Find the product AB, if possible.
115)
A =3–2 1
0 4 –3, B =5 0
–2 2
115)
A)
15 –10 5
–612 –8
B)
15 –6
–10 12
5–8
C)
15 0
0 8
D)
AB is not defined.
116)
A =–2 2 6 , B =
7
0
–3
116)
A)
–14 0–18
B)
–14
0
–18
C)
–32
D)
78
Use Cramer’s rule to determine if the system is inconsistent system or contains dependent equations.
117)
4x +y=14
12x+ 3y=42
117)
A)
system is inconsistent
B)
system contains dependent equations
Encode or decode the given message, as requested, numbering the letters of the alphabet 1 through 26 in their usual order.
118)
Use the coding matrix A =
1 0 –2
1 2 3
1 1 1
to encode the message COME_HERE.
118)
A)
–7–21 3
28 69 44
13 33 26
B)
19 27 –21
–14 –30 39
8 11 –13
C)
3 5 5
15 018
13 8 5
D)
–23 –11 –5
72 29 56
31 13 28
39
Solve the matrix equation for X.
119)
Let A =
3 5
0 1
7–9
and B =
6–6
–3 5
0 6
;B – X = 3A
119)
A)
X =
–3–21
3 2
–21 33
B)
X =
–3–21
–3 2
–21 33
C)
X =
15 9
3 8
7–21
D)
X =
15 9
3 8
21 –21
Solve the problem.
120)
Let A =
2–6 3
4–8–2
–5 5 6
and B =
7 6 2
–5 0 –3
4 7 –5
. Find A – B.
120)
A)
–512 1
9–8 1
–9 2 11
B)
9 0 5
–1–8–5
–112 1
C)
–5–12 1
9–8 1
–9–211
D)
–5 0 1
9–8 1
1–2 1
Solve the matrix equation for X.
121)
Let A =4 2
3 5 and B =3–4
4–5;X + A = B
121)
A)
X =–10 –1
–6 1
B)
X =–6–1
–10 1
C)
X =–1–6
1–10
D)
X =1–10
–1–6
40
Solve the system using the inverse that is given for the coefficient matrix.
122)
x+2y +3z = – 4
x+y+z=1
2x +2y +z= – 5
The inverse of
1 2 3
1 1 1
2 2 1
is
–1 4 –1
1 –5 2
0 2 –1
.
122)
A)
{(–5, –9, –3)}
B)
{(–16, –2, –5)}
C)
{(5, –31, 3)}
D)
{(13, –19, 7)}
Solve the system of equations using matrices. Use Gaussian elimination with back–substitution.
123)
3x + 5y – 2w = – 13
2x + 7z – w = – 1
4y + 3z + 3w =1
–x + 2y + 4z = – 5
123)
A)
{(3
4, –2, 0, 3
4)}
B)
{(4
3, –13
20 , 0, 5
2)}
C)
{(–1, –20
13 , 0, 2
5)}
D)
{(1, –2, 0, 3)}
Write a system of linear equations in three variables, and then use matrices to solve the system.
124)
Ron attends a cocktail party (with his graphing calculator in his pocket). He wants to limit his food
intake to 133 g protein, 120 g fat, and 165 g carbohydrate. According to the health conscious
hostess, the marinated mushroom caps have 3 g protein, 5 g fat, and 9 g carbohydrate; the spicy
meatballs have 14 g protein, 7 g fat, and 15 g carbohydrate; and the deviled eggs have 13 g protein,
15 g fat, and 6 g carbohydrate. How many of each snack can he eat to obtain his goal?
124)
A)
3 mushrooms; 8 meatballs; 5 eggs
B)
8 mushrooms; 5 meatballs; 3 eggs
C)
5 mushrooms; 3 meatballs; 8 eggs
D)
9 mushrooms; 6 meatballs; 4 eggs
41
Find the inverse of the matrix, if possible.
125)
A =–6–2
0 5
125)
A)
01
5
–1
6–1
15
B)
–1
6–1
15
01
5
C)
–1
6
1
15
01
5
D)
1
5–1
15
0–1
6
Encode or decode the given message, as requested, numbering the letters of the alphabet 1 through 26 in their usual order.
126)
Use the coding matrix A =3 7
2 5 to encode the message LIFE.
126)
A)
99 53
69 37
B)
78 62
54 43
C)
–3–5
3 3
D)
54 28
129 67
Write the matrix equation as a system of linear equations without matrices.
