Solve the problem.
114)
Let A =
–1 4
0 4
9 –4
and B =
7 2
17 4
4 2
. Find A – B.
114)
A)
–8 2
–17 0
5 –6
B)
3–3
7 0
–5 6
C)
1 5
7 8
13 0
D)
1 2
7 0
5 –2
Solve the matrix equation for X.
115)
Let A =
6–6
–8 0
11 7
and B =
11 0
0–2
6 7
;4X + A = B
115)
A)
X =
–5 6
–8 2
5 0
B)
X =
–5
4–3
2
–21
2
5
40
C)
X =
5
4
3
2
2–1
2
–5
40
D)
X =
5 6
8–2
–5 0
Evaluate the determinant.
116)
2 3
1–1
116)
A)
–5
B)
7
C)
1
D)
5
Solve the matrix equation for X.
117)
Let A =
–6 8
5–3
4–2
and B =
9 2
–8 5
–1 8
;X – B = A
117)
A)
310
–3 2
3 6
B)
3–3
–3 2
3 6
C)
–15 6
13 –8
5–6
D)
310
3–3
3–6
Solve the system of equations using matrices. Use Gauss–Jordan elimination.
118)
5x + 8y–z=36
x– 2y+ 7z=18
3x +y+z=21
118)
A)
{(12, 1, –6)}
B)
{(6, 2, 1)}
C)
{(–6, 1, 12)}
D)
{(6, 1, 2)}
Solve the system using the inverse that is given for the coefficient matrix.
119)
x+2y +3z =6
x+y+z= – 5
–x+y+2z = – 9
The inverse of
1 2 3
1 1 1
–1 1 2
is
1 –1–1
–3 5 2
2 –3–1
.
119)
A)
{(20, –61, 36)}
B)
{(6, –20, 18)}
C)
{(3, –4, –7)}
D)
{(–8, –25, –12)}
Encode or decode the given message, as requested, numbering the letters of the alphabet 1 through 26 in their usual order.
120)
Use the coding matrix A =3 7
2 5 to encode the message LIFE.
120)
A)
78 62
54 43
B)
54 28
129 67
C)
–3–5
3 3
D)
99 53
69 37
Write the augmented matrix for the system of equations.
121)
7x + 7y + 8z= 25
9x + 3y + 3z= 15
3x + 3y – 2z= – 11
121)
A)
7 7 8 25
9 3 3 15
3 3 –2–11
B)
25 877
15 339
–11 –233
C)
7 7 8
9 3 3
3 3 –2
D)
7 9 3 25
7 3 3 15
8 3 –2–11
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.
122)
Adjust the contrast by changing the black to white and the light grey to dark grey. Use matrix
addition to accomplish this.
122)
A)
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
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
+
1 –3 1
1 –3 1
–3–3–3
=
2 0 2
2 0 2
0 0 0
D)
131
131
333
+
–1–1–1
–1–1–1
–1–1–1
=
0 2 0
0 2 0
2 2 2
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)
{(4
3, –13
20 , 0, 5
2)}
B)
{(–1, –20
13 , 0, 2
5)}
C)
{(1, –2, 0, 3)}
D)
{(3
4, –2, 0, 3
4)}
Evaluate the determinant.
124)
1
5
1
6
–3
8
8
7
124)
A)
163
560
B)
–223
840
C)
11
1680
D)
93
560
Find the products AB and BA to determine whether B is the multiplicative inverse of A.
125)
A =5 3
3 2 ,B =2–3
–3 5
125)
A)
B =A–1
B)
B A–1
Evaluate the determinant.
126)
1
6–1
12
12 6
126)
A)
5
2
B)
0
C)
2
D)
–2
Find the products AB and BA to determine whether B is the multiplicative inverse of A.
127)
A =
1 0 0 –1
2 1 0 0
1 1 1 –2
0 0 0 1
,B =
1 0 0 1
–2 1 0–2
1 –1 1 3
0 0 0 1
127)
A)
B =A–1
B)
B A–1
Find values for the variables so that the matrices are equal.
