C H A P T E R 7Linear Systems and Matrices
Section 7.1 Solving Systems of Equations . . . . . . . . . . . . . . . 549
Section 7.2 Systems of Linear Equations in Two Variables . . . . . . 564
Section 7.3 Multivariable Linear Systems . . . . . . . . . . . . . . . 578
Section 7.4 Matrices and Systems of Equations . . . . . . . . . . . . 603
Section 7.5 Operations with Matrices . . . . . . . . . . . . . . . . . . 620
Section 7.6 The Inverse of a Square Matrix . . . . . . . . . . . . . . 633
Section 7.7 The Determinant of a Square Matrix . . . . . . . . . . . . 645
Section 7.8 Applications of Matrices and Determinants . . . . . . . . 654
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662
Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 687
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549
C H A P T E R 7Linear Systems and Matrices
Section 7.1 Solving Systems of Equations
■ You should be able to solve systems of equations by the method of substitution.
1. Solve one of the equations for one of the variables.
2. Substitute this expression into the other equation and solve.
3. Back-substitute into the first equation to find the value of the other variable.
4. Check your answer in each of the original equations.
■ You should be able to find solutions graphically. (See Example 5 in textbook.)
1. (a)
No, is not a solution.
(b)
No, is not a solution.��1, �5�
�11 � �6
1 � 1
6��1� � ��5� �?
�6
4��1� � ��5� �?
1
�0, �3�
�3 � �6
3 � 1
6�0� � ��3� �?
�6
4�0� � ��3� �?
1
Vocabulary Check
1. system, equations 2. solution 3. method, substitution
4. point, intersection 5. break-even point
(c)
No, is not a solution.
(d)
Yes, is a solution.��12, �3�
�6 � �6
1 � 1
6��12� � ��3� �
?�6
4��12� � ��3� �
?1
��32, 3�
�6 � �6
�9 � 1
6��32� � �3� �
?�6
4��32� � �3� �
?1
2. (a)
Yes, is a solution.�2, �13�
11 � 11
3 � 3
�2 � ��13� �?
11
4�2�2 � ��13� �?
3 (b)
No, is not a solution.
––CONTINUED––
��2, �9�
11 � 11
7 � 3
���2� � ��9� �?
11
4��2�2 � ��9� �?
3
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550 Chapter 7 Linear Systems and Matrices
3. (a)
No, is not a solution.
(b)
Yes, is a solution.
(c)
No, is not a solution.
(d)
No, is not a solution.��1, �5�
2 � 2
�5 � �2e�1
3��1� � ��5� �?
2
�5 �?
�2e�1
�0, �3�
3 � 2
�3 � �2
3�0� � ��3� �?
2
�3 �?
�2e0
�0, �2�
2 � 2
�2 � �2
3�0� � ��2� �?
2
�2 �?
�2e0
��2, 0�
�6 � 2
0 � �2e�2
3��2� � 0 �?
2
0 �?
�2e�2 4. (a)
No, is not a solution.
(b)
Yes, is a solution.
(c)
Yes, is a solution.
(d)
No, is not a solution.�1, 1�
109 �
289
3 � 1
19 �1� � 1 �? 28
9
�log10 1 � 3 �?
1
�1, 3�
289 �
289
3 � 3
19 �1� � 3 �? 28
9
�log10 1 � 3 �?
3
�10, 2�
289 �
289
2 � 2
19 �10� � 2 �? 28
9
�log10 10 � 3 �?
2
�100, 1�
1099 �
289
1 � 1
19 �100� � 1 �? 28
9
�log10�100� � 3 �?
1
5.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Back-substitute
Answer: �2, 2�
x � 2: y � 6 � 2(2) � 2
�3x � 6 � 0 ⇒ x � 2x
�x � (6 � 2x) � 0y
y � 6 � 2xy
Equation 1Equation 2� 2x � y � 6
�x � y � 06.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for
Back-substitute
Answer: ��1, 3�
x � 3 � 4 � �1y � 3:
3y � 4 � 5 ⇒ y � 3y:
� y � 4� � 2y � 5x
x � y � 4x
Equation 1Equation 2�x � y
x � 2y�
�
�45
2. ––CONTINUED––
(c)
No, is not a solution.��32, 6�
�92 � 11
15 � 3
���32� � �6� �
?11
4��32�2
� �6� �?
3 (d)
Yes, is a solution.��74, �37
4 � 11 � 11
3 � 3
���74� � ��37
4 � �?
11
4��74�2
� ��374 � �
?3
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Section 7.1 Solving Systems of Equations 551
9.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Back-substitute:
Answer: �0, 2�, ��3, 2 � 3�3 �, ���3, 2 � 3�3 �x � ��3: y � 2 � 3�3
x � �3: y � 2 � 3�3
x � 0: y � 2
x � 0, ±�3 ⇒ x�x2 � 3� � 0 ⇒ x3 � 3x � 0x
x3 � 2 � �2 � 3x� � 0y
y � 2 � 3xy
Equation 1Equation 2�3x � y � 2
x3 � 2 � y � 0
10.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for
Back-substitute
Back-substitute
Back-substitute
Answer: �0, 0�, �2, �2�, ��2, 2�
y � ���2� � 2x � �2:
y � �2x � 2:
y � �0 � 0x � 0:
x3 � 4x � 0 ⇒ x�x2 � 4� � 0 ⇒ x � 0, ±2x:
x3 � 5x � ��x� � 0y
y � �xy
Equation 1Equation 2� x � y
x3 � 5x � y�
�
00
11.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Hence, and Answer: �4, 4�x � �27�4� �
367 � 4.y � 4
�y � 4��49y2 � 180y � 1296� � 0
49y3 � 16y2 � 576y � 5184 � 0
�2y3 � 8� 449 y2 �
14449 y �
362
49 � � 0x
8��27y �
367 �2
� 2y3 � 0x
x � �27 y �
367⇒�
72x � y � 18x
Equation 1Equation 2��
72x � y � �18
8x2 � 2y3 � 0
8.
Solve for y in Equation 1:
Substitute for y in Equation 2:
Solve for x:
Back-substitute
Back-substitute
Answer: �0, �5�, �4, 3�
x � 4: y � 3
x � 0: y � �5
x � 0, 4
5x�x � 4� � 0
5x2 � 20x � 0
x2 � 4x2 � 20x � 25 � 25
x2 � �2x � 5�2 � 25
y � 2x � 5
Equation 1Equation 2��2x � y
x2 � y2
�
�
�525
7.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Back-substitute
Back-substitute
Answer: ��1, 3�, �2, 6�
x � 2: y � 2 � 4 � 6
x � �1: y � �1 � 4 � 3
⇒ x � �1, 2
⇒ �x � 1��x � 2� � 0
x2 � x � 2 � 0 x
x2 � (x � 4) � �2y
y � x � 4y
Equation 1Equation 2� x
x2
�
�
yy
�
�
�4�2
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552 Chapter 7 Linear Systems and Matrices
12.
Substitute for in Equation 1:
Solve for
Back-substitute
Back-substitute
Back-substitute
Answer: �0, 4�, �1, 2�, �2, 0�
y � �2�2� � 4 � 0x � 2:
y � �2�1� � 4 � 2x � 1:
y � �2�0� � 4 � 4x � 0:
0 � x�x � 2��x � 1� ⇒ x � 0, 1, 2
0 � x�x2 � 3x � 2�
0 � x3 � 3x2 � 2xx:
�2x � 4 � x3 � 3x2 � 4y
Equation 1Equation 2�y �
y �
x3 � 3x2 � 4�2x � 4
13.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Back-substitute in Equation 1:
Answer: �5, 5�
y � x � 5
2x � 10 ⇒ x � 5x
5x � 3x � 10y
y � xy
Equation 1Equation 2� x
5x�
�
y3y
�
�
010
14.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for
Back-substitute
Answer: ��3, 2�
x � 1 � 2y � 1 � 4 � �3y � 2:
�14y � �28 ⇒ y � 2y:
5�1 � 2y� � 4y � �23x
x � 1 � 2yx
Equation 1Equation 2� x � 2y
5x � 4y�
�
1�23
15.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Back-substitute
Answer: �12, 3�
x �12: y � 2x � 2 � 2� 1
2� � 2 � 3
4x � �2x � 2� � 5 � 0 ⇒ 6x � 3 � 0 ⇒ x �12
x
4x � �2x � 2� � 5 � 0y
y � 2x � 2y
Equation 1Equation 2�2x � y � 2 � 0
4x � y � 5 � 0
16.
Solve for in Equation 2:
Substitute for in Equation 1:
Solve for
Back-substitute
Answer: �43, 43�
x � 4 � 2y � 4 � 2�43� �
43y �
43:
24 � 12y � 3y � 4 � 0 ⇒ �15y � �20 ⇒ y �43y:
6�4 � 2y� � 3y � 4 � 0x
x � 4 � 2yx
Equation 1Equation 2�6x � 3y � 4
x � 2y � 4�
�
00
17.
Solve for in Equation 2:
Substitute for in Equation 1:
Then, Answer: �1, 1�y �3x � 1
2�
3�1� � 1
2� 1.
x � 1
27x � 27
15x � 12x � 4 � 23
15x � 8�3x � 1
2 � � 23y
�2y �3x � 1
2
�2y � 1 � 3xy
�1.5x � 0.8y � 2.3 ⇒ 0.3x � 0.2y � 0.1 ⇒
15x3x
�
�
8y2y
�
�
231
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Section 7.1 Solving Systems of Equations 553
18.
Multiply both equations by 10:
Solve for in Equation 2: x � �18 � 8yx
2x � 16y � �36
5x � 32y � 90
Equation 1Equation 2�0.5x � 3.2y
0.2x � 1.6y�
�
9.0�3.6
19.
Solve for in Equation 2:
Substitute for in Equation 1:
Solve for :
Back-substitute
Answer: �203 , 40
3 � � 20 �
403 �
203
x � 20 � yy �403 :
4 �310 y � 8 ⇒ y �
403y
15�20 � y� �
12 y � 8x
x � 20 � yx
Equation 1Equation 2�
15xx
�
�
12yy
�
�
820
20.
Solve for in Equation 2:
Substitute for in Equation 1:
Solve for
Back-substitute
Answer: �20817 , 88
17�y �
34 �208
17 � � 4 �8817x �
20817 :
12 x �
916 x � 3 � 10 ⇒ 17
16 x � 13 ⇒ x �20817
x:
12 x �
34 �3
4 x � 4� � 10y
y �34 x � 4y
Equation 1
Equation 2�12 x �
34 y
34x � y
�
�
10
4
21.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Inconsistent
No solution
15 � 6
�5x � 15 � 5x � 6x
�5x � 3�5 �53x� � 6y
y � 5 �53xy
Equation 1Equation 2��
53x
�5x�
�
y3y
�
�
56
22.
From Equation 1,
In Equation 2,
No solution
2x � 3�2 �23 x� � �6 � 6
y � 2 �23 x
Equation 1Equation 2��
23 x �
y2x � 3y
�
�
26
23.
Solve for in Equation 2:
Substitute for in Equation 1:
Answer: �3, 5�, ��2, 0�
x � �2 ⇒ y � 0
x � 3 ⇒ y � 5
x � 3, �2
�x � 3��x � 2� � 0
x2 � x � 6 � 0
x2 � 2x � �x � 2� � 8
y
y � x � 2y
Equation 1Equation 2�x2 � 2x � y
x � y�
�
8�2
24.
Solve for in Equation 2:
Substitute for in Equation 1:
Answer: �4, 10�, ��2, �2�
x � �2 ⇒ y � �2
x � 4 ⇒ y � 10
x � 4, �2
�x � 4��x � 2� � 0
x2 � 2x � 8 � 0
2x2 � 4x � 16 � 0
2x2 � 2x � �2x � 2� � 14
y
y � 2x � 2y
Equation 1Equation 2�2x2 � 2x � y
2x � y�
�
14�2
Substitute for in Equation 1:
Solve for
Back-substitute
Answer: (2, 2.5�
y � 2.5: x � �18 � 8�2.5� � 2
y �18072 � 2.5
72y � 180
�90 � 40y � 32y � 90y:
5��18 � 8y� � 32y � 90x
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554 Chapter 7 Linear Systems and Matrices
25.
Solve for in Equation 2:
Substitute for in Equation 1:
No real solution
2x2 � x � 1 � 0
2x2 � �x � 2� � 1y
y � x � 2y
Equation 1Equation 2�2x2
x�
�
yy
�
�
12
26.
Solve for in Equation 2:
Substitute for in Equation 1:
No real solution
2x2 � x � 1 � 0
2x2 � ��x � 4� � 3y
y � �x � 4y
Equation 1Equation 2�2x2
x�
�
yy
�
�
34
27.
Solve for in Equation 2:
Substitute for in Equation 1:
Solve for
Back-substitute:
Answer: �0, 0�, �1, 1�, ��1, �1�
x � �1 ⇒ y � �1
x � 1 ⇒ y � 1
x � 0 ⇒ y � 0
x�x � 1��x � 1� � 0 ⇒ x � 0, 1, �1x:
x3 � x � 0y
y � xy
Equation 1Equation 2�x3
x�
�
yy
�
�
00
28.
Answer: �0, 0�
x � 0 ⇒ y � 0
x�x2 � 3x � 3� � 0
x3 � 3x2 � 3x � 0
y � �x � x3 � 3x2 � 2x
29.
Point of intersection: �4, 3�
x−2 4 6−1 1 2 3
−2
3
5
2
4
6
3x + y = 15
−x + 2y = 2
(4, 3)
y
��x3x
�
�
2yy
�
�
215
30.
Point of intersection: �2, �2�
x
−1
1
−1 1 2 4
−2
−3
−4
x y+ = 0
3 2 = 10x y−
y
(2, −2)
� x � y3x � 2y
�
�
010
31.
Point of intersection: � 52, 32�
( (,
y
x
x − 3y = −2
5x + 3y = 17
−1−2 1 2 3 5 6
−2
−3
1
2
3
4
5
52
32
� x5x
�
�
3y3y
�
�
�217
32.
Point of intersection: �5, 3�
x
(5, 3)2
2 4 6
4
6
x y− = 2
−x y+ 2 = 1
y
��xx
�
�
2yy
� 1� 2
33.
No solution
−2−3−4 2 3 4
−2
−1
−3
−4
−5
2
3
y
x
x + y = 2x2 + y = 1
�x2 � y � 1x � y � 2
34.
Points of intersection:�2, 0�, ��1, �3�
−1−3−4 3 4
−2
−5
2
1
3
y
x
x − y = 2
x2 − y = 4
�x2 � y � 4x � y � 2
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Section 7.1 Solving Systems of Equations 555
35.
Points of intersection: (�3, 0�, �3, 6�
y3 � ��6x � x2 � 27
14126 8 1042
10
86
y
x−6
−6
−8
−10
��x � y � 3 ⇒ y1 � x � 3x2 � 6x � 27 � y2 � 0 ⇒ y2 � �6x � x2 � 27
36.
Points of intersection: �15, 7�, �3, 1�
16 181486 10 122
10
86
4
2x
y
−4
−6
−8−10
�y2 � 4x � 11 � 0�
12 x � y � �
12
⇒ ⇒
y � ±�4x � 11y �
12 x �
12
38.
Point of intersection: �2, 2�
� x � y � 0
5x � 2y � 6
⇒ ⇒
y1 � x
y2 �52x � 3
37.
Point of intersection: �4, �12�
−4
−4
8
4
�7xx
�
�
8y8y
�
�
248 ⇒ ⇒
y1
y2
�
�
�78 x18 x
�
�
31
39.
Points of intersection: �8, 3�, �3, �2�
x � y � 5 ⇒ y3 � x � 5
y2 � ��x � 1−5 13
−6
6 x � y2 � �1 ⇒ y2 � x � 1 ⇒ y1 � �x � 1
40.
Points of intersection: �2, �2�, �14, 4�
x � 2y � 6 ⇒ y3 �12
�x � 6�−6 18
−8
8x � y2 � �2 ⇒ y1 � �x � 2, y2 � ��x � 2
41.
Points of intersection: �1.540, 2.372�, ��1.540, 2.372�
y � x2 ⇒ y3 � x2 −6
−4
6
4x2 � y2 � 8 ⇒ y1 � �8 � x2, y2 � ��8 � x2
600
4
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556 Chapter 7 Linear Systems and Matrices
42.
Points of intersection: �3, 4�, �3, �4�
−8
−8
16
8
y4 � ��41 � �x � 8�2
�x � 8�2 � y2 � 41 ⇒ y3 � �41 � �x � 8�2
y2 � ��25 � x2
x2 � y2 � 25 ⇒ y1 � �25 � x2 43.
Point of intersection: �0, 1�
−3
−1
3
3
� y � ex
x � y � 1 � 0 ⇒ y � x � 1
44.
Point of intersection: ��0.490, �6.530�
−9
−10
9
2
�y � �4e�x
y � 3x � 8 � 0 ⇒ y � �3x � 845.
Point of intersection: �2.318, 2.841�
−1
−1
8
5
�x � 2yy
�
�
82 � ln x
⇒ ⇒
y1 � 4 � x2y2 � 2 � ln x
46.
Point of intersection: �5.309, �0.539�
−6
−6
12
6
�y � �2 � ln�x � 1�3y � 2x � 9 ⇒ y �
13 �9 � 2x�
47.
Point of intersection: �94, 11
2 �
−3
−3
15
9
�y � �x � 4y � 2x � 1
48.
Point of intersection: �4, 1�
−2
−4
10
4
� x � y � 3�x � y � 1
⇒ ⇒
y � x � 3 y � �x � 1
49.
Points of intersection: �0, �13�, �±12, 5�
−27
−18
27
18
�x2
x2
�
�
y2
8y
�
�
169
104
⇒
⇒
y1 � �169 � x2 andy2 � ��169 � x2
y3 �18x2 � 13
50.
Points of intersection:or �0, �2�, �±1.323, 1.500��0, �2�, �1
2�7, 32�, ��12�7, 32�
2x2 � y � 2 ⇒ y3 � 2x2 � 2
y2 � ��4 � x2
−6
−4
6
4 x2 � y2 � 4 ⇒ y1 � �4 � x2
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Section 7.1 Solving Systems of Equations 557
51.
Substitute for in Equation 2:
Solve for :
Back-substitute
Answer: �1, 2�
x � 1 in Equation 1: y � 2x � 2
x2 � 2x � 1 � �x � 1�2 � 0 ⇒ x � 1x
2x � x2 � 1y
Equation 1Equation 2�y � 2x
y � x2 � 152.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for
No real solutions because the discriminant in the Quadratic Formula is negative.
Inconsistent. No solution
x2 � x � 2 � 0x:
x2 � �4 � x� � 2y
y � 4 � xy
Equation 1Equation 2�x � y
x2 � y�
�
42
53.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Back-substitute
Back-substitute
Answers: �2910, 21
10�, ��2, 0�
x � �2: y �3x � 6
7� 0
x �29
10: y �
3x � 6
7�
3(29�10) � 6
7�
21
10
10x2 � 9x � 58 � 0 ⇒ x �9 ± �81 � 40�58�
20 ⇒ x �
29
10, �2
40x2 � 36x � 232 � 0
49x2 � �9x2 � 36x � 36� � 196
x2 � �9x2 � 36x � 36
49 � � 4x
x2 � �3x � 6
7 �2
� 4y
y �3x � 6
7y
Equation 1Equation 2�3x � 7y � 6 � 0
x2 � y2 � 4
54.
Solve for in Equation 2:
Substitute for in Equation 1:
Solve for
Back-substitute
Back-substitute
Answer: �3, 4�, �5, 0�
y � 10 � 2�5� � 0x � 5:
y � 10 � 2�3� � 4x � 3:
⇒ �x � 5��x � 3� � 0 ⇒ x � 3, 5
x2 � 100 � 40x � 4x2 � 25 ⇒ x2 � 8x � 15 � 0x:
x2 � �10 � 2x�2 � 25y
y � 10 � 2xy
Equation 1Equation 2� x2
2x �
�
y2
y �
�
2510
55.
Graphing and you see that there are no points of
intersection. No solution
y3 � 4 � x,y1 � �1 � x2, y2 � ��1 � x2
x � y � 4
x2 � y2 � 1 56.
Graphing and you see that there are no points of
intersection.No solution
y3 � x � 5,y1 � �4 � x2, y2 � ��4 � x2
x � y � 5
x2 � y2 � 4
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558 Chapter 7 Linear Systems and Matrices
57.
Point of intersection: �14, 32�
−5 7
−2
6
�y � 2x � 1y � �x � 2
58.
extraneous
Answer: or �1.25, 1.5��54, 32�
x �54 ⇒ y �
32
x � 0
x�4x � 5� � 0
4x2 � 5x � 0
4x2 � 4x � 1 � x � 1
2x � 1 � �x � 1
�yy
�
�
2x � 1�x � 1
59.
Point of intersection: Approximately �0.287, 1.751�
6−3
−1
5
� y � e�x � 1 ⇒ y � e�x � 1y � ln x � 3 ⇒ y � ln x � 3
60. Graph and
Point of intersection: �1.262, 3.534�
6−30
6
y � exy � 4 � 2 ln x
61.
Substitute for in Equation 2:
Solve for :
Back-substitute:
Answer: �0, 1�, �1, 0�
x � 1 ⇒ y � 0
x � 0 ⇒ y � 1
x2�x � 1� � 0 ⇒ x � 0, 1
x3 � x2 � 0x
x3 � 2x2 � 1 � 1 � x2y
Equation 1Equation 2�y � x3 � 2x2 � 1
y � 1 � x262.
Substitute for in Equation 1:
Solve for
Back-substitute in Equation 2:
Back-substitute in Equation 2:
Back-substitute in Equation 2:
Answer: �0, �1�, �2, 1�, ��1, �5�
y � ���1�2 � 3��1� � 1 � �5
x � �1
y � �22 � 3�2� � 1 � 1
x � 2
y � �02 � 3�0� � 1 � �1
x � 0
0 � x�x � 2��x � 1� ⇒ x � 0, 2, �1
0 � x�x2 � x � 2�
0 � x3 � x2 � 2xx:
�x2 � 3x � 1 � x3 � 2x2 � x � 1
y
Equation 1Equation 2�y
y�
�
x3 � 2x2 � x � 1�x2 � 3x � 1
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Section 7.1 Solving Systems of Equations 559
63.
Solve for in Equation 1:
Substitute for in Equation 2:
Solve for :
Back-substitute
Back-substitute
Two points of intersection: �1
2, 2�, ��4, �
1
4�
x � �4: y �1
�4� �
1
4
x �1
2: y �
1
12� 2
⇒ �2x � 1��x � 4� � 0 ⇒ x �1
2, �4
2x2 � 4 � 7x � 0x
2x � 4�1
x� � 7 � 0y
y �1
xy
Equation 1Equation 2
� xy � 1 � 02x � 4y � 7 � 0
65.
(Answers will vary.)R $1,910,400
x 192 units
1300x � 250,000
9950x � 8650x � 250,000
00
400
3,500,000
C
R
R � C
C � 8650x � 250,000, R � 9950x
66.
R $968,333 x 233,333 units,
1.50x � 350,000
4.15x � 2.65x � 350,000
R � C
500,00000
2,500,000
R
C
C � 2.65x � 350,000, R � 4.15x
67.
In order for the revenue to break even with the cost, 3133 units must be sold, R $10,308.
x 3133 units
10.8241x2 � 65,830.25x � 100,000,000 � 0
10.8241x2 � 65,800x � 100,000,000 � 30.25x
3.29x � 10,000 � 5.5�x
3.29x � 5.5�x � 10,000
R � C
00
5,000
15,000
C
R
R � 3.29xC � 5.5�x � 10,000,
64.
Solve for in Equation 1:
Substitute for in Equation 2:
(x cannot be 0.)
Two points of intersection: �23
, 3�, ��2, �1�
x � �2 ⇒ y � �1
x �23
⇒ y � 3
�3x � 2��x � 2� � 0
3x2 � 4x � 4 � 0
3x � 2�2
x� � 4 � 0
y
y �2
xy
Equation 1Equation 2
� xy � 2 � 03x � 2y � 4 � 0
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560 Chapter 7 Linear Systems and Matrices
68.
Quadratic in
units, R $18,798x 1464
�x12.84x � 7.8�x � 18,500 � 0
12.84x � 7.8�x � 18,500
R � C
5,00000
25,000
R
C
C � 7.8�x � 18,500, R � 12.84x
69. Animated film
Horror film
(a)
(b) For
(c)
(d) The answers are the same.
(e) During week 8 the same number (168) were rented.
N � 24 � 18�8� � 168
x � 8
336 � 42x
360 � 24x � 24 � 18x
x � 8, N � 168
N � 24 � 18x
N � 360 � 24x
Week 1 2 3 4 5 6 7 8 9 10 11 12
Animated 336 312 288 264 240 216 192 168 144 120 96 72
Horror 42 60 78 96 114 132 150 168 186 204 222 240
x
70. Sofia
Paige
(a)
(b) In the table, at
(c)
Answer:
(d) The solutions are the same.
(e) Both girls scored 18 points in game 3.
�3, 18�
Ps � Pp � 18
3 � x
12 � 4x
24 � 2x � 12 � 2x
x � 3.Ps � Pp � 18
Pp � 12 � 2x
Ps � 24 � 2x
1 2 3 4 5 6 7 8 9 10 11 12
Sofia 22 20 18 16 14 12 10 8 6 4 2 0
Paige 14 16 18 20 22 24 26 28 30 32 34 36
x
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Section 7.1 Solving Systems of Equations 561
73.
To make the straight commission offer better, youwould have to sell more than $11,666.67 per week.
x $11,666.67
0.03x � 350
0.06x � 0.03x � 350 74.
For the second offer to be better, you would haveto sell more than $500,000 per year.
500,000 � x
5000 � 0.01x
25,000 � 0.01x � 20,000 � 0.02x
75. (a)
(b)
As increases, decreases and the amount ofinterest decreases.
(c) The curves intersect at Thus, $5000should be invested at 6.5%.
x � 5000.
yx
00
25,000
20,000 0.065x + 0.085y = 1600
x + y = 20,000
� x � y � 20,0000.065x � 0.085y � 1600
76.
(a)
(b) The two graphs intersect at Algebraically:
Since the scales agree when inches.
(c) V is larger using the Scribner Log Rule whenV is larger using the Doyle Log
Rule when Therefore, for largediameters, you would use the Doyle Log Rule.
24.7 ≤ D ≤ 40.5 ≤ D ≤ 24.7.
D 24.72 5 ≤ D ≤ 40,
D 24.72, 3.9
0.21D2 � 6D � 20 � 0
D2 � 8D � 16 � 0.79D2 � 2D � 4
�D � 4�2 � 0.79D2 � 2D � 4
D � 24.72.
4050
750
ScribnerDoyle
V � 0.79D2 � 2D � 4, 5 ≤ D ≤ 40
V � �D � 4�2, 5 ≤ D ≤ 40
71. (a)
(b)
for units
Algebraically,
x �16,000
20.5 780 units
16,000 � 20.5x
35.45x � 16,000 � 55.95x
x 780C � R
020,000
2,000
100,000
C
R
R � 55.95x
C � 35.45x � 16,000 72. (a)
(b)
3759 units to break even.
Algebraically,
x 3759.4 � 3759 units
5000 � 1.33x
5000 � 2.16x � 3.49x
5,00000
20,000
R
C
R � 3.49x
C � 5000 � 2.16x
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562 Chapter 7 Linear Systems and Matrices
77.
(a)
(b) From 1998 to 2010
(c)
Point of intersection: �7.87, 5477.0�
45000 20
6500
P � 76.5t � 4875
M � 47.4t � 5104
Year 1990 1994 1998 2002 2006 2010
Missouri 5104 5294 5483 5673 5862 6052
Tennessee 4875 5181 5487 5793 6099 6405
78. (a)
(b)
(c) when or late 2023.
t 23.9,Tpublic � Tprivate 33,495
−1 50
20,000
Tpublic
Tprivate
Tprivate � 810.5x � 14,109
Tpublic � 55.21x2 � 25.4x � 2516
79.
Dimensions: 6 meters � 9 meters
l � w � 3 � 9
w � 6
2w � 12
l � w � 3 ⇒ �w � 3� � w � 15
2l � 2w � 30 ⇒ l � w � 15 80.
Dimensions: centimeters60 � 80
w � l � 20 � 80 � 20 � 60
l � 80
2l � 160
w � l � 20 ⇒ l � �l � 20� � 140
2l � 2w � 280 ⇒ l � w � 140
81.
If the length is supposed to be greater than thewidth, we have miles and miles.w � 8 l � 12
l � 12, w � 8
l � 8 or l � 12
0 � �l � 8��l � 12�
0 � l2 � 20l � 96
20l � l2 � 96
lw � 96 ⇒ l�20 � l� � 96
2l � 2w � 40 ⇒ l � w � 20 ⇒ w � 20 � l 82.
The dimensions areinches and
hypotenuse inches.� 2b � h � �2
a � �2
a2 � 2
1 �12a2
a 2
a
A �12bh
(d)
Using the positive value, or late 2023.
(e) The results are the same.
x 23.9,
�835.9 ± 1805.2
110.42
x �835.9 ± �835.92 � 4�55.21���11,593�
2�55.21�
55.21x2 � 835.9 � 11,593 � 0
55.21x2 � 25.4x � 2516 � 810.5x � 14,109
(d)
(e) The population of Tennessee surpassed that ofMissouri during 1997.
t � 22929.1 7.87
229 � 29.1t
47.4t � 5104 � 76.5t � 4875
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Section 7.1 Solving Systems of Equations 563
90. (a)
(b) Based on the graphs in part (a) it appears thatfor there are three points of intersection for the graphs of and when b isan even number.
y � xby � bxb > 1,
−3
−30
5
300
y = 4x
y = x4
−5
−2
5
20
y = x2 y = 2x
88. Answers will vary. For example,
(a)
(b)
(c)
2x � y � 3
6x � 3y � 9
2x � y � 2
3x � y � 4
3x � y � 5
3x � y � 3
89. Answers will vary. For example,
2y � x � 4
y � x � 3
91.
2x � 7y � 45 � 0
7y � 49 � �2x � 4
y � 7 � �2
7�x � ��2��
m �5 � 7
5 � ��2�� �
2
7
��2, 7�, �5, 5�
92.
2x � 7y � 22 � 0
7y � 28 � 2x � 6
y � 4 �2
7�x � 3�
m �6 � 4
10 � 3�
2
7
�3, 4�, �10, 6� 93.
The line is horizontal.
y � 3 ⇒ y � 3 � 0
m �3 � 3
10 � 6� 0
�6, 3�, �10, 3� 94.
x � 4
�4, �2�, �4, 5�
83. False. You could solve for first.x
85. The system has no solution if you arrive at a falsestatement, ie. , or you have a quadraticequation with a negative discriminant, which wouldyield imaginary roots.
4 � 8
87. (a) The line intersects the parabola at two points, and
(b) The line intersects at only.
(c) The line does not intersect
(Other answers possible.)
y � x2.y � x � 2
�0, 0�y � x2y � 0
�2, 4�.�0, 0�y � x2y � 2x
84. False. There could be four points of intersection.For example, and y � x2 � 3.x2 � y2 � 4
86. The advantage of the method of substitution overthe graphical method is that substitution gives anexact answer.
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564 Chapter 7 Linear Systems and Matrices
Section 7.2 Systems of Linear Equations in Two Variables
■ You should be able to solve a linear system by the method of elimination.
1. Obtain coefficients for either or that differ only in sign. This is done by multiplying all the terms ofone or both equations by appropriate constants.
2. Add the equations to eliminate one of the variables and then solve for the remaining variable.
3. Use back-substitution into either original equation and solve for the other variable.
4. Check your answer.
■ You should know that for a system of two linear equations, one of the following is true.
(a) There are infinitely many solutions; the lines are identical. The system is consistent.
(b) There is no solution; the lines are parallel. The system is inconsistent.
(c) There is one solution; the lines intersect at one point. The system is consistent.
yx
Vocabulary Check
1. method, elimination 2. equivalent 3. consistent, inconsistent
99. Domain: all
Vertical asymptotes:
Horizontal asymptote: y � 1
x � ±4
x � ±4 100.
Domain: all
Vertical asymptote:
Horizontal asymptote: y � 3
x � 0
x � 0
f �x� � 3 �2x2 �
3x2 � 2x2
95.
0 � 30x � 17y � 18
17y � 102 � 30x � 120
y � 6 �30
17�x � 4�
m �6 � 0
4 � 35
�6175
�30
17
�3
5, 0�, �4, 6� 96.
45x � 29y � 127 � 0
29y �29
2� �45x �
225
2
y �1
2� �
45
29�x �5
2�
�152
�296
� �45
29 m �
8 �12
�73 �
52
�� 7
3, 8�, �5
2,
1
2�
97. Domain: all
Vertical asymptotes:
Horizontal asymptote: y � 0
x � 6
x � 6 98. Domain: all
Vertical asymptote:
Horizontal asymptote: y �23
x � �23
x � �23
101. Domain: all real numbers x
Horizontal asymptote: y � 0
102. Domain: all real numbers x
Horizontal asymptote: y � 0
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Section 7.2 Systems of Linear Equations in Two Variables 565
4.
Multiply Equation 1 by 3:
Add this to Equation 2 to eliminate
Substitute in Equation 1:
Answer: �3, 3�
2�3� � y � 3 ⇒ y � 3x � 3
10x � 30 ⇒ x � 3y:
6x � 3y � 9
−4
−8 10
82x − y = 3
4x + 3y = 21
Equation 1Equation 2�2x
4x�
�
y3y
�
�
321
5.
Multiply Equation 1 by 2:
Add this to Equation 2:
There are no solutions.
0 � 9
2x � 2y � 4
Equation 1Equation 2� x
�2x�
�
y2y
�
�
25
−4
−6 6
4x − y = 2−2x + 2y = 5
6.
Multiply Equation 1 by 2 and add to Equation 2:
There are infinitely many solutions.
All points on line .3x � 2y � 5
0 � 0 � 0
−4
−5 7
4
3x − 2y = 5
−6x + 4y = −10Equation 1Equation 2� 3x
�6x�
�
2y4y
�
�
5�10
1.
Add to eliminate
Substitute in Equation 2:
Answer: �2, 1�
2 � y � 1 ⇒ y � 1x � 2
y: 3x � 6 ⇒ x � 2
Equation 1Equation 2�2x
x�
�
yy
�
�
51
−3
−5 7
5x − y = 1
2x + y = 5
2.
Add to eliminate
Substitute in Equation 1:
Answer: ��2, 1�
x � 3�1� � 1 ⇒ x � �2y � 1
5y � 5 ⇒ y � 1x:
−3
−8 4
5x + 3y = 1 −x + 2y = 4Equation 1
Equation 2� x�x
�
�
3y2y
�
�
14
3.
Multiply Equation 1 by
Add this to Equation 2 to eliminate
Substitute in Equation 1:
Answer: �1, �1�
1 � y � 0 ⇒ y � �1x � 1
y: x � 1
�2: �2x � 2y � 0
Equation 1Equation 2� x
3x�
�
y2y
�
�
01
−4
−6 6
4 x + y = 0
3x + 2y = 1
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566 Chapter 7 Linear Systems and Matrices
7.
Add to eliminate
Substitute in Equation 1:
Answer: �52, 34�
52 � 2y � 4 ⇒ y �
34
x �52
x �52
2x � 5
y:
Equation 1Equation 2�x
x�
�
2y2y
�
�
41
8.
Adding the equations,
Then
Answer: �3, 2�
3 � 2y � 7 ⇒ y � 2
4x � 12 ⇒ x � 3.
Equation 1Equation 2�3x
x�
�
2y2y
�
�
57
9.