127)
6 3
5 4
x
y
=–3
4
127)
A)
6x + 3y =3
5x + 4y = – 4
B)
6x + 3y = – 3
5x + 4y = 4
C)
3x + 6y = – 3
5x + 4y = 4
D)
6x + 3y = – 3
4x + 5y = 4
42
Solve the matrix equation for X.
128)
Let A =
6–6
–8 0
11 7
and B =
11 0
0–2
6 7
;4X + A = B
128)
A)
X =
–5
4–3
2
–21
2
5
40
B)
X =
–5 6
–8 2
5 0
C)
X =
5
4
3
2
2–1
2
–5
40
D)
X =
5 6
8–2
–5 0
Find the products AB and BA to determine whether B is the multiplicative inverse of A.
129)
A =–2 4
4–4,B =
1
2
1
4
1
2
1
4
129)
A)
B =A–1
B)
B A–1
43
The shape in the figure below is shown using 9 pixels in a 3 ×
3 grid. The color levels are given to the right of the
figure. Use the matrix
1 3 1
1 3 1
3 3 3
that represents a digital photograph of the shape to solve the problem.
130)
Adjust the contrast by changing the black to white and the light grey to dark grey. Use matrix
addition to accomplish this.
130)
A)
1 3 1
1 3 1
3 3 3
+
–1–1–1
–1–1–1
–1–1–1
=
0 2 0
0 2 0
2 2 2
B)
131
131
333
+
–1 3 –1
–1 3 –1
3 3 3
=
2 0 2
2 0 2
0 0 0
C)
1 3 1
1 3 1
3 3 3
+
0 –3 0
0 –3 0
–3–3–3
=
1 0 1
1 0 1
0 0 0
D)
131
131
333
+
1 –3 1
1 –3 1
–3–3–3
=
2 0 2
2 0 2
0 0 0
Solve the problem.
131)
Let A =–9 1
2 5 and B =6 2
1 1 . Find A + B.
131)
A)
–3 3
3 6
B)
9
C)
–3–7
–8–8
D)
3 4
2 6
44
Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.
132)
x + y + z = 9
2x – 3y + 4z = 7
132)
A)
{(27
5, 13
5, 1)}
B)
C)
{(–7
5z +34
5, 2
5z +11
5, z)}
D)
{(3
5z +16
5, –8
5z +29
5, z)}
Find the product AB, if possible.
133)
A = [6–1–1], B =
9–4 6
6–9–4
–1 5 –9
133)
A)
6–1 0 –1
9–4 6
6–9–4
–1 5 –9
B)
49
–20
49
C)
49 –20 49
D)
54 4–6
36 9 4
–6–5 9
Solve the matrix equation for X.
134)
Let A =1 4
3 1 and B =3–5
1–5;X + A = B
134)
A)
X =–6–2
–9 2
B)
X =–9 2
–6–2
C)
X =–2–6
2–9
D)
X =2–9
–2–6
45
Write the linear system as a matrix equation in the form AX = B, where A is the coefficient matrix and B is the constant
matrix.
135)
–2x + 3y = 10
5x – 2y = 19
135)
A)
–2 3
–2 5
x
y
=19
10
B)
10 3
19 5
x
y
=–2
–2
C)
–2 3
5–2
x
y
=10
19
D)
–2 5
3–2
x
y
=10
19
Give the order of the matrix, and identify the given element of the matrix.
136)
615 –4 3 –10
–11 –e–5 2
112 –812 0
1
3115 11 8
; a34
136)
A)
4 × 4;–8
B)
4 × 5;12
C)
5 × 4;15
D)
20; 0
Write the augmented matrix for the system of equations.
137)
x –9y + z =15
y +6z =17
z =11
137)
A)
0–9 0 15
0 0 6 17
0 0 0 11
B)
1 9 1 15
0 1 6 17
0 0 1 11
C)
1–9 1 15
1 1 6 17
1 1 1 11
D)
1–9 1 15
0 1 6 17
0 0 1 11
46
Solve the problem.
138)
Let A =
1
–3
2
and B =
–1
3
–2
. Find A – 2B.
138)
A)
–3
9
–6
B)
3
–9
6
C)
–1
3
–2
D)
3
–6
4
139)
Let A =
–1 4
0 4
9 –4
and B =
7 2
17 4
4 2
. Find A – B.
139)
A)
1 5
7 8
13 0
B)
3–3
7 0
–5 6
C)
–8 2
–17 0
5 –6
D)
1 2
7 0
5 –2
Solve the problem using matrices.