128)
x y +8
2z 1
=10 13
18 1
128)
A)
x =1; y =21; z =36
B)
x =10; y =5; z =9
C)
x =10; y =13; z =18
D)
x =13; y =1; z =10
Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.
129)
x – y + z – w = 10
–2x + 3y + 5w = – 28
x + 2y + 8z + 3w = – 10
x – 4y – 6z – 5w = 30
129)
A)
{(3w – 2, –8w + 3, 4w + 9, w)}
B)
{(24, 10, –6, –2)}
C)
D)
{(–17w – 10, –13w – 16, 5w + 4, w)}
130)
x + y + z = 7
x – y + 2z =7
2x + 3z =14
130)
A)
{(–3z
2– 7, 2z, z)}
B)
{(–3z
2+ 7, z
2, z)}
C)
{(–3z
2+ 7, 2z, z)}
D)
{(–3z
2– 7, z
2, z)}
42
Find the inverse of the matrix, if possible.
131)
A =2–6
–2 0
131)
A)
–1
6–1
6
0–1
2
B)
–1
6–1
2
–1
60
C)
01
2
1
6–1
6
D)
0–1
2
–1
6–1
6
Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.
132)
5x – y + z = 8
7x + y + z = 6
132)
A)
{(–z + 3, 4z + 7, z)}
B)
{(1
6z +7
6, 1
6z, z)}
C)
D)
{(–1
6z +7
6, 1
6z –13
6, z)}
Evaluate the determinant.
133)
–8–4
–4 6
133)
A)
64
B)
–32
C)
56
D)
–64
Find the products AB and BA to determine whether B is the multiplicative inverse of A.
134)
A =10 1
–1 0 ,B =0 1
–110
134)
A)
B A–1
B)
B =A–1
Find the product AB, if possible.
135)
A =
2 –7–9
–6–9–1
1 4 1
, B =
–6–5–3
5 4 4
5 2 7
135)
A)
–92 –14 19
–56 –8 13
–97 –25 20
B)
2–7–9
–6–9–1
1 4 1
–6–5–3
5 4 4
5 2 7
C)
–12 35 27
–30 –36 –4
587
D)
–92 –56 –97
–14 –8–25
19 13 20
Solve the matrix equation for X.
136)
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
136)
A)
X =
–20 5 5
15 20 –5
–25 –25 0
B)
X =
–25 –25 0
–20 1 1
15 20 –5
C)
X =
–20 1 1
15 20 –5
–25 –25 0
D)
X =
–25 –25 0
–20 5 5
15 20 –5
Find the product AB, if possible.
137)
A =–2 2 6 , B =
7
0
–3
137)
A)
–32
B)
–14 0–18
C)
78
D)
–14
0
–18
Solve the system of equations using matrices. Use Gaussian elimination with back–substitution.
138)
x + y + z – w =6
2x – y + 3z + 4w = – 4
4x + 2y – z – w = – 13
–x – 2y + 4z + 3w =12
138)
A)
{(–1
4, 1
3, 1
5, –1
2)}
B)
{(4, –3, –5, 2)}
C)
{(–4, 3, 5, –2)}
D)
{(1
4, –1
3, –1
5, 1
2)}
Write a system of linear equations in three variables, and then use matrices to solve the system.
139)
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?
139)
A)
5 mushrooms; 3 meatballs; 8 eggs
B)
3 mushrooms; 8 meatballs; 5 eggs
C)
8 mushrooms; 5 meatballs; 3 eggs
D)
9 mushrooms; 6 meatballs; 4 eggs
Find the product AB, if possible.
140)
A =1–7 8
5–1–4, B =
–7
6
–1
140)
A)
–57
–37
B)
AB is not defined.
C)
1–7 8
5–1–4
–7 6 –1
D)
–57 –37
Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.