Multiply Equation 2 by 3:
Add this to Equation 1 to eliminate
Substitute in Equation 1:
Answer: �3, 4�
6 � 3y � 18 ⇒ y � 4
x � 3
17x � 51 ⇒ x � 3
y:
15x � 3y � 33
Equation 1Equation 2�2x
5x�
�
3yy
�
�
1811
10.
Multiply Equation 1 by
Add this to Equation 2 to eliminate
Substitute in Equation 1:
Answer: �5, 1�
x � 7 � 12 ⇒ x � 5
y � 1
�26y � �26 ⇒ y � 1
x:
�3x � 21y � �36�3:
Equation 1Equation 2� x
3x�
�
7y5y
�
�
1210
11.
Multiply Equation 1 by 2 and Equation 2 by
Add to eliminate
Substitute in Equation 1:
Answer: �4, �1�
3r � 2��1� � 10 ⇒ r � 4
s � �1
�11s � 11 ⇒ s � �1
r:
�6r � 15s � �9
6r � 4s � 20
�3:
Equation 1Equation 2�3r
2r�
�
2s5s
�
�
103
12.
Multiply Equation 1 by
Add this to Equation 2:
Substitute in Equation 1:
Answer: �r, s� � �56, 56�
2r � 4�56� � 5 ⇒ r �
56
s �56
18s � 15 ⇒ s �56
�16r � 32s � �40�8:
Equation 1Equation 2� 2r
16r�
�
4s50s
�
�
555
13.
Multiply Equation 1 by 3 and Equation 2 by
Add to eliminate
Substitute in Equation 2:
Answer: �127 , 18
7 �3u � 5�18
7 � � 18 ⇒ u �127v �
187
�7v � �18 ⇒ v �187u:
�15u � 25v � �90
15u � 18v � 72��5�:
Equation 1Equation 2�5u
3u�
�
6v5v
�
�
2418
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Section 7.2 Systems of Linear Equations in Two Variables 567
22.
Multiply Equation 1 by 5 and Equation 2 by
Adding,
Substituting into Equation 1,
Answer: ��149 , 20
9 �⇒ x � �
149 2x � 5�20
9 � � 8 ⇒ 2x � 8 �1009
9y � 20 ⇒ y �209
�10x � 16y � �20
10x � 25y � 40
�2:
Equation 1Equation 2�2x
5x�
�
5y8y
�
�
810
14.
Multiply Equation 1 by 2 and Equation 2 by 3:
Adding,
Then,
Answer: ��17, 5�
3u � 11�5� � 4 ⇒ u � �17
7v � 35 ⇒ v � 5
� 6u�6u
�
�
22v15v
�
�
827
Equation 1Equation 2� 3u
�2u�
�
11v5v
�
�
49
15.
Multiply Equation 1 by
Add this to Equation 2:
Inconsistent; no solution
0 � �17
�9x � 6y � �20��5�:
Equation 1Equation 2�1.8x
9x�
�
1.2y6y
�
�
43
16.
Multiply Equation 1 by
Add this to Equation 2:
Substituting this value into Equation 2:
Answer: �13031 , 8�
31x � 12�8� � 34 ⇒ 31x � 130 ⇒ x �13031
17y � 136 ⇒ y � 8.
�31x � 29y � 102�10:
Equation 1Equation 2�3.1x
31x�
�
2.9y12y
�
�
�10.234
17.
Matches (b)
One solution; consistent
x � 5, y � 2
�3x � �15
2x � 5�x � 3� � 0
2x � 5y � 0 x � y � 3 ⇒ y � x � 3
18.
Lines coincide.
Matches (a)
Infinite number of solutions; consistent
14x � 12y � 8
�7x � 6y � �4 19.
One solution; consistent
Matches (c)
y � �2, x � �52x � 3y � �4
�2y � 42x � 5y � 0 �
20.
Parallel lines
Inconsistent; no solution
�7x � 6y � �4
7x � 6y � �6 21.
Multiply Equation 1 by 3 and Equation 2 by
Add to eliminate
Substitute into Equation 1:
Answer: �� 635, 43
35�4x � 3�43
35� � 3 ⇒ x � �635
y �4335
�35y � �43 ⇒ y � 4335x:
� 12x�12x
�
�
9y44y
�
�
9�52
�4:
Equation 1Equation 2�4x
3x�
�
3y11y
�
�
313
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568 Chapter 7 Linear Systems and Matrices
23.
Multiply Equation 2 by and add to Equation 1:
Inconsistent; no solution
0 � 8
�2
Equation 1Equation 2�
25 x15 x
�
�
32 y34 y
�
�
4
�224.
Multiply Equation 1 by and add to Equation 2:
There are an infinite number of solutions. All pointson the line 4x � y � 4.
0 � 0
�6
Equation 1Equation 2�
23 x4x
�
�
16 yy
�
�
23
4
25.
Multiply Equation 1 by
Add this to Equation 2:
There are an infinite number of solutions.
The solutions consist of all satisfying or 6x � 8y � 1.34 x � y �
18,�x, y�
0 � 0
�94x � 3y � �
38�3:
Equation 1Equation 2�
34 x94 x
�
�
y
3y
�
�
1838
26.
Multiply Equation 1 by 12 and add to Equation 2:
Inconsistent; no solution
0 � 12
Equation 1
Equation 2� 1
4
�3
x
x
�
�
16
2
y
y
�
�
1
0
27.
Multiply Equation 1 by 12 and Equation 2 by 4:
Add to eliminate
Substitute into Equation 2:
Answer: �5, �2�
2�5� � y � 12 ⇒ y � �2
x � 5
11x � 55 ⇒ x � 5y:
�3x8x
�
�
4y4y
�
�
748
Equation 1
Equation 2�x � 3 4
�y � 1 � 1
3 2x � y � 12
28.
Multiply Equation 1 by 4:
Add to eliminate
Substitute into Equation 2:
Answer: �72, �1
2�
7
2� y � 4 ⇒ y � �
1
2
x �7
2
2x � 7 ⇒ x �7
2y:
�xx
�
�
yy
�
�
34
Equation 1
Equation 2�x � 2 4
�y � 1 � 1
4 x � y � 4
29.
Multiply Equation 1 by 6:
Add this to Equation 2 to eliminate
Substitute in Equation 2:
Answer: �7, 1�
7 � 2y � 5 ⇒ y � 1
x � 7
4x � 28 ⇒ x � 7
y:
3�x � 1� � 2�y � 2� � 24 ⇒ 3x � 2y � 23
Equation 1
Equation 2�x � 1
2x
�
�
y � 2
32y
�
�
4
5
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Section 7.2 Systems of Linear Equations in Two Variables 569
34.
Multiply Equation 1 by 10 and Equation 2 by 2:
Subtract to eliminates
Hence,
Answer: �1, �2�
x � 2 � 0.5�y� � 2 � 0.5��2� � 1
7y � �14 ⇒ y � �2.
x:
�2x2x
�
�
6yy
�
�
�104
Equation 1Equation 2�0.2x
x�
�
0.6y0.5y
�
�
�12
35.
Multiply Equation 1 by 200 and Equation 2 by 300:
Add to eliminate y:
Substitute in Equation 2:
Answer: �9031, �67
31� y � �
6731
0.07�9031� � 0.02y � 0.16
x �9031
31x �9031
31x � 90
�10x21x
�
�
6y6y
�
�
4248
Equation 1Equation 2�0.05x
0.07x�
�
0.03y0.02y
�
�
0.210.16
30.
Multiply Equation 1 by 2:
Add to eliminate
Substitute into Equation 2:
Answer: �72, 32�
7
2� y � 2 ⇒ y �
3
2
x �72
2x � 7 ⇒ x �7
2y:
�xx
�
�
yy
�
�
52
Equation 1
Equation 2�x � 1 2
�y � 2 � 1
2 x � y � 2
31.multiplied by 5
Multiply Equation 1 by
Add this to Equation 2 to eliminate
The solution set consists of all points lying on the line
All points on the line
Let
Answer: where is any real number. a�a, 56a �12�,
x � a, then y �56a �
12.
5x � 6y � 3.
10x � 12y � 6.
0 � 0
x:
�10x � 12y � �6
��4�:
Equation 1Equation 2�2.5x
10x�
�
3y12y
�
�
1.56
32.
There are an infinite number of solutions. All points on the line 7x � 8y � 6.
�Divide by 0.9��Divide by 0.8��7x
7x�
�
8y8y
�
�
66
Equation 1Equation 2�6.3x
5.6x�
�
7.2y6.4y
�
�
5.44.8
33.
Multiply Equation 1 by 40 and Equation 2 by 50:
Adding the equation eliminates
Substitute into Equation 1:
Answer: �101, 96�
8�101� � 20y � �1112 ⇒ y � 96
x � 101
23x � 2323 ⇒ x � 101
y:
� 8x15x
�
�
20y20y
�
�
�11123435
Equation 1Equation 2�0.2x
0.3x�
�
0.5y0.4y
�
�
�27.868.7
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570 Chapter 7 Linear Systems and Matrices
36.
Multiply Equation 1 by 10 and Equation 2 by 2:
Subtract to eliminates
Hence,
Answer: ��3, 1�
x � �2.5 � 0.5�y� � �2.5 � 0.5�1� � �3
3y � 3 ⇒ y � 1.
x:
�2x2x
�
�
4yy
�
�
�2�5
Equation 1Equation 2�0.2x
x�
�
0.4y0.5y
�
�
�0.2�2.5
37. Let and .
Multiply Equation 1 by 4:
Subtract to eliminates
Hence,
Answer: ��1, 1�
x �1X
� �1, y �1Y
� 1
X � 2 � 3Y � 2 � 3�1� � �1
13Y � 13 ⇒ Y � 1.
X:
�4X4X
�
�
12YY
�
�
8�5
Equation 1Equation 2� X
4X�
�
3YY
�
�
2�5
Y �1y
X �1x
39. Let and .
Multiply Equation 1 by 2:
Adding the equations eliminates
Hence,
Answer: �1, 12�
x �1X
� 1, y �1Y
�12
2Y � 5 � X � 4 ⇒ Y � 2.
5X � 5 ⇒ X � 1.
Y:
�2X3X
�
�
4Y4Y
�
�
10�5
Equation 1Equation 2� X
3X�
�
2Y4Y
�
�
5�5
Y �1y
X �1x
38. Let and .
Multiply Equation 1 by 2:
Subtract to eliminates
Hence,
Answer: �2, 1�
x �1X
� 2, y �1Y
� 1
X �12
.
Y � 1X:
�4X4X
�
�
2Y3Y
�
�
0�1
Equation 1Equation 2�2X
4X�
�
Y3Y
�
�
0�1
Y �1y
X �1x
40. Let and .
Adding the equations,
Hence,
Answer: �12, �1�
x �1X
�12
, y �1Y
� �1
Y � 11 � 6X � 11 � 12 � �1.
8X � 16 ⇒ X � 2.
Equation 1Equation 2�2X
6X�
�
YY
�
�
511
Y �1y
X �1x
41.
The system is consistent. There is one solution,
−5
−6
13
6
�5, 2�.
�2x � 5y � 0 ⇒ y �25 x
x � y � 3 ⇒ y � x � 3©
Hou
ghto
n M
ifflin
Com
pany
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right
s re
serv
ed.
Section 7.2 Systems of Linear Equations in Two Variables 571
48.
Answer: �3, �2�
−7
−8
11
4
� 4y � �8
7x � 2y � 25
⇒ ⇒
y �
y �
�2
�7x � 25��249.
Answer: �6, 5�
−3
−2
15
10
�32 x
�2x�
�
15 y3y
�
�
8 ⇒ y � 5�32x � 8�
3 ⇒ y �13�3 � 2x�
44.
The system is consistent. The solution set consists of all points on the line or 4x � 6y � 9 � 0.y �
23 x �
32,
−5
−4
7
4� 4x � 6y � 9163 x � 8y � 12
⇒ ⇒
y � �4x � 9��6
y � �163 x � 12��8
�
�
23 x �
32
23 x �
32
45.
The system is consistent. The solution set consists ofall points on the line or 8x � 14y � 5.y �
47 x �
514,
−6
−4
6
4
�8x2x
�
�
14y3.5y
�
�
51.25
⇒ ⇒
yy
�
�
(8x(2x
�
�
5)1.25)
�14 �47 x �
514
�3.5 �47 x �
514
46.
Inconsistent; no solution
−9
−4
9
8
� y �17�x � 3�
y � 5 �17x
47.
Answer: �236 , 7� � �3.833, 7�
903
9
�6x �
6y y
�
�
4216
⇒ ⇒
yy
�
�
76x � 16
43.
The lines are parallel. The system is inconsistent.
−9
−6
9
6
� 35 x � y � 3 ⇒ y �35 x � 3
�3x � 5y � 9 ⇒ y �15 �3x � 9� �
35 x �
95
42.
The system is consistent. There is one solution,
−8
−5
10
7
�1.8, 1.4�
�2xx
�
�
y2y
�
�
5�1
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572 Chapter 7 Linear Systems and Matrices
50.
Answer: �8, 6�
−4
−2
14
10
�x � 6y � 28 ⇒ y � �x � 28��6 � x6
�143
34
x �52
y � �9 ⇒ y �25 �3
4x � 9� �
310
x �185� 51.
Answer: �1, �0.667�
−4
−6 6
4
5x � 3y � 7 ⇒ y �13
�5x � 7�
13
x � y � �13
⇒ y � �13
�13
x
52.
Answer: ��0.4, 2�
−4 2
−1
3
2x �35 y �
25 ⇒ y �
53��2x �
25�
5x � y � �4 ⇒ y � 5x � 4 53.
Answer: ��4, 5�
−12
−2
6
10
�0.5x6x
�
�
2.2y0.4y
�
�
9 ⇒ y � 1�2.2�9 � 0.5x� �22 ⇒ y � 1�0.4��22 � 6x�
54.
Answer: ��1, �4�
−6
−7
6
1
�2.4x � 3.8y4x � 0.2y
�
�
�17.6�3.2
⇒ ⇒
y �
y �
��2.4x � 17.6��3.8�4x � 3.2��0.2
55.
Multiply Equation 2 by 5:
Add this to Equation 1:
Back-substitute into Equation 2:
Answer: �4, 1�
2�4� � y � 9 ⇒ y � 1
x � 4
13x � 52 ⇒ x � 4
10x � 5y � 45
Equation 1Equation 2�3x
2x�
�
5yy
�
�
79
56.
Subtract Equation 2 from Equation 1 to eliminate
Substitute in Equation 1:
Answer: ��2, 5�
���2� � 3y � 17 ⇒ y � 5
x � �2
�5x � 10 ⇒ x � �2
y:
Equation 1Equation 2��x
4x�
�
3y3y
�
�
177
57.
The lines intersect at
��53, �11
3 � � ��1.667, �3.667�
−10 8
−8
4
�yy
�
�
4x�5x
�
�
312
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Section 7.2 Systems of Linear Equations in Two Variables 573
66. There are infinitely many systems that have thesolution One possible system:
6��23� � 1��10� � 6 ⇒ 6x � y � 6
3��23� � 1��10� � �12 ⇒ 3x � y � �12
��23, �10�.
67.
Answer: �80, 10�
p � $10
x � 80 units
50 � 0.625x
50 � 0.5x � 0.125x
Demand � Supply
58.
Answer: �1310, 23
10�y �
1310 � 1 �
2310
x �1310
10x � 13
7x � 3�x � 1� � 16
�7x � 3yy
�
�
16 x � 1
59.
Adding the equations,
Back-substituting,
Answer: �6, �3�
�5y � 15 ⇒ y � �3
x � 5y � 6 � 5y � 21 ⇒ 7x � 42 ⇒ x � 6.
� x6x
�
�
5y5y
�
�
2121
60.
Answer: ��23, 61�
x � �23 ⇒ y � �3��23� � 8 � 61
�x � 23
�3x � 8 � 15 � 2x
�yy
�
�
�3x � 815 � 2x
61.
Substitute into Equation 1,
Back-substituting,
Answer: �436 , 25
6 �y � x � 3 �
436 � 3 �
256
⇒ x �436
6x � 43 ⇒ �2x � 8�x � 3� � 19
Equation 1Equation 2��2x � 8y � 19
y � x � 3
62.
Multiply Equation 1 by 5 and Equation 2 by 4.
Adding,
Then,
Answer: �3, 2�
4x � 3�2� � 6 ⇒ x � 3
13y � 26 ⇒ y � 2
�20x � 28y � �4
20x � 15y � 30
Equation 1Equation 2� 4x
�5x�
�
3y7y
�
�
6�1
64. There are infinitely many systems that have as the solution. For example,
� x2x
�
�
yy
�
�
�110
�3, �4�
63. There are infinitely many systems that have thesolution One possible system is
� x�x
�
�
yy
�
�
88
�0, 8�.
65. There are infinitely many systems that have thesolution One possible system:
3 � 4�52� � �7 ⇒ x � 4y � �7
2�3� � 2�52� � 11 ⇒ 2x � 2y � 11
�3, 52�.
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574 Chapter 7 Linear Systems and Matrices
68.
Answer: �500, 75�
p � 75
x � 500
0.15x � 75
25 � 0.1x � 100 � 0.05x
Supply � Demand 69.
Answer: �2,000,000, 100�
p � $100.00
x � 2,000,000 units
60 � 0.00003x
140 � 0.00002x � 80 � 0.00001x
Demand � Supply
70.
Answer: �250,000, 350�
p � 350
x � 250,000
0.0007x � 175
225 � 0.0005x � 400 � 0.0002x
Supply � Demand 71. Let the ground speed and the wind speed.
Substituting in Equation 2:
Answer: x � 550 mph, y � 50 mph
y � 50
550 � y � 600
x � 550
Equation 1
Equation 2�3.6�x � y� � 1800
3�x � y� � 1800
y �x �
x � 550
2x � 1100
x � y � 600
x � y � 500
72. Let the speed of the plane that leaves first and the speed of the plane that leaves second.
Answer: First plane: 880 kilometers per hour; Second plane: 960 kilometers per hour
x � 880
960 � x � 80
y � 960
72 y � 3360
��2x2x
�
�
2y32 y
�
�
1603200
Equation 1Equation 2� y
2x�
�
x32 y
�
�
803200
y �x �
73. (a)
(b) Multiply Equation 1 by 5:
Subtracting eliminates A:
Hence,
650 adult tickets and 525 child tickets
A � 1175 � 525 � 650.
C � 525
1.5C � 787.5
5A � 3.5C � 5087.5
5A � 5C � 5875
Equation 1
Equation 2� A � C � 1175
5A � 3.5C � 5087.5
(c)
Let
Point of intersection: �A, C� � �650, 525�
C � y2 �1
3.5�5087.5 � 5x�
C � y1 � 1175 � x
120000
1200
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Section 7.2 Systems of Linear Equations in Two Variables 575
78. Let number of pairs of $75.50 shoes.
Let number of pairs of $89.95 shoes.
Solving for y in the first equation and substitutinginto the second equation gives:
170 of $75.50 shoes, 80 of $89.95 shoes
y � 250 � x � 80
x � 170
2456.5 � 14.45x
75.50x � 89.95250 � x � 20,031
� x75.50x
�
�
y89.95y
�
�
25020,031
y �
x �
74. (a)
(b) Multiply Equation 1 by 4 and Equation 2 by 3:
Subtracting,
Hence,
(c) Let
Point of intersection: �C, S� � �1.25, 0.75�
C � y2 � �10.5 � 4S��6
C � y1 � �8.5 � 3S��5
C � $1.25, S � $0.75
S � �8.50 � 5C��3 � 0.75
2C � 2.5 ⇒ C � 1.25:
18C � 12S � 31.5
20C � 12S � 34
Equation 1
Equation 2�5C � 3S � 8.50
6C � 4S � 10.50
76. Family Dollar
Dollar General
(a) Using a graphing utility, the solution is
(b) In 1994, both stores had sales of $905.59 million.
�S, T � � �905.59, 4.38�.
S � 691.48t � �2122.8
S � 440.36t � �1023.0
75. Let number of oranges and let number of grapefruit.
Solving for R in Equation 1:
Substituting into Equation 2:
Hence,
9 oranges and 7 grapefruit
R � 16 � 9 � 7.
M � 9
0.9 � 0.1M
0.95M � 16.8 � 1.05M � 15.9
0.95M � 1.05�16 � M� � 15.9
R � 16 � M.
Equation 1Equation 2� M
0.95M�
�
R1.05R
�
�
16 15.90
R �M �
77. Let number of movies and let number of videos.
Solve for m in Equation 1:
Substitute for m Equation 2:
185 movies and 125 videos
m � 310 � 125 � 185
v � 125
62.5 � 0.5v
3�310 � v� � 2.5v � 867.5
m � 310 � v.
Equation 1Equation 2� m
3m�
�
v2.5v
�
�
310 867.5
v �m �
79.
Least squares regression line: y � 0.97x � 2.10
b � 2.10
a � 0.97
10a � 9.7
� 5b10b
�
�
10a30a
�
�
20.250.1
⇒ ⇒
�10b10b
�
�
20a30a
�
�
�40.450.1
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576 Chapter 7 Linear Systems and Matrices
80.
Least squares regression line: y � 0.22x � 1.9
b � 1.9
5b � 10�0.22� � 11.7
a � 0.22
10a � 2.2
� 5b � 10a10b � 30a
� 11.7 ⇒ � 25.6 ⇒
�10b � 20a10b � 30a
�
�
�23.425.6
81.
Multiply Equation 1 by
Adding,
y � �2.5x � 5.54
b � �2.7 � 10�2.5���5 � 5.54
a � �2.5
10a � �25.0
10b � 30a � �19.6
�10b � 20a � �5.4
�2:
Equation 1Equation 2� 5b
10b�
�
10a30a
�
�
2.7�19.6
82.
Multiply Equation 1 by
Adding,
y � �1.48x � 2.5
b � ��2.3 � 10�1.48���5 � 2.5
a � �1.48
10a � �14.8
10b � 30a � �19.4
�10b � 20a � 4.6�2:
Equation 1Equation 2� 5b
10b�
�
10a30a
�
�
�2.3�19.4
83. (a)
Adding,
Thus,
(b) Using a graphing utility, you obtain y � 14x � 19.
y � 14x � 19
a � 14, b � 19. ⇒ �5a � �70
�4b7b
�
�
7a13.5a
�
�
174322
⇒ ⇒
28b�28b
�
�
49a54a
�
�
1218�1288
(c)
(d) If bushels per acre.y � 14�1.6� � 19 � 41.4
�160 pounds�acre�,x � 1.6,
300
60
84. (a)
Solving this system, you obtain and
(b) y � �44.21x � 89.53
y � �44.21x � 89.53
b � 89.53a � �44.21
�3.00b3.70b
�
�
3.70a4.69a
�
�
105123.9
(c)
(d) For x � 1.75, y � 12
200
50
85. True. A consistent linear system has either one solution or an infinite number of solutions.
86. True
©H
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Section 7.2 Systems of Linear Equations in Two Variables 577
93.
Multiply the first equation by the second byand add the equations:
v � �tan x.
Hence, v cos x � �u sin x � �sin x
u � 1
u sin2 x � u cos2 x � sec x � cos x
cos x,sin x,
u cos x � v sin x � sec x
u sin x � v cos x � 0 94.
Multiply the first equation by the secondby and add the equations:
u � �12.
Hence, u cos 2x �12 cot 2x � sin 2x � 0
v �12 cot 2x
2v � cot 2x
2v sin2 2x � 2v cos2 2x � csc 2x � cos 2x
cos 2x,2 sin 2x,
u��2 sin 2x� � v�2 cos 2x� � csc 2x
u cos 2x � v sin 2x � 0
88.
Multiply Equation 2 by
Add this to Equation 1 to eliminate
Substitute in Equation 1:
Answer:
The lines are not parallel. The scale on the axes must be changed to see the point of intersection.
�300, 315�
21�300� � 20y � 0 ⇒ y � 315x � 300
� 23x � �200 ⇒ x � 300y:
� 653 x � 20y � �200�� 5
3�:
Equation 1Equation 2�21x
13x�
�
20y12y
�
�
0120
89. No, it is not possible for a consistent system oflinear equations to have exactly two solutions.Either the lines will intersect once or they willcoincide and then the system would haveinfinite solutions.
90. (a)
Subtract Equation 2 from Equation 1:
System is inconsistent no solution
(b)
Multiply Equation 1 by
Add this to Equation 2: (dependent)
The system has an infinite number of solutions.
0 � 0
�2x � 2y � �6��2�:
Equation 1Equation 2� x
2x�
�
y2y
�
�
36
⇒
0 � �10
Equation 1Equation 2�x
x�
�
yy
�
�
1020
92.
Multiply Equation 1 by
Add this to Equation 2:
The system is inconsistent if k � �2.
�k � 2�y � 13
ky � 2y � 13
10x � 2y � 423:
Equation 1Equation 2� 15x
�10x�
�
3yky
�
�
69
91.
Multiply Equation 2 by
Add this to Equation 1:
The system is inconsistent if
This occurs when Note that for the two original equations represent parallel lines.
k � �4,k � �4.
�8y � 2ky � 0.
�8y � 2ky � �35
�2: �4x � 2ky � �32
Equation 1Equation 2�4x
2x�
�
8yky
�
�
�316
87.
Subtract Equation 2 from Equation 1 to eliminate x:
Substitute into Equation 1:
Answer:
The lines are not parallel. The scale on the axes must be changed to see the point of intersection.
�39,600, 398�
100�398� � x � 200 ⇒ x � 39,600y � 398
y � 398
Equation 1Equation 2�100y
99y�
�
xx
�
�
200�198
©H
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578 Chapter 7 Linear Systems and Matrices
Section 7.3 Multivariable Linear Systems
■ You should know the operations that lead to equivalent systems of linear equations:
(a) Interchange any two equations.
(b) Multiply all terms of an equation by a nonzero constant.
(c) Replace an equation by the sum of itself and a constant multiple of any other equation in the system.
■ You should be able to use the method of elimination.
95.
−10 −9 −8 −7 −6 −5
3−
x
22
x ≤ �446 � �
223
�6x ≥ 44
�11 � 6x ≥ 33 97.
42 6 8 10 12 14 16 180−2x
�2 < x < 18
�10 < x � 8 < 10
�x � 8� < 10
99.
Critical numbers:Testing the three intervals,
43210−1−2−3−4−5−6x
72
�5 < x < 72.
72, �5.
�2x � 7��x � 5� < 0
2x2 � 3x � 35 < 0
101. ln x � ln 6 � ln 6x
103. log9 12 � log9 x � log9 12x
96.
1 2 3 4 5 6
3 3
x
4 16
43 ≤ x < 163
4 ≤ 3x < 16
�6 ≤ 3x � 10 < 6
98. is true for all x,since
x43210−1−2−3−4
�x � 10� ≥ 0.�x � 10� ≥ �3 100.
Critical numbers:Checking the three intervals,you obtain and
210−1−2−3−4−5−6
x
x > 0.x < �4
0, �4.
3x�x � 4� > 0
3x2 � 12x > 0
102.
� ln x
�x � 3�5
ln x � 5 ln�x � 3� � ln x � ln�x � 3�5
104.
� log6�3x�1�4
14
log6 3 �14
log6 x �14
log6�3x�
105.
� ln� x2
x � 2� 2 ln x � ln�x � 2� � ln x2 � ln�x � 2� 106.
� ln� �x2 � 4
x 12
ln�x2 � 4� � ln x � ln�x2 � 4�1�2 � ln x
107. Answers will vary.
©H
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Section 7.3 Multivariable Linear Systems 579
1. (a) Yes
No
No
No, is not a solution.
(c) No
Yes
No
No, is not a solution.��4, 1, 2�
5�1� � 2�2� �?
8
2��4� � 3�2� �?
�14
3��4� � 1 � 2 �?
1
�3, 5, �3�
5�5� � 2��3� �?
8
2�3� � 3��3� �?
�14
3�3� � 5 � ��3� �?
1 (b) Yes
Yes
Yes
Yes, is a solution.
(d) No
No
Yes
No, is not a solution.�1, 0, 4�
5�0� � 2�4� �?
8
2�1� � 3�4� �?
�14
3�1� � 0 � 4 �?
1
��1, 0, 4�
5�0� � 2�4� �?
8
2��1� � 3�4� �?
�14
3��1� � �0� � 4 �?
1
2. (a) Yes
No
No
No, is not a solution.
(c) Yes
Yes
Yes
Yes, is a solution.�1, 3, �2�
2�1� � 3�3� � 7��2� �?
�21
5�1� � 3 � 2��2� �?
�2
3�1� � 4�3� � ��2� �?
17
�1, 5, 6�
2�1� � 3�5� � 7�6� �?
�21
5�1� � 5 � 2�6� �?
�2
3�1� � 4�5� � 6 �?
17 (b) No
Yes
No
No, is not a solution.
(d) No
No
Yes
No, is not a solution.�0, 7, 0�
2�0� � 3�7� � 7�0� �?
�21
5�0� � 7 � 2�0� �?
�2
3�0� � 4�7� � 0 �?
17
��2, �4, 2�
2��2� � 3��4� � 7�2� �?
�21
5��2� � ��4� � 2�2� �?
�2
3��2� � 4��4� � 2 �?
17
3. (a) Yes
No
No
No, is not a solution.
(c) Yes
Yes
Yes
Yes, is a solution.��12, 34, �5
4� 3��1
2� �34 �
?�
94
�8��12� � 6�3
4� �54 �
?�
74
4��12� �
34 � ��5
4� �?
0
�0, 1, 1�
3�0� � 1 �?
�94
�8�0� � 6�1� � 1 �?
�74
4�0� � 1 � 1 �?
0 (b) No
No
No
No, is not a solution.
(d) Yes
No
No
No, is not a solution.��12, 2, 0�
3��12� � 2 �
?�
94
�8��12� � 6�2� � 0 �
?�
72
4��12� � 2 � 0 �
?0
��32, 54, �5
4� 3��3
2� � �54� �
?�
94
�8��32� � 6�5
4� � ��54� �
?�
74
4��32� �
54 � ��5
4� �?
0
Vocabulary Check
1. row-echelon 2. ordered triple 3. Gaussian
4. independent, dependent 5. nonsquare 6. three-dimensional
7. partial fraction decomposition
©H
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580 Chapter 7 Linear Systems and Matrices
4. (a)
is a solution.
(b)
is not a solution.��332 , �10, 10�
4��332 � � 7��10� �
? 6 No
� 10 � 10 �?
0 Yes
�4��332 � � ��10� � 8�10� �? �6 No
��2, �2, 2�
4��2� � 7��2� �?
6 Yes
� 2 � 2 �?
0 Yes
�4��2� � ��2� � 8�2� �? �6 Yes (c)
is not a solution.
(d)
is a solution.��112 , �4, 4�
4��112 � � 7��4� �
? 6 Yes
�4 � 4 �?
0 Yes
�4��112 � � ��4� � 8�4� �? �6 Yes
�18, �1
2, 12� 4�1
8� � 7��12� �
? 6 No
�12 �
12 �
? 0 Yes
�4�18� � ��1
2� � 8�12� �
? �6 No
5.
Back-substitute into Equation 2:
Back-substitute and into Equation 1:
Answer: �2, �2, 2�
x � 2
2x � 4
2x � ��2� � 5�2� � 16
y � �2z � 2
y � �2
y � 2�2� � 2
z � 2
Equation 1Equation 2Equation 3
2x � yy
�
�
5z2zz
�
�
�
1622
7.
Back-substitute into Equation 2:
Back-substitute and into Equation 1:
Answer: �3, 10, 2�
x � 3
2x � 6
2x � 10 � 3(2) � 10
z � 2y � 10
y � 10
y � 2 � 12
z � 2
Equation 1Equation 2Equation 3
2x � yy
�
�
3zzz
�
�
�
10122
6.
Back-substitute into Equation 2:
Back-substitute and intoEquation 1:
Answer: ��8, �11�3, �2� x � �8
4x � �32
4x � 3��11�3� � 2��2� � �17
y � �11�3z � �2
y � �11�3
6y � �22
6y � 5��2� � �12
z � �2
Equation 1Equation 2Equation 3
4x � 3y6y
�
�
2z5zz
�
�
�
�17�12�2
8.
Answer: �17, �11, �3�
x � �11 � 2��3� � 22 � 17
3y � 8��3� � 9 � �33 ⇒ y � �11
z � �3
x � y3y
�
�
2z8zz
�
�
�
22�9�3
©H
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.
Section 7.3 Multivariable Linear Systems 581
9.
Back-substitute into Equation 2:
Back-substitute and into Equation 1:
Answer: �12, �2, 2�
x �12
4x � 2
4x � 2��2� � 2 � 8
z � 2y � �2
y � �2
�y � 2 � 4
z � 2
Equation 1Equation 2Equation 3
4x �
�
2yy
�
�
zzz
�
�
�
842
10.
Back-substitute in Equation 2:
Back-substitute in Equation 1:
Answer: ��2, � 103 , �4�
5x � 8��4� � 22 ⇒ x � �2
z � �4
3y � 5��4� � 10 ⇒ y � � 103
z � �4
5x 3y
�
�
8z5zz
�
�
�
2210
�4
11.
Add Equation 1 to Equation 2.
New Equation 2
This is the first step in putting the system inrow-echelon form.
x
2x
� 2yy
�
�
�
3z2z3z
�
�
�
590
y � 2z � 9
Equation 1Equation 2Equation 3
x�x2x
�
�
2y3y
�
�
�
3z5z3z
�
�
�
540
12.
Add times Equation 1 to Equation 3.
This is a step in putting the system in row-echelon form.
4y � 9z � �10
�2
x�x2x
�
�
2y3y
�
�
�
3z5z3z
�
�
�
540
13.
Add times Equation 1 to Equation 2.
New Equation 2
This is the first step in putting the system in row-echelon form.
3y � z � �2
�2
Equation 1Equation 2Equation 3
x2x3x
�
�
�
2yyy
�
�
�
z3z4z
�
�
�
101
14.
Add times Equation 1 to Equation 3.
New Equation 3
This eliminates the term 3 in Equation 3.x
�10y � 13z � �5
�3
Equation 1Equation 2Equation 3
x�x3x
�
�
�
2y4y4y
�
�
�
4zzz
�
�
�
2�2
1
15.
Answer: �1, 2, 3�
x � 2 � 3 � 6 ⇒ x � 1
�3y � 3 � �9 ⇒ y � 2
�3z � �9 ⇒ z � 3
��1� Eq. 2 � Eq. 3x �
�
y3y
�
�
�
zz
3z
�
�
�
6�9�9
��2� Eq. 1 � Eq. 2��3� Eq. 1 � Eq. 3
x �
�
�
y3y3y
�
�
�
zz
4z
�
�
�
6�9
�18
Equation 1Equation 2Equation 3
x2x3x
�
�
yy
�
�
�
zzz
�
�
�
630
©H
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.
582 Chapter 7 Linear Systems and Matrices
16.
Answer: �0, 4, �2�
x � 4 � 2 � 2 ⇒ x � 0
12y � 9��2� � 30 ⇒ y � 4
�7z � 14 ⇒ z � �2
Eq. 2 � Eq. 3x � y
12y�
�
z9z
�7z
�
�
�
23014
3 Eq. 24 Eq. 3
x � y12y
�12y
�
�
�
z9z
16z
�
�
�
230
�16
Eq. 1 � Eq. 2�4 Eq. 1 � Eq. 3
x � y4y
�3y
�
�
�
z3z4z
�
�
�
210
�4
Equation 1Equation 2Equation 3
x�x4x
�
�
�
y3yy
�
�
z2z
�
�
�
284
17.
Answer: ��4, 8, 5�
x � 5 � 1 ⇒ x � �4
3y � 5�5� � �1 ⇒ y � 8
��1� Eq. 2 � Eq. 3x
3y
�
�
z5zz
�
�
�
1�1
5
��5� Eq. 1 � Eq. 2x
3y3y
�
�
�
z5z4z
�
�
�
1�1
4
�12� Eq. 1 x
5x � 3y3y
�
�
z
4z
�
�
�
144
Equation 1Equation 2Equation 3
2x5x � 3y
3y
�
�
2z
4z
�
�
�
244
18.
