1.15
$35,075 $2,560 $1, 070 $640
$39,045 $1,840 $770 $460
$41, 667 $2,890 $1, 200 $725
The new markup is:
Basic AM/FM Cruise
car Air radio control
$3,141 $484 $211 $124 Model
A
70. (A) [2 3 5 0 6]
1, 800
2, 400
= $48,480 (B) [10 4 3 4 3]
1, 400
1, 800
= $39,300
(C)
33, 400 42,160
35,600 48, 480
MN
The entries are the wholesale and retail values of the inventories at each
Round Wrinkled
(B) [1 1](M + N)1
1
= [1 1] 685 229
218 68
1
1
= [1 1] 914
286
= 914 + 286 = 1200
1200 (M + N) = 1
1200
218 68
18% 6% Green
4-34 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
74. (A) M
0.25
=
91658492 0.25
=
83
Bob
(B) M
0.2
=
84.8
Bob
Class Average
Test 1 Test 2 Test 3 Test 4
EXERCISE 4-5
6. No, the matrix is not a square matrix. 8. Yes, the matrix is the identity matrix.
16.
100
010
340
125
=
340
125
18.
340
125
100
010
=
340
125
24. 7434 10
Yes 26.
1 0 1 10 1 100
31 2 31 1 010
Yes
28.
10 1 1 0 1
31 1, 31 2
30. Only square matrices have inverses. 32. Only square matrices have inverses.
EXERCISE 4-5 4-35
40. 1510
0101
~ 1510
0101
~ 101 5
010 1
01
01
(0)(1) (1)(0) (0)(5) (1)(1)
01
42. 2110
5301
~
11
22
10
5301
~
11
22
5
1
22
10
01
~
11
22
10
01 52
1
52
53
(5)2 25 (5)1 23
01
44. 2110
1101
~ 101 1
110 1
~ 1011
0112
12
11
(1)2 21 (1)1 21
01
46.
2 3 0100 1 2 3010 1 2 30 1 0
1 2 3010 2 3 0100 0 1 61 20
~
~
~
4-36 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
Copyright © 2019 Pearson Education, Inc.
Thus, M-1 =
715 9
5106
12 1
1
M
M
715 9 23 0 100
5106123 010
12 1015 001
48.
1 0 1 100 1 0 11 00 10 11 0 0
2 1 0 010 0 1 2 210~~012210
~
M
311 1 1 2 001
50. 4310
~
31
10
44
~
31
10
44
~
31
10
44
01 54
54
52. 2410
~
1
2
12 0
~
1
2
12 0
EXERCISE 4-5 4-37
54. 5310
2201
~
31
55
10
2201
~
31
55
42
55
10
01
~
31
55
5
1
24
10
01
5
1
24
01
0.5 1.25
56. 51010
~ 22401
~
1
2
1120
~
1
2
5
1120
58.
10
5
60. 10
62.
2 2 4 1 0 0 1 0 10 0 1 1 0 10 0 1 1 0 10 0 1
1 1 1 0 1 0 1 1 10 1 0 0 1 00 1 1 0 1 00 1 1
~~
~
1
64.
110100 110100 110100
211010 01121~~0 011210
4-38 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
66.
11
55
4
5
4
422100 505001 10100 10 100
420010 420010 420010 02 401
505001 422100 422100 02 21
~
0
~~
41111
52222
02210 002110001 0001 0
25
11
22
0
68.
11 4100
3322010
2119001
~
2119001
3322010
11 4100
1
22 2
111 1
13 2
33
01 0 1
~
13 2
33
01 0 1
~
EXERCISE 4-5 4-39
Copyright © 2019 Pearson Education, Inc.
35 1
33
10 0 1
100 3.5 1.5 1
13
3
R3 + R2 → R2
0.3 0.1 0
70. A = 43
32
; 4310
3201
~
31
44
10
3201
~
31
44
3
1
44
10
01
~
4
R2 + R1 → R1
51
10
51
10
22
10
~ 10 7 5
Thus, B-1 = 75
.
32
37
12 29
17 41 1 0
41 1
17 17
10
41 1
17 17
10
17 17
~ 1 0 29 41
Thus, (AB)-1 = 29 41
.
17
4-40 CHAPTER 4: SYSTEMS OF LINEAR EQUATIONS; MATRICES
Copyright © 2019 Pearson Education, Inc.
We have B-1A-1 = 75
32
23
34
= 29 41
12 17
.
Therefore, (AB)-1 = B-1A-1.
72. 10
001
ab
d
1
010
d
1
0
d
0
a
ad
For an
n × n upper triangular matrix, the inverse exists if and only if all the elements on the main diagonal
are nonzero.
74. A = 54
A-1: 5410
~
~
6
1
~
76. If A = A-1, then A2 = AA = AA-1 = A-1A = I.
78. A = 12
13
13 9 9 5 9 314 1 5 0 334135272343233513