140)
A company that manufactures products A, B, and C does both assembly and testing. The hours
needed to assemble and test each product are shown in the table below.
Hours needed
weekly to assemble
Hours needed
weekly to test
Product A 1 4
Product B 1 5
Product C 2 10
The company has exactly 21 hours per week available for assembly and 95 hours per week
available for testing. If the company must produce t units of Product C this week, how many units
of Products A and B can they produce?
140)
A)
10 of Product A; 11 of Product B
B)
t +10 of Product A; t +11 of Product B
C)
10t of Product A; 2t +11 of Product B
D)
10 of Product A; –2t +11 of Product B
47
Find values for the variables so that the matrices are equal.
141)
x
7
=4
y
141)
A)
x =4; y =4
B)
x =7; y =7
C)
x =4; y =7
D)
x =7; y =4
Solve the system using the inverse that is given for the coefficient matrix.
142)
x+2y +3z =4
x+y+z=7
x–2z =11
The inverse of
1 2 3
1 1 1
1 0 –2
is
–2 4 –1
3 –5 2
–1 2 –1
.
142)
A)
{(2, 3, –1)}
B)
{(4, 0, 0)}
C)
{(47, 69, 29)}
D)
{(9, –1, –1)}
Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.
143)
3x – 2y + 2z – w =2
4x + y + z + 6w =8
–3x + 2y – 2z + w =5
5x + 3z – 2w =1
143)
A)
{(1, –1
3, 4
9, 6)}
B)
{(2, 0, –3
37 , 9
37 )}
C)
{(1
2, 0, –37
3, 37
9)}
D)
Solve the problem.
144)
Let A =–3 3
0 2 . Find 2A.
144)
A)
–6 3
0 2
B)
–6 6
0 2
C)
–1 5
2 4
D)
–6 6
0 4
48
Evaluate the determinant.
145)
3 4
6 5
145)
A)
39
B)
9
C)
–18
D)
–9
146)
0 0 0 7
4 4 4 9
6 1 3 8
2 5 5 9
146)
A)
168
B)
–168
C)
–35
D)
–21
Find the product AB, if possible.
147)
A =–2 3
3 2 , B =–2 0
–1 1
147)
A)
1 3
–8 2
B)
4–6
–4–1
C)
4 0
–3 2
D)
3 1
2–8
Find the inverse of the matrix, if possible.
148)
A =5 0
–3–2
148)
A)
No inverse
B)
1
50
3
10 –1
2
C)
1
50
–3
10 –1
2
D)
–1
20
–3
10
1
5
49
Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.
149)
4x – y + 3z =12
x + 4y + 6z = – 32
5x + 3y + 9z =20
149)
A)
{(–8, –7, 9)}
B)
C)
{(8, –7, –2)}
D)
{(2, –7, –1)}
Evaluate the determinant.
150)
1
6–1
12
12 6
150)
A)
2
B)
0
C)
5
2
D)
–2
Solve the matrix equation for X.
151)
Let A =
4 1 –5
4 0 0
1–4 5
and B =
–1–4–5
0 1 1
4 0 4
; 5B –5A = X
151)
A)
X =
–20 1 1
15 20 –5
–25 –25 0
B)
X =
–25 –25 0
–20 1 1
15 20 –5
C)
X =
–20 5 5
15 20 –5
–25 –25 0
D)
X =
–25 –25 0
–20 5 5
15 20 –5
50
The shape in the figure below is shown using 9 pixels in a 3 ×
3 grid. The color levels are given to the right of the
figure. Use the matrix
1 3 1
1 3 1
3 3 3
that represents a digital photograph of the shape to solve the problem.
152)
Using the same color levels from the instructions, write a 3 × 3 matrix A that represents the letter L
in dark grey on a white background. Then find a 3 × 3 matrix B so that A + B lightens only the letter
L from dark grey to light grey.
152)
A)
A =
1 0 0
1 0 0
1 1 1
; B =
–1 0 0
–1 0 0
–1–1–1
B)
A =
2 0 0
2 0 0
2 2 2
; B =
–1 0 0
–1 0 0
–1–1–1
C)
A =
2 1 1
2 1 1
2 2 2
; B =
1 0 0
1 0 0
1 1 1
D)
A =
3 1 1
3 1 1
3 3 3
; B =
–2–1–1
–2–1–1
–2–2–2
51
Answer Key
Testname: C6
Answer Key
Testname: C6
53
Answer Key
Testname: C6
54
Answer Key
Testname: C6
55