Answer: �6, �3, 0�
x � 6⇒x � ��3� � 3
y � �3⇒�2y � 3�0� � 6
z � 0⇒�5z � 0
3 Eq. 2 � Eq. 3x �
�
y2y �
�
3z5z
�
�
�
360
Interchangethe equations.x �
�
y2y6y
�
�
3z4z
�
�
�
36
�18
��2� Eq. 1 � Eq. 3x �
�
y6y2y
�
�
4z3z
�
�
�
3�18
6
13 �New Eq. 1� x
2x
� y6y �
�
4z3z
�
�
�
3�18
12
Interchange equations1 and 2.3x
2x
� 3y6y �
�
4z3z
�
�
�
9�18
12
Equation 1Equation 2Equation 3
3x2x
�
6y3y
�
�
4z
3z
�
�
�
�189
12
19.
Answer: �2, �3, �2�
x � ��3� � ��2� � �3 ⇒ x � 2
⇒ z � �2
⇒ �7z � 14
y � �3 ⇒ �3��3� � 7z � 23
��2� Eq. 1 � Eq. 2��4� Eq. 1 � Eq. 3
x � y�5y�3y
�
�
z
7z
�
�
�
�31523
x2x4x
�
�
�
y3yy
�
�
�
z2z3z
�
�
�
�39
11
4x2xx
�
�
�
y3yy
�
�
�
3z2zz
�
�
�
119
�3
Equation 1Equation 2Equation 3
Interchange equations1 and 3
©H
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.
Section 7.3 Multivariable Linear Systems 583
20.
Answer: �12, � 3
2, 1�2x � 4�� 3
2� � 1 � �4 ⇒ x �12
�40y � 25�1� � 85 ⇒ y � � 32
�29z � �29 ⇒ z � 1
�Eq. 2 � Eq. 32x � 4y
�40y�
�
z25z
�29z
�
�
�
�485
�29
5 Eq. 24 Eq. 3
2x � 4y�40y�40y
�
�
�
z25z4z
�
�
�
�48556
�Eq. 1 � Eq. 2�2 Eq. � Eq. 3
2x � 4y�8y
�10y
�
�
�
z5zz
�
�
�
�41714
Equation 1Equation 2Equation 3
2x2x4x
�
�
�
4y4y2y
�
�
�
z6zz
�
�
�
�4136
21.
Inconsistent; no solution.
�Eq. 2 � Eq. 3x �
�
y5y
�
�
2z10z
0
�
�
�
3�810
�3 Eq. 1 � Eq. 2�2 Eq. 1 � Eq. 3
x �
�
�
y5y5y
�
�
�
2z10z10z
�
�
�
3�8
2
Interchangethe equations. x
3x2x
�
�
�
y2y3y
�
�
�
2z4z6z
�
�
�
318
22.
Inconsistent; no solution.
�12 Eq. 2 � Eq. 3
x � 11y52y
�
�
4z18z
0
�
�
�
3�12
7
�5 Eq. 1 � Eq. 2�2 Eq. 1 � Eq. 3
x � 11y52y26y
�
�
�
4z18z9z
�
�
�
3�12
1
Interchange rows x5x2x
�
�
�
11y3y4y
�
�
�
4z2zz
�
�
�
337
Equation 1Equation 2Equation 3
5x2xx
�
�
�
3y4y
11y
�
�
�
2zz
4z
�
�
�
373
23.
Answer: �1, �32, 12�
x � 3��32� � 7�1
2� � 2 ⇒ x � 1
84y � 180� 12� � �36 ⇒ y � �
32
2z � 1 ⇒ z �12
�Eq. 2 � Eq. 3x � 3y
84y�
�
7z180z
2z
�
�
�
2�36
1
6 Eq. 23.5 Eq. 3
x � 3y84y84y
�
�
�
7z180z182z
�
�
�
2�36�35
�3 Eq. 1 � Eq. 2�5 Eq. 1 � Eq. 3
x � 3y14y24y
�
�
�
7z30z52z
�
�
�
2�6
�10
�Eq. 3 � Eq. 1 x3x5x
�
�
�
3y5y9y
�
�
�
7z9z
17z
�
�
�
200
2 Eq. 16x3x5x
�
�
�
6y5y9y
�
�
�
10z9z
17z
�
�
�
200
3x3x5x
�
�
�
3y5y9y
�
�
�
5z9z
17z
�
�
�
100
©H
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Miff
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ny. A
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rese
rved
.
584 Chapter 7 Linear Systems and Matrices
24.
Answer: � 310, 25, 0�
2x �25 � 3�0� � 1 ⇒ x �
310
5y � 5�0� � 2 ⇒ y � 25
4z � 0 ⇒ z � 0
�Eq. 2 � Eq. 32x � y
5y�
�
3z5z4z
�
�
�
120
�Eq. 1 � Eq. 2�3 Eq. 1 � Eq. 3
2x � y5y5y
�
�
�
3z5z9z
�
�
�
122
Equation 1Equation 2Equation 3
2x2x6x
�
�
�
y6y8y
�
�
�
3z8z
18z
�
�
�
135
25.
Answer: ��3a � 10, 5a � 7, a�
x � �3a � 10
y � 5a � 7
Let z � a, then:
�2 Eq. 2 � Eq. 1x y
�
�
3z5z
�
�
10�7
�13 Eq. 2x � 2y
y�
�
7z5z
�
�
�4�7
Eq. 2 � Eq. 3x �
�
2y3y
�
�
7z15z
0
�
�
�
�4210
�2 Eq. 1 � Eq. 2�3 Eq. 1 � Eq. 3
x �
�
2y3y3y
�
�
�
7z15z15z
�
�
�
�421
�21
Equation 1Equation 2Equation 3
x2x3x
�
�
�
2yy
9y
�
�
�
7zz
36z
�
�
�
�413
�33
26.
Answer: �� 12a �
52, 4a � 1, a�
x � � 12a �
52
y � 4a � 1
z � a
�Eq. 2 � Eq. 12x y
�
�
z4z
�
�
5�1
�12 Eq. 2
2 Eq. 2 � Eq. 32x � y
y�
�
3z4z0
�
�
�
4�1
0
�2 Eq. 1 � Eq. 2Eq. 1 � Eq. 3
2x � y�2y
4y
�
�
�
3z8z
16z
�
�
�
42
�4
Equation 1Equation 2Equation 3
2x4x
�2x
�
�
y
3y
�
�
�
3z2z
13z
�
�
�
410
�8
27.
Answer: �1, �1, 2�
x � y � 2z � 6 ⇒ x � 6 � ��1� � 2�2� � 1
y � z � �3 ⇒ y � 2 � 3 � �1
��2� Eq. 2 � Eq. 3x � yy
�
�
2zzz
�
�
�
6�3
2
�13� Eq. 2 x � y
y2y
�
�
�
2zzz
�
�
�
6�3�4
��2� Eq. 1 � Eq. 2��1� Eq. 1 � Eq. 3x � y
3y2y
�
�
�
2z3zz
�
�
�
6�9�4
Equation 1Equation 2Equation 3
x2xx
�
�
�
yyy
�
�
�
2zzz
�
�
�
632
28.
Answer: �0, 1, �2�
x � y � z � 3 ⇒ x � 3 � 1 � 2 � 0
�3y � z � �5 ⇒ �3y � 2 � 5 � �3 ⇒ y � 1
z � �2
��2� Eq. 1 � Eq. 2��1� Eq. 1 � Eq. 3x � y
�3y�
�
zz
�z
�
�
�
3�5
2
Equation 1Equation 2Equation 3
x2xx
�
�
�
yyy
�
�
�
zz
2z
�
�
�
315
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.3 Multivariable Linear Systems 585
29.
Let then:
Answer: ��a � 3, a � 1, a�
x � �a � 3
y � a � 1
z � a,
x y
�
�
zz
�
�
31
x � yy
�
�
2zz0
�
�
�
210
��1� Eq. 1 � Eq. 2��5� Eq. 1 � Eq. 3
x �
�
y3y3y
�
�
�
2z3z3z
�
�
�
23
�3
�13� Equation 1 x
x5x
�
�
�
y2y8y
�
�
�
2zz
13z
�
�
�
257
Equation 1Equation 2Equation 3
3xx
5x
�
�
�
3y2y8y
�
�
�
6zz
13z
�
�
�
657
31.
No solution; inconsistent.
�Eq. 2 � Eq. 3x � 2y
5y�
�
3z7z0
�
�
�
4�12
6
�3 Eq. 1 � Eq. 2�1 Eq. 1 � Eq. 3
x � 2y5y5y
�
�
�
3z7z7z
�
�
�
4�12�6
Equation 1Equation 2Equation 3
x3xx
�
�
�
2yy
3y
�
�
�
3z2z4z
�
�
�
40
�2
33.
No solution; inconsistent.
Eq. 2 � Eq. 3x
y�
�
4z6z0
�
�
�
192
�Eq. 1 � Eq. 2�2 Eq. � Eq. 3
x
�
yy
�
�
�
4z6z6z
�
�
�
19
�7
xx
2x�
�
yy
�
�
�
4z10z2z
�
�
�
110
�5
30.
Answer: ��4a � 13, � 152 a �
452 , a�
x � �4a � 13
y � � 152 a �
452
z � a
�Eq. 2 � Eq. 3x
�2y
�
�
4z15z
0
�
�
�
13�45
0
�4 Eq. 1 � Eq. 2�2 Eq. 1 � Eq. 3
x
�2y�2y
�
�
�
4z15z15z
�
�
�
13�45�45
Equation 1Equation 2Equation 3
x4x2x
�
�
2y2y
�
�
�
4zz
7z
�
�
�
137
�19
32.
No solution; inconsistent.
�Eq. 2 � Eq. 3�x � 3y
10y�
�
zz0
�
�
�
49
11
4 Eq. 1 � Eq. 22 Eq. 1 � Eq. 3
�x � 3y10y10y
�
�
�
zzz
�
�
�
49
20
Equation 1Equation 2Equation 3
�x4x2x
�
�
�
3y2y4y
�
�
�
z5z3z
�
�
�
4�712
34.
No solution; inconsistent.
Eq. 2 � Eq. 3x � y
y�
�
5z9z0
�
�
�
�35
�3
�3 Eq. 1 � Eq. 23 Eq. 1 � Eq. 3
x � yy
�y
�
�
�
5z9z9z
�
�
�
�35
�8
Interchange theequations x
3x�3x
�
�
�
y2y2y
�
�
�
5z6z6z
�
�
�
�3�4
1
Equation 1Equation 2Equation 3
3x�3x
x
�
�
�
2y2yy
�
�
�
6z6z5z
�
�
�
�41
�3
©H
ough
ton
Miff
lin C
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ny. A
ll rig
hts
rese
rved
.
586 Chapter 7 Linear Systems and Matrices
35.
Answer: ��1, 1, 0�
x � 2y � z � 1 ⇒ x � 1 � 2 � �1
y � 1 � 0 � 1
z � 0
��3� Eq. 2 � Eq. 3x � 2y
y�
�
z12z52z
�
�
�
11
0
��14� Eq. 2
��1� Eq. 3 x � 2y
y3y
�
�
�
z12zz
�
�
�
113
��1� Eq. 1 � Eq. 2��2� Eq. 1 � Eq. 3x � 2y
�4y�3y
�
�
�
z2zz
�
�
�
1�4�3
Equation 1Equation 2Equation 3
xx
2x
�
�
�
2y2yy
�
�
�
z3zz
�
�
�
1�3�1
36.
Answer: �0, �12, 1�
x � 2y � z � 2 ⇒ x � 2 � 2��12� � 1 � 0
6y � 5z � �8 ⇒ 6y � �8 � 5 � �3 ⇒ y � �12
z � 1
��53� Eq. 2 � Eq. 3
x � 2y
6y
�
�
z
5z133 z
�
�
�
2
�8133
��2� Eq. 1 � Eq. 2��5� Eq. 1 � Eq. 3x � 2y
6y10y
�
�
�
z5z4z
�
�
�
2�8�9
Equation 1Equation 2Equation 3
x2x5x
�
�
2y2y
�
�
�
z3zz
�
�
�
2�4
1
37.
Let Then and
Answer: �14 a, 21
8 a � 1, a�x �
14 ay �
218 a � 1z � a.
2 Eq. 2 � Eq. 1x
y
�
�
14 z
218 z
�
�
0
�1
18 Eq. 2x � 2y
y�
�
5z218 z
�
�
2�1
�4 Eq. 1 � Eq. 2x � 2y8y
�
�
5z21z
�
�
2�8
x4x
� 2y �
�
5zz
�
�
20
39.
Answer: ��3
2a �
1
2, �
2
3a � 1, a
x � �32
a �12
y � �23
a � 1
Let z � a, then:
Eq. 2 � Eq. 12x 3y
�
�
3z2z
�
�
13
2 Eq. 1 � Eq. 22x � 3y3y
�
�
z2z
�
�
�23
2x�4x
�
�
3y9y
� z �
�
�27
38.
To avoid fractions, let then:
Answer: �9a, �35a, 67a�
x � 9a
x � 6��35a� � 3�67a� � 0
y � �35a
�67y � 35�67a� � 0
z � 67a,
2 Eq. 2 � Eq. 1�12 Eq. 1 � Eq. 2x � 6y
�67y�
�
3z35z
�
�
00
Interchangethe equations.23x
12x�
�
4y5y
�
�
zz
�
�
00
40.
Infinite number of solutions. Let Then
Answer: or �137
a, 487
a, a�13a, 48a, 7a�
x � y � 5z � 48a � 5�7a� � 13a
y �96z14
�96�7a�
14� 48a
z � 7a.
�19 Eq. 1 � Eq. 2x � y14y
�
�
5z96z
�
�
00
2 Eq. 1 � Eq. 2 x19x
�
�
y5y
�
�
5zz
�
�
00
Equation 1Equation 210x
19x�
�
3y5y
�
�
2zz
�
�
00
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.3 Multivariable Linear Systems 587
41.
Let then and
Answer: ��5a � 3, �a � 5, a�
x � �5a � 3.y � �a � 5,z � a,
3 Eq. 2 � Eq. 1x y �
�
5zz
�
�
3�5
12 Eq. 2x � 3y
y�
�
2zz
�
�
18�5
�5 Eq. 1 � Eq. 2x � 3y2y
�
�
2z2z
�
�
18�10
Equation 1Equation 2 x
5x�
�
3y13y
�
�
2z12z
�
�
1880
42.
Let then:
Answer: ��38a �
14, �3
4a �52, a�
2x �34a � �1
2 ⇒ x � �38a �
14
12y � 9a � 30 ⇒ y � �34 a �
52
z � a,
�14 Eq. 2 � Eq. 12x
12y�
�
34z9z
�
�
�12
30
�2 Eq. 1 � Eq. 22x � 3y12y
�
�
3z9z
�
�
730
Equation 1Equation 22x
4x�
�
3y18y
�
�
3z15z
�
�
744
43.
Answer: �12, 0, 12, 32�
x � y � 2z � w � 0 ⇒ x � �2�12� �
32 �
12
2y � 2z � �1 ⇒ 2y � �1 � 2�12� � 0 ⇒ y � 0
�2z � 2w � 2 ⇒ �2z � 2 � 2�32� � �1 ⇒ z �
12
w �32
��32� Eq. 2 � Eq. 3
��1� Eq. 2 � Eq. 4
x � y2y
�
�
2z2z
�2z
�
�
w
w2w
�
�
�
�
0�1
32
2
InterchangeEq. 2 and 3
x � y2y3y2y
�
�
�
�
2z2z3z4z
�
�
�
w
w2w
�
�
�
�
0�1
01
��2� Eq. 1 � Eq. 2��1� Eq. 1 � Eq. 3��1� Eq. 1 � Eq. 4
x � y
3y2y2y
�
�
�
�
2z3z2z4z
�
�
�
ww
2w
�
�
�
�
00
�11
Equation 1Equation 2Equation 3Equation 4
x
2xxx
�
�
�
�
yyyy
�
�
�
2zz
2z
�
�
�
�
wwww
�
�
�
�
00
�11
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
588 Chapter 7 Linear Systems and Matrices
44.
Answer: �1, 0, �1, 2�
x � 2y � z � 2w � 6 ⇒ x � 6 � 2�0� � ��1� � 2�2� � 1
y � 3z � 3w � �9 ⇒ y � �9 � 3��1� � 3�2� � 0
2z � w � �4 ⇒ 2z � �4 � 2 � �2 ⇒ z � �1
w � 2
�2� Eq. 3 � Eq. 4
x � 2yy
�
�
z3z2z
�
�
�
2w3ww
2w
�
�
�
�
6�9�4
4
��1� Eq. 2 � Eq. 4
x � 2yy
�
�
z3z2z
�4z
�
�
�
�
2w3ww
4w
�
�
�
�
6�9�412
��2� Eq. 1 � Eq. 2��1� Eq. 1 � Eq. 3
x � 2yy
y
�
�
�
z3z2zz
�
�
�
�
2w3www
�
�
�
�
6�9�4
3
Equation 1Equation 2Equation 3Equation 4
x
2xx
�
�
�
2y3y2yy
�
�
�
�
zzzz
�
�
�
�
2wwww
�
�
�
�
6323
45. Let
Answer: �1, �2, �1�
x �1X
� 1, y �1Y
� �2, z �1Z
� �1
X � 2Y � 3Z � 3 ⇒ X � 3 � 2��1
2 � 3��1� � 1
2Y � 2Z � 1 ⇒ 2Y � 1 � 2Z � �1 ⇒ Y � �1
2
Eq. 2 � Eq. 3X � 2Y
2Y�
�
3Z2ZZ
�
�
�
31
�1
��12� Eq. 2X � 2Y
2Y�2Y
�
�
�
3Z2Z3Z
�
�
�
31
�2
��1� Eq. 1 � Eq. 2��2� Eq. 1 � Eq. 3
X � 2Y�4Y�2Y
�
�
�
3Z4Z3Z
�
�
�
3�2�2
Equation 1Equation 2Equation 3
XX
2X
�
�
�
2Y2Y2Y
�
�
�
3ZZ
3Z
�
�
�
314
X �1x, Y �
1y, Z �
1z. 46. Let
Answer: ��1, �1, �1�
x �1X
� y � z � �1
Z � �1, Y � �1, X � �1
�� 310� Eq. 2 � Eq. 3
X � 2Y�10Y
�
�
2Z9Z
�1710Z
�
�
�
�11
1710
��4� Eq. 1 � Eq. 2��2� Eq. 1 � Eq. 3
X � 2Y�10Y�3Y
�
�
�
2Z9ZZ
�
�
�
�112
InterchangeEq. 1 and 2 X
4X2X
�
�
�
2Y2YY
�
�
�
2ZZ
3Z
�
�
�
�1�3
0
Equation 1Equation 2Equation 3
4XX
2X
�
�
�
2Y2YY
�
�
�
Z2Z3Z
�
�
�
�3�1
0
X �1x, Y �
1y, Z �
1z.
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.3 Multivariable Linear Systems 589
47. Let
Answer: �1, �1, 2�
x �1X
� 1, y �1Y
� �1, z �1Z
� 2
X � 2Y � 2Z � �2 ⇒ X � �2 � 2��1� � 2�1
2 � 1
�5Y � 6Z � 8 ⇒ �5Y � 8 � 6�1
2 � 5 ⇒ Y � �1
Z �12
��35�Eq. 2 � Eq. 3
X � 2Y�5Y
�
�
2Z6Z
325 Z
�
�
�
�28
165
��2� Eq. 1 � Eq. 2��3� Eq. 1 � Eq. 3
X � 2Y�5Y�3Y
�
�
�
2Z6Z
10Z
�
�
�
�288
InterchangeEq. 1 and 2 X
2X3X
�
�
�
2YY
3Y
�
�
�
2Z2Z4Z
�
�
�
�242
Equation 1Equation 2Equation 3
2XX
3X
�
�
�
Y2Y3Y
�
�
�
2Z2Z4Z
�
�
�
4�2
2
X �1x, Y �
1y, Z �
1z.
48. Let
Answer: �1, �1, 2�
x � 1, y � �1, z � 2
Z �12
, Y � �1, X � 1
��1� Eq. 2�4� Eq. 2 � Eq. 3
X � YY
�
�
2Z6Z
22Z
�
�
�
12
11
��2� Eq. 1 � Eq. 2��3� Eq. 1 � Eq. 3
X � Y�Y
�4Y
�
�
�
2Z6Z2Z
�
�
�
1�2
3
Equation 1Equation 2Equation 3
X2X3X
�
�
�
YYY
�
�
�
2Z2Z4Z
�
�
�
106
X �1x, Y �
1y, Z �
1z. 49. There are an infinite number of linear systems that
have as their solution. One such system isas follows:
3�4�
�4�
��4�
�
�
�
��1�
2��1�
��1�
�
�
�
�2�
�2�
3�2�
�
�
�
9
0
1
⇒
⇒
⇒
3x
x
�x
�
�
�
y
2y
y
�
�
�
z
z
3z
� 9
� 0
� 1
�4, �1, 2�
50. There are an infinite number of linear systems thathave as their solution. One such system is:
2�1� � 2
2��2� � 1 � �3
1��5� � 1��2� � 1 � �6 ⇒
��5, �2, 1�51. There are an infinite numbers of linear systems that
have as their solution. One such system is:
4z � 7 ⇒ 4�74� � 7
4y � 8z � 12 ⇒ 4��12� � 8�7
4� � 12
x � 2y � 4z � 9 ⇒ 1�3� � 2��12� � 4�7
4� � 9
�3, �12, 74�
x � y � z2y � z
2z
�
�
�
�6�3
2
©H
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lin C
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ny. A
ll rig
hts
rese
rved
.
590 Chapter 7 Linear Systems and Matrices
52. There are an infinite number of linear systems that have as their solution. One such system is:
�2�� 32� � 5�4� � 3��7� � 44 ⇒
4��32� � 2�4� � ��7� � �5 ⇒
2�� 32� � 4 � ��7� � 8 ⇒
�� 32, 4, �7�
2x
4x
�2x
�
�
�
y
2y
5y
�
�
�
z
z
3z
�
�
�
8
�5
44
53.
x y
4
22
6 64
6
z
�6, 0, 0�, �0, 4, 0�, �0, 0, 3�, �4, 0, 1�
2x � 3y � 4z � 12
55.
x y
4
6 644
6
z
�2, 0, 0�, �0, 4, 0�, �0, 0, 4�, �0, 2, 2�
2x � y � z � 4
57.7
x2 � 14x�
7
x�x � 14��
A
x�
B
x � 14
59.12
x3 � 10x2�
12
x2�x � 10��
A
x�
B
x2�
C
x � 10
61.4x2 � 3�x � 5�3 �
A�x � 5� �
B�x � 5�2 �
C�x � 5�3
54.
x y
4
2
22
6 644
6
z
�6, 0, 0�, �0, 6, 0�, �0, 0, 6�, �1, 1, 4�
x � y � z � 6
56.
x y
4
6 64
6
2
z
�6, 0, 0�, �0, 3, 0�, �0, 0, 3�, �2, 1, 1�
x � 2y � 2z � 6
58.x � 2
x2 � 4x � 3�
A
x � 3�
B
x � 1
60.x2 � 3x � 24x3 � 11x2 �
x2 � 3x � 2x2�4x � 11� �
Ax
�Bx2 �
C4x � 11
62.6x � 5
�x � 2�4 �A
x � 2�
B�x � 2�2 �
C�x � 2�3 �
D�x � 2�4
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.3 Multivariable Linear Systems 591
65.
1x2 � x
�1x
��1
x � 1�
1x
�1
x � 1
A � BA
�
�
01 ⇒ B � �1
1 � A�x � 1� � Bx � �A � B�x � A
1
x2 � x�
1x�x � 1� �
Ax
�B
x � 1
63.
1x2 � 1
��1�2x � 1
�1�2
x � 1�
12�
1x � 1
�1
x � 1�2B � 1 ⇒ B �
12 ⇒ A � �
12
A�A
�
�
BB
�
�
01
1 � A�x � 1� � B�x � 1� � �A � B�x � �B � A�
1
x2 � 1�
Ax � 1
�B
x � 164.
Let :
1
4x2 � 9�
1
6�1
2x � 3�
1
2x � 3�
Let x �3
2: 1 � 6B ⇒ B �
1
6
1 � �6A ⇒ A � �1
6x � �
3
2
1 � A�2x � 3� � B�2x � 3�
1
4x2 � 9�
A
2x � 3�
B
2x � 3
66.
Let :
Let :
3
x2 � 3x�
1
x � 3�
1
x
3 � �3B ⇒ B � �1x � 0
3 � 3A ⇒ A � 1x � 3
3 � Ax � B�x � 3�
3
x2 � 3x�
A
x � 3�
B
x
67.
12x2 � x
��2
2x � 1�
1x
�1x
�2
2x � 1
A � 2BB
�
�
01 ⇒ A � �2
1 � Ax � B�2x � 1� � �A � 2B�x � B
12x2 � x
�1
x�2x � 1� �A
2x � 1�
Bx
68.
Let :
Let :
5
x2 � x � 6�
1
x � 2�
1
x � 3
5 � 5B ⇒ B � 1x � 2
5 � �5A ⇒ A � �1x � �3
5 � A�x � 2� � B�x � 3�
5
x2 � x � 6�
A
x � 3�
B
x � 2
69.
and
5 � x2x2 � x � 1
�3
2x � 1�
�2x � 1
A � 3��1 � 2B� � B � 5 ⇒ B � �2
AA
�
�
2BB
�
�
�15 ⇒ A � �1 � 2B
� �A � 2B�x � �A � B�
5 � x � A�x � 1� � B�2x � 1�
�A
2x � 1�
Bx � 1
5 � x2x2 � x � 1
�5 � x
�2x � 1��x � 1� 70.
Solving for A and B,
x � 2x2 � 4x � 3
�5�2
x � 3�
3�2x � 1
A �52, B � �
32
A � BA � 3B
�
�
1�2
(A � B�x � �A � 3B� � x � 2
A(x � 1� � B�x � 3� � x � 2
x � 2x2 � 4x � 3
�A
x � 3�
Bx � 1
©H
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592 Chapter 7 Linear Systems and Matrices
71.
x2 � 12x � 12x3 � 4x
��3x
��1
x � 2�
5x � 2
2C � 10 ⇒ C � 5 ⇒ B � �1
�
BB
�
�
CC
�
�
46
�
A
4A
�
�
B2B
�
�
C2C
�
�
�
11212 ⇒ A � �3
� �A � B � C�x2 � ��2B � 2C�x � ��4A�
x2 � 12x � 12 � A�x � 2��x � 2� � Bx�x � 2� � Cx�x � 2�
x2 � 12x � 12
x3 � 4x�
x2 � 12x � 12x�x � 2��x � 2� �
Ax
�B
x � 2�
Cx � 2
72.
Solving,
x2 � 12x � 9x3 � 9x
�1x
�2
x � 3�
2x � 3
A � 1, B � 2 and C � �2
A
�9A
� B3B
�
�
C3C
�
�
�
112
�9
A�x2 � 9� � Bx�x � 3� � Cx�x � 3� � x2 � 12x � 9
�Ax
�B
x � 3�
Cx � 3
x2 � 12x � 9x3 � 9x
�x2 � 12x � 9
x�x � 3��x � 3�
73.
4x2 � 2x � 1x2�x � 1� �
3x
��1x2 �
1x � 1
B � �1 ⇒ A � 3 ⇒ C � 1
AA � B
B
� C �
�
�
42
�1
� �A � C�x2 � �A � B�x � B
4x2 � 2x � 1 � Ax�x � 1� � B�x � 1� � Cx2
4x2 � 2x � 1
x2�x � 1� �Ax
�Bx2 �
Cx � 1
74.
Let :
2x � 3
�x � 1�2�
2
x � 1�
1
�x � 1�2
2 � A
�3 � �A � 1
Let x � 0: �3 � �A � B
�1 � Bx � 1
2x � 3 � A�x � 1� � B
2x � 3
�x � 1�2�
A
x � 1�
B
�x � 1�2
75.
27 � 7xx�x � 3�2 �
3x
��3
x � 3�
2�x � 3�2
A � 3 ⇒ B � �3 ⇒ C � �7 � 18 � 9 � 2
�
A6A9A
�
�
B3B � C
�
�
�
0�727
� �A � B�x2 � ��6A � 3B � C�x � 9A 27 � 7x � A�x � 3�2 � Bx�x � 3� � Cx
27 � 7xx�x � 3�2 �
Ax
�B
x � 3�
C�x � 3�2
©H
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Miff
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.
Section 7.3 Multivariable Linear Systems 593
76.
Solving,
x2 � x � 2x�x � 1�2 �
2x
�1
x � 1�
2�x � 1�2
A � 2, B � �1 and C � 2
A�2A
A
�
�
BB � C
�
�
�
1�1
2
A�x � 1�2 � Bx�x � 1� � Cx � x2 � x � 2
x2 � x � 2
x�x � 1�2�
A
x�
B
x � 1�
C
�x � 1�277.
2x3 � x2 � x � 5x2 � 3x � 2
� 2x � 7 �1
x � 1�
17x � 2
A � 1 ⇒ B � 17
A2A
�
�
BB
�
�
1819
� �A � B�x � �2A � B�
18x � 19 � A�x � 2� � B�x � 1�
18x � 19
�x � 1��x � 2� �A
x � 1�
Bx � 2
2x3 � x2 � x � 5
x2 � 3x � 2� 2x � 7 �
18x � 19�x � 1��x � 2�
78.
Let
Let
x3 � 2x2 � x � 1x2 � 3x � 4
� x � 1 �27
5�x � 4� �3
5�x � 1�
x � �4: �27 � �5A ⇒ A �275
x � 1: 3 � 5B ⇒ B �35
6x � 3 � A�x � 1� � B�x � 4�
6x � 3�x � 4��x � 1� �
Ax � 4
�B
x � 1
x3 � 2x2 � x � 1x2 � 3x � 4
� x � 1 �6x � 3
�x � 4��x � 1�
79.
x4
�x � 1�3 �6
x � 1�
4�x � 1�2 �
1�x � 1�3 � x � 3
A � 6 ⇒ B � �8 � 2�6� � 4 ⇒ C � 3 � 6 � 4 � 1
A�2A
A�
�
BB � C
�
�
�
6�8
3
� Ax2 � ��2A � B�x � �A � B � C� 6x2 � 8x � 3 � A�x � 1�2 � B�x � 1� � C
6x2 � 8x � 3
�x � 1�3 �A
x � 1�
B�x � 1�2 �
C�x � 1�3
x4
�x � 1�3 � x � 3 �6x2 � 8x � 3
�x � 1�3
©H
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.
594 Chapter 7 Linear Systems and Matrices
80.
Answer:x2
�34
�3
2�2x � 1� �1
�2x � 1�2 �1
4�2x � 1�3
A � B � C � 3 ⇒ C � 1
�4A � 2B � �16 ⇒ 2B � 8 ⇒ B � 4
4A � 24 ⇒ A � 6
24x2 � 16x � 3 � A�2x � 1�2 � B�2x � 1� � C � A4x2 � ��4A � 2B�x � �A � B � C�
24x2 � 16x � 34�2x � 1�3 �
14�
A2x � 1
�B
�2x � 1�2 �C
�2x � 1�3�
4x4
�2x � 1�3 �x2
�34
�24x2 � 16x � 3
4�2x � 1�3
81.
Vertical asymptotes: Vertical asymptotes:and and
The combination of the vertical asymptotes of theterms of the decompositions are the same as thevertical asymptotes of the rational function.
x � 4x � 0x � 4x � 0
–6 2 8 10
–8
2
8
x
y = 3x
y = 3x
y = 2x − 4
−
y = 2x − 4
−
y
–6 –4 2 8 10
–8
2
4
6
8
x
y
y �3
x, y � �
2
x � 4y �
x � 12
x�x � 4�
x � 12
x�x � 4��
3
x�
2
x � 4
⇒ A � 3, B � �2 A�4A
� B �
�
1�12
x � 12 � A(x � 4� � Bx
x � 12
x�x � 4��
A
x�
B
x � 482.
Let
Let
Vertical asymptotes: Vertical asymptotes:
The combination of the vertical asymptotes of the terms of the decompositions are the same as the vertical asymptotes of the rational function.
x � 3, x � �3x � ±3
–4 2 4 6 8
–8
–6
–4
6
8
x
y = 5x + 3
y = 5x + 3
y = 3x − 3
y = 3x − 3
y
–4 4 6 8
–8
–6
–4
4
6
8
x
y
y �3
x � 3, y �
5
x � 3y �
2�4x � 3�x2 � 9
2�4x � 3�x2 � 9
�3
x � 3�
5
x � 3
x � �3: 1�30 � �6B ⇒ B � 5
x � 3: 18 � 6A ⇒ A � 3
2�4x � 3� � A(x � 3� � B�x � 3�
2�4x � 3�x2 � 9
�A
x � 3�
B
x � 3
83.
Solving the system,
Thus,
� �16t2 � 144.
s �12��32�t2 � �0�t � 144
a � �32, v0 � 0, s0 � 144.
128800
�
�
�
12 a2a92 a
�
�
�
v0
2v0
3v0
�
�
�
s0
s0
s0
⇒ ⇒ ⇒
a2a9a
�
�
�
2v0
2v0
6v0
�
�
�
2s0
s0
2s0
�
�
�
256800
�1, 128�, �2, 80�, �3, 0�
s �12at2 � v0t � s0 84.
Solving the system,
Thus,
� �16t2 � 64t.
s �12��32�t2 � 64t � 0
s0 � 0.a � �32, v0 � 64,
486448
�
�
�
12a2a92a
�
�
�
v0
2v0
3v0
�
�
�
s0
s0
s0
⇒ ⇒ ⇒
a2a9a
�
�
�
2v0
2v0
6v0
�
�
�
2s0
s0
2s0
�
�
�
966496
�1, 48�, �2, 64�, �3, 48�
s �12at2 � v0t � s0
©H
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Miff
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ny. A
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hts
rese
rved
.
Section 7.3 Multivariable Linear Systems 595
87. passing through
Answer:
The equation of the parabola is y �12x2 � 2x.
a �12, b � �2, c � 0
� �0, 0�: 0�2, �2�: �2
�4, 0�: 0
�
�
�
4a4a
16a
�
�
�
2b2b4b
�
�
�
ccc
⇒ ⇒ ⇒
c�1
0
�
�
�
�4a � 2b2a � b4a � b
−15
−10
15
10�0, 0�, �2, �2�, �4, 0�y � ax2 � bx � c
88. passing through
Answer:
The equation of the parabola is y � �x2 � 2x � 3.
a � �1, b � 2, c � 3
��0, 3�: 3�1, 4�: 4�2, 3�: 3
�
�
�
a4a
�
�
b2b
�
�
ccc
⇒ ⇒
10
�
�
a � b2a � b
−15
−10
15
10�0, 3�, �1, 4�, �2, 3�y � ax2 � bx � c
89. passing through
Answer:
The equation of the parabola is y � x2 � 6x � 8.
a � 1, b � �6, c � 8
� �2, 0�: 0�3, �1�: �1
�4, 0�: 0
�
�
�
4a9a
16a
�
�
�
2b3b4b
�
�
�
ccc
⇒ ⇒ ⇒
c�1
0
�
�
�
�4a � 2b5a � b
12a � 2b
−15
−10
15
10�2, 0�, �3, �1�, �4, 0�y � ax2 � bx � c
90. passing through
Answer:
The equation of the parabola is y � �2x2 � 5x.
a � �2, b � 5, c � 0
��1, 3�: 3�2, 2�: 2�3, �3�: �3
�
�
�
a4a9a
�
�
�
b2b3b
�
�
�
ccc
⇒ ⇒
�1�6
�
�
3a � b 8a � 2b
−15
−10
15
10�1, 3�, �2, 2�, �3, �3�y � ax2 � bx � c
85.