141)
x + y + z = 7
x – y + 2z = 7
141)
A)
{(–3
2z + 7, 1
2z, z)}
B)
{(4, 1, 2)}
C)
{(–3z + 14, 2z – 7, z)}
D)
{(8, –3, 2)}
Use Cramer’s rule to determine if the system is inconsistent system or contains dependent equations.
142)
–2x – 7y= – 47
–4x – 14y= – 49
142)
A)
system contains dependent equations
B)
system is inconsistent
Evaluate the determinant.
143)
900
744
695
143)
A)
504
B)
–139
C)
144
D)
–144
Solve the problem.
144)
The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is
Area = ± 1
2
x1y1 1
x2y2 1
x3y3 1
,
where the symbol ± indicates that the appropriate sign should be chosen to yield a positive area.
Use this formula to find the area of a triangle whose vertices are (10, 4), (3, –5), and (–7, –4).
144)
A)
97
B)
3
2
C)
97
2
D)
3
Solve the system using the inverse that is given for the coefficient matrix.
145)
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
.
145)
A)
{(5, –31, 3)}
B)
{(–16, –2, –5)}
C)
{(–5, –9, –3)}
D)
{(13, –19, 7)}
Find the inverse of the matrix, if possible.
146)
A =5 0
–3–2
146)
A)
–1
20
–3
10
1
5
B)
No inverse
C)
1
50
–3
10 –1
2
D)
1
50
3
10 –1
2
Write the system of linear equations represented by the augmented matrix. Use x, y, z, and, if necessary, w for the
variables.
147)
9 1 0 5 –5
–1 4 1 0 4
6 0 0 7 9
0 8 0 –812
147)
A)
9x +y+z+5w = – 5
–x+4y +z+w=4
6x +y+z+7w =9
x+8y +z–8w =12
B)
9x +y+5z = – 5
–x+4y +z=4
6x +7y =9
8x –8y =12
C)
9x +y+5w = – 5
x+4y +z=4
6x +7w =9
8y +8w =12
D)
9x +y+5w = – 5
–x+4y +z=4
6x +7w =9
8y –8w =12
Use Gaussian elimination to find the complete solution to the system of equations, or state that none exists.
148)
3x – 2y + 2z – w =2
4x + y + z + 6w =8
–3x + 2y – 2z + w =5
5x + 3z – 2w =1
148)
A)
{(2, 0, –3
37 , 9
37 )}
B)
{(1, –1
3, 4
9, 6)}
C)
D)
{(1
2, 0, –37
3, 37
9)}
Encode or decode the given message, as requested, numbering the letters of the alphabet 1 through 26 in their usual order.
149)
Use the coding matrix A =
1 0 –2
1 2 3
1 1 1
to encode the message COME_HERE.
149)
A)
–7–21 3
28 69 44
13 33 26
B)
19 27 –21
–14 –30 39
8 11 –13
C)
–23 –11 –5
72 29 56
31 13 28
D)
3 5 5
15 018
13 8 5
Find the products AB and BA to determine whether B is the multiplicative inverse of A.
150)
A =–5 1
–7 1 ,B =
1
2–1
2
7
2–5
2
150)
A)
B =A–1
B)
B A–1
Use Cramer’s rule to solve the system.
151)
–4x – 4y – 3z = – 22
6x + 2y – 3z =14
–9x – 5y + 8z = – 16
151)
A)
{(3, –1, –2)}
B)
{(1, 2, 1)}
C)
{(3, 1, 2)}
D)
{(4, –1, 2)}
Use Cramer’s rule to solve the system.
152)
11x +3y=1
5x +2y =4
152)
A)
{(–7
10 , 7
39 )}
B)
{(14
37 , 49
37 )}
C)
{(39
7, –10
7)}
D)
{(–10
7, 39
7)}
49
Answer Key
Testname: C6
Answer Key
Testname: C6
Answer Key
Testname: C6
Answer Key
Testname: C6