Solving the system,
Thus,
� �16t2 � 32t � 500
s �12 ��32�t2 � 32t � 500
a � �32, v0 � �32, s0 � 500
�452372260
�
�
�
12 a2a92 a
�
�
�
v0
2v0
3v0
�
�
�
s0
s0
s0
⇒ ⇒ ⇒
a2a9a
�
�
�
2v0
2v0
6v0
�
�
�
2s0
s0
2s0
�
�
�
904372520
�1, 452�, �2, 372�, �3, 260�
s �12 at2 � v0t � a0 86.
Solving the system
Thus,
� �16t2 � 16t � 132.
s �12��32�t2 � 16t � 132
s0 � 132.a � �32, s0 � 16,
�13210036
�
�
�
12a2a92a
�
�
�
v0
2v0
3v0
�
�
�
s0
s0
s0
⇒ ⇒ ⇒
a2a9a
�
�
�
2v0
2v0
6v0
�
�
�
2s0
s0
2s0
�
�
�
26410072
�1, 132�, �2, 100�, �3, 36�
s �12at2 � v0t � s0
©H
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Miff
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rved
.
596 Chapter 7 Linear Systems and Matrices
91. passing through
The equation of the circle is
To graph, let y1 � �4x � x2 and y2 � ��4x � x2.
x2 � y2 � 4x � 0.
�4, 0�: 16 � 4D � 2E � F � 0 ⇒ D � �4 and E � 0
�2, 2�: 18 � 2D � 2E � F � 0 ⇒ D � E � �4
�0, 0�: 16 � 4D � 2E � F � 0
�0, 0�, �2, 2�, �4, 0�x2 � y2 � Dx � Ey � F � 0
−6
−4
6
4
92. passes through
The equation of the circle is
To graph, complete the square first, then solve for
Let and y2 � 3 � �9 � x2.y1 � 3 � �9 � x2
y � 3 ± �9 � x2
y � 3 � ±�9 � x2
�y � 3�2 � 9 � x2
x2 � �y � 3�2 � 9
−9
−2
9
10x2 � y2 � 6y � 9 � 9
y.
x2 � y2 � 6y � 0.
��0, 0�:�0, 6�:�3, 3�: 18 �
363D
�
�
6E3E
�
�
F � 0F � 0F � 0
⇒ ⇒
E � �6D � 0
�0, 0�, �0, 6�, �3, 3�.x2 � y2 � Dx � Ey � F � 0
93. passes through
Answer:
The equation of the circle is To graph, complete the squares first, then solve for
Let y1 � 4 � �25 � �x � 3�2 and y2 � 4 � �25 � �x � 3�2.
y � 4 ±�25 � �x � 3�2
y � 4 � ±�25 � �x � 3�2
�y � 4�2 � 25 � �x � 3�2
�x � 3�2 � �y � 4�2 � 25
�x2 � 6x � 9� � �y2 � 8y � 16� � 0 � 9 � 16
y.x2 � y2 � 6x � 8y � 0.
D � 6, E � �8, F � 0
��6, �8�: 100 � 6D � 8E � F � 0 ⇒ 100 � �6D � 8E � F
��2, �4�: 120 � 2D � 4E � F � 0 ⇒ 20 � �2D � 4E � F
��3, �1�: 110 � 3D � 8E � F � 0 ⇒ 10 � �3D � 8E � F
−9
−2
9
10��3, �1�, �2, 4�, ��6, 8�.x2 � y2 � Dx � Ey � F � 0
©H
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Miff
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rved
.
Section 7.3 Multivariable Linear Systems 597
95. Let amount at 8%.
Let amount at 9%.
Let amount at 10%.
Solving the system, at 8%,at 9%, at 10%.z � $91,666.67y � $316,666.67
x � $366,666.67
� x8xx
�
�
y9y
�
�
�
z10z4z
�
�
�
775,0006,700,000
0
� x0.08x
x
�
�
y0.09y
�
�
z0.10z
�
�
�
775,00067,000
4z
z �
y �
x � 96. Let amount at 8%.
Let amount at 10%.
Let amount at 12%.
Solving the system,z � $316,000 at 12%.y � $228,000 at 10%,
x � $456,000 at 8%,
�0.08xxx
�
�
�
0.10yy
2y
�
�
0.12zz
�
�
�
97,200 1,000,0000
z �
y �
x �
98. Let amount in certificates of deposit.
Let amount in municipal bonds.
Let amount in blue-chip stocks.
Let amount in growth or speculative stocks.
The system has infinitely many solutions.
Let then
Answer:
One possible solution is to let $100,000.
Certificates of deposit: $356,250
Municipal bonds: $18,750
Blue-chip stocks: $25,000
Growth or speculative stocks: $100,000
s �
�406,250 �12s, �31,250 �
12s, 125,000 � s, s�
C � 406,250 �12s.
M �12 s � 31,250
B � 125,000 � sG � s,
� C � M � B � G0.09C � 0.05M � 0.12B � 0.14G
B � G
�
�
�
500,000 0.10�500,000�14�500,000�
G �
B �
M �
C �
94. passes through
Solving the system and the circle is
y2 � �1 � �25 � �x � 1�2
y1 � �1 � �25 � �x � 1�2
�x � 1�2 � � y � 1�2 � 25
�x2 � 2x � 1� � �y2 � 2y � 1� � 23 � 1 � 1−9
−7
9
5 x2 � y2 � 2x � 2y � 23 � 0
D � 2, E � 2, F � �23,
���6, �1�:��4, 3�:�2, �5�:
36164
�
�
�
19
25
�
�
�
6D4D2D
�
�
�
E � F � 03E � F � 05E � F � 0
⇒ ⇒ ⇒
6D4D2D
�
�
�
E3E5E
� F� F� F
� 37� 25� �29
��6, �1�, ��4, 3�, �2, �5�.x2 � y2 � Dx � Ey � F � 0
97. Let amount in certificates of deposit.
Let amount in municipal bonds.
Let amount in blue chip stocks.
Let amount in growth or speculative stocks.
Solving the system
G � s
B � 218,750 � 1.75s
M � 125,000
C � 156,250 � 0.75s
� C � M � B � G0.08C � 0.09M � 0.12B � 0.15G
M
�
�
�
500,000 0.10�500,000�14�500,000�
G �
B �
M �
C �
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.
598 Chapter 7 Linear Systems and Matrices
99. Let number of 1-point free throws.
Let number of 2-point field goals.
Let number of 3-point basket.
Solving the system,
18 free throws, 24 2-point field goals,6 3-point baskets.
x � 18, y � 24, z � 6.
� x�x
�
�
2yyy
�
�
3z
4z
�
�
�
8460
z �
y �
x � 100. Let number of 1-point free throws.
Let number of 2-point field goals.
Let number of 3-point baskets.
Solving the system,
12 free throws, 18 2-point field goals,9 3-point baskets.
x � 12, y � 18, z � 9.
�xxx
�
�
2y
y
�
�
3zz
�
�
�
753
�6
z �
y �
x �
101. Let number of touchdowns.
Let number of extra-point kicks.
Let number of field goals.
Solving the system,
4 touchdowns, 4 extra-points and 1 field goal
x � 4, y � 4, z � 1.
�x
6xxx
�
�
�
yy
y
�
�
�
z3z4z
�
�
�
�
93100
z �
y �
x � 102. Let number of touchdowns.
Let number of extra points.
Let number of field goals.
Solving the system, touchdowns,extra points, field goalsz � 4y � 3
x � 4
� x6xx
�
�
yy
�
�
�
z3zz
�
�
�
11390
z �
y �
x �
103.
Answer: ampere, amperes,ampereI3 � 1
I2 � 2I1 � 1
I1 � 2 � 1 � 0 ⇒ I1 � 1
10I2 � 6�1� � 14 ⇒ I2 � 2
26I3 � 26 ⇒ I3 � 1
�Eq. 2 � Eq. 3�I1 � I2
10I2
�
�
I3
6I3
26I3
�
�
�
01426
2 Eq. 25 Eq. 3
�I1 � I2
10I2
10I2
�
�
�
I3
6I3
20I3
�
�
�
01440
�3 Eq. 1 � Eq. 2�I1 � I2
5I2
2I2
�
�
�
I3
3I3
4I3
�
�
�
078
Equation 1Equation 2Equation 3
� I1
3I1
�
�
I2
2I2
2I2
�
�
I3
4I3
�
�
�
078
104.
Answer: a � �16 ft�sec2t2 � 48 lb,t1 � 96 lb,
t1 � 96
t2 � 48
a � �16
�4a � 64
�t1t1
� 2t2
t2
�
�
2aa
�
�
�
012832
⇒ ⇒ ⇒
2t2�2t2
�
�
2a2a
�
�
128�64
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.
Section 7.3 Multivariable Linear Systems 599
107. Least squares regression parabola through
Solving the system,
Thus, y � �524x2 �
310x �
416 .
a � �524, b � �
310, and c �
416 .
� 4c
40c
40b
�
�
40a
544a
�
�
�
19�12160
��4, 5�, ��2, 6�, �2, 6�,�4, 2�108. Least squares regression parabola through
Solving the system,
Thus, y �37 x2 �
65x �
2635.
c �2635.b �
65,a �
37,
� 5c
10c
�
�
10a10b34a
�
�
�
81222
�2, 5��1, 2�,�0, 1�,��1, 0�,��2, 0�,
109. Least squares regression parabola through
Solving the system,Thus, y � x2 � x.
a � 1, b � �1, and c � 0.
� 4c9c
29c
�
�
�
9b29b99b
�
�
�
29a99a
353a
�
�
�
2070
254
�0, 0�, �2, 2�, �3, 6�, �4, 12�110. Least squares regression parabola through
Solving the system,
Thus, y � � 54x2 �
920x �
19920 .
a � � 54, b �
920, c �
19920 .
� 4c6c
14c
�
�
�
6b14b36b
�
�
�
14a36a98a
�
�
�
252133
�0, 10�, �1, 9�, �2, 6�, �3, 0�
111. (a)
Solving the system,and
(b)
(c) For feet.y � 453x � 70,
6000
500
y � 0.165x2 � 6.55x � 103
c � 103.b � �6.55a � 0.165,
�a�30�2
a�40�2
a�50�2
�
�
�
b�30�b�40�b�50�
�
�
�
ccc
�
�
�
55105188
112. (a)
Solving the system,and
Parabola:
(b)
(c) For x � 170, y � 13%
1701100
100
y � �0.015x2 � 3.25x � 106
c � �106.a � �0.015, b � 3.25
�a�120�2
a�140�2
a�160�2
�
�
�
b�120�b�140�b�160�
�
�
�
ccc
�
�
�
685530
105. Let number of par-3 holes.
Let number of par-4 holes.
Let number of par-5 holes.
Solving the system,
4 par-3 holes, 10 par-4 holes, and 4 par-5 holes
x � 4, y � 10, z � 4.
�3x
x
� 4yy
�
�
�
5z2zz
�
�
�
7220
z �
y �
x � 106. Let number of par-3 holes.
Let number of par-4 holes.
Let number of par-5 holes.
Solving the system,
2 par-3 holes, 14 par-4 holes, and 2 par-5 holes
x � 2, y � 14, z � 2.
�3x
x
� 4yy
�
�
�
5z7zz
�
�
�
7204
z �
y �
x �
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Miff
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rved
.
600 Chapter 7 Linear Systems and Matrices
113. (a)
�2000
7 � 4x�
2000
11 � 7x
2000�4 � 3x�
�11 � 7x��7 � 4x��
�2000
11 � 7x�
2000
7 � 4x
A � �2000B � 2000
⇒ ��60008000
�
�
�4A7A
�
�
7B11B
2000�4 � 3x� � A�7 � 4x� � B�11 � 7x�
2000�4 � 3x��11 � 7x��7 � 4x�
�A
11 � 7x�
B
7 � 4x, 0 ≤ x ≤ 1 (b)
100
700 20007 − 4x
200011 − 7x
y2 �2000
11 � 7x
y1 �2000
7 � 4x
114.
Hence, and 60
100 � p�
60100 � p
�120p
10,000 � p2.A � 60, B � �60
� A100A
�
�
B100B
�
�
1200
A�100 � p� � B�100 � p� � 120p
120p�100 � p��100 � p� �
A100 � p
�B
100 � p
0 ≤ p < 100C �120p
10,000 � p2 �120p
�100 � p��100 � p� ,
115. False. The coefficient of in the second equationis not 1.
y
117. False. The correct form is
Ax � 10
�B
x � 10�
C�x � 10�2.
116. True. A common point would be a solution.
118. No, the problem was notworked correctly. You must first divide theimproper fraction.
A � �1 ⇒ B � 1.
119.
1a2 � x2 �
1�2aa � x
�1�2aa � x
�12a�
1a � x
�1
a � x�
A �12a
, B �12a
⇒ ��AAa
�
�
BBa
�
�
01
1 � A�a � x� � B�a � x� � ��A � B�x � �Aa � Ba�
1
a2 � x2 �1
�a � x��a � x� �A
a � x�
Ba � x
120.
Let :
Let :
1
�x � 1��a � x��
1
a � 1 � 1
x � 1�
1
a � x�
B�a � 1� ⇒ B �1
a � 1x � a
1 � A�a � 1� ⇒ A �1
a � 1x � �1
1 � A�a � x� � B�x � 1�
1
�x � 1��a � x��
A
x � 1�
B
a � x121.
1
y�a � y��
1
a�1
y�
1
a � y
A �1a
, B �1a
1 � A�a � y� � By � ��A � B�y � aA
1
y�a � y��
A
y�
B
a � y
©H
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Miff
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.
Section 7.3 Multivariable Linear Systems 601
122.
Let :
Let :
1
x�x � a��
1
a �1
x�
1
x � a�
1 � �aB ⇒ B � �1
ax � �a
1 � aA ⇒ A �1
ax � 0
1 � A�x � a� � Bx
1
x�x � a��
A
x�
B
x � a, a is a constant. 123. No, they are not equivalent. In the second system,
the constant in the second equation should be and the coefficient of in the third equationshould be 2.
z�11
124. When using Gaussian elimination to solve a system of linear equations, a system has no solution when there is a row representing a contradictory equation such as where
is a nonzero real number.
For instance:
No solution
Eq. 1 � Eq. 2�x � y0
�
�
36
Equation 1Equation 2� x
�x�
�
yy
�
�
33
N0 � N,
125.
� � �5
y � 5
x � 5
⇒ x � y � ��
⇒ 2x � 10 � 0 �y
x
�
�
xy
�
�
�
�
10
�
�
�
126.
� � �4
y � 2
x � 2
2x � 4
� 2x � � � 02y � � � 0
x � y � 4 � 0
⇒
⇒
x � y � �
2x � 4 � 0
��2 127.
From the first equation,
If then and
If then
Thus, the solutions are:
(1)
(2)
(3) x � ��2
2, y �
1
2, � � 1
x ��2
2, y �
1
2, � � 1
x � y � � � 0
� � 1 ⇒ y �12 and x � ±�1
2.x � 0,
� � 0.y � 02 � 0x � 0,
x � 0 or � � 1.
�2x � 2x� � 0�2y � � � 0
y � x2 � 0
⇒ ⇒ ⇒
x � x�
2y � �
y � x2
128.
Then, and
Answer: x � 0, y � 100, � � �1.
y � 100 � 2x � 100� � �1
x � 0
�2x � 0
2 � 2x � 2��2x � 1� � 0
� 2 � 2x � 2� � 02x � 1 � � � 0
2x � y � 100 � 0
⇒
� � �2x � 1
129.
7654321−1−2−3
5
4
3
2
1
6
7
x
y
y � �3x � 7
©H
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.
602 Chapter 7 Linear Systems and Matrices
130.
9854321−1
9
8
5
4
3
2
1x
y132.
9
8
5
4
3
2
6
7
54321−1−2−3−4−5x
y
y �14x2 � 1131.
5432−2−3−4−5
1
−5
−6
−7
−8
−9
x
y
y � �2x2
133.
65421−1−2−3−4
7
6
5
−2
−3
x
y
y � �x2�x � 3� 134.
5432−1−2−3−4−5
5
4
3
2
1
−3
−4
−5
x
y
y �12x3 � 1
135. (a)
(b)25
20
−5
−10
−15
642−2−6x
y
⇒ x � 0, �4, 3
� x�x2 � x � 12� � x�x � 4��x � 3�
f�x� � x3 � x2 � 12x
137. (a)
(b)
642−2−6
20
10
−30
−40
−50
−60
x
y
⇒ x � �32, �4, 3
� �2x � 3��x � 4��x � 3�
f �x� � 2x3 � 5x2 � 21x � 36
136. (a)
(b)
5431−1−3−4−5
35
x
y
⇒ x � 0, 0, �2, 2
� 8x2��x2 � 4� � 8x2�2 � x��2 � x� f �x� � �8x4 � 32x2
138. (a)
(b)
8642−2−4
10x
y
⇒ x � 5, �12, 13
� �x � 5��2x � 1��3x � 1� f �x� � 6x3 � 29x2 � 6x � 5
©H
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Miff
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ny. A
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rved
.
Section 7.4 Matrices and Systems of Equations 603
Section 7.4 Matrices and Systems of Equations
■ You should be able to use elementary row operations to produce a row-echelon form (or reducedrow-echelon form) of a matrix.
1. Interchange two rows
2. Multiply a row by a nonzero constant
3. Add a multiple of one row to another row
■ You should be able to use either Gaussian elimination with back-substitution or Gauss-Jordan eliminationto solve a system of linear equations.
139.
−1−2 1 2 3 4 5 6 7 8
1
2
3
4
8
9
x
yy � �12�x�2
x 0 1 2
y 16 8 4 2 1
�1�2 140.
−2−4−6−8−10 2 4 6 8 10
2
16
18
x
yy � �13�x
� 1
x 0 1
y 10 4 2 1.33
�1�2
141.
−1
−2
1 2 3
1
2
3
4
5
6
7
8
x
yy � 2x�1
� 1
x 0 1 2
y 0 1 3 7�12
�1�2142.
−1−2−3−4−5 1 2 3 4 5
1
3
4
56
7
8
9
x
yy � 3x�1
� 2
x 0 1 2
y 2.037 2.33 3 5
�2
143. Answers will vary.
Vocabulary Check
1. matrix 2. square 3. row matrix, column matrix
4. augmented matrix 5. coefficient matrix 6. row-equivalent
7. reduced row-echelon form 8. Gauss-Jordan elimination
©H
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.
604 Chapter 7 Linear Systems and Matrices
1. Since the matrix has one rowand two columns, its order is1 � 2.
2. Since the matrix has one rowand four columns, its order is1 � 4.
3. Since the matrix has threerows and one column, its orderis 3 � 1.
5. Since the matrix has two rowsand two columns, its order is2 � 2.
4. Since the matrix has threerows and four columns, itsorder is 3 � 4.
6. Since the matrix has two rowsand three columns, its order is2 � 3.
10. �100
�340
107
���
10
�5�
14.
�6x
�x4x
�
�
2y
y8y
�
�
�
�
z7z
10zz
�
�
�
�
5w3w6w
11w
�
�
�
�
�257
23�21
16.
13R1→
�1
4
2
�3
83
6�
�3
4
6
�3
8
6�
18.
�R1 � R2→�2R1 � R2→ �
1
0
0
2
�3
2
4
�7
�4
3212
6
�
12R1→�
1
1
2
2
�1
6
4
�3
4
32
2
9
�
�2
1
2
4
�1
6
8
�3
4
3
2
9�
7.
� 6
�2
�7
5��
11
�1�
6x�2x
�
�
7y5y
�
�
11�1
8.
�7
5
4
�9��
22
15�
�7x5x
�
�
4y9y
�
�
2215
9.
�1
5
2
10
�3
1
�2
4
0
���
2
0
6�
� x5x2x
�
�
�
10y3yy
�
�
2z4z
�
�
�
206
11.
�314
�1��
9
�3�
3xx
�
�
4yy
�
�
9�3
12.
�7
8
�5
3��
0
�2�
�7x8x
�
�
5y3y
�
�
0�2
13.
� 9x�2x
x
�
�
�
12y18y7y
�
�
�
3z5z8z
�
�
�
010
�4
�9
�21
12187
35
�8
���
010
�4�
15.
�2R1 � R2→ �1
0
4
2
3
�1�
�1
2
4
10
3
5�
17.
15R2→�1
0
0
1
1
3
4
�25
20
�165
4�
�3R1 � R2→2R1 � R3→
�1
0
0
1
5
3
4
�2
20
�1
6
4�
�1
3
�2
1
8
1
4
10
12
�1
3
6�
©H
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Miff
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.
Section 7.4 Matrices and Systems of Equations 605
30.
(a) (b) (c) (d)
(e) (f) This matrix is in reduced row-echelon form.�1
0
0
0
0
1
0
0��
1
0
0
0
5
1
19
�34�
�1
0
0
0
5
2
19
�34��
1
0
0
7
5
2
19
1��
1
0
�3
7
5
2
4
1��
7
0
�3
1
1
2
4
5�
�7
0
�3
4
1
2
4
1�
19. Add times Row 2 to Row 1.�3 20. 3 times Row 1 added to Row 2.
21. Interchange Rows 1 and 2. 22. 5 times Row 1 added to Row 3.
23.
This matrix is in reduced row-echelon form.
�1
0
0
0
1
0
0
1
0
0
5
0� 24.
This matrix is in reduced row-echelon form.
�1
0
0
3
0
0
0
1
0
0
8
0�
25.
The first nonzero entries in rows one and two are not one. The matrix is not in row-echelon form.
�3
0
0
0
�2
0
3
0
1
7
4
5� 26.
This matrix is in row-echelon form, but not reducedrow-echelon form.
�1
0
0
0
1
0
2
�3
1
1
10
0�
27.
The first nonzero entry in row three is two, not one.The matrix is not in row-echelon form.
�1
0
0
0
1
0
0
0
0
1
�1
2� 28.
The matrix is in row-echelon form, but not reducedrow-echelon form. There is a one above the leadingone of row three.
�1
0
0
0
1
0
1
0
1
0
2
0�
29.
(a) (b) (c)
(d) (e) This matrix is in reduced row-echelon form.�1
0
0
0
1
0
�1
2
0��
1
0
0
2
1
0
3
2
0�
�1
0
0
2
�5
0
3
�10
0��
1
0
0
2
�5
�5
3
�10
�10��
1
0
3
2
�5
1
3
�10
�1�
�1
2
3
2
�1
1
3
�4
�1�
©H
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.
606 Chapter 7 Linear Systems and Matrices
36.
�2R2 � R3→�1
0
0
�3
1
0
0
1
0
�7
2
0�
3R1 � R2→�4R1 � R3→
�1
0
0
�3
1
2
0
1
2
�7
2
4�
�1
�3
4
�3
10
�10
0
1
2
�7
23
�24�
33.
(Answers may vary.)
�14R3→ �
1
0
0
2
1
0
312
1
0
�56
�3512�
�2R2 � R3→�1
0
0
2
1
0
312
�4
0
�56
353
�
16R2→�1
0
0
2
1
2
312
�3
0
�56
10�
R1 � R2→�2R1 � R3→
�1
0
0
2
6
2
3
3
�3
0
�5
10�
�1
�1
2
2
4
6
3
0
3
0
�5
10
� 34.
�3R2 � R3 → �
1
0
0
2
1
0
�1
�2
1
3
5
�1�
�3R1 � R2 →2R1 � R3 →
�1
0
0
2
1
3
�1
�2
�5
3
5
14�
�1
3
�2
2
7
�1
�1
�5
�3
3
14
8�
31. (See Exercise 29.) (Answer is a series of screens.)
(a) (b)
(c) (d)
(e)
32. (a) (b)
(c) (d)
(e) (f)
35.
�2R2 � R3→
�1
0
0
�1
1
0
�1
6
0
1
3
0�
�5R1 � R2→ 6R1 � R3→
�1
0
0
�1
1
2
�1
6
12
1
3
6�
�1
5
�6
�1
�4
8
�1
1
18
1
8
0�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.4 Matrices and Systems of Equations 607
40.
5R2 � R1 → �1
001
02
2�6�
��1�R1 →
126R2 →�1
0�5
1�10
232
�6�
5R1 � R2 → ��10
526
1052
�32�156�
� 5�1
15
210
4�32�
R1
R2 ��1
551
102
�324�
38.
�1R2→ �1
0
0
3
0
0
0
1
0�
2R2 � R1→
6R2 � R3→ �1
0
0
3
0
0
0
�1
0�
�5R1 � R2→�2R1 � R3→ �
1
0
0
3
0
0
2
�1
6�
�1
5
2
3
15
6
2
9
10�37.
�4R3 � R1 →
3R3 � R2 → �100
010
001�
12R3 →�
100
010
4�3
1�
�R2 � R1 →
�2R2 � R3 → �100
010
4�3
2�
R1 � R2 →�2R1 � R3 → �
100
112
1�3�4�
13R1 →
�1
�12
104
1�4�2�
�3
�1
2
3
0
4
3
�4
�2�
39.
2R2 � R1 → �1
0
0
1
�37
�127
�87
107�
�
17R2 →�1
0�2
13
�127
�4107�
4R1 � R2 → �1
0�2�7
312
�4�10�
R1 →R2 → � 1
�4�2
130
�46�
��41
1�2
03
6�4�
41.
Answer: �x, y� � ��2, �3�
x � 2y � 4 � 2��3� � 4 � �2
y � �3
x � 2y � 4 42.
Answer: ��12, 3�
x � �12
x � 8�3� � 12
y � 3
x � 8y � 12
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
608 Chapter 7 Linear Systems and Matrices
44.
Answer: �x, y, z� � ��31, 12, �3�
x � �2y � 2z � 1 � �24 � 6 � 1 � �31
y � �z � 9 � 3 � 9 � 12
z � �3
y � z � 9
x � 2y � 2z � �1
50.
Answer: ��1, 3�
y � 3, x � �1
�20
01
��
�23�
�20
6�3
��
16�9�
�22
63
��
167�
43.
Answer: �8, 0, �2�
x � 8
x � 0 � 2��2� � 4
y � 0
y � ��2� � 2
�x � yy
�
�
2zzz
�
�
�
42
�2
45.
Answer: �7, �5�
y � �5
x � 7
�1
0
0
1��
7
�5� 46.
Answer: ��2, 4�
y � 4
x � �2
�10
01
��
�24�
47.
Answer: ��4, �8, 2�
z � 2
y � �8
x � �4
�1
0
0
0
1
0
0
0
1
���
�4
�8
2� 48.
Answer: �3, �1, 0�
z � 0
y � �1
x � 3
�1
0
0
0
1
0
0
0
1
���
3
�1
0�
49.
Answer: �3, 2�
x � 2�2� � 7 ⇒ x � 3
y � 2
�1
3R2→�1
0
2
1��
7
2� �2R1 � R2→
�1
0
2
�3��
7
�6�
�1
2
2
1��
7
8�� x2x
�
�
2yy
�
�
78
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.4 Matrices and Systems of Equations 609
54.
Answer: ��1, 0, 6, 4�
x � 9 � 4�0� � 3�6� � 2�4� � �1
y � �4 �45�6� �
15�4� � 0
w �16040
� 4, z � 50 � 4�11� � 6
�1000
�4100
3�
45
10
�215
1140
����
9�450
160�
�1
0
0
0
�4
1
0
0
3
�45
�25
�185
�215
�22525
����
9
�4
�20
�20�
�1000
�410
�13�7
3�8102
�22
�7�1
����
9�40
328�
�13
�4�2
�4�2
31
31
�2�4
�2�4
13
����
9�13�4
�10�
52.
No solution, inconsistent
�R2→�8R2 � R3→ �
1
0
0
2
1
0
���
0
6
�40�
�R1 � R2→�3R1 � R3→ �
1
0
0
2
�1
�8
���
0
6
8�
�1
1
3
2
1
�2
���
0
6
8�
� xx
3x
�
�
�
2yy
2y
�
�
�
068
51.
No solution, inconsistent
�14R2 →
�7R2 � R3 → ��1
00
110
���
�2214
�160�
R2 →R3 → �
�100
1�4
7
���
�22�56�62�
3R1 � R2 → 4R1 � R3 → �
�100
17
�4
���
�22�62�56�
��1
34
14
�8
���
�224
32���x
3x4x
�
�
�
y4y8y
�
�
�
�224
32
53.
Answer: �3, �2, 5, 0�
x � 25 � ��2� � 4�5� � 3
w � 0, z �18537 � 5, y � �52 � 10�5� � �2
�1000
�1100
4�10
370
2�3101537
����
25�52185
0�
�1000
�1100
4�10
3717
2�3105
����
25�5218585
��1000
�15
�12
4�13
10�3
2�5
3�1
����
25�75
52�19
��
31
�21
2�1
11
�1421
12
�11
����
02526�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
610 Chapter 7 Linear Systems and Matrices
58.
Answer: �� 32a �
32, 13a �
13, a�
x � � 32a �
32
y �13a �
13
z � a
12R1 →
�13R2 → �1
0
0
0
1
0
32
�13
0
�
�
�
3213
0�
�3R2 � R3 → �
200
0�3
0
310
���
3�1
0�
�2R1 � R2 →�4R1 � R3 → �
200
0�3�9
313
���
3�1�3�
�248
0�3�9
37
15
���
359�
�2x4x8x
�
�
�
3y9y
�
�
3z7z
15z
�
�
�
359
56.
Answer: �8, 10, 6�
�5R3 � R1 →
�100
010
001
���
8106�
12R1 →
�100
010
501
���
38106�
R2 � R1 →
�7R3 � R2 → �200
010
1001
���
76106�
17R1 →
�115R3→ �
200
�110
371
���
24526�
�1400
�710
217
�15
�
�
�
16852
�90��1400
�72
�3
21�1
�21
�
�
�
16814
�156��140
14
�72
�10
21�1
0
�
�
�
1681412�
�207
�12
�5
3�1
0
���
24146�55.
Answer: �4, �3, 2�
3R3 � R1→�7R3 � R2→�
1
0
0
0
1
0
0
0
1
���
4
�3
2�
�17R3→
�1
0
0
0
1
0
�3
7
1
���
�2
11
2�
�2R2 � R3→�1
0
0
0
1
0
�3
7
�7
���
�2
11
�14�
�3R1 � R2→�2R1 � R3→
�1
0
0
0
1
2
�3
7
7
���
�2
11
8�
�1
3
2
0
1
2
�3
�2
1
���
�2
5
4�
� x3x2x
�
�
y2y
�
�
�
3z2zz
�
�
�
�254
57.
Let any real number
Answer: �2a � 1, 3a � 2, a�
x � 2a � 1 ⇒ x � 2a � 1
y � 3a � 2 ⇒ y � 3a � 2
z � a,
R2 � R1→�R2→�
1
0
0
0
1
0
�2
�3
0
���
1
2
0�
�3R2 � R3→�1
0
0
1
�1
0
�5
3
0
���
3
�2
0�
�R1 � R2→�2R1 � R3→
�1
0
0
1
�1
�3
�5
3
9
���
3
�2
�6�
�1
1
2
1
0
�1
�5
�2
�1
���
3
1
0�
� xx
2x
�
�
y
y
�
�
�
5z2zz
�
�
�
310
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.4 Matrices and Systems of Equations 611
59.
Answer: �7, �3, 4�
R2 � R1→
�100
010
001
���
7�3
4�
� R3 � R1 →
R3 � R2 → �100
�110
001
���
10�3
4�
13 R3 →�
100
�110
1�1
1
���
14�7
4�
�R1 →
�5R2 � R3 → �100
�110
1�1
3
���
14�712�
2R1 � R2 →3R1 � R3 → �
�100
115
�1�1�2
���
�14�7
�23�
��1
23
1�1
2
�111
���
�142119�
��x2x3x
�
�
�
yy
2y
�
�
�
zzz
�
�
�
�142119
61.
Let any real number
Answer: �0, �4a � 2, a�
x � 0
y � �4a � 2
z � a,
�3
1
2
�1
3
1
5
2
12
4
20
8
����
6
2
10
4� ⇒ �
1
0
0
0
0
1
0
0
0
4
0
0
����
0
2
0
0�
�3xx
2x�x
�
�
�
�
3yy
5y2y
�
�
�
�
12z4z
20z8z
�
�
�
�
62
104
60.
Answer: �22, 36, 114�
3R2 � R1 →
�12R2 →�
100
010
001
�
�
�
2236
114�
�R3 � R1 →�R3 � R2 → �
100
�3�2
0
001
�
�
�
�86�72114�
�100
�3�2
0
111
�
�
�
2842
114��100
�38
�2
1�3
1
�
�
�
28�54
42��
21
�1
2�3
1
�110
���
22814�
62.
Let any real number
Answer: ��2a, a, a�
x � �2a
y � a
z � a,
�123
135
111
���
000� ⇒ �
100
010
2�1
0
���
000�
� x2x3x
�
�
�
y3y5y
�
�
�
zzz
�
�
�
000
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
612 Chapter 7 Linear Systems and Matrices
66. The solutions are not the same.
(a)
Answer:
(b)
Answer: ��3, �2, 2�
x � �4��2� � 11 � �3.y � �3�2� � 4 � �2,z � 2,
��25, �2, 2�
x � 3��2� � 4�2� � 11 � �25.y � 2 � 4 � �2,z � 2,
68. The solutions are not the same.
(a)
Answer:
(b)
Answer: �3, 6, �4�
x � 6 � 3��4� � 15 � 3.y � 2��4� � 14 � 6,z � �4,
��3, 6, �4�
x � �3�6� � ��4� � 19 � �3.y � �6��4� � 18 � 6,z � �4,
64. row reduces to
Answer: �1, 0, 4, �2�
�1000
0100
0010
0001
����
104
�2��
2315
1452
�102
�1
216
�1
����
�61
�33�
63.
Answer: ��5a, a, 3�
x � �5ay � a,z � 3,
�2x
xx
�3x
�
�
�
�
10y5y5y
15y
�
�
�
�
2z2zz
3z
�
�
�
�
663
�9�
2
1
1
�3
10
5
5
�15
2
2
1
�3
����
6
6
3
�9� ⇒ �
1
0
0
0
5
0
0
0
0
1
0
0
����
0
3
0
0�
65. Yes, the systems yield the same solutions.
(a)
Answer:
(b)
Answer: ��1, 1, �3�
x � �1 � 2��3� � 6 � �1y � �3��3� � 8 � 1,z � �3,
��1, 1, �3�
x � 2�1� � ��3� � 6 � �1y � 5��3� � 16 � 1;z � �3;
67. No, solutions are different.
(a)
Answer:
(b)
Answer: �19, 2, 8�
x � 6�2� � 8 � 15 � 19y � �5�8� � 42 � 2,z � 8,
��5, 2, 8�
x � 4�2� � 5�8� � 27 � �5y � 7�8� � 54 � 2,z � 8,
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.4 Matrices and Systems of Equations 613
70.
Answer: y � �x2 � 2x � 8
a � 8 � 2 � 9 ⇒ a � �1
b �32�8� � 14 ⇒ b � 2
c � 8
�12R2→
�3R2 � R3→�1
0
0
1
1
0
132
1
���
9
14
8�
�4R1 � R2→�9R1 � R3→
�1
0
0
1
�2
�6
1
�3
�8
���
9
�28
�76�
�1
4
9
1
2
3
1
1
1
���
9
8
5�
� f�1�f�2�f�3�
�
�
�
a4a9a
�
�
�
b2b3b
�
�
�
ccc
�
�
�
985
f�x� � ax2 � bx � c69.
Answer: y � x2 � 2x � 5
a � 2 � 5 � 8 ⇒ a � 1
b �32 �5� �
192 ⇒ b � 2
c � 5
�12R2 →
�3R2 � R3 → �1
0
0
1
1
0
132
1
���
8192
5�
�4R1 � R2 →�9R1 � R3 → �
1
0
0
1
�2
�6
1
�3
�8
���
8
�19
�52�
�1
4
9
1
2
3
1
1
1
���
8
13
20�
� f �1� � a � b � c � 8f �2� � 4a � 2b � c � 13f �3� � 9a � 3b � c � 20
f �x� � ax2 � bx � c
71.
Answer: y � 2x2 � x � 1
a � b � c � 2 ⇒ a � 2 � 1 � 1 � 2
�6b � 3c � 3 ⇒ �6b � 3 � 3 � 6 ⇒ b � �1
�5c � �5 ⇒ c � 1
��1�R2 � R3→ �1
0
0
1
�6
0
1
�3
�5
���
2
3
�5�
�4R1 � R2 →�9R1 � R3 → �
1
0
0
1
�6
�6
1
�3
�8
���
2
3
�2�
�1
4
9
1
�2
3
1
1
1
���
2
11
16�
� f �1� � a � b � c � 2f ��2� � 4a � 2b � c � 11f �3� � 9a � 3b � c � 16
f�x� � ax2 � bx � c 72.
Answer: y � �x2 � 2x � 2
a � b � c � �1 ⇒ a � �1 � 2 � 2 � �1
2b � c � �2 ⇒ 2b � �2 � 2 � �4 ⇒ b � �2
�2c � �4 ⇒ c � 2
R2 � R3→ �
1
0
0
1
2
0
1
1
�2
���
�1
�2
�4�
�13R2→�
1
0
0
1
2
�2
1
1
�3
���
�1
�2
�2�
�4R1 � R2→�4R1 � R3→
�1
0
0
1
�6
�2
1
�3
�3
���
�1
6
�2
�
�1
4
4
1
�2
2
1
1
1
���
�1
2
�6�
� f �1� � a � b � c � �1f ��2� � 4a � 2b � c � 2f �2� � 4a � 2b � c � �6
f �x� � ax2 � bx � c
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
614 Chapter 7 Linear Systems and Matrices
74.
Solving the system,
f �x� � �2x2 � 2x � 1
a � �2, b � �2, c � 1.
f ��2�f �1�f �2�
�
�
�
4aa
4a
�
�
�
2bb
2b
�
�
�
ccc
�
�
�
�3�3
�11
f �x� � ax2 � bx � c
76.
Solving the system,
f �x� � x3 � x2 � 2x � 1
a � 1, b � �1, c � 2, d � �1.
f ��2�f ��1�
f �1�f �2�
�
�
�
�
�8a�a
a8a
�
�
�
�
4bbb
4b
�
�
�
�
2ccc
2c
�
�
�
�
dddd
�
�
�
�
�17�5
17
f �x� � ax3 � bx2 � cx � d
73.
Solving the system,
f�x� � �9x2 � 5x � 11
a � �9, b � �5, c � 11.
f ��2�f ��1�
f �1�
�
�
�
4aaa
�
�
�
2bbb
�
�
�
ccc
�
�
�
�157
�3
f�x� � ax2 � bx � c
75.
Solving the system,
f�x� � x3 � 2x2 � 4x � 1
a � 1, b � �2, c � �4, d � 1.
f ��2�f ��1�
f �1�f �2�
�
�
�
�
�8a�a
a8a
�
�
�
�
4bbb
4b
�
�
�
�
2ccc
2c
�
�
�
�
dddd
�
�
�
�
�72
�4�7
f�x� � ax3 � bx2 � cx � d
77. amount at 7%, amount at 8%, amount at 10%
Answers: $150,000 at 7%, $750,000 at 8%, $600,000 at 10%
x � 750,000 � 600,000 � 1,500,000 ⇒ x � 150,000
y � 3�600,000� � 2,550,000 ⇒ y � 750,000
7z � 4,200,000 ⇒ z � 600,00
100R2→4R2 � R3→ �
100
110
137
���
1,500,0002,550,0004,200,000�
�0.07R1 � R2→�4R1 � R3→ �
100
10.01�4
10.03�5
���
1,500,00025,500
�6,000,000�
�1
0.074
10.08
0
10.1�1
���
1,500,000130,500
0�� x
0.07x4x
�
�
y0.08y
�
�
�
z0.1z
z
�
�
�
1,500,000130,500
0
z �y �x �
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Section 7.4 Matrices and Systems of Equations 615
78. amount at 9%, amount at 10%, amount at 12%
Answers: $100,000 at 9%, $250,000 at 10%, $150,000 at 12%
x � 100,000y � 250,000,z � 150,000,
116R3→
�1
0
0
0
1
0
�2
3
1
���
�200,000
700,000
150,000�
�R2 � R1→
7R2 � R3→ �1
0
0
0
1
0
�2
3
16
���
�200,000
700,000
2,400,000�
100R2→2R3→ �
1
0
0
1
1
�7
1
3
�5
���
500,000
700,000
�2,500,000�
�0.09R1 � R2→�2.5R1 � R3→ �
1
0
0
1
0.01
�3.5
1
0.03
�2.5
���
500,000
7,000
�1,250,000�
�1
0.09
2.5
1
0.10
�1
1
0.12
0
���
500,000
52,000
0�
� x0.09x2.5x
�
�
�
y0.10y
y
�
�
z0.12z
�
�
�
500,00052,000
0
z �y �x �
79.
amperes; amperes; amperesI1 �115 �
910 � 0 ⇒ I1 �
1310I2 � 2� 9
10� � 4 ⇒ I2 �115I3 �
910
� 1
10R3→�1
0
0
�1
1
0
1
2
1
���
0
4910�
�4R2 � R3→ �1
0
0
�1
1
0
1
2
�10
���
0
4
�9�
12R2→�1
0
0
�1
1
4
1
2
�2
���
0
4
7�
R3→R2→ �
1
0
0
�1
2
4
1
4
�2
���
0
8
7�
�2R1 � R2→ �1
0
0
�1
4
2
1
�2
4
���
0
7
8�
�1
2
0
�1
2
2
1
0
4
���
0
7
8�
� I1
2I1
�
�
I2
2I2
2I2
�
�
I3
4I3
�
�
�
078
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616 Chapter 7 Linear Systems and Matrices
80. (a)
(b)
(c) Maximum height feet
Strikes ground when feet.
(d) Complete the square:
Maximum height feet
Range: � 103.793 feety � 0 ⇒ x ��0.367 ± �0.3672 � 4�0.004�5
�0.008
� 13.418
�0.004�x2 � 91.75x � 2104.5� � 5 � 8.418
x � 104�y � 0�
� 13
12000
18
y � �0.004x2 � 0.367x � 5
�0
225900
01530
111
���
5.09.6
12.4� ⇒ 1
�00
010
001
���
�0.0040.367
5�� f�0�
f�15�f�30�
�
�
�
c225a900a
�
�
�
5.015b30b
�
�
cc
�
�
9.612.4
f�x� � ax2 � bx � c
81. (a)
Solving the system using matrices,
you obtain
(b)
(c) For 2005, and dollars
For 2010, and dollars
For 2015, and dollarsy � 103.08t � 15
y � 86.33t � 10
y � 67.58t � 5
0 50
80
y � �0.04t2 � 4.35t � 46.83
a � �0.04, b � 4.35, c � 46.83.
�49
16
234
111
���
55.3759.5263.59� ⇒
1
�00
010
001
���
�0.044.35
46.83�
� 4a9a
16a
�
�
�
2b3b4b
�
�
�
ccc
�
�
�
55.3759.5263.59
�2, 55.37��3, 59.52��4, 63.59�
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Section 7.4 Matrices and Systems of Equations 617
82. (a)
Solving the system using matrices,
you obtain
(b)
(c) For 2005, and thousand
For 2010, and thousand
For 2015, and thousand
(d) Answers will vary.
y � 12.6t � 15
y � 36.6t � 10
y � 45.6t � 5
0 50
60
y � �0.3t2 � 2.7t � 39.6
a � �0.3, b � 2.7, c � 39.6.
�49
16
234
111
���
43.845.045.6� ⇒
1
�00
010
001
���
�0.32.7
39.6�
� 4a9a
16a
�
�
�
2b3b4b
�
�
�
ccc
�
�
�
43.845.045.6
�2, 43.8��3, 45.0��4, 45.6�
84. Let number of adults
Let number of students
Let number of children
Solving the system,
120 adults, 80 students and 60 children
x � 120, y � 80 and z � 60.
� 5xx
�12x
�
�
3.5y
y
�
�
2.5z2z
�
�
�
10300
20
z �
y �
x �
83. Let number of pounds of glossy.
Let number of pounds of semi-glossy
Let number of pounds of matte
Solving the system,
50 pounds of glossy, 35 pounds of semi-glossy and 15 pounds of matte
x � 50, y � 35 and z � 15.
� x5.5x
�
�
y4.25y
y
�
�
�
z3.75z
z
�
�
�
10048050
z �
y �
x �
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618 Chapter 7 Linear Systems and Matrices
x1 � x3 � 600x1 � x2 � x4 ⇒ x1 � x2 � x4 � 0x2 � x5 � 500x3 � x6 � 600x4 � x7 � x6 ⇒ x4 � x6 � x7 � 0x5 � x7 � 500
85. (a)
�R3 � R2→�R4 � R3→
�R4→
1
0�0
0
0
0
0
�1
0
0
0
0
1
0
�1
0
0
0
0
0
0
1
0
0
0
�1
0
�1
1
0
0
0
�1
�1
0
0
0
0
0
0
1
0
������
600
�500
�600
�500
500
0
�
�R1 � R2→R2 � R3→R3 � R4→R4 � R5→
�R5 � R6→�
1
0
0
0
0
0
0
�1
0
0
0
0
1
�1
�1
0
0
0
0
�1
�1
�1
0
0
0
0
1
1
1
0
0
0
0
1
0
0
0
0
0
0
1
0
������
600
�600
�100
500
500
0
�
�1
1
0
0
0
0
0
�1
1
0
0
0
1
0
0
1
0
0
0
�1
0
0
1
0
0
0
1
0
0
1
0
0
0
1
�1
0
0
0
0
0
1
1
������
600
0
500
600
0
500
�
Let then:
(b) If
x6 � x7 � 0
x5 � 500
x4 � 0
x3 � 600
x2 � 0
x1 � 0
x6 � x7 � 0, then s � t � 0, and
x1 � 600 � �600 � s� � s
x2 � 500 � �500 � t� � t
x3 � 600 � s
x4 � �500 � s � �500 � t� � s � t
x5 � 500 � t
x7 � t and x6 � s,
(c) IfThus:
x7 � �500
x6 � 0
x5 � 1000
x4 � 500
x3 � 600
x2 � �500
x1 � 0
x5 � 1000 and x6 � 0, then s � 0 and t � �500.
©H
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.
Section 7.4 Matrices and Systems of Equations 619
86. (a)
Let
x1 � 200 � s � t � 300 ⇒ x1 � 500 � s � t
x2 � s � 350 � t � 150 ⇒ x2 � �200 � s � t
Let x3 � s.
x4 � t � 350 ⇒ x4 � 350 � t
x5 � t.
�R2→�R3→
R3 � R4→�1
0
0
0
1
1
0
0
0
�1
0
0
0
1
1
0
0
0
1
0
����
300
150
350
0�
R2 � R3→ �
1
0
0
0
1
�1
0
0
0
1
0
0
0
�1
�1
1
0
0
�1
1
����
300
�150
�350
350�
�R1 � R2→ �
1
0
0
0
1
�1
1
0
0
1
�1
0
0
�1
0
1
0
0
�1
1
����
300
�150
�200
350�
�1
1
0
0
1
0
1
0
0
1
�1
0
0
�1
0
1
0
0
�1
1
����
300
150
�200
350�
�x1
x1
x2
x4
�
�
�
�
x2
x3
200x5
�
�
�
�
300150
x3
350
� x4
� x5
⇒ x1 � x3 � x4 � 150 ⇒ x2 � x3 � x5 � �200
(b) When and
(c) When and
x1 � 150
x2 � 150,x3 � 0,x4 � 0,x5 � 350,
150 � �200 � 0 � t ⇒ t � 350.
x2 � �200 � s � t
x3 � 0:x2 � 150
x1 � 100
x2 � 200,x3 � 50,x4 � 0,x5 � 350,
200 � �200 � 50 � t ⇒ t � 350.
x2 � �200 � s � t
x3 � 50:x2 � 200
87. False. It is a matrix.2 � 4 88. False. Gauss-Jordan elimination reduces a matrixto reduced row-echelon form.
89.
� xx
2x
�
�
�
y2yy
�
�
�
7z11z10z
�
�
�
�10
�3
2�Equation 1� � �Equation 2� → new Equation 3
�Equation 1� � 2�Equation 2� → new Equation 2
�Equation 1� � �Equation 2� → new Equation 1
Equation 1Equation 2�x
y�
�
3z4z
�
�
�21
90. (a) In the row-echelon form of an augmentedmatrix that corresponds to an inconsistentsystem of linear equations, there exists a rowconsisting of all zeros except for the entry inthe last column.
(b) In the row-echelon form of an augmentedmatrix that corresponds to a system with aninfinite number of solutions, there are fewerrows with nonzero entries than there are variables. Nor does the last row consist of allzeros except for the entry in the last column.
91. The row operation was not performedon the last column. Nor was �R2 � R1.
�2R1 � R2 92. Answers will vary.
©H
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620 Chapter 7 Linear Systems and Matrices
93.
Asymptotes: x � �1, y � 0
128−4−8−12
4
−4
−12
−8
x
y
f �x� �7
�x � 1
95.
Asymptotes: x � 4, y � x � 2
4
−4
−8 128
12
16
8
x
y
f �x� �x2 � 2x � 3
x � 4� x � 2 �
5x � 4
94.
Horizontal asymptote: y � 0
2−4
0.4
−0.8
−2 4 6 8
0.2
0.6
0.8
x
y
f �x� �4x
5x2 � 2
96.
Vertical asymptote:
Slant asymptote: y � x � 1
x � �1,
8 1−4− 28
4
8
−12
−8
−12
x
y
f �x� �x2 � 36x � 1
■ if and only if they have the same order and
■ You should be able to perform the operations of matrix addition, scalar multiplication, and matrixmultiplication.
■ Some properties of matrix addition and scalar multiplication are:
(a) (b)
(c) (d)
(e) (f)
■ Some properties of matrix multiplication are:
(a) (b)
(c) (d)
■ You should remember that in general.AB � BA
c�AB� � �cA�B � A�cB��A � B�C � AC � BC
A�B � C� � AB � ACA�BC� � �AB�C
�c � d�A � cA � dAc�A � B� � cA � cB
1A � A�cd�A � c�dA�A � �B � C� � �A � B� � CA � B � B � A
aij � bij.A � B
Section 7.5 Operations with Matrices
Vocabulary Check
1. equal 2. scalars 3. zero, 0 4. identity
5. (a) iii (b) i (c) iv (d) v (e) ii 6. (a) ii (b) iv (c) i (d) iii
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.
Section 7.5 Operations with Matrices 621
1. x � �4, y � 22
3.
3z � 14 � 4 ⇒ z � 6
3y � 12 ⇒ y � 4
2x � 7 � 5 ⇒ x � �1
5. (a)
(b)
(c)
(d) 3A � 2B � �3
6
�3
�3� �2 � 2
�1
�1
8� � �3
6
�3
�3� � ��4
2
2
�16� � ��1
8
�1
�19�
3A � 3�1
2
�1
�1� � �3�1�3�2�
3��1�3��1�� � �3
6
�3
�3�
A � B � �1
2
�1
�1� � � 2
�1
�1
8� � �1 � 2
2 � 1
�1 � 1
�1 � 8� � ��1
3
0
�9�
A � B � �1
2
�1
�1� � � 2
�1
�1
8� � �1 � 2
2 � 1
�1 � 1
�1 � 8� � �3
1
�2
7�
2. x � 13, y � 12
4.
z � 2 � 11 ⇒ z � 9
2y � �8 ⇒ y � �4
x � 4 � 2x � 9 ⇒ x � �5
6. (a)
(b)
(c)
(d) 3A � 2B � �3
6
6
3� � 2��3
4
�2
2� � �3 � 6
6 � 8
6 � 4
3 � 4� � � 9
�2
10
�1�
3A � 3�1
2
2
1� � �3�1�3�2�
3�2�3�1�� � �3
6
6
3�
A � B � �1
2
2
1� � ��3
4
�2
2� � �1 � 3
2 � 4
2 � 2
1 � 2� � � 4
�2
4
�1�
A � B � �1
2
2
1� � ��3
4
�2
2� � �1 � 3
2 � 4
2 � 2
1 � 2� � ��2
6
0
3�
7.
(a) (b)
(c) (d) 3A � 2B � �246
�12
�39
15� � �2
�22
12�10
20� � �228
�14
�1519
�5�3A � �246
�12
�39
15�
A � B � �73
�5
�78
�5�A � B � �91
�3
5�215�
A � �82
�4
�135�, B � �
1�1
1
6�510�
8. (a)
(b)
(c)
(d) 3A � 2B � �30
�318
927� � ��4
�608
�10�14� � �7
6�310
1941�
3A � �30
�318
927�
A � B � �33
�12
816�
A � B � �10
�16
39� � ��2
�304
�5�7� � ��1
�3�110
�22�
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622 Chapter 7 Linear Systems and Matrices
9.
(a)
(b)
(c)
(d)
� �1015
15�10
�1�10
73
1214�
3A � 2B � �123
156
�3�6
9�3
120� � � 2
�120
16 �2
42
�60
�14�
3A � �123
156
�3�6
9�3
120�
A � B � �37
5�6
0�4
22
47�
A � B � � 5�5
510
�20
4�4
4�7�
B � � 1�6
08
�12
1�3
0�7�A � �4
152
�1�2
3�1
40�,
10. (a)
(c) 3A � ��3
9150
�12
12�61224
�3
06
�3�18
0�
A � B � ��4
5153
�4
9�6�5100
1�5�2
�10�2
� (b)
(d) 3A � 2B � �35
�5�6
�12
22
3020
�5
�220
�1�10
4�
A � B � �21
�5�3�4
�12
136
�2
�190
�22�
11.
(a) is not possible.
(b) is not possible.
(c)
(d) is not possible.3A � 2B
3A � � 18�3
0�12
90�
A � B
A � B
B � �84
�1�3�A � � 6
�10
�430�, 12. (a) is not defined.
(b) is not defined.
(c)
(d) is not defined.3A � 2B
3A � 3�32
�1� � �96
�3�A � B
A � B
13. � ��53
0�6� � ��3
12�7
5� � ��815
�7�1���5
30
�6� � � 7�2
1�1� � ��10
14�8
6�
14. ��6
�17
901� � �
0�2
3
5�1�6�� � �
�134
�6
�7�1
0� � �6
�310
14�1�5� � �
�134
�6
�7�1
0� � ��7
14
7�2�5�
15. ��24�12
�432
1212�4���4
002
13� � �2
31
�6�2
0�� � 4��6�3
�18
33� �
16. �12 �19 4 �14 9� � �19
2 2 �7 92�1
2�5 �2 4 0� � �14 6 �18 9��
17. � 2�1
5�4� � ��3
202� � ��1
15
�2� 18. ��8�9
925�
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Section 7.5 Operations with Matrices 623
19. �12�
3.211�1.004
0.055
6.8294.914
�3.889� � 8�1.6305.256
�9.768
�3.0908.3354.251� � �
�14.645�41.546
78.117
21.305�69.137�32.064�
20.
� ��4
2�9
�111
�3� � �13
�1
1
2312
2� � ��113
1
�8
�31332
�1�� �
�42
�9
�111
�3� �16�
2�6
6
43
12���1��4
�29
11�1
3� �16��
�530
�14
13� � �7
�96
5�1�1��
21. X � 3��2
1
3
�1
0
�4� � 2 �
0
2
�4
3
0
�1� � �
�6
3
9
�3
0
�12� � �
0
4
�8
6
0
�2� � �
�6
�1
17
�9
0
�10�
22.
X � A �12B � �
�2
1
3
�1
0
�4� �
12�
0
2
�4
3
0
�1� � �
�2
1
3
�1
0
�4� � �
0
1
�2
32
0� 1
2� � �
�2
0
5
� 52
0� 7
2�
2X � 2A � B
23. X � �32 A �
12 B � �
32�
�2
1
3
�1
0
�4� �
12 �
0
2
�4
3
0
�1� � �
3
�12
�132
3
0112
�24.
X � �A � 2B � �1��2
1
3
�1
0
�4� � 2�
0
2
�4
3
0
�1� � �
2
�1
�3
1
0
4� � �
0
�4
8
�6
0
2� � �
2
�5
5
�5
0
6�
2A � 4B � �2X
25. is and is is not defined.3 � 3 ⇒ ABB3 � 2A 26. AB � �067
�10
�1
238� �
241
�1�5
6� � ��21518
171246�
27. AB � ��1�4
0
653� �2
039� � �
�2�8
0
513327� 28. is is is
�1
0
0
0
4
0
0
0
�2� �
3
0
0
0
�1
0
0
0
5� � �
3
0
0
0
�4
0
0
0
�10�
3 � 3.3 � 3 ⇒ AB3 � 3, BA
29.
AB � �5
0
0
0
�8
0
0
0
7� �
15
0
0
0
�18
0
0
012� � �
1
0
0
0
1
0
0
072�
A is 3 � 3, B is 3 � 3 ⇒ AB is 3 � 3.
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
624 Chapter 7 Linear Systems and Matrices
30. �0
0
0
0
0
0
5
�3
4� �
6
8
0
�11
16
0
4
4
0� � �
0
0
0
0
0
0
0
0
0�
31. AB � �56���3 �1 �5 �9� � ��15
�18�5 �6
�25�30
�45�54�
32. is is is not defined.2 � 2 ⇒ AB2 � 4, BA
33. (a)
(b)
(c) A2 � �15
22��
15
22� � �1 � 10
5 � 102 � 4
10 � 4� � �1115
614�
BA � � 2�1
�18��
15
22� � � 2 � 5
�1 � 404 � 2
�2 � 16� � ��339
214�
� �08
1511�AB � �1
522��
2�1
�18� � � 2 � 2
10 � 2�1 � 16�5 � 16�
34. (a) (b)
(c) A2 � AA � � 30�4
610�
BA � ��124
�6�10�AB � � 6
�23
�4���2
204� � ��6
�412
�16�
35. (a)
(b)
(c) A2 � �3
1
�1
3� �3
1
�1
3� � �9 � 1
3 � 3
�3 � 3
�1 � 9� � �8
6
�6
8�
BA � �1
3
�3
1� �3
1
�1
3� � �3 � 3
9 � 1
�1 � 9
�3 � 3� � � 0
10
�10
0�
AB � �3
1
�1
3� �1
3
�3
1� � �3 � 3
1 � 9
�9 � 1
�3 � 3� � � 0
10
�10
0�
36. (a)
(b)
(c) A2 � �1
1
�1
1� �1
1
�1
1� � �1�1� � ��1��1�1�1� � �1��1�
1��1� � ��1��1�1��1� � 1�1�� � �0
2
�2
0�
BA � � 1
�3
3
1� �1
1
�1
1� � � 1�1� � �3�1�3�1� � �1��1�
1��1� � 3�1��3��1� � 1�1�� � � 4
�2
2
4�
AB � �1
1
�1
1� � 1
�3
3
1� � �1�1� � ��1���3�1�1� � 1��3�
1�3� � ��1��1�1�3� � 1�1�� � � 4
�2
2
4�
37. (a)
(b)
(c) is not defined.A2
BA � �1 1 2� �78
�1� � �7 � 8 � 2� � �13�
AB � �78
�1� �1 1 2� � �78
�1
78
�1
1416
�2�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.5 Operations with Matrices 625
38. (a)
(b)
(c) The number of columns of does not equal the number of rows of the multiplication is not possible.A;A
BA � �2
3
0� �3 2 1� � �
2�3�3�3�0�3�
2�2�3�2�0�2�
2�1�3�1�0�1�� � �
6
9
0
4
6
0
2
3
0�
AB � �3 2 1� �2
3
0� � �3�2� � 2�3� � 1�0�� � �12�
39. AB � �703216
�1711
�38
736
70� 40. AB � �124228192
�70452
�72�
41. ��3
�12
5
8
15
�1
�6
9
1
8
6
5� �
3
24
16
8
1
15
10
�4
6
14
21
10� � �
151
516
47
25
279
�20
48
387
87�
42. is is is not defined.
4 � 2 ⇒ AB3 � 3, BA 43.is not defined.A is 2 � 4 and B is 2 � 4 ⇒ AB 44. AB � �
�238119210
50115135
�484342555�
45. ��30
1�2��
1�2
02���
12
04� � �1
42
�4��12
04� � � 5
�48
�16�
46. � 27�6
�6�27�
47. � ��43
1014��0
421
�22���
40
�1
0�1
2� � ��2�3
0
35
�3�� � �04
21
�22��
2�3�1
34
�1�
48. �3
�157��4 2� � �
12�42028
6�21014
�49.
(a) is not a solution.
(b) is a solution.
(c) is not a solution.
(d) is not a solution.� 2�3� ⇒ �1
322��
2�3� � ��4
0�
��44� ⇒ �1
322��
�44� � � 4
�4�
��23� ⇒ �1
322��
�23� � �4
0�
�21� ⇒ �1
322��
21� � �4
8�
�13
22
��
40�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
626 Chapter 7 Linear Systems and Matrices
50. Let
(a) is a solution.
(b) is not a solution.
(c) is not a solution.
(d) is not a solution.A��39� � � 0
48� ⇒ ��39�
A� 3�9� � � 0
�48� ⇒ � 3�9�
A� 2�6� � � 0
�32� ⇒ � 2�6�
A��13� � � 0
16� ⇒ ��13�
A � � 6�1
25�� 6
�125
��
016�. 51.
(a) is not a
solution.
(b) is a
solution.
(c) is not
a solution.
(d) is not
a solution.
�42� ⇒ ��2
4�3
2��42� � ��14
20�
��66� ⇒ ��2
4�3
2���6
6� � � �6�12�
� 6�2� ⇒ ��2
4�3
2��6
�2� � ��620�
�30� ⇒ ��2
4�3
2��30� � ��6
12�
��24
�32
��
�620�
52. Let
(a) is a solution.
(b) is not a solution.
(c) is not a solution.
(d) is not a solution.A� 211� � ��67
17� ⇒ � 211�
A��4�5� � � 15
�17� ⇒ ��4�5�
A�52� � �11
17� ⇒ �52�
A�45� � ��15
17� ⇒ �45�
A � �53
�71��5
3�7
1��
�1517�. 53. (a)
(b) By Gauss-Jordan elimination on
we have
Thus, X � �48�.x1 � 4 and x2 � 8.
�R2 � R1→
�R2→ �1
0
0
1��
4
8�,
�R1→
2R1 � R2→ �1
0
�1
�1��
�4
�8�
��1
�2
1
1 ��
4
0�
A � ��1
�2
1
1�, X � �x1
x2�, B � �4
0�
54. (a)
(b) By Gauss-Jordan elimination on
we have
Answer: ��23�
x1 � �2 and x2 � 3.
�4R2 � R1→
�15R2→ �
1
0
0
1
..
.
..
.�2
3�,
�2R1 � R2→ �1
0
4
�5
..
.
..
.10
�15�
�1
2
4
3
..
.
..
.10
5�
A � �2
1
3
4�, X � �x1
x2�, B � � 5
10� 55. (a)
(b)
Answer: X � ��76�
x1 � �7, x2 � 6.
�
12R1 �1
001
��
�76�
3R2 � R1 ��2
001
��
146�
��18�R2
��20
�31
��
�46�
3R1 � R2
��20
�3�8
��
�4�48�
��26
�31
��
�4�36�
A � ��26
�31�, X � �x1
x2�, B � � �4
�36�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.5 Operations with Matrices 627
56. (a)
(b)
Answer: ��23�
353�
x2 � �353x1 � �23,
�13 R2 →�1
001
��
�23�
353�
��1�R2 � R1 → �1
00
�3��
�2335�
4R1 � R2 → �10
�3�3
��
1235�
��41
9�3
�1312�
R1
R2 � 1
�4�3
912
�13�
B � ��1312�x � �x1
x2�,A � ��4
19
�3�,57. (a)
(b)
Answer: X � �1
�12�.
x1 � 1, x2 � �1, x3 � 2.
�7R3 � R1 →�2R3 � R2 → �
1
0
0
0
1
0
0
0
1
���
1
�1
2�
2R2 � R1 →
R2 � R3 → �
1
0
0
0
1
0
7
2
1
���
15
3
2�
R1 � R2 →�2R1 � R3 →
�1
0
0
�2
1
�1
3
2
�1
���
9
3
�1�
�1
�1
2
�2
3
�5
3
�1
5
���
9
�6
17�
A � �1
�1
2
�2
3
�5
3
�1
5�, X � �
x1
x2
x3�, B � �
9
�6
17�
58. (a)
(b) row reduces to
Answer: �23232�
�100
010
001
21.51.5��
1�1
1
12
�1
�301
���
�112�
A � �1
�11
12
�1
�301�, X � �
x1
x2
x3�, B � �
�112�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
628 Chapter 7 Linear Systems and Matrices
60. (a)
(b)
Answer: �4
�52�x1 � 4x2 � �5,x3 � 2,
R2 � R1 →
�100
010
001
���
4�5
2�
4R3 � R1 →
�R3 � R2 →�R3 → �
100
�110
001
���
9�5
2�
6R2 � R3 → �100
�110
4�1�1
���
17�7�2�
14 R2 →�100
�11
�6
4�1
5
���
17�740�
�110
�13
�6
405
���
17�11
40�
��1�R1 � R2 → �100
�14
�6
4�4
5
���
17�28
40�B � �
17�11
40�x � �x1
x2
x3�,A � �
110
�13
�6
405�,
59. (a)
(b)
Answer: X � ��1
3�2�x1 � �1, x2 � 3, x3 � �2
5R2 � R1
�100
010
001
���
�13
�2� ��1
2�R2 �100
�510
001
���
�163
�2� �2R3 � R1
�5R3 � R2 �100
�5�2
0
001
���
�16�6�2�
�
130R3
�100
�5�2
0
251
���
�20�16�2�
�7R2 � R3
�100
�5�2
0
25
�30
���
�20�16
60� R2
R3
�100
�5�2
�14
255
���
�20�16�52�
3R1 � R2 �100
�5�14�2
255
���
�20�52�16��
1�3
0
�51
�2
2�1
5
���
�208
�16�A � �
1�3
0
�51
�2
2�1
5�, X � �x1
x2
x3�, B � �
�208
�16�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.5 Operations with Matrices 629
61. (a)
(b)
The answers are the same.
AB � AC � �7
�616
�21311
5�8�3�
A�B � C� � �7
�616
�21311
5�8�3� 62. (a)
(b)
The answers are the same.
BA � CA � �94
�10
5�5�8
05
13�
�B � C�A � �94
�10
5�5�8
05
13�
63. (a)
(b)
The answers are the same.
A2 � AB � BA � B2 � �261111
112014
0�3
0�
�A � B�2 � �261111
112014
0�3
0� 64. (a)
(b)
The answers are the same.
A2 � AB � BA � B2 � �0
4125
9�12
0
�6�1
�26�
�A � B�2 � �0
4125
9�12
0
�6�1
�26�
65. (a)
(b)
The answers are the same.
�AB�C � �25
�53�76
�343430
28�721�
A�BC� � �25
�53�76
�343430
28�721� 66. (a)
(b)
The answers are the same.
�cA�B � �33
�33�18
91239
�121218�
c�AB� � �33
�33�18
91239
�121218�
67. (a)
(b) � ��1�5
10�1
�40�A � cB � � 1
�121
�20� � 2��1
�24
�1�1
0�
A � cB � ��1�5
10�1
�40�
68. Not possible
B and C have different orders.
�a�, �b� A�B � C� 69. Not possible
Number of columns of A (3) does not equal thenumber of rows of B (2).
�a�, �b� c�AB�
70. (a)
(b) � ��41
�2�4
50�B � dA � ��1
�24
�1�1
0� � ��3�� 1�1
21
�20�
B � dA � ��41
�2�4
50�
71. Not possible
is is 2 � 2.BC3 � 3,CA
�a�, �b� CA � BC 72. Not possible
not defined, nor B2AB
�a�, �b� dAB2
73. (a)
(b) � ��66
�12�6
120�cd A � 2��3�� 1
�121
�20�
cd A � ��66
�12�6
120�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
630 Chapter 7 Linear Systems and Matrices
74. (a) cA � dB � �54
�85
�10�
75.
f�A� � A2 � 5A � 2I � �2
4
0
5 ��2
4
0
5� � 5 �2
4
0
5� � 2 �1
0
0
1� � ��4
8
0
2�
A � �2
4
0
5�
76.
f�A� � A2 � 7A � 6 � �5
1
4
2� �5
1
4
2� � 7�5
1
4
2� � 6�1
0
0
1� � �0
0
0
0�
A � �5
1
4
2�
77. 1.20�7035
50100
2570� � �84
4260
1203084� 78. 1.10�100
40
90
20
70
60
30
60� � �110
44
99
22
77
66
33
66�
79.
The entries in the last matrix BA represent the profit for both crops at each of the three outlets.
� �1037.50 1400 1012.50�BA � �3.50 6.00� �125100
100175
75125�
80.
The entries represent the costs of the three models of the product at each of the two warehouses.
�39.50 44.50 56.50� �500060008000
400010,000
5000� � �916,500 885,500�
81.
The entries represent the wholesale and retail prices of the inventory at each outlet.
ST � �3
0
4
2
2
2
2
3
1
3
4
3
0
3
2� �
840
1200
1450
2650
3050
1100
1350
1650
3000
3200� � �
$15,770
$26,500
$21,260
$18,300
$29,250
$24,150�
82.
This represents the labor cost for each boat size at each plant.
ST � �1.0
1.6
2.5
0.5
1.0
2.0
0.2
0.2
0.4� �
12
9
6
10
8
5� � �
$17.70
$29.40
$50.40
$15.00
$25.00
$43.00�
83.
This product represents the changes in party affiliation after two elections.
P2 � �0.6
0.2
0.2
0.1
0.7
0.2
0.1
0.1
0.8� �
0.6
0.2
0.2
0.1
0.7
0.2
0.1
0.1
0.8� � �
0.40
0.28
0.32
0.15
0.53
0.32
0.15
0.17
0.68�
(b)
� �54
�85
�10�
� � 2�2
42
�40� � �3
6�12
33 0 �
cA � dB � 2� 1�1
21
�20� � ��3���1
�24
�1�1
0�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.5 Operations with Matrices 631
84.
As is raised to higher and higher powers, the resulting matrices appear to be approaching the matrix
�0.2
0.3
0.5
0.2
0.3
0.5
0.2
0.3
0.5�.
P
P8 � �0.203
0.305
0.492
0.199
0.309
0.492
0.199
0.292
0.508�
P7 � �0.206
0.308
0.486
0.198
0.316
0.486
0.198
0.288
0.514�
P6 � �0.213
0.311
0.477
0.197
0.326
0.477
0.197
0.280
0.523�
P5 � P4P � �0.250
0.315
0.435
0.188
0.377
0.435
0.188
0.248
0.565� �
0.6
0.2
0.2
0.1
0.7
0.2
0.1
0.1
0.8� � �
0.225
0.314
0.461
0.194
0.345
0.461
0.194
0.267
0.539�
P4 � P3P � �0.300
0.308
0.392
0.175
0.433
0.392
0.175
0.217
0.608� �
0.6
0.2
0.2
0.1
0.7
0.2
0.1
0.1
0.8� � �
0.250
0.315
0.435
0.188
0.377
0.435
0.188
0.248
0.565�
P3 � P2P � �0.4
0.28
0.32
0.15
0.53
0.32
0.15
0.17
0.68� �
0.6
0.2
0.2
0.1
0.7
0.2
0.1
0.1
0.8� � �
0.300
0.308
0.392
0.175
0.433
0.392
0.175
0.217
0.608�
85. True 86. False.
�40
0�1��
�62
�2�6� � ��24
�2�8
6�
��62
�2�6��
40
0�1� � ��24
826�
For 87–93, A is of order B is of order C is of order and D is of order 2 � 2.3 � 22 � 3,2 � 3,
87. is not possible. and are not of thesame order.
CAA � 2C 88. Not possible
89. AB is not possible. The number of columns of Adoes not equal the number of rows of B.
90. Possible. Order 2 � 2
91. is possible. The resulting order is 2 � 2.BC � D 92. Not possible
93. is possible. The resulting order is 2 � 3.D�A � 3B� 94. Possible.Order 2 � 3
95.
A2 � 2AB � B2 � �03
02�
�A � B�2 � �12
01� 96.
A2 � 2AB � B2 � �87
�1622�
�A � B�2 � �78
�1623�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
632 Chapter 7 Linear Systems and Matrices
97.
A2 � B2 � �25
�24�
�A � B��A � B� � �34
�23� 98. �A � B�2 � �1
201� � A2 � AB � BA � B2
99.
AC � BC, but A � B.
BC � �1
1
0
0� �2
2
3
3� � �2
2
3
3�
AC � �0
0
1
1� �2
2
3
3� � �2
2
3
3� 100.
but and B � 0.A � 0AB � 0
AB � �3
4
3
4� � 1
�1
�1
1� � �0
0
0
0�
A � �3
4
3
4�, B � � 1
�1
�1
1�
101. (a)
(b)
The identity matrix
B2 � �0
i
�i
0� �0
i
�i
0� � �1
0
0
1�,
A4 � A3A � ��i
0
0
�i� � i
0
0
i� � �1
0
0
1� and i4 � 1
A3 � A2A � ��1
0
0
�1� � i
0
0
i� � ��i
0
0
�i� and i3 � �i
A2 � � i
0
0
i� � i
0
0
i� � ��1
0
0
�1� and i2 � �1
102. The product of two diagonal matrices of the same order is a diagonal matrix whoseentries are the products of the corresponding diagonal entries of A and B.
103. (a)
(b) and are both zero matrices.
(c) If A is then will be the zero matrix.
(d) If A is then is the zero matrix.Ann � n,
A44 � 4,
B3A2
A � �0
0
2
0�, B � �0
0
0
2
0
0
3
4
0� 104. Matrix multiplication can be used to find the
number of subscribers each company will haveone year later. Multiply a matrix containingthe current number of subscribers per company on the left by the given matrix.
3 � 1
105. 3 ln 4 �13
ln�x2 � 3� � ln 43 � ln�x2 � 3�13 � ln� 64�x2 � 3�13�
106.
� ln� x�x2 � 36�3�
� ln x � ln�x2 � 36�3
ln x � 3�ln�x � 6� � ln�x � 6�� � ln x � 3�ln�x � 6��x � 6��
107.
� ln��x � 5�xx � 8 �
12
�2 ln�x � 5� � ln x � ln�x � 8�� � ln�x � 5� � ln x12 � ln�x � 8�12
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.6 The Inverse of a Square Matrix 633
108.
� ln�73�2t6
t3 � � ln�73�2t3�
� ln�73�2t6� � ln t3
32
ln 7t4 �35
ln t5 � ln�7t4�3�2 � ln�t5�3�5
Section 7.6 The Inverse of a Square Matrix
■ You should be able to find the inverse, if it exists, of a square matrix.
(a) Write the matrix that consists of the given matrix on the left and the identity matrix on the right to obtain Note that we separate the matrices and by a dotted line. We callthis process adjoining the matrices and
(b) If possible, row reduce to using elementary row operations on the entire matrix The result will be the matrix If this is not possible, then is not invertible.
(c) Check your work by multiplying to see that
■ You should be able to use inverse matrices to solve systems of equation.
■ You should be able to find inverses using a graphing utility.
AA�1 � I � A�1A.
A�I � A�1�.�A � I�.IA
I.AIA�A � I�.
In � nAn � 2n
1.
BA � � 3
�5
�1
2� �2
5
1
3� � �3�2� � ��1��5��5�2� � 2�5�
3�1� � ��1��3��5�1� � 2�3�� � �1
0
0
1�
AB � �2
5
1
3� � 3
�5
�1
2� � �2�3� � 1��5�5�3� � 3��5�
2��1� � 1�2�5��1� � 3�2�� � �1
0
0
1�
2.
BA � �2
1
1
1� � 1
�1
�1
2� � �2 � 1
1 � 1
�2 � 2
�1 � 2� � �1
0
0
1�
AB � � 1
�1
�1
2� �2
1
1
1� � � 2 � 1
�2 � 2
1 � 1
�1 � 2 � � �1
0
0
1�
3.
BA � ��232
1
�12� �1
3
2
4� � ��2 � 332 �
32
�4 � 4
3 � 2� � �1
0
0
1�
AB � �1
3
2
4� ��232
1
�12� � ��2 � 3
�6 � 6
1 � 1
3 � 2� � �1
0
0
1�
4.
BA � �35
� 25
1515� �1
2
�1
3� � �35 �
25
�25 �
25
�35 �
35
25 �
35� � �1
0
0
1�
AB � �1
2
�1
3� �35
� 25
1515� � �
35 �
25
65 �
65
15 �
15
25 �
35� � �1
0
0
1�
Vocabulary Check
1. square 2. inverse 3. nonsingular, singular
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
634 Chapter 7 Linear Systems and Matrices
5.
� �100
010
001�� �
2 � 14 � 46 � 6
�17 � 11 � 6�34 � 44 � 9
�51 � 66 � 15
11 � 7 � 422 � 28 � 6
33 � 42 � 10�BA � �123
146
2�3�5��
2�1
0
�17113
11�7�2�
� �100
010
001� � �
2 � 34 � 33�1 � 22 � 21
6 � 6
2 � 68 � 66�1 � 44 � 42
12 � 12
4 � 51 � 55�2 � 33 � 35
�9 � 10�
AB � �2
�10
�17113
11�7�2��
123
146
2�3�5�
6.
BA � �100
010
001�
AB � �1
�11
012
�100� 13 �
00
�3
�21
�2
111� � �
100
010
001�
7. BA � �10
01�AB � ��1
1�4
2��1
�12
2�
12� � �1
001�;
9. BA � �10
01�AB � � 1.6
�3.52
�4.5��22.5
�17.510
�8� � �10
01�;
8. BA � �10
01�AB � �11
2�12�2��
�1�1
6112� � �1
001�;
10.
BA � �100
010
001�
AB � �410
023
�2�4
1��0.28
�0.020.06
�0.120.08
�0.24
0.080.280.16� � �
100
010
001�
11.
A�1 � �12
0
013� �
16 �3
0
0
2�
12R1 →13R2 →�1
001
��
12
0013� � �I � A�1�
�A � I� � �20
03
��
10
01�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.6 The Inverse of a Square Matrix 635
12.
A�1 � � 7
�3
�2
1�
�2R2 � R1→ �1
0
0
1��
7
�3
�2
1� � �I � A�1�
�3R1 � R2→ �1
0
2
1��
1
�3
0
1�
�A � I� � �1
3
2
7��
1
0
0
1�
14.
A�1 � ��19
�4
�33
�7�
�7R2 → �1
0
0
1
��
�19
�4
�33
�7�
�33R2 � R1→ �1
00
�17
��
�1947
�331�
�4R1 � R2 → �1
0
�337
�17
��
�1747
0
1�
�
17R1→ �1
4�
337
�19��
�17
001�
�A � I� � ��74
33�19
��
10
01� 15.
A has no inverse because it is not square.
A � � 2
�3
7
�9
1
2�
16.
A has no inverse because it is not square.
A � ��2
6
0
5
�15
1�
13.
A�1 � ��3
�2
2
1�
2R2 � R1 → �1
001
��
�3�2
21�
�2R1 � R2 → �1
0�2
1��
1�2
01�
�A � I� � �12
�2�3
��
10
01�
17.
A�1 � �1
�3
3
1
2
�3
�1
�1
2�
� �I � A�1�
�R3 � R1→�R3 � R2→
2R3→�1
0
0
0
1
0
0
0
1
�
�
�
1
�3
3
1
2
�3
�1
�1
2�
�R2 � R1→12R2→
�3R2 � R3→�1
0
0
0
1
0
12
12
12
�
�
�
52
�32
32
�12
12
�32
0
0
1�
�3R1 � R2→�3R1 � R3→
�1
0
0
1
2
3
1
1
2
�
�
�
1
�3
�3
0
1
0
0
0
1�
�A � I� � �1
3
3
1
5
6
1
4
5
���
1
0
0
0
1
0
0
0
1�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
636 Chapter 7 Linear Systems and Matrices
18.
A�1 � ��13
12
�5
6
�5
2
4
�3
1�
� �I � A�1�
4R3 � R1→�3R3 � R2→ �
1
0
0
0
1
0
0
0
1
�
�
�
�13
12
�5
6
�5
2
4
�3
1�
�2R2 � R1→
2R2 � R3→�1
0
0
0
1
0
�4
3
1
�
�
�
7
�3
�5
�2
1
2
0
0
1�
�3R1 � R2→R1 � R3→
�1
0
0
2
1
�2
2
3
�5
�
�
�
1
�3
1
0
1
0
0
0
1�
�A � I� � �1
3
�1
2
7
�4
2
9
�7
���
1
0
0
0
1
0
0
0
1�
20.
Since the first 3 entries of row 3 are all zeros, the inverse does not exist.
→ �100
050
000
���
1�3
1
01
�1
001�
→ �100
055
000
���
1�3�2
010
001�
�A � I� � �132
055
000
���
100
010
001� 21. Not invertible.
does not exist.A�1
22. A�1 � ��175
9514
37�20�3
�1371� 23.
A�1 � ��12�4�8
�5�2�4
�9�4�6�
A � ��
12
1
0
34
0
�1
14
�3212
� 24.
does not exist.A�1
A � ��56
0
1
1323
�12
116
2
�52�
25.
A�1 �511�
0
�22
22
�4
11
�6
2
11
�8�
A � �0.1
�0.3
0.5
0.2
0.2
0.4
0.3
0.2
0.4� 26.
A�1 � �3.75
3.45834.1667
0�1
0
�1.25�1.375
�2.5�
A � �0.60.7
1
0�1
0
�0.30.2
�0.9�
19.
Not invertible (row of zeros)
does not exist.A�1
��1
5�R1 →��2�R1 � R2 → �
10
�1
005
007
���
�1525
0
010
001�
�A � I� � ��5
2�1
005
007
���
100
010
001�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.6 The Inverse of a Square Matrix 637
29. � 5�2
1�2�
�1
�1
5��2� � ��2��1� ��2
2�1
5� �1
�8��2
2�1
5� � �14
�14
18
�58�
36. �23
�51�
�1
�1
2�1� � ��5��3� �1
�352� �
117�
1�3
52� � � 1
17
�317
517
217�
37. � 1�2
20�
�1
� �012
�1214� ⇒ k � 0 38. ��1
211�
�1
� ��1323
1313� ⇒ k �
23
40. ��10
�22�
�1
� ��10
�112� ⇒ k �
12
30. �180
��10�11
0�8���8
110
�10��1
�1
��8���10� � 11�0� ��10�11
0�8�
32. Inverse does not exist.⇒�10 �
89
�13
23
�14� ��
1413
�2389�
�1
� 1
��14��8
9� � �13���2
3� �
89
�13
23
�14�
33. � 2�1
35�
�1
�1
2�5� � �3���1� �51
�32� �
113�
51
�32� � �
513
113
�313
213�
34. � 1�3
�22�
�1
�1
1�2� � ��2���3� �23
21� �
�14 �2
321� � ��
12
�34
�12
�14�
27.
A�1 � �1
0
2
0
0
1
0
1
1
0
1
0
0
1
0
2�
A � ��1
0
2
0
0
2
0
�1
1
0
�1
0
0
�1
0
1� 28.
A�1 � ��24
�10
�29
12
7
3
7
�3
1
0
3
�1
�2
�1
�2
1�
A � �1
3
2
�1
�2
�5
�5
4
�1
�2
�2
4
�2
�3
�5
11�
31. �2059
�45
�15
3472� �
159 �
16�4
1570� �
7215
�3445�
�1
� 1
�72��4
5� � ��34��1
5� �45
�15
3472�
35. ��13
0�2�
�1
�1
��1���2� � 0�3� ��2�3
0�1� �
12�
�2�3
0�1� � ��1
�32
0�
12�
39. ��1�3
21�
�1
� �1535
�25
�15� ⇒ k �
35
©H
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ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
638 Chapter 7 Linear Systems and Matrices
41.
Answer: �5, 0�
�x
y� � ��3
�2
2
1� � 5
10� � �5
0� 42.
Answer: �6, 3�
�xy� � ��3
�2
2
1� �0
3� � �6
3�
43.
Answer: ��8, �6�
�x
y� � ��3
�2
2
1��4
2���8
�6� 44.
Answer: ��7, �4�
�xy� � ��3
�2
2
1� � 1
�2� � ��7
�4�
45.
Answer: �3, 8, �11�
�x
y
z� � �
1
�3
3
1
2
�3
�1
�1
2� �
0
5
2� � �
3
8
�11� 46.
Answer: �1, 7, �9�
�x
y
z� � �
1
�3
3
1
2
�3
�1
�1
2� �
�1
2
0� � �
1
7
�9�
47.
Answer: �2, 1, 0, 0�
�x1
x2
x3
x4
� � ��24
�10
�29
12
7
3
7
�3
1
0
3
�1
�2
�1
�2
1� �
0
1
�1
2� � �
2
1
0
0� 48.
Answer: ��32, �13, �37, 15�
�x
y
z
w� � �
�24
�10
�29
12
7
3
7
�3
1
0
3
�1
�2
�1
�2
1� �
1
�2
0
�3� � �
�32
�13
�37
15�
49.
Answer: �2, �2�
� � 2�2� �x
y� � �1
11 � 3
�5
�4
3� ��2
4� � �1
11 ��22
22�
A�1 �1
9 � 20 � 3
�5
�4
3�
A � �3
5
4
3�
50.
Answer: �12, 13�
�xy� � A�1b �
112�
4�5
�23��
1323� � �
1213�
A�1 �1
18�24� � 12�30� �24
�30�12
18� �112�
4�5
�23�
A � �1830
1224�
51.
does not exist.
[The system actually has no solution.]
A�1
A�1 �1
1.6 � 1.6 ��4
�2
�0.8
�0.4�
A � ��0.4
2
0.8
�4� 52.
Answer: �6, �2�
�xy� �
�18 �35
25155��
2.4�8.8� � � 6
�2�
A�1 ��18 �35
25155�
A � � 0.2�1
�0.61.4�
©H
ough
ton
Miff
lin C
ompa
ny. A
ll rig
hts
rese
rved
.
Section 7.6 The Inverse of a Square Matrix 639
53.
Answer: ��4, �8�
�xy� � A�1b � ��1
2
1213�� �2
�12� � ��4�8�
A�1 � ��1
2
1213�
A � ��1432
3834�
54.
Answer: ��12, 10�
�xy� �
119�
4216
�12�10��
�20�51� � ��12
10�
�119�
4216
�12�10�A�1 �
1
�56��
72 � �4
3��1���
72
�43
156� �
�1219 ��
72
�43
156�
A � �5643
�1
�72�
56.
Answer: �5, 8, �2�
�x
y
z� �
1
82 �
�21
�44
26
19
32
�4
16
14
�12� �
�2
16
4� � �
5
8
�2�
A�1 �1
82 �
�21
�44
26
19
32
�4
16
14
�12�
A � �4
2
8
�2
2
�5
3
5
�2�
57.
does not exist.
The system actually has an infinite number of solutions of the form
where is any real number.t
z � t
y � 1.1875t � 0.6875
x � 0.3125t � 0.8125
A�1
A � �5
2
�1
�3
2
7
2
�3
�8� 58.
Answer: ��1, 2, 0�
�xyz� � A�1B � �
�120�
B � �47
13�A � �235
359
5�917�
55.
Answer: ��1, 3, 2�
� ��1
32� �
155
�55
� 165
110
�
�x
y
z� �
155 �
18
3
�14
4
19
3
�5
�10
10��
�5
10
1
�
A�1 �155�
18
3
�14
4
19
3
�5
�10
10�
A � �4
2
5
�1
2
�2
1
3
6�
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.
640 Chapter 7 Linear Systems and Matrices
59. row reduces to
Answer: �5, 0, �2, 3�
�1000
0100
0010
0001
����
50
�23��
7�2
4�1
�3101
0010
2�1�2�1
����
41�13
12�8
�
60.
Answer: �6.21, �0.77, �2.67, 2.40�
�0.338
0.042
�0.141
0.113
�0.352
0.164
0.230
�0.117
0.141
�0.066
0.108
0.047
0.394
�0.117
�0.164
�0.202� �
11
�7
3
�1� � �
6.21
�0.77
�2.67
2.40��
x
y
z
w� �
A�1 �0.338
0.042
�0.141
0.113
�0.352
0.164
0.230
�0.117
0.141
�0.066
0.108
0.047
0.394
�0.117
�0.164
�0.202�
A � �2
1
2
1
5
4
�2
0
0
2
5
0
1
�2
1
�3�
61. (a) (b)
(c) (d) (e)
The point has been translated back to ��3, 2�.��1, 1�
A�1B���3
21�A�1��
100
010
�211�B � AX � �
�111�
�x � h, y � k� � ��3 � 2, 2 � ��1�� � ��1, 1��x, y� � ��3, 2�
62. (a) (b)
(c) (d) (e)
The point has been translated back to �1, �2�.��1, 1�
A�1B��1
�21�A�1��
100
010
2�3
1�B � AX � ��1
11�
�x � h, y � k� � �1 � 2, �2 � 3� � ��1, 1��x, y� � �1, �2�
63. (a) (b)
(c) (d) (e)
The point has been translated back to �2, �4�.��1, 0�
A�1B��2
�41�A�1��
100
010
3�4
1�B � AX � ��1
01�
�x � h, y � k� � �2 � 3, �4 � 4� � ��1, 0��x, y� � �2, �4�
64. (a) (b)
(c) (d) (e)
The point has been translated back to �0, �3�.�3, �5�
A�1B��0
�31�A�1��
100
010
�321�B � AX � �
3�5
1��x � h, y � k� � �0 � 3, �3 � 2� � �3, �5��x, y� � �0, �3�
©H
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.
Section 7.6 The Inverse of a Square Matrix 641
For 65–68 use Using the methods of this section, we have A�1 �111�
50�13�26
�600200400
�45
�1�.A � �
10.065
0
10.07
2
10.09�1
�.
66.
Answer: 4000 in AAA bonds, 2000 in A bonds, 4000 in B bonds.
X � A�1B �1
11 �
50
�13
�26
�600
200
400
�4
5
�1� �
10,000
760
0� � �
4000
2000
4000�
68.
Answer: $200,000 in AAA bonds, $100,000 in A bonds, and $200,000 in B bonds.
X � A�1B �1
11 �
50
�13
�26
�600
200
400
�4
5
�1� �
500,000
38,000
0� � �
200,000
100,000
200,000�
70.
Answer:I3 � 15�7 ampsI1 � 5�7 amps, I2 � 10�7 amps,
�I1
I2
I3� �
114 �
5
�4
1
�4
6
2
4
8
�2� �
10
10
0� � �
5�7
10�7
15�7�
A�1 �114 �
5
�4
1
�4
6
2
4
8
�2�
A � �2
0
1
0
1
1
4
4
�1�
65.
Answer: $10,000 in AAA bonds, $5000 in A-bonds and $10,000 in B-bonds.
X � A�1 B �111�
50�13�26
�600200400
�45
�1� �25,000
19000� � �
10,0005000
10,000�
67.
Answer: $20,000 in AAA bonds, $15,000 in A-bonds and $30,000 in B-bonds.
X � A�1 B �111�
50�13�26
�600200400
�45
�1� �65,000
50500� � �
20,00015,00030,000�
69.
Answer: I1 � �3 amps, I2 � 8 amps, I3 � 5 amps
�I1
I2
I3� �
114 �
5
�4
1
�4
6
2
4
8
�2� �
14
28
0� � �
�3
8
5�
A�1 �114 �
5
�4
1
�4
6
2
4
8
�2�
A � �2
0
1
0
1
1
4
4
�1�
©H
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.
642 Chapter 7 Linear Systems and Matrices
For Exercises 71–74, let number of muffins, number of bones, number of cookies.
A X
A�1 � [ 1�5
2
1�2
0
�28
�2]
[232
111
21
1.5][xyz] � [ Beef
ChickenLiver]
z �y �x �
71.
300 units of bones
200 units of cookies
A�1�700500600� � �
0300200� 72.
5 units of muffins
415 units of bones
50 units of cookies
A�1�525480500� � �
541550�
73.
100 units of muffins
300 units of bones
150 units of cookies
A�1�800750725� � �
100300150� 74.
150 units of muffins
300 units of bones
200 units of cookies
A�1�1000950900� � �
150300200�
75. (a)
(b)
(c)
2 pounds French vanilla
4 pounds hazelnut
4 pounds Swiss chocolate
X � A�1B � �113
�43
�43
�432323
�1323
�13��10
260� � �
244�
B�XA
�120
12.5
1
13
�1��fhs � � �
10260�
f2f
�
�
h2.5h
h
�
�
�
s3ss
�
�
�
10260
76. (a) Let number of roses.
Let number of lilies.
Let number of irises.
(b)
(c)
80 roses, 10 lilies, 30 irises
X � A�1B � �23
�7632
012
�12
13
�112
�14��120
3000� � �
801030�
B�XA
�1
2.51
14
�2
12
�2��xyz � � �
120300
0�
x2.5x
x
�
�
�
y4y2y
�
�
�
z2z2z
�
�
�
120300
0
z �
y �
x �
©H
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.
Section 7.6 The Inverse of a Square Matrix 643
77. (a)
(b)
(c)
(d) For 2005, and thousand.
For 2010, and thousand.
For 2015, and thousand.
(e) Answers will vary.
y � 45,513t � 15
y � 20,583t � 10
y � 7503t � 5
0 50
8,000
y � 237t2 � 939t � 6273
A�1�534355896309� � �
abc� � �
237�9396273�
A � �49
16
234
111�, A�1 � �
12
�72
6
�1
6
�8
12
�52
3�� 4a
9a16a
�
�
�
2b3b4b
�
�
�
ccc
�
�
�
534355896309
���
�2, 5343��3, 5589��4, 6309�
78. (a)
(b)
(c)
(d) For 2005, and millions.
For 2010, and millions.
For 2015, and millions.
(e) Answers will vary.
y � 87,929t � 15
y � 37,199t � 10
y � 11,744t � 5
0 50
15,000
y � 505.5t2 � 2491.5t � 11,564
A�1�860386399686� � �
abc� � �
505.5�2491.5
11,564�
A � �49
16
234
111�, A�1 � �
12
�72
6
�1
6
�8
12
�52
3�� 4a
9a16a
�
�
�
2b3b4b
�
�
�
ccc
�
�
�
860386399686
���
�2, 8603��3, 8639��4, 9686�
79. True. AA�1 � A�1A � I 80. True
81.
A�1A �1
ad � bc � d
�c
�b
a� �a
c
b
d� �1
ad � bc �ad � bc
0
0
ad � bc � � �1
0
0
1�
�1
ad � bc �ad � bc
0
0
ad � bc�� �1
0
0
1�
AA�1 � �a
c
b
d� � 1
ad � bc�d
�c
�b
a� �1
ad � bc �a
c
b
d� � d
�c
�b
a�
82. (a)
—CONTINUED—
Given A � �a11
0
0
0
a22
0
0
0
a33�, A�1 �
1 a11
0
0
0
1 a22
0
0
0
1 a33
, a11, a22, a33, � 0
, a11 � 0, a22 � 0Given A � �a11
0
0
a22�, A�1 � �
1a11
0
0
1a22
�
©H
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.
644 Chapter 7 Linear Systems and Matrices
82. —CONTINUED—
(b) In general, the inverse of the diagonal matrix A is
�assuming aii � 0�
1 a11
0
0
�
0
0
1 a22
0
�
0
0
0
1 a33
�
0
. . .
. . .
. . .
�. . .
0
0
0
� 1 ann
83.�9
x�6
x� 2
��9
x�6 � 2x
x �
9x
�x
6 � 2x�
96 � 2x
, x � 0
84.1 �
2x
1 �4x
�x � 2x � 4
, x � 0 85.
�x2 � 2x � 13
x�x � 2� , x � ±3
�2x2 � 4x � 26
2x2 � 4x
�4�x � 2� � 2�x2 � 9�
�x � 3��x � 2� � �x � 3��x � 2�
4x2 � 9
�2
x � 21
x � 3�
1x � 3
��x2 � 9��x � 2��x2 � 9��x � 2�
86. �x2 � 4x � 3
3, x � �1�
�x � 3��x � 1�3
1x � 1
�12
32x2 � 4x � 2
�2 � x � 12�x � 1� �
2�x � 1�2
3
87. e2x � 2ex � 15 � �ex � 5��ex � 3� � 0 ⇒ ex � 3 ⇒ x � ln 3 1.099
88.
ex � 4 ⇒ x � ln 4 1.386
ex � 6 ⇒ x � ln 6 1.792
e2x � 10ex � 24 � �ex � 6��ex � 4� � 0 89.
x �13 e12�7 1.851
3x � e12�7
ln 3x �127
7 ln 3x � 12
90.
x � e2 � 9 �1.611
x � 9 � e2
ln�x � 9� � 2 91. Answers will vary.
©H
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.
Section 7.7 The Determinant of a Square Matrix 645
Section 7.7 The Determinant of a Square Matrix
■ You should be able to determine the determinant of a matrix of order by using the products of thediagonals.
■ You should be able to use expansion by cofactors to find the determinant of a matrix of order 3 or greater.
■ The determinant of a triangular matrix equals the product of the entries on the main diagonal.
■ You should be able to calculate determinants using a graphing utility.
2 � 2
1. �4� � 4 2. det� ��10�� � �10
3. �82 43� � 8�3� � 4�2� � 24 � 8 � 16 4. ��9
602� � ��9��2� � 0 � �18
5. � 18 � 10 � 28� 6�5
23� � 6�3� � �2���5� 6. �34 �3
�8� � 3��8� � ��3��4� � �24 � 12 � �12
7. ��712
6
3� � �7�3� � 6�12� � �21 � 3 � �24 8. �40 �3
0� � �4��0� � �0���3� � 0
9. �244 �1
2
2
0
1
1� � 2�22 1
1� � 4��1
2
0
1� � 4��1
2
0
1� � 2�0� � 4��1� � 4��1� � 0
10. ��2
1
0
2
�1
1
3
0
4� � 0� 2
�1
3
0� � 1��2
1
3
0� � 4��2
1
2
�1� � 0�3� � 1��3� � 4�0� � 3
11. (Upper Triangular)��1
0
0
2
3
0
�5
4
3� � ��1��3��3� � �9
12. (Lower Triangular)� 1
�4
5
0
�1
1
0
0
5� � �1���1��5� � �5
13. � 0.3
0.2
�0.4
0.2
0.2
0.4
0.2
0.2
0.3� � �0.002 14. � 0.1
�0.3
0.5
0.2
0.2
0.4
0.3
0.2
0.4� � �0.022
Vocabulary Check
1. determinant 2. minor 3. cofactor
4. expanding by cofactors 5. triangular 6. diagonal
©H
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.
646 Chapter 7 Linear Systems and Matrices
15.
(a) (b)
C22 � �M22 � �3M22 � �3
C21 � �M21 � �4M21 � �4
C12 � �M12 � �2M12 � �2
C11 � �M11 � �5M11 � �5
�3
2
4
�5� 16.
(a)
M22 � 11
M21 � 0
M12 � �3
M11 � 2
� 11
�3
0
2�(b)
C22 � M22 � 11
C21 � �M21 � 0
C12 � �M12 � 3
C11 � M11 � 2
17.
(a)
M22 � ��4
1
3
�5� � 17
M21 � �60 3
�5� � �30
M13 � �71 �2
0� � 2
M12 � �71 8
�5� � �43
M11 � ��2
0
8
�5� � 10
��4
7
1
6
�2
0
3
8
�5�
M33 � ��4
7
6
�2� � �34
M32 � ��4
7
3
8� � �53
M31 � � 6
�2
3
8� � 54
M23 � ��4
1
6
0� � �6 (b)
C33 � �34
C32 � 53
C31 � 54
C23 � 6
C22 � 17
C21 � 30
C13 � 2
C12 � 43
C11 � 10
18.
(a)
M22 � ��2
6
4
�6� � �12
M33 � ��2
7
9
�6� � �51M21 � �97 4
�6� � �82
M32 � ��2
7
4
0� � �28M13 � �76 �6
7� � 85
M31 � � 9
�6
4
0� � 24M12 � �76 0
�6� � �42
M23 � ��2
6
9
7� � �68M11 � ��6
7
0
�6� � 36
��2
7
6
9
�6
7
4
0
�6�
(b)
C33 � ��1�6M33 � �51
C32 � ��1�5M32 � 28
C31 � ��1�4M31 � 24
C23 � ��1�5M23 � 68
C22 � ��1�4M22 � �12
C21 � ��1�3M21 � 82
C13 � ��1�4M13 � 85
C12 � ��1�3M12 � 42
C11 � ��1�2M11 � 36
19. (a)
(b) ��3
4
2
2
5
�3
1
6
1� � �2�42 6
1� � 5��3
2
1
1� � 3��3
4
1
6� � �2��8� � 5��5� � 3��22� � �75
��3
4
2
2
5
�3
1
6
1� � �3� 5
�3
6
1� � 2�42 6
1� � �42 5
�3� � �3�23� � 2��8� � 22 � �75
©H
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.
Section 7.7 The Determinant of a Square Matrix 647
20. (a)
(b) ��3
6
4
4
3
�7
2
1
�8� � 2�64 3
�7� � ��3
4
4
�7� � 8��3
6
4
3� � 2��54� � �5� � 8��33� � 151
��3
6
4
4
3
�7
2
1
�8� � �6� 4
�7
2
�8� � 3��3
4
2
�8� � 1��3
4
4
�7� � �6��18� � 3�16� � �5� � 151
21. (a)
(b)
� 0 � 13��298� � 0 � 6�674� � 170
� 6
4
�1
8
0
13
0
6
�3
6
7
0
5
�8
4
2� � 0� 4
�1
8
6
7
0
�8
4
2� � 13� 6
�1
8
�3
7
0
5
4
2� � 0�648 �3
6
0
5
�8
2� � 6� 6
4
�1
�3
6
7
5
�8
4� � �4��282� � 13��298� � 6��174� � 8��234� � 170
� 6
4
�1
8
0
13
0
6
�3
6
7
0
5
�8
4
2� � �4�006 �3
7
0
5
4
2� � 13� 6
�1
8
�3
7
0
5
4
2� � 6� 6
�1
8
0
0
6
5
4
2� � 8� 6
�1
8
0
0
6
�3
7
0�
22. (a)
(b)
� 10�24� � 4�245� � 0��64� � 1�427� � �1167
�10
4
0
1
8
0
3
0
3
5
2
�3
�7
�6
7
2� � 10�030 5
2
�3
�6
7
2� � 4�830 3
2
�3
�7
7
2� � 0�800 3
5
�3
�7
�6
2� � 1�803 3
5
2
�7
�6
7� � 0��64� � 3��3� � 2��112� � 7�136� � �1167
�10
4
0
1
8
0
3
0
3
5
2
�3
�7
�6
7
2� � 0�800 3
5
�3
�7
�6
2� � 3�10
4
1
3
5
�3
�7
�6
2� � 2�10
4
1
8
0
0
�7
�6
2� � 7�10
4
1
8
0
0
3
5
�3�
23. Expand by Column 3.
� 1
3
�1
4
2
4
�2
0
3� � �2� 3
�1
2
4� � 3�13 4
2� � �2�14� � 3��10� � �58
24. (Lower Triangular)��3
7
1
0
11
2
0
0
2� � �3�11
2
0
2� � �3�22� � �66
25. Expand by Column 3.
�2213 6
7
5
7
6
3
0
0
2
6
1
7� � 6�213 7
5
7
6
1
7� � 3�213 6
5
7
2
1
7� � 6��20� � 3�16� � �168©H
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ll rig
hts
rese
rved
.
648 Chapter 7 Linear Systems and Matrices
26. Expand by Row 2.
� 3
�2
1
0
6
0
1
3
�5
6
2
�1
4
0
2
�1� � ���2��613 �5
2
�1
4
2
�1� � 6�310 6
1
3
4
2
�1� � 2��63� � 6��3� � �108
27. Expand by Column 2.
�� 3
�2
1
6
3
2
0
0
0
0
4
1
0
2
5
�1
3
4
�1
1
5
2
0
0
0�� � �2��2
1
6
3
1
0
2
5
3
4
�1
1
2
0
0
0� � ��2���2��163 0
2
5
4
�1
1� � 4�103� � 412
28. Expand by Column 1.
�
5
0
0
0
0
2
1
0
0
0
0
4
2
3
0
0
3
6
4
0
�2
2
3
1
2� � 5�1000 4
2
3
0
3
6
4
0
2
3
1
2� � 5 � 1�230 6
4
0
3
1
2� � 5��20� � �100
29. (Lower Triangular)�4611 0�5
3�2
0017
0003� � �4���5��1��3� � �60
30.
(Upper Triangular)
�A� � �5��6���2���1� � 60 31.
(Upper Triangular)
det�A� � ��6���1���7���2���2� � �168
32.
(Lower Triangular)
�A� � ��2��4��1��10���3� � 240
33. �1220 �1
6
0
2
8
0
2
8
4
�4
6
0� � �336 34. � 0
8
�4
�7
�3
1
6
0
8
�1
0
0
2
6
9
14� � 7441
35. �� 3
�1
5
4
1
�2
0
�1
7
2
4
2
0
�8
3
3
1
3
0
0
1
0
2
0
2�� � 410 36. �
�2
0
0
0
0
0
3
0
0
0
0
0
�1
0
0
0
0
0
2
0
0
0
0
0
�4� � �48
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.
Section 7.7 The Determinant of a Square Matrix 649
37. (a)
(c) ��1
0
0
3� �2
0
0
�1� � ��2
0
0
�3���1
0
0
3� � �3 (b)
(d) [Note: �AB� � �A� �B����2
0
0
�3� � 6
�20 0
�1� � �2
38. (a)
(c) AB � ��41
4�1�
�A� � �8 (b)
(d) �AB� � 0
�B� � 0
39. (a) (b)
(c) (d) � 1
�1
0
4
0
2
3
3
0� � �12��1
1
0
2
0
1
1
1
0� �
�1
0
0
0
2
0
0
0
3� � �
1
�1
0
4
0
2
3
3
0�
��1
0
0
0
2
0
0
0
3� � �6��1
1
0
2
0
1
1
1
0� � 2
40. (a)
(c) AB � �2
1
3
0
�1
1
1
2
0� �
2
0
3
�1
1
�2
4
3
1� � �
7
8
6
�4
�6
�2
9
3
15�
�A� � �213 0
�1
1
1
2
0� � 0 (b)
(d) �AB� � �786 �4
�6
�2
9
3
15� � 0
�B� � �203 �1
1
�2
4
3
1� � �7
[Note:
�AB� � �A� �B��
41. (a)
(b)
(c)
(d) Note: �AB� � �A� �B����AB� � 5500
AB � ��7�413
�2
�16�14
43
�1�11
42
�288
�42�
�B� � �220
�A� � �25 42. (a)
(b)
(c)
(d) �AB� � �4094
AB � �53
�1�29
35
�102
1816
105
�6�1
221
�1312
��B� � 89
�A� � �46
43.
Thus, �wy x
z� � �� y
w
z
x�.�� y
w
z
x� � ��xy � wz� � wz � xy
�wy x
z� � wz � xy 44.
Thus, �wy cx
cz� � c�wy x
z�.c�wy x
z� � c�wz � xy�
�wy cx
cz� � cwz � cxy � c�wz � xy�
45.
Thus, �wy x
z� � �wy x � cw
z � cy�.�wy x � cw
z � cy� � w�z � cy� � y�x � cw� � wz � xy
�wy x
z� � wz � xy 46.
Thus, � w
cw
x
cx� � 0.
� w
cw
x
cx� � cxw � cxw � 0
©H
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.
650 Chapter 7 Linear Systems and Matrices
47.
� �y � x��z � x��z � y�
� �y � x��z�z � x� � y�z � x��
� �y � x��z2 � zx � zy � xy�
� �y � x��z2 � zy � zx � xy�
� �y � x��z2 � z�y � x� � xy�
� z2�y � x� � z�y � x��y � x� � xy�y � x�
� z2�y � x� � z�y2 � x2� � xy�y � x�
� yz2 � xz2 � y2z � x2z � xy�y � x�
� �yz2 � y2z� � �xz2 � x2z� � �xy2 � x2y�
�111 x
y
z
x2
y2
z2� � �yz y2
z2� � �xz x2
z2� � �xy x2
y2�
48.
� 3ab2 � b3 � b2�3a � b�
� a3 � 3a2b � 3ab2 � b3 � 3a3 � 3a2b � 2a3
� �a � b�3 � 3a2�a � b� � 2a3
� �a � b�3 � a2�a � b� � a2�a � b� � a3 � a3 � a2�a � b�
� �a � b���a � b�2 � a2� � a�a�a � b� � a2� � a�a2 � a�a � b��
�a � b
a
a
a
a � b
a
a
a
a � b� � �a � b��a � b
a
a
a � b� � a�aa a
a � b� � a� a
a � b
a
a �
49.
x � ±2
x2 � 4
x2 � 2 � 2
�x1
2x� � 2 50.
x � ± 4
x2 � 16
x2 � 4 � 20
� x�1
4x� � 20 51.
x � ± 32
x2 �94
4x2 � 9
4x2 � 6 � 3
� 2x�2
�32x� � 3
52.
x � ± 43
x2 �169
9x2 � 16
9x2 � 8 � 8
�x4
29x� � 8 54.
x � �2, 1
�x � 2��x � 1� � 0
x2 � x � 2 � 0
x2 � x � 2 � 4
�x � 1�1
2x� � 453.
x � 1 ± �2
x �2 ± �4 � 4
2
x2 � 2x � 1 � 0
x2 � 2x � 2 � �1
�x2
1x � 2� � �1
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.
Section 7.7 The Determinant of a Square Matrix 651
55.
x � �4, �1
�x � 4��x � 1� � 0
x2 � 5x � 4 � 0
�x � 3��x � 2� � 2 � 0
�x � 31
2x � 2� � 0 56.
x � �1 or x � 3
�x � 1��x � 3� � 0
x2 � 2x � 3 � 0
x�x � 2� � ��3���1� � 0
�x � 2
�3
�1
x� � 0
57.
x � 1, 12
�x � 1��2x � 1� � 0
2x2 � 3x � 1 � 0
2x2 � 2x � 1 � x
� 2x�1
1x � 1� � x 58.
x � 3, �2
�x � 3��x � 2� � 0
x2 � x � 6 � 0
�x2 � x � 6 � 0
2x � 2 � x2 � x � �8
�x � 1x � 1
x2� � �8
60.
x � 8
2x � 16
�12 � ��2x � 4� � 0
1�32 3�2��1�x
2�2�2� � 0 �Expand first column�
�110
x32
�23
�2� � 0
61. � 4u
�1
�1
2v� � 8uv � 1 62. �3x2
1
�3y2
1 � � 3x2 � ��3y2� � 3x2 � 3y2
63. � e2x
2e2x
e3x
3e3x � � 3e5x � 2e5x � e5x 64.
� e�2x
� e�2x � xe�2x � xe�2x
� e�x
�e�x
xe�x
�1 � x�e�x � � �1 � x�e�2x � ��xe�2x�
65. �x
1
ln x1x� � 1 � ln x
66.
� x � x ln x � x ln x � x
�x
1
x ln x
1 � ln x� � x�1 � ln x� � x ln x
67. True. Expand along the row of zeros. 68. True
59.
x � 3
21 � 7x
7 � 2��7� � x��7� � 0
1� 3�2
21� � 2��1
321� � x��1
33
�2� � 0
� 1�1
3
23
�2
x21� � 0
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.
652 Chapter 7 Linear Systems and Matrices
69. Let
Thus, Your answer may differ, depending on how you choose and B.A�A � B� � �A� � �B�. A � B � ��3
1
3
9�, �A � B� � ��3
1
3
9� � �30
�A� � � 1
�2
3
4� � 10, �B� � ��4
3
0
5� � �20
A � � 1
�2
3
4� and B � ��4
3
0
5�.
70.
For an matrix with consecutive integer entries, the determinant appears to be 0.
� �3x � 6x � 6 � 3x � 6 � 0
� �3x � �x � 1���6� � �x � 2���3�
� �x2 � 11x � 30�� � �x � 2���x2 � 10x � 21� � �x2 � 10x � 24��
� x��x2 � 12x � 32� � �x2 � 12x � 35�� � �x � 1���x2 � 11x � 24�
� �x � 6��x � 5�� � �x � 2���x � 3��x � 7� � �x � 6��x � 4��
� x��x � 4��x � 8� � �x � 7 ��x � 5�� � �x � 1���x � 3��x � 8�
� x
x � 3
x � 6
x � 1
x � 4
x � 7
x � 2
x � 5
x � 8� � x�x � 4
x � 7
x � 5
x � 8� � �x � 1��x � 3
x � 6
x � 5
x � 8� � �x � 2��x � 3
x � 6
x � 4
x � 7��n > 2�n � n
�57
61
65
69
58
62
66
70
59
63
67
71
60
64
68
72� � 0�19
23
27
31
20
24
28
32
21
25
29
33
22
26
30
34� � 0
��5
�2
1
�4
�1
2
�3
0
3� � 0�33
36
39
34
37
40
35
38
41� � 0
�10
13
16
11
14
17
12
15
18� � 0� 4
7
10
5
8
11
6
9
12� � 0
71. (a)
(b)
(c)
(d) In general, det�A�1� �1
det A.
det�A�1� �16
A�1 � �1313
�1316�
�A� � 6 72. (a)
(b)
(c)
(d) In general, det�A�1� �1
det A.
det�A�1� � �13
A�1 � �1323
�13
�53�
�A� � �3
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.
Section 7.7 The Determinant of a Square Matrix 653
80. x2 � 5x � 6 � �x � 2��x � 3�
81. 4y2 � 12y � 9 � �2y � 3�2 82. 4y2 � 28y � 49 � �2y � 7�2
83.
Answer: �2, �4�
y � �2 � 2 � �4
x � 2
13x � 26
3x � 10��x � 2� � 46
y � �x � 2
x � y � �2
3x � 10y � 46
84.
Answer: ��1, 4�
x � �1, y � 2 � 2��1� � 4
�9x � 9
5x � 7�2 � 2x� � 23
5x�4x
�
�
7y2y
�
�
23�4 ⇒ 2x � y � 2 ⇒ y � 2 � 2x
78. Answers will vary.
79. x2 � 3x � 2 � �x � 2��x � 1�
73. (a)
(b)
(c)
(d) In general, det�A�1� �1
det A.
det�A�1� �12
A�1 � ��4�1�1
�5�1�1
1.50.5
0��A� � 2 74. (a)
(b)
(c)
(d) In general, det�A�1� �1
det A.
det�A�1� �14
A�1 � ��1.25
0.25�0.5
201
�2.250.25
�1.5��A� � 4
75. (a) Columns 2 and 3 are interchanged.
(b) Rows 1 and 3 are interchanged.
76. (a) times Row 1 is added to Row 2.
(b) times Row 2 is added to Row 1.��2�
��5�
77. (a) 5 is factored out of the first row of
(b) 4 and 3 are factored out of columns 2 and 3.
A.
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654 Chapter 7 Linear Systems and Matrices
Section 7.8 Applications of Matrices and Determinants
■ You should be able to find the area of a triangle with vertices
The symbol indicates that the appropriate sign should be chosen so that the area is positive.
■ You should be able to test to see if three points, are collinear.
if and only if they are collinear.
■ You should be able to use Cramer’s Rule to solve a system of linear equations.
■ Now you should be able to solve a system of linear equations by substitution, elimination, elementary rowoperations on an augmented matrix, using the inverse matrix, or Cramer’s Rule.
■ You should be able to encode and decode messages by using an invertible matrix.n � n
�x1
x2
x3
y1
y2
y3
1
1
1� � 0,
�x1, y1�, �x2, y2�, and �x3, y3�,±
Area � ±12�x1
x2
x3
y1
y2
y3
1
1
1��x1, y1�, �x2, y2�, and �x3, y3�.
1. Vertices:
Area �52 square units
� 12 ��2��2� � 4�3� � 13� �
12 �5� �
52
12��22
�1
435
111� �
12��2�35 1
1� � 4� 2�1
11� � � 2
�135��
��2, 4�, �2, 3�, ��1, 5�
2. Vertices:
Area square units� 28
12��3
23
56
�5
111� �
12 ���3�11 � 5��1� � 1��28�� �
12 ��56�
��3, 5�, �2, 6�, �3, �5�
3. Vertices:
Area �338 square units
12�0524 1
2
03
111� �
12 ��1
2��32� �
152 � �
12 �33
4 � �338 .
�0, 12�, �52, 0�, �4, 3�
Vocabulary Check
1. collinear 2. Cramer’s Rule 3. cryptogram 4. uncoded, coded
©H
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.
Section 7.8 Applications of Matrices and Determinants 655
5.
Area rhombus square units� ��24� � 24
��3
1
�1
2
2
�4
1
1
1� � �3 �6� � 2�2� � 1��2� � �24
6.
Area rhombus square units� ��54� � 54
��4
6
2
4
8
1
1
1
1� � �4 �7� � 4�4� � 1��10� � �54
7.
or
or
x � 0, �165
x � �165x � 0
5x � 8 � �85x � 8 � 8
8 � ± �5x � 8�
8 � ± ���1���2� � 5��2 � x� � 1��4��
4 � ± 12��1
�2
x
5
0
2
1
1
1� 8.
or
or
x � 3, 19
x � 3x � 19
x � 11 � �8x � 11 � 8
8 � ± �x � 11�
8 � ± ��4�5 � x� � 2��2� � 1��3x � 5��
4 � ± 12��4
�3
�1
2
5
x
1
1
1�
9. Points:
The points are collinear.
� 3
0
12
�1
�3
5
1
1
1� � 3��8� � 12�2� � 0
�3, �1�, �0, �3�, �12, 5� 10. Points:
The points are not collinear.
� 8 � 26 � 33 � 15 � 0
�364 �5
1
2
1
1
1� � �64 1
2� � �34 �5
2� � �36 �5
1��3, �5�, �6, 1�, �4, 2�
11. Points:
The points are not collinear.
� �3 � 0
� 2
�4
6
�12
4
�3
1
1
1� � 2�7� �12 ��10� � 1��12�
�2, �12�, ��4, 4�, �6, �3� 12. Points:
The points are collinear.
� �3 � 3 � 0
� 02
�4
12
�172
111� � �
12 �2 � 4� � 1�7 � 4�
�0, 12�, �2, �1�, ��4, 72�
4. Vertices:
.
Area square units�1238
� 12 �30.75� � 15.375 �
1238
12� 9
2
20
06
�32
111� �
12 �9
2�6 �32� � 1��3��
�92, 0�, �2, 6�, �0, �3
2�©
Hou
ghto
n M
ifflin
Com
pany
. All
right
s re
serv
ed.
656 Chapter 7 Linear Systems and Matrices
15.
Answer: ��3, �2�
y ���7
3�1
9���7
311
�9� ��6030
� �2
x ���1
911
�9���7
311
�9� ��9030
� �3
��7x3x
�
�
11y9y
�
�
�19
16.
Answer: ��1, 2�
y ��46 �10
12��46 �3
9� �10854
� 2
x ���10
12�3
9��46 �3
9� ��5454
� �1
�4x6x
�
�
3y9y
�
�
�1012
17.
Cramer’s rule cannot be used.
(In fact, the system is inconsistent.)
�36 24� � 12 � 17 � 0
�3x6x
�
�
2y4y
�
�
�24
18.
Answer: �7, 5�
y �� 6�13
17�76�
� 6�13
�53� �
�235�47
� 5
x �� 17�76
�53�
� 6�13
�53� �
�329�47
� 7
� 6x�13x
�
�
5y3y
�
�
17�76
19.
Answer: 32
7,
30
7
y ���0.4
0.2
1.6
2.2��0.28
��1.20
�0.28�
30
7
x ��1.6
2.2
0.8
0.3��0.28
��1.28
�0.28�
32
7
D � ��0.4
0.2
0.8
0.3� � �0.28
��0.4x0.2x
�
�
0.8y0.3y
�
�
1.62.2
20.
Answer: �5, 1.5�
y ��2.44.6
10.824.8�
6.56�
9.846.56
� 1.5
x ��10.824.8
�0.81.2�
6.56�
32.86.56
� 5
D � �2.44.6
�0.81.2� � 6.56
�2.4x4.6x
�
�
0.8y1.2y
�
�
10.824.8
13.
x � 3
8x � 24 � 0
1��4� � 2�x � 5� � 1�6x � 10� � 0
�1x5 �2
2
6
1
1
1� � 0 14.
x � 3
�3x � �9
�25 � 3x � 24 � 6x � 10 � 0
��5
�3
x
5� � ��6
�3
2
5� � ��6
�5
2
x� � 0
��6
�5
�3
2
x
5
1
1
1� � 0
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.
Section 7.8 Applications of Matrices and Determinants 657
21.
Answer: ��1, 3, 2�
z ��425 �1
2
�2
�5
10
1�55
�110
55� 2y �
�425 �5
10
1
1
3
6�55
�165
55� 3,x �
��5
10
1
�1
2
�2
1
3
6�55
��55
55� �1,
D � �425 �1
2
�2
1
3
6� � 55�4x
2x
5x
�
�
�
y
2y
2y
�
�
�
z
3z
6z
�
�
�
�5
10
1
22.
Answer: �5, 8, �2�
z ��428 �2
2
�5
�2
16
4��82
�164
�82� �2 y �
�428 �2
16
4
3
5
�2��82
��656
�82� 8 x �
��2
16
4
�2
2
�5
3
5
�2��82
��410
�82� 5
D � �428 �2
2
�5
3
5
�2� � �82�4x2x8x
�
�
�
2y2y5y
�
�
�
3z5z2z
�
�
�
�2164
23. (a)
Answer: �0, �12, 12�
3x � 3��12� � 5�1
2� � 1 ⇒ x � 0
2y � 4�12� � 1 ⇒ y � �
12
z �12
�3x � 3y2y
�
�
5z4z23z
�
�
�
1113
�3x � 3y2y4y
�
�
�
5z4z
263 z
�
�
�
1173
�3x3x5x
�
�
�
3y5y9y
�
�
�
5z9z
17z
�
�
�
124
(b)
Answer: �0, �12, 12�
z ��335 3
5
9
1
2
4�4
�1
2
y ��335 1
2
4
5
9
17�4
� �1
2
x ��124 3
5
9
5
9
17�4
� 0
D � �335 3
5
9
5
9
17� � 4
©H
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Miff
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.
658 Chapter 7 Linear Systems and Matrices
24. (a)
Answer: �1, �2, �1�
x � �1 � 5 � 6��2 � 1
y � �35 � 33 � �2
z � �1
�2x � 3yy
�
�
5z33z
�20z
�
�
�
1�35
20
�2x � 3y
12y32y
�
�
�
5z332 z592 z
�
�
�
1
�352
�652
�2x3x5x
�
�
�
3y5y9y
�
�
�
5z9z
17z
�
�
�
1�16�30
(b)
Answer: �1, �2, �1�
z ��235 3
5
9
1
�16
�30��20
�20
�20� �1
y ��235 1
�16
�30
�5
9
17��20
�40
�20� �2
x �� 1
�16
�30
3
5
9
�5
9
17��20
��20
�20� 1
D � �235 3
5
9
�5
9
17� � �20
25. (a)
Answer: �1, �1, 2�
x � 2��1� � 2 � 1 ⇒ x � 1
y � 2 � 1 ⇒ y � �1
z � 2
�x � 2yy
�
�
zz
�3z
�
�
�
11
�6
�x � 2y3y7y
�
�
�
z3z4z
�
�
�
131
�2xx
3x
�
�
�
y2yy
�
�
�
zzz
�
�
�
514
(b)
Answer: �1, �1, 2�
z ��213 �1
�2
1
5
1
4�9
� 2
y ��213 5
1
4
1
�1
1�9
� �1
x ��514 �1
�2
1
1
�1
1�9
� 1
D � �213 �1
�2
1
1
�1
1� � 9
©H
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.
Section 7.8 Applications of Matrices and Determinants 659
(b)
z ��321 �1
1
1
1
�4
5��16
� �21
8
y ��321 1
�4
5
�3
2
�1��16
�7
2
x �� 1
�4
5
�1
1
1
�3
2
�1��16
� �9
8
D � �321 �1
1
1
�3
2
�1� � �1626. (a)
Answer: ��98, 72, �21
8 �
x �72
�218 � 5 ⇒ x � �
98
�72
� 4z � �14 ⇒ z � �218
y �72
x � y�y
�4y
�
�
z4z
�
�
�
5�14�14
�3x2xx
�
�
�
yyy
�
�
�
3z2zz
�
�
�
1�4
5
27. (a)
(b)
(c) No, because of the negative coefficient.t2
−1 50
1,500
y � �26.36t2 � 241.5t � 784
a �� 5
10
30
10
30
100
5543
12,447
38,333�700
��18,450
700� �26.36
b �� 5
10
30
5543
12,447
38,333
30
100
354�700
�169,070
700� 241.5
c �� 5543
12,447
38,333
10
30
100
30
100
354�700
�548,580
700� 784
D � � 5
10
30
10
30
100
30
100
354� � 700 28. (a)
(b)
(c) for t � 4.3, or 2004.y � 1500
−1 50
15,000
y � �689.21t2 � 2349.6t � 10,182
a �� 5
10
30
10
30
100
53,729
103,385
296,433�700
��482,450
700� �689.21
b �� 5
10
30
53,729
103,385
296,433
30
100
354�700
�1,644,690
700� 2349.6
c �� 53,729
103,385
296,433
10
30
100
30
100
354�700
�7,127,380
700� 10,182
D � � 5
10
30
10
30
100
30
100
354� � 700
©H
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Miff
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ny. A
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.
660 Chapter 7 Linear Systems and Matrices
31.
Cryptogram: 38 63 51 58 119 133 44 88 95�32�14�1
GEII
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[8 1 16] [16 25 0] [2 9 18] [20 8 4] [1 25 0]
Cryptogram: �5 �41 �87 91 207 257 11 �5 �41 40 80 84 76 177 227
� 11 25 10�A � � 76 177 227�
� 20 28 14�A � � 40 80 84�
� 12 29 18�A � � 11 �5 2�41�
� 16 25 10�A � � 91 207 257�� 18 21 16�A � ��5 �41 �87�
33.
Message: HAPPY NEW YEAR �11642529234055
211125053467592
� ��5
3
2
�1� � �
816251423251
1160505
18
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HPYNWYA
AP
E
ER
A�1 � �1
3
2
5��1
� ��5
3
2
�1�
29. The uncoded row matrices are the rows of the matrix on the left.
Answer:
��62, 15, 32�, ��54, 12, 27�, �23, �23, 0�
��68, 21, 35�, ��66, 14, 39�, ��115, 35, 60�
CLEOR
W
A
MR
LMTOO
3125�15
1823
100
13180
12132015150� �
11
�6
�102
0�1
3� � ��68�66
�115�62�54
23
2114351512
�23
35396032270�
6 � 3
30. The uncoded row matrices are the rows of the matrix on the left.
Answer: �43 6 9�, ��38 �45 �13�, ��42 �47 �14�, �44 16 10�, �49 9 12�, ��55 �65 �20�
43
�38
�42
44
49
�55
6
�45
�47
16
9
�65
9
�13
�14
10
12
�20
�4
�3
3
2
�3
2
1
�1
1� �
P
A
N
M
E
L
S
S
D
O
Y
E
E
E
N
16
1
0
14
13
5
12
19
19
4
15
25
5
5
5
0
14
0
6 � 3
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Miff
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ny. A
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.
Section 7.8 Applications of Matrices and Determinants 661
34.
TO BE OR NOT TO BE
T—EO
—O
—OB
OB
—RNTT
—E
2005
150
150
152
1520
1814202005
856
10844290603019
1208
1511756
125804526
A � �23
34�, A�1 � ��4
33
�2�
��43
3�2� �
35.
YANKEES WIN WORLD SERIES
381142
�5268
3223
36171518282420
�1
�111
�27371720
�7�19
23�013
1�1
1
43
�123� �
2511199
2312199
150
1415455
145
230
180
1819
YKSI
WLSI
AE_
NODEE
NE
W_
R_
RS
� �23
013
1�1
1
43
�123�A�1 � �
1�1
1
20
�1
12
�2��1
36. Let A be the matrix needed to decode the message.
Message: MEET ME TONIGHT RON�
8
�15
�13
5
5
5
�1
20
�18
1
21
�10
�13
10
25
19
6
40
�18
16
� ��1
1
�2
1� � �
13
5
0
5
20
14
7
20
0
15
5
20
13
0
15
9
8
0
18
14
�
M
E
E
T
N
G
T
O
E
T
M
O
I
H
R
N
A � ��18
1
�18
16��1
� 0
15
18
14� � ��8
1351
270
�115115� � 0
15
18
14� � ��1
1
�2
1�
��18
1
�18
16�A � � 0
15
18
14�
O
R
N
2 � 2
37. True. Cramer’s Rule requires that the determinantof the coefficient matrix be nonzero.
38. False. The system has solutions,
yet det�12
12� � 0.
� x2x
�
�
y2y
�
�
12©
Hou
ghto
n M
ifflin
Com
pany
. All
right
s re
serv
ed.
662 Chapter 7 Linear Systems and Matrices
39. Answers will vary. 40. Answers will vary.
41.
x � 4y � 19 � 0
4y � x � 19
4y � 20 � �x � 1
y � 5 �5 � 3
�1 � 7�x � 1� �
�14
�x � 1�
43.
2x � 7y � 27 � 0
7y � 2x � �27
7y � 21 � 2x � 6
y � 3 ��3 � 13 � 10
�x � 3� �27
�x � 3�
45.
Horizontal asymptote:
1
2 31−1−2−3
3
−1
−2
−3
x
y
y � 2
f �x� �2x2
x2 � 4.
42.
8x � y � 6 � 0
y � �8x � 6
y � 6 �10 � ��6�
�2 � 0�x � 0� � �8x
44.
5x � 4y � 28 � 0
4y � 5x � 28
4y � 48 � �5x � 20
y � 12 �12 � 2�4 � 4
�x � 4� � �54
�x � 4�
46.
Vertical asymptotes:
Horizontal asymptote:
4
6−4
2
−2 4
6
−6
−4
−8
8
x
y
y � 0
x � �6, 3
f �x� �2x
x2 � 3x � 18�
2x�x � 6��x � 3�
Review Exercises for Chapter 7
2.
Answer: ��3, �3�
x � �3 ⇒ y � �3
�x � 3
2x � 3�x� � 3
�2xx
�
�
3yy
�
�
3 0 ⇒ y � x
1.
Answer: �1, 1�
y � 2 � 1 � 1
x � 1
2x � 2 � 0
x � y � 0 ⇒ x � �2 � x� � 0
x � y � 2 ⇒ y � 2 � x�©
Hou
ghto
n M
ifflin
Com
pany
. All
right
s re
serv
ed.
Review Exercises for Chapter 7 663
4.
Answer: �13, 0�, �5, 12�
y � 12: x �13�39 � 2�12�� � 5
y � 0: x �13�39 � 2�0�� � 13
133 y�1
3y � 4� � 0 ⇒ y � 0, 12
139 y2 �
523 y � 0
169 �523 y �
49y2 � y2 � 169
19�1521 � 156y � 4y2� � y2 � 169
�13�39 � 2y��2
� y2 � 169
�x2
3x�
�
y2
2y�
�
169 39 ⇒ x �
13�39 � 2y�
3.
Answer: �5, 4�
x � 5
y � 4
2y � 1 � 9
�y � 1�2 � y2 � 9
x � y � 1 ⇒ x � y � 1
x2 � y2 � 9 �
5.
Answer: �0, 0�, ��2, 8�, �2, 8�
y � 0, y � 8, y � 8
x � 0, x � �2, x � 2
0 � x2�x � 2��x � 2� 0 � x2�x2 � 4� 0 � x4 � 4x2
y � x4 � 2x2 ⇒ 2x2 � x4 � 2x2
y � 2x2
� 6.
Answer: �5, 2�, �2, �1�
y � �1: x � �1 � 3 � 2
y � 2: x � 2 � 3 � 5
0 � � y � 2��y � 1� ⇒ y � 2, �1
0 � y2 � y � 2
y � 3 � y2 � 1
�xx
�
�
yy2
�
�
31
7.
Answer: �2, �12�
−9
−6
9
6
y2 � �x4
⇒ �x � 4y � 0
y1 �16
�7 � 5x� ⇒ 5x � 6y � 7
� 8.
Answer: �1.5, 5�
−9
−2
9
10
�8x2x
�
�
3y5y
�
�
�328
⇒ ⇒
y �
y �
13�8x � 3�
15�28 � 2x�
�83 x � 1
9.
Points of intersection: �0, 0�, �4, �4�
−6 12
−6
6
�y2
x�
�
4xy
�
�
00
⇒ ⇒
y2 �
y �
4x�x
⇒ y � ± 2�x 10.
Points of intersection: �2, 1�, �3.25, �1.5�
−2 10
−4
4
�y2
y�
�
x2x
�
�
�15
⇒ ⇒
y2 �
y �
x � 15 � 2x
⇒ y � ±�x � 1
©H
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Miff
lin C
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ny. A
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rese
rved
.
664 Chapter 7 Linear Systems and Matrices
14.
Point of intersection: �2, 10�
y � 16 � 3x
−2 70
12�3x � yy
�
�
161 � 3x
15.
Point of intersection: ��1, 1� −4 5
−3
3�y � ln�x � 2� � 1y � �x
16.
Point of intersection: �2, 3�−1 8
−1
5�y � ln�x � 1� � 3y � 4 �
12x
13.
Point of intersection: �4, 4�−2
−2
10
6�y � 2�6 � x�y � 2x�2
17. Revenue
Cost
Break even when Revenue Cost
x � 4762 units
2.10x � 10,000
4.95x � 2.85x � 10,000
�
� 2.85x � 10,000
� 4.95x 18.
Answer: More than $500,000
$500,000 � x
2500 � 0.005x
22,500 � 0.015x � 20,000 � 0.02x
�yy
�
�
22,50020,000
�
�
0.015x0.02x
19.
The dimensions are meters.96 � 144
l � 144
w � 96
5w � 480
2�1.50w� � 2w � 480
l � 1.50w
2l � 2w � 480�20.
The dimensions are 16 � 18 feet.
w � 16
l � 18
34l9
� 68
2l � 2�89
l � 68
w �89l
2l � 2w � 68
21.
Answer: �52, 3�
y � 3
x �5522 �
52
22x � 55
6x � 8y � 39 ⇒ 6x � 8y � 39
2x � y � 2 ⇒ 16x � 8y � 16� 22.
Answer: � 310, 25�
x �310
y �25
130y � 52
�40x20x
�
�
30y50y
�
�
24�14
⇒ ⇒
40x�40x
�
�
30y100y
�
�
2428
11.
Points of intersection:
�0.67, 2.56�, ��1, 2�−4 5
−2
4� y � 3 � x2
y � 2x2 � x � 112.
Points of intersection:
or
�1.41, �0.66�, ��1.41, 10.66�
��2, 5 � 4�2�, ���2, 5 � 4�2�
−2
−2
4
14�yy
�
�
2x2
x2
�
�
4x � 14x � 3
�
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Miff
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rved
.
Review Exercises for Chapter 7 665
24. Interchange the equations:
times Eq. 1 added to Eq. 2 produces:
Then
Answer: �35, �8�
�x �78��8� � �
385 ⇒ x �
35
�3796 y �
3712 ⇒ y � �85
12
Equation 1
Equation 2��x �78 y
512 x �
34 y
�
�
�385254
25.
Answer: �0, 0�
y � 0
x � 0
6x � 0
3x � 2�y � 5� � 10 ⇒ 3x � 2y � 0
3x � 2y � 0 ⇒ 3x � 2y � 0� 26.
Answer: ��3, 7�
y � 7
x � �3
�7x2x
�
�
12y3y
�
�
6315
⇒ ⇒
�7x8x
�
�
12y12y
�
�
�6360
28.
Inconsistent; no solution
0 � 5
�1.5x6x
�
�
2.5y10y
�
�
8.524
⇒ ⇒
3x�3x
�
�
5y5y
�
�
17�12
29.
Consistent.
Answer: �1.6, �2.4�
−4
−6
8
2
y � x � 4 x � y � 4 ⇒ y � �
32x 3x � 2y � 0 ⇒ � 30.
Consistent. Infinite number of solutions of formor All points on line
y � 6 � x.�6 � y, y�.�x, 6 � x�
−3
−1
9
7
�x � y�2y
�
�
6�12
⇒ � 2x ⇒
yy
�
�
66
�
�
xx
23.
Answer: ��12, 45� or ��0.5, 0.8�
x � �12
y � 45
5y � 4
25x � 12y � 15 ⇒ 4x � 5y � 2 ⇒ �20x � 25y � �10
15x � 310y � 750 ⇒ 20x � 30y � 14 ⇒ 20x � 30y � 14�
27.
Infinite number of solutions
Let then
Answer: �145 �
85a, a�5x � 8a � 14 ⇒ x �
145 �
85 a.y � a,
0 � 0
5x � 8y � 14 ⇒ �5x � 8y � �14
1.25x � 2y � 3.5 ⇒ 5x � 8y � 14�
©H
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Miff
lin C
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ny. A
ll rig
hts
rese
rved
.
666 Chapter 7 Linear Systems and Matrices
33.
Consistent
Answer: ��4.6, �8.6�
4x � 1.5y � �5.5 ⇒ y � 83 x � 113
−11
−11
7
1 2x � 2y � 8 ⇒ y � x � 4
35.
Point of equilibrium: �500,0007
, 1597
x �500,000
7, p �
159
7
15 � 0.00021x
37 � 0.0002x � 22 � 0.00001x
Demand � Supply
�
34.
Consistent
Answer: ��7.52, 0.9�
�2x � 9.6y � 6.4 ⇒ y ��19.6
�2x � 6.4� −15
−7
6
7�x � 3.2y � 10.4 ⇒ y �1
3.2�x � 10.4�
36.
Points of equilibrium: �250,000, 95�
p � $95.00
x � 250,000 units
0.0003x � 75
45 � 0.0002x � 120 � 0.0001x
Supply � Demand
�
37. Let speed of the slower plane.
Let speed of the faster plane.
Then, distance of first plane distance of second plane 275 miles.
y � x � 25 � 218.75 mph
x � 193.75 mph
4x � 775
4x � 50 � 825
23x �
23�x � 25� � 275
y � x � 25
x�4060� � y�40
60� � 275
�rate of first plane��time� � �rate of second plane��time� � 275 miles
��
y �
x �
�
31.
Inconsistent; lines are parallel.
−7
−6
11
6
y �14�8 � 5x� �
54x � 2 �5x � 4y � 8 ⇒
y �54x � 10 14x � 15 y � 2 ⇒ � 32.
Inconsistent; lines are parallel.
−3
−1
3
3
�x � 2y � 4 ⇒ y � x2
� 2
72
x � 7y � �1 ⇒ y � �72�x � 1
7�
x2
�17�
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.
Review Exercises for Chapter 7 667
40.
Answer: �3, �1, �3�
x � 7y � 8z � �14 ⇒ x � �14 � 7��1� � 8��3� � 3
y � 9z � 26 ⇒ y � 9��3� � 26 � �1
z � �3
41.
Answer: �3817, 40
17, �6317�
x � 3�4017� � ��63
17� � 13 ⇒ x �3817
�6y � 3��6317� � �3 ⇒ y �
4017
172 z � �
632 ⇒ z � �
6317
�x �
�
3y6y
�
�
z3z
172 z
�
�
�
13�3�
632
�x �
�
�
3y6y
13y
�
�
�
z3z2z
�
�
�
13�3
�38
� x2x4x
�
�
3y
y
�
�
�
z5z2z
�
�
�
132314
42.
Answer: ��0.8, �2.4, 1.6�
x � 4 � 6�1.6� � 2��2.4� � �0.8
y � ��8 � 17�1.6����8� � �2.4
z �85 � 1.6
�x � 2y�8y
�
�
6z17z
�5z
�
�
�
4�8�8
�x � 2y�8y�8y
�
�
�
6z17z22z
�
�
�
4�8
�16
� x3x4x
�
�
2y2y
�
�
�
6zz
2z
�
�
�
440
38. Let amount invested in 6.75% bond.
Let amount invested in 7.25% bond.
Solving the system,
At most $18,000 can be invested in the 6.75% bond.
x � 18,000, y � 28,000.
�x � y � 46,0000.0675x � 0.0725y � 3245
y �
x � 39.
Answer: �4, 3, �2�
⇒ x � 4�3� � 3��2� � 14 � 4
x � 4y � 3z � �14 �y � z � �5 ⇒ �y � 2 � �5 ⇒ y � 3
z � �2
43.
Let then
Answer: where is any real number.
a�3a � 4, 2a � 5, a�
x � 3a � 4
x � 3a � 10 � �6
x � 2�2a � 5� � a � �6
y � 2a � 5.z � a,
�Eq. 2 � Eq. 3�x � 2y
y�
�
z2z0
�
�
�
�650
�2 Eq.1 � Eq. 2Eq. 1 � Eq. 3
�x � 2yyy
�
�
�
z2z2z
�
�
�
�655
� x2x
�x
�
�
�
2y3y3y
�
�
z
3z
�
�
�
�6�711
44.
Answer: �� 34, 0, � 5
4��x � 2�0� � 5�� 5
4� � 7 ⇒ x � �34
4�0� � 4z � 5 ⇒ z � �54
�3y � 0 ⇒ y � 0
�2 Eq. 2 � Eq. 3��x � 2y
4y�3y
�
�
5z4z
�
�
�
750
3 Eq. 1 � Eq. 23 Eq. 1 � Eq. 3
��x � 2y4y5y
�
�
�
5z4z8z
�
�
�
75
10
�Eq. 2 � Eq. 1��x3x3x
�
�
�
2y2yy
�
�
�
5z11z7z
�
�
�
7�16�11
Equation 1Equation 2Equation 3
�2x3x3x
�
�
2yy
�
�
�
6z11z7z
�
�
�
�9�16�11
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.
668 Chapter 7 Linear Systems and Matrices
45.
Answer: ��196 , 17
12, 13�x � 2�17
12� � 3�13� � 5 � �
196
4y � 5 � 2�13� �
173 ⇒ y �
1712
z �13
�x � 2y4y
�
�
3z2z3z
�
�
�
�551
�x � 2y8y4y
�
�
�
3zz
2z
�
�
�
�5115
� x2xx
�
�
�
2y4y2y
�
�
�
3z5zz
�
�
�
�510
46.
Adding the second and third equations,
Answer: �72, � 7
10, 110�
x � 2y � z � 5 � 2�� 710� �
110 � 5 �
72
z � 7y � 5 � 7�� 710� � 5 �
110
10y � �7 ⇒ y � �7
10
�x � 2y7y3y
�
�
�
zzz
�
�
�
5�5�2
� x2xx
�
�
�
2y3yy
�
�
�
zz
2z
�
�
�
553
47.
3 times Eq. 1 and times Eq. 2:
Adding,
Let then and
Answer: where is any real number.
a�a � 4, a � 3, a�
y � a � 3.x � a � 4z � a,
x � z � 4
5x � 5z � 20
5x � 5z � 36 � 16
5x � 12�z � 3� � 7z � 16
�y � z � 3 ⇒ y � z � 3.
� 15x�15x
�
�
36y35y
�
�
21z20z
�
�
48�45
��5�
Equation 1Equation 2�5x
3x�
�
12y7y
�
�
7z4z
�
�
169
48.
Let
Answer: �2 � 3a, 5a � 6, a�
x � �102 � 57a � 15�5a � 6��6 � �3a � 2
y � 5a � 6
z � a
�6x � 15yy
�
�
57z5z
�
�
1026
�6x6x
�
�
15y16y
�
�
57z62z
�
�
102108
�2x3x
�
�
5y8y
�
�
19z31z
�
�
3454
49.
�1, 0, 6��0, 0, 8�,�4, 0, 0�, �0, �2, 0�,
x y
22
64
64
8
z 50.
�0, 0, �9�, �0, 3, 0�, �3, 0, 0�, �1, 1, �3�
x y
z
−8
−10
6 64
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.
Review Exercises for Chapter 7 669
55.
x2 � 3x � 3x3 � 2x2 � x � 2
��1
x � 2�
2x � 1x2 � 1
C � �1, B � 2��1� � 4 � 2, A � 1 � 2 � �1
�A � BB � 2C
5C
�
�
�
14
�5
�A � B2B
�B�
�
C2C
�
�
�
13
�4
�A
A
� B2B �
�
C2C
�
�
�
13
�3
x2 � 3x � 3 � A�x2 � 1� � �Bx � C��x � 2�
�A
x � 2�
Bx � Cx2 � 1
x2 � 3x � 3x3 � 2x2 � x � 2
�x2 � 3x � 3
�x � 2��x2 � 1�
51.
4 � x
x2 � 6x � 8�
3
x � 2�
4
x � 4
⇒ A � 3, B � �4� A4A
�
�
B2B
�
�
�14
� �A � B�x � �4A � 2B�
4 � x � A�x � 4� � B�x � 2�
4 � x
x2 � 6x � 8�
A
x � 2�
B
x � 452.
Let :
Let :
�x
x2 � 3x � 2�
1
x � 1�
2
x � 2
2 � �B ⇒ B � �2x � �2
1 � Ax � �1
�x � A�x � 2� � B�x � 1�
�x
x2 � 3x � 2�
A
x � 1�
B
x � 2
53.
x2 � 2x
x3 � x2 � x � 1�
32
x � 1�
��12�x � 32
x2 � 1�
1
2�3
x � 1�
x � 3
x2 � 1
⇒ A �32, B � �
12, C �
32� A
�BA
�
�
�
BCC
�
�
�
120
x2 � 2x � A�x2 � 1� � �Bx � C��x � 1� � �A � B�x2 � �C � B�x � �A � C�
x2 � 2x
x3 � x2 � x � 1�
A
x � 1�
Bx � C
x2 � 1
54.
3x3 � 4xx4 � 2x2 � 1
�3x
x2 � 1�
x�x2 � 1�2
D � 0
C � 1
B � 0
A � 3
�A
AB
B� C
� D
�
�
�
�
3 0 4 0
3x3 � 4x � �Ax � B��x2 � 1� � Cx � D
3x3 � 4xx4 � 2x2 � 1
�3x3 � 4x�x2 � 1�2 �
Ax � Bx2 � 1
�Cx � D
�x2 � 1�2
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.
670 Chapter 7 Linear Systems and Matrices
56.
2x2 � x � 7x4 � 8x2 � 16
�2
x2 � 4�
�x � 1�x2 � 4�2
A � 0, B � 2, C � �1, D � �1
�A
4AB
4B� C
� D
�
�
�
�
0 2
�1 7
2x2 � x � 7 � �Ax � B��x2 � 4� � Cx � D
2x2 � x � 7x4 � 8x2 � 16
�2x2 � x � 7
�x2 � 4�2 �Ax � Bx2 � 4
�Cx � D
�x2 � 4�2
57.
Solving the system,
y � 2x2 � x � 5
a � 2, b � 1, c � �5.
��1, �4��1, �2�
�2, 5�
���
aa
4a
�
�
�
bb
2b
�
�
�
ccc
�
�
�
�4�2
5
y � ax2 � bx � c 58.
Solving the system,
y � �x2 � 2x � 3
a � �1, b � 2, c � 3.
��1, 0��1, 4��2, 3�
���
aa
4a
�
�
�
bb
2b
�
�
�
ccc
�
�
�
043
y � ax2 � bx � c
59. Let gallons of spray X
Let gallons of spray Y
Let gallons of spray Z
Subtracting gives
Then and
Answer: 10 gallons of spray X
5 gallons of spray Y
12 gallons of spray Z
y � 5.z � 12
15 x � 2 ⇒ x � 10.Eq. 2 � Eq. 1
�Chemical A: Chemical B: Cehmical C:
15x25x25x
�
�
� y �
13z13z13z
�
�
�
6
8
13
z �
y �
x � 60. Let be amount invested at 7%.
Let be amount invested at 9%.
Let be amount invested at 11%.
Solving the system,x � $8000, y � $5000, z � $7000.
�x
0.07xxx
�
�
�
y0.09y
y
�
�
�
z0.11z
z
�
�
�
�
20,000178030001000
z
y
x
62. Order 2 � 461. Order 3 � 1 63. Order 1 � 1 64. Order 1 � 5
65. �3
5
�10
4��
15
22� 66. ��110
1�4
��
12�90�
68. �3
�46
�501
1�2
0
���
25�14
15�
67. �8
3
5
�7
�5
3
4
2
�3
���
12
20
26�
69.
�5x � 2y � 7z � �9
4x � 2y � 7z � 10
9x � 4y � 2z � �3
�5
4
9
1
2
4
7
0
2
���
�9
10
3�
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.
Review Exercises for Chapter 7 671
75. �1.5
0.2
2.0
3.6
1.4
4.4
4.2
1.8
6.4� ⇒ �
1
0
0
0
1
0
0
0
1� 76. �
1
0
0
0
1
0
0
0
1�
78.
Answer: ��9, �4�
x � 2��4� � 1 � �9
y � �4
�3R1 � R2 → �10
�2�1
��
�14�
R2 � R1 → �13
�2�7
��
�11�
�23
�5�7
��
21�
80.
Answer: �0.6, 0.5�
x � 0.5�0.5� � 0.35 � 0.6
y � 0.5
5R1 →�2R1 � R2 → �1
0�0.5�0.3
��
0.35�0.15�
�0.20.4
�0.1�0.5
��
0.07�0.01�
70.
�13xx
4x
�
�
�
16y21y10y
�
�
�
7z8z4z
�
�
�
3w5w3w
�
�
�
212
�1
�13
1
4
16
21
10
7
8
�4
3
5
3
���
2
12
�1�
72.
Other answers possible
�31
�2
5�2
0
245� ⇒ �
100
�211
�4
4�10
13� ⇒ �100
�210
4�29
1�
73. �3
4
�2
�3
1
0
0
1� ⇒ �1
0
0
1
3
4
�2
�3� 74. �100
010
001
�611
�2
�46
�1
3�5
1�
71.
3R2 � R1→�R2→
�4R2 � R3→�1
0
0
0
1
0
1
1
�2�
R1 � R2→�R1 � R2→
�2R1 � R3→�1
0
0
3
�1
�4
4
�1
�6�
�0
1
2
1
2
2
1
3
2�
77.
Answer: �10, �12�
x � �8��12� � 86 � 10
y � �12
9y � �108
4R2 � R1→R1 � R2→
�1
0
8
9��
�86
�108�
� 5
�1
4
1��
2
�22�
79.
Answer: ��0.2, 0.7�
x � �0.2, y � 0.7
x � 2�0.7� � 1.6 � �0.2
5y � 3.5 ⇒ y � 0.7
�2R1 � R2→
�1
0
�2
5��
�1.6
3.5�
�1
2
�2
1��
�1.6
0.3�
�R1 � R2→
�2
1
1
�2��
0.3
�1.6�
�2
3
1
�1��
0.3
�1.3�
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.
672 Chapter 7 Linear Systems and Matrices
82.
Inconsistent, no solution
0 � 1
�2R2 � R1→
5R2 � R3→ �
1
0
0
0
1
0
0
3
0
���
�3
2
1�
�2R1 � R2→�3R1 � R3→
�1
0
0
2
1
�5
6
3
�15
���
1
2
�9�
�1
2
3
2
5
1
6
15
3
���
1
4
�6�
83.
Answer: a is a real number.�1 � 3a, �a, a�,
x � 1 � 3z
y � �z
�10
21
�11
��
10� ⇒ �1
001
�31
��
10�
�x � 2y � z � 1y � z � 0
84.
Answer: �1, �1, 0, �2�
�1102
�1310
4�2�1
1
�1111
����
4�4�3
0� ⇒ �
1000
0100
0010
0001
����
1�1
0�2
��
xx
2x
�
�
y3yy
�
�
�
�
4z2zzz
�
�
�
�
wwww
�
�
�
�
4�4�3
0
81.
Answer: �12, � 1
3, 1�x � 3�1� �
72 ⇒ x �
12
y � 1 � � 43 ⇒ y � � 1
3
z � 1
12R1→
� 13R2→
� 128R3→
�1
0
0
0
1
0
3
�1
1
�
��
72
� 43
1�
R2 � R1→
�3R2 � R3→ �2
0
0
0
�3
0
6
3
�28
���
7
4
�28�
�3R1 � R2→�6R1 � R3→ �
2
0
0
3
�3
�9
3
3
�19
���
3
4
�16�
�2
6
12
3
6
9
3
12
�1
���
3
13
2�
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.
Review Exercises for Chapter 7 673
88.
Answer: �15, �3
5, 110�
�1
2
�1
1
1
1
4
2
�2
���
0
0
�1� ⇒ �1
0
0
0
1
0
0
0
1
���
15
�35110�
� x2x
�x
�
�
�
yyy
�
�
�
4z2z2z
�
�
�
00
�1
90.
�x, y, z� � ��2, 1, 1�
�3
�2
0
0
1
1
6
0
2
���
0
5
3� ⇒ �
1
0
0
0
1
0
0
0
1
���
�2
1
1�
86.
Answer: �3142, 5
14, 1384�
x �3142, y �
514, z �
1384
8R3 � R1→12R3 � R2→
184R3→
�1
0
0
0
1
0
0
0
1
..
.
..
.
..
.
31425141384�
R2 � R1→12R2→
�8R2 � R3→ �
1
0
0
0
1
0
�8
�12
84
..
.
..
.
..
.
�12
�32
13�
R3 � R1→�4R1 � R2→�5R1 � R3→
�1
0
0
�1
2
8
4
�24
�12
..
.
..
.
..
.
1
�3
1�
�4
4
5
4
�2
3
4
�8
8
..
.
..
.
..
.
5
1
6� 85.
Answer: �2, �3, 3�
x � 2, y � �3, z � 3
R3 � R1→�R3 � R2→ �
1
0
0
0
1
0
0
0
1
���
2
�3
3�
13R3→
�1
0
0
0
1
0
�1
1
1
���
�1
0
3�
R2 � R1→
�9R2 � R3→�1
0
0
0
1
0
�1
1
3
���
�1
0
9�
15 R2→ �1
0
0
�1
1
9
�2
1
12
���
�1
0
9�
�R1→2R1 � R2→5R1 � R3→
�1
0
0
�1
5
9
�2
5
12
���
�1
0
9�
��1
2
5
1
3
4
2
1
2
���
1
�2
4�
87.
Answer: �1, 2, 12�
�1
1
2
1
�1
�1
2
4
2
���
4
1
1� ⇒ �
1
0
0
0
1
0
0
0
1
���
1
212�� x
x2x
�
�
�
yyy
�
�
�
2z4z2z
�
�
�
411
89. reduces to
Answer: �3, 0, �4�
�100
010
001
���
30
�4��104
2�1
0
�1�1�1
���
74
16�
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.
674 Chapter 7 Linear Systems and Matrices
92.
Inconsistent. No solution �0 � 1�
�411
�2
1266
�10
241
�2
����
20128
�10� ⇒ �
1000
0100
0010
����
0001�
94.
y � 0
x � 8
96.
Answer: x � 12, y � �2
6 �12 x ⇒ x � 12
�4 � 2y ⇒ y � �2
2 � x � 10 ⇒ x � 12
93.
y � �7
x � 12
95.
Answer: x � 1, y � 11
6x � 6 ⇒ x � 1
y � 5 � 16 ⇒ y � 11
�4y � �44 ⇒ y � 11
x � 3 � 5x � 1 ⇒ x � 1
98. (a) not possible (b) not possible
(c) (d) not possibleA � 3B4A � ��44�28
64�8
764�
A � BA � B
97. (a)
(b) (c)
(d) A � 3B � � 7�1
35� � �30
42�60�9� � �37
41�57�4�
4A � � 28�4
1220�A � B � � �3
�15238�
A � B � � 7�1
35� � �10
14�20�3� � �17
13�17
2�
99. (a)
(b)
(c)
(d) A � 3B � �653
0�1
2
723� � �
0�12
6
1524
�3
3183� � �
6�7
9
1523
�1
10206�
4A � �242012
0�4
8
288
12�
A � B � �691
�5�9
3
6�4
2�
A � B � �653
0�1
2
723� � �
0�4
2
58
�1
161� � �
615
571
884�
91. reduces to
Answer: �2, 6, �10, �3�
�1000
0100
0010
0001
����
26
�10�3
��3150
�16
�14
541
�1
�2�1
3�8
����
�441
�1558
�
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.
Review Exercises for Chapter 7 675
105. �32
6
�172
46
�32
33� 106. �4.1
�10.1�49
�10.260.95.1�
107. X � 3A � 2B � 3��4
1
�3
0
�5
2� � 2�
1
�2
4
2
1
4� � �
�14
7
�17
�4
�17
�2�
108. �1
6��13
�2
0
6
�17
20�X �
1
6�4A � 3B� �
1
6�4��4
1
�3
0
�5
2� � 3�
1
�2
4
2
1
4�
109. X �1
3 �B � 2A� �
1
3 ��
1
�2
4
2
1
4� � 2 �
�4
1
�3
0
�5
2� �
1
3 �
9
�4
10
2
11
0�
110. �1
3��13
12
�26
�10
�15
�16�X �
1
3�2A � 5B� �
1
3�2��4
1
�3
0
�5
2� � 5�
1
�2
4
2
1
4�
100. (a)
(c) 4A � �80
�1216
244�
A � B � ��11
25
112� (b)
(d) A � 3B � ��73
127
214�
A � B � � 5�1
�83
10�
101.
� ��13
0
�8
11
18
�19�
�2
0
1
5
0
�4� � 3�5
0
3
�2
�6
5� � �2
0
1
5
0
�4� � �15
0
9
�6
�18
15�
102. �2�1
5
6
2
�4
0� � 8�
7
1
1
1
2
4� � �
�2
�10
�12
�4
8
0� � �
56
8
8
8
16
32� � �
54
�2
�4
4
24
32�
103.
� � 9�9
�74�
� ��82
1�4� � ��10
150
�5� � �74
�83��� 8
�2�1
4� � 5 ��23
0�1� � �7
4�8
3�
104.
� ��3096
0�48
6090
�18�66�
� 6��516
0�8
1015
�3�11�6���4
2�1�5
�37
4�10� � ��1
141
�3138
�7�1�
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676 Chapter 7 Linear Systems and Matrices
111. � �14
14
36
�2
�10
�12
8
40
48��
1
5
6
2
�4
0� �6
4
�2
0
8
0� � �1�6� � 2�4�
5�6� � ��4��4�6�6� � �0��4�
1��2� � 2�0�5��2� � ��4��0�
6��2� � �0��0�
1�8� � 2�0�5�8� � ��4��0�
6�8� � �0��0��112. is undefined.�1
2
5
�4
6
0� �7
0
5
1
2
0� 113. AB � �31
�24
09��
75
�1
033� � �11
18�639�
114. AB � �101
32
�1
2�4
3��400
�336
2�1
2� � �404
18�18
12
3�10
9�115. �
4
11
12
1
�7
3� �3
2
�5
�2
6
�2� � �14
19
42
�22
�41
�66
22
80
66� 116. ��2
4
3
�2
10
2� �1
�5
3
1
2
2� � �13
20
24
4�
117.
� � 1
12
17
36�
� �2�2� � 1��3�6�2� � 0
2�6� � 1�5�6�6� � 0�
�2
6
1
0� �� 4
�3
2
1� � ��2
0
4
4� � �2
6
1
0� � 2
�3
6
5�
118. �14
�12���
01
32��
15
0�3� � �1
4�1
2��1511
�9�6� � � 4
82�3
�48�
119. (a)
This is the sales and profit for milk on Friday,Saturday and Sunday.
(b) Profit � 47.4 � 66.2 � 77.2 � $190.80
� �438.2612.1
717.28
47.466.277.2�
AB � �406076
648296
527684��
2.652.812.93
0.250.300.35� 120. (a)
(b)
(c) This gives the total number of calories burnedby a 120 pound person and a 150 pound personwho perform the given exercises for the givenamount of time.
A � �2 12 3��
10912764
13615979� � �473.5 588.5�
A � �2 12 3�
121. BA � I2AB � ��47
�12��
�27
�14� � �1
001�;
122. BA � I3AB � �116
102
013��
�232
�334
1�1�1� � �
100
010
001�;
123. row reduces to
��6�5
54�
�1
� �45
�5�6�
�10
01
��
45
�5�6���6
�554
��
10
01�
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Review Exercises for Chapter 7 677
129. �1
�10
21
�1
010�
�1
� �101
001
2�1
3� 130. �101
�112
�2�2�4�
�1
� � 0
�12
�14
�2
�12
�34
11214�
131. ��7�8
22�
�1
�1
��7��2� � �2���8� �28
�2�7� � �1
4�1�
72�
132. �107
43�
�1
�1
�10��3� � 4�7��3
�7�410� � �
32
�72
�2
5�133. ��1
21020�
�1
�1
��1��20� � 10�2��20
�2�10�1� �
1�40�
20
�2�10
�1� � ��12120
14140�
134. ��63
�53�
�1
�1
��6��3� � ��5��3��3
�35
�6� �1
�3�3
�35
�6� � ��11
�53
2�136.
Answer: ��12, �3�
�xy� � �
11427
314
�17���10
1� � ��12
�3��2
43
�1��1
� �11427
314
�17�
124. reduces to
��32
�53�
�1
� � 3�2
5�3�
�10
01
��
3�2
5�3���3
2�5
3��
10
01�
125. row reduces to
��1
31
�274
�297�
�1
� �13
�125
6�5
2
�43
�1��100
010
001
���
13�12
5
6�5
2
�43
�1���1
31
�274
�297
���
100
010
001�
126. reduces to
�0
�57
�2�2
3
1�3
4��1
� �1
�1�1
11�7
�14
8�5
�10��100
010
001
���
1�1�1
11�7
�14
8�5
�10��0
�57
�2�2
3
1�3
4
���
100
010
001�
127. �2
3
6
�6��1
� �151
10
15
�115� 128. �3
4
�10
2��1
� �123
� 223
523346�
135.
Answer: �1, �25�
�xy� � �
14320
14
�120���1
5� � � 1
�25�
�13
5�5�
�1� �
14320
14
�120�
� x3x
�
�
5y5y
�
�
�15
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.
678 Chapter 7 Linear Systems and Matrices
138.
Answer: �2, �4, 3�
�xyz� � �
�11�3�1
�6�2�1
�2�1�1��
12�25
10� � �2
�43�
��1
2�1
4�9
5
�25
�4��1
� ��11�3�1
�6�2�1
�2�1�1�
140.
Answer: �1, 1, �2, 1�
�xyz
w� � �
232313
�23
�13
�131313
�1
0
0
1
23
�13
�2313
�� 1�3
22� � �
11
�21�
�1101
1�1
10
1200
1111��1
� �232313
�23
�13
�131313
�1
0
0
1
23
�13
�2313
��1
141.
Answer: ��3, 1�
x � �3, y � 1
�1
3
2
4��1
� ��232
1
�12� ⇒ �x
y� � ��232
1
�12� ��1
�5� � ��3
1�
� x3x
�
�
2y4y
�
�
�1�5
142.
Answer: �5, 6�
x � 5, y � 6
� 1
�6
3
2��1
� �0.1
0.3
�0.15
0.05� ⇒ �x
y� � �0.1
0.3
�0.15
0.05� � 23
�18� � �5
6�
� x�6x
�
�
3y2y
�
�
23�18
139.
Answer: �1, �2, 1, 0�
�xyz
w� � �
�1
2
�1
1
43
�53
1
�1
�2343
�1
1
�73
113
�2
3� ��2
101� � �
1�2
10�
�1210
21
�10
11
�31
�1101�
�1
� ��1
2
�1
1
43
�53
1
�1
�2343
�1
1
�73
113
�2
3�
�1
137.
Answer: �2, �1, �2�
�xyz� � �
�1
3
2
�18373
1
�73
�53
� � 6�1
7� � �2
�1�2�
�315
2�1
1
�121�
�1
� ��1
3
2
�18373
1
�73
�53
�
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.
Review Exercises for Chapter 7 679
146.
� 36 � 77 � �41
��97
11�4� � ��9���4� � �11��7� 147. �50
10
�30
5� � 50�5� � ��30��10� � 550
148. �1412
�24�15� � 14��15� � ��24��12� � 78
149.
Minors:
Cofactors:
C22 � 2C12 � �7
C21 � 1C11 � 4
M22 � 2M12 � 7
M21 � �1M11 � 4
A � �27
�14� 150.
Minors:
Cofactors:
C22 � 3C21 � �6
C12 � �5C11 � �4
M22 � 3M21 � 6
M12 � 5M11 � �4
A � �35
6�4�
143.
Answer: �1, 1, �2�
x � 1, y � 1, z � �2
��3
0
4
�3
1
3
�4
1
4�
�1
� �1
4
4
0
4
�3
1
3
�3� ⇒ �
x
y
z� � �
1
4
�4
0
4
�3
1
3
�3� �
2
�1
�1� � �
1
1
�2�
��3x
4x
�
�
3yy
3y
�
�
�
4zz
4z
�
�
�
2�1�1
144.
does not exist.
Inconsistent; no solution.
�2
1
3
3
�1
7
�4
2
�10�
�1
�2xx
3x
�
�
�
3yy
7y
�
�
�
4z2z
10z
�
�
�
1�4
0
145. �82 5
�4� � 8��4� � 2�5� � �42
151.
Minors:
Cofactors:
C31 � 5, C32 � 2, C33 � 19
C21 � �20, C22 � 19, C23 � �22
C11 � 30, C12 � 12, C13 � �21
M31 � �25 �10� � 5, M32 � � 3
�2�1
0� � �2, M33 � � 3�2
25� � 19
M21 � �28 �16� � 20, M22 � �31 �1
6� � 19, M23 � �31 28� � 22
M11 � �58 06� � 30, M12 � ��2
106� � �12, M13 � ��2
158� � �21
A � �3
�21
258
�106�
©H
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rved
.
680 Chapter 7 Linear Systems and Matrices
152. Minors:
Cofactors:
C31 � �47, C32 � 96, C33 � 22
C21 � �2, C22 � 32, C23 � �20
C11 � 19, C12 � 24, C13 � 26
M33 � �86 35� � 22M32 � �86 4
�9� � �96,M31 � �35 4�9� � �47,
M23 � � 8�4
31� � 20M22 � � 8
�442� � 32,M21 � �31 4
2� � 2,
M13 � � 6�4
51� � 26M12 � � 6
�4�9
2� � �24,M11 � �51 �92� � 19,
153. ��2�6
5
403
124� � 6�43 1
4� � 2��25
43� � 6�13� � 2��26� � 130
154. �4 � 126 � 13 � �117� 42
�5
7�3
1
�14
�1� � 4�3 � 4� � 7��2 � 20� � 1�2 � 15� �
155. � 1
0
�2
0
1
0
�2
0
1� � 1� 1
�2
�2
1� � 1�1� � ��2���2� � 1 � 4 � �3
156. � �12 � 32 � 20�051 3
�2
6
1
1
1� � �3�5 � 1� � 1�30 � 2�
157.
� 279
� 3�128 � 5 � 56� � 4��12�
� 3�8�8 � ��8�� � 1�1 � 6� � 2��4 � 24�� � 4�0 � 6�8 � 6� � 0�
�3060 0
8
1
3
�4
1
8
�4
0
2
2
1� � 3�813 1
8
�4
2
2
1� � ��4��060 8
1
3
2
2
1� �Expansion along Row 1�
158. (Expansion along Row 1.)
� �255
� �5�54 �3� � 6�9 �9�
� �5�6��1 � 10� � 3��5 � 4�� � 6���1 � 10� � 3�0 � 3��
��5
0
�3
1
6
1
4
6
0
�1
�5
0
0
2
1
3� � �5�146 �1
�5
0
2
1
3� � 6� 0
�3
1
�1
�5
0
2
1
3�159. (Upper Triangular)det�A� � 8��1��4��3� � �96 160. det (Lower Triangular)� ��5���2��2��14� � 280
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.
Review Exercises for Chapter 7 681
167. The figure is a rhombus.
Area square units� 48
��24
�2
�19
�9
111� � �48
168. The figure is a rhombus.
Area square units� 64
��44
�4
800
111� � �64
169.
Not collinear
��124
751
111� � �8 � 0 170.
Collinear
�024 �517
111� � �2��12� � 4��6� � 0
171.
Answer: �1, 2�
y �� 1
�1
5
1�� 1
�1
2
1� �6
3� 2
x ��51 2
1�� 1
�1
2
1� �3
3� 1 172.
Answer: �x, y� � ��3, 4�
y ��23 �10
�1��23 �1
2� �28
7� 4
x ���10
�1
�1
2��23 �1
2� ��21
7� �3
162.
Area
Area � 24 square units
�12��4
4
0
0
0
6
1
1
1� �12 �48� � 24
��4, 0�, �4, 0�, �0, 6�161.
Area � 16 square units
12�155 0
0
8
1
1
1� �12 �32� � 16
�1, 0�, �5, 0�, �5, 8�
163.
Area �74 square units
12� 1
2
232
1�
52
1
111� �
12�7
2� �74 164.
Area square units�12� 3
2
4
4
1� 1
2
2
1
1
1� �12�25
4 � �258
�32, 1�, �4, �1
2�, �4, 2�
165.
Area square units�132
12�254 4
61
111� �
12��13� 166.
Area square units�352
12��3
2�4
2�3�4
111� �
12��35�
©H
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.
682 Chapter 7 Linear Systems and Matrices
175.
Answer: ��1, 4, 5�
z ���2
4�1
3�1�4
�11�315�
14�
7014
� 5
y ���2
4�1
�11�315
�516�
14�
5614
� 4
x ���11
�315
3�1�4
�516�
��24
�1
3�1�4
�516
� ��1414
� �1 176.
Answer: �6, 8, 1�
z � �532 �2�3�1
15�7�3�
65�
6565
� 1
y � �532 15�7�3
1�1�7�
65�
52065
� 8
x � � 15�7�3
�2�3�1
1�1�7�
�532 �2�3�1
1�1�7� �
39065
� 6
173.
Answer: �4, 7�
y �� 5�11
6�23�
� 5�11
�23� �
�49�7
� 7
x �� 6�23
�23�
� 5�11
�23� �
�28�7
� 4 174.
Answer: �3, �2�
y ��39 �7
37��39 8
�5� �174�87
� �2
x ���7
378
�5��39 8
�5� ��261�87
� 3
177.
Answer: �0, �2.4, �2.6�
z � �5220 � �2.6y � �
4820 � �2.4,x � 0,
�A3� � �52
�A2� � �48
�A1� � 0
�121 �32
�7
2�3
8� � 20 178.
Answer: �x, y, z� � 3
4,
25
28, �
73
28
z �� 14
�4
56
�21
2
�21
10
4
5�� 14
�4
56
�21
2
�21
�7
�2
7� ��4088
1568� �
73
28
y �� 14
�4
56
10
4
5
�7
�2
7�� 14
�4
56
�21
2
�21
�7
�2
7� �1400
1568�
25
28
x ��10
4
5
�21
2
�21
�7
�2
7�� 14
�4
56
�21
2
�21
�7
�2
7� �1176
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3
4
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Review Exercises for Chapter 7 683
(b)
Answer: �815
, �12
5,
�73
z � �124 2�1�2
�3�1
5�15
��3515
��73
y � �124 �3�1
5
�11
�1�15
��3615
��12
5
x � ��3�1
5
2�1�2
�11
�1�15
��815
D � �124 2�1�2
�11
�1� � 15180. (a)
Answer: �815
, �12
5,
�73
x � 2�125 � �
73 � �3 ⇒ x � �
815
�5y � 3�73 � 5 ⇒ y � �
125
z � �73
�x �
�
2y5y
�
�
�
z3z3z
�
�
�
�357
�x � 2y�5y
�10y
�
�
�
z3z3z
�
�
�
�35
17
� x2x4x
�
�
�
2yy
2y
�
�
�
zzz
�
�
�
�3�1
5
179. (a)
Answer: 5333
, �1733
, 6166
x � 3�1733 � 261
66 � 5 ⇒ x �5333
7y � 86166 � �11 ⇒ y � �
1733
z �6166
�x � 3y7y
�
�
2z8z
667 z
�
�
�
5�11
617
�x � 3y7y
10y
�
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�
2z8z2z
�
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�
5�11�7
� x2x2x
�
�
�
3yy
4y
�
�
�
2z4z2z
�
�
�
5�1
3
(b)
Answer: 5333
, �1733
, 6166
z � �122 �314
5�1
3�66
�6166
y � �122 5�1
3
2�4
2�66
��3466
��1733
x � � 5�1
3
�314
2�4
2�66
�10666
�5333
D � �122 �314
2�4
2� � 66
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684 Chapter 7 Linear Systems and Matrices
181. I _ H A V E _ A _ D R E A M
Cryptogram: 24 38 8 3 0
32 2 41 �39�2�39
�3�51�2�30
�1 13 0�A � �41 �2 �39�
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182. J U S T _ D O _ I T
Cryptogram: 52 28 4
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183.
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Review Exercises for Chapter 7 685
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187. (a)
Using Cramer’s Rule, you obtain,
and
(c) when or 2010t � 20.15,y � 20
y � 0.41t � 11.74
b � 11.74a � 0.41
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(d)
That is, year 2010.
t > 20.15
0.41t > 8.26
0.41t � 11.74 > 20
0 500
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188. False
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686 Chapter 7 Linear Systems and Matrices
190. The row operations on matrices are equivalent tothe operations used in the method of elimination.
189. True. Expansion by Row 3 gives
Note: Expand each of these matrices by Row 3 to see the previous step.
� �a11
a21
a31
a12
a22
a32
a13
a23
a33� � �a11
a21
c1
a12
a22
c2
a13
a23
c3�� c1�a12
a22
a13
a23� � c2�a11
a21
a13
a23� � c3�a11
a21
a12
a22�� a32�a11
a21
a13
a23� � a33�a11
a21
a12
a22�� a31�a12
a22
a13
a23�� �a32 � c2��a11
a21
a13
a23� � �a33 � c3��a11
a21
a12
a22�� a11
a21
a31 � c1
a12
a22
a32 � c2
a13
a23
a33 � c3� � �a31 � c1��a12
a22
a13
a23�
191. A square matrix has an inverse if det�A� � 0.
n � n
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Practice Test for Chapter 7 687
Chapter 7 Practice Test
For Exercises 1–3, solve the given system by the method of substitution.
1.
3x � y � 15
3x � y � 51 2.
x2 � 6y � �5
2x � 3y � �3 3.
5x � 2y � z � �3
2x � y � 3z � 0
x � y � z � 6
4. Find the two numbers whose sum is 110 andproduct is 2800.
5. Find the dimensions of a rectangle if its perimeter is170 feet and its area is 2800 square feet.
For Exercises 6–7, solve the linear system by elimination.
6.
x � 3y � 23
2x � 15y � 24 7.
38x � 19y � 7
x � y � 2
8. Use a graphing utility to graph the two equations.Use the graph to approximate the solution of thesystem. Verify your answer analytically.
0.3x � 0.7y � �0.131
0.4x � 0.5y � �0.112
9. Herbert invests $17,000 in two funds that pay 11%and 13% simple interest, respectively. If he receives$2080 in yearly interest, how much is invested ineach fund?
10. Find the least squares regression line for the points and ��2, �1�.
�4, 3�, �1, 1�, ��1, �2�,
11.
x � 4y � 3z � �20
2x � y � z � �11
x � y � a � 1�2 12.
2x � 4y � 8z � 0
2x � y � z � 0
4x � y � 5z � 4 13.
6x � y � 5z � 2
3x � 2y � z � 5
For Exercises 11–13, solve the system of equations.
14. Find the equation of the parabola passing through the points �0, �1�, �1, 4� and �2, 13�.y � ax2 � bx � c
15. Find the position equation given that feet after 1 second, feetafter 2 seconds, and feet after 3 seconds.s � 4
s � 5s � 12s �12at2 � v0t � s0
16. Write the matrix in reduced row-echelon form.
�1
3
�2
�5
4
9�
For Exercises 17–19, use matrices to solve the system of equations.
17.
2x � 5y � �11
3x � 5y � �13 18.
3x � 2y � �1
3x � 2y � �8
2x � 3y � �3 19.
3x � y � 3z � �3
2x � y � 3z � �0
2x � y � 3z � �5
©H
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688 Chapter 7 Linear Systems and Matrices
20. Multiply �1
2
4
0
5
�3� �1
0
�1
6
�7
2�
21. Given find 3A � 5B.A � � 9
�4
1
8� and B � �6
3
�2
5�,
22. Find
f�x� � x2 � 7x � 8, A � �3
7
0
1�f�A�: 23. True or false:
where andare matrices.
(Assume that exist.)A2, AB, and B2
BA�A � B��A � 3B� � A2 � 4AB � 3B2
For Exercises 24 and 25, find the inverse of the matrix, if it exists.
24. �1
3
2
5� 25. �1
3
6
1
6
10
1
5
8�
For Exercises 27 and 28, find the determinant of the matrix.
27. �6
3
�1
4� 28. �1
5
6
3
9
2
�1
0
�5�
29. Use a graphing utility to find the determinant of the matrix.
�1
0
3
2
4
1
5
0
2
�2
�1
6
3
0
1
1�
30. Evaluate ��60000 4
5
0
0
0
3
1
2
0
0
0
4
7
9
0
6
8
3
2
1�.
31. Use a determinant to find the area of the triangle with vertices and �3, 9�.�0, 7�, �5, 0�,
32. Use a determinant to find the equation of the line through �2, 7� and ��1, 4�.
For Exercises 33–35, use Cramer’s Rule to find the indicated value.
33. Find
2x � 5y � 11
6x � 7y � 14
x. 34. Find
3x � y � 4z � 2
3x � y � 4z � 3
3x � y � 4z � 1
z. 35. Find
745.9x � 105.6y � 19.85
721.4x � 129.1y � 33.77
y.
26. Use an inverse matrix to solve the systems.
(a) (b)
3x � 5y � �23x � 5y � 1
3x � 2y � �33x � 2y � 4
©H
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