Question 1

Find $A B$ . 
 $A = \left[ \begin{array} { c c } 
- 2 & 0 \
0 & 4
\end{array} \right] , B = \left[ \begin{array} { c c } 
3 & 0 \
0 & - 2
\end{array} \right]$

Accepted Answer

A)  $\left[ \begin{array} { l l } 
0 & - 6 \
0 & - 8
\end{array} \right]$ 
B)  $\left[ \begin{array} { c c } 
- 2 & 0 \
0 & 4
\end{array} \right]$ 
C)  $\left[ \begin{array} { c c } 
- 6 & 0 \
0 & - 8
\end{array} \right]$ 
D)  $\left[ \begin{array} { c c } 
0 & - 8 \
- 6 & 0
\end{array} \right]$ 
E)  $\left[ \begin{array} { l l } 
- 6 & 0 \
- 8 & 0
\end{array} \right]$ 
A)  $\left[ \begin{array} { l l } 
0 & - 6 \
0 & - 8
\end{array} \right]$ 
B)  $\left[ \begin{array} { c c } 
- 2 & 0 \
0 & 4
\end{array} \right]$ 
C)  $\left[ \begin{array} { c c } 
- 6 & 0 \
0 & - 8
\end{array} \right]$ 
D)  $\left[ \begin{array} { c c } 
0 & - 8 \
- 6 & 0
\end{array} \right]$ 
E)  $\left[ \begin{array} { l l } 
- 6 & 0 \
- 8 & 0
\end{array} \right]$ C

Question 2

Use the matrix capabilities of a graphing utility to find AB, if possible. $$A = \left[ \begin{array} { c c } 
4 & - 5 \
- 4 & 2 \
- 3 & - 9 \
- 5 & 8
\end{array} \right] , B = \left[ \begin{array} { c c c c } 
- 1 & - 9 & 8 & - 7 \
- 9 & - 1 & - 1 & 9
\end{array} \right]$$

Accepted Answer

A)  $\left[ \begin{array} { c } 
41 \
34 \
- 32 \
107
\end{array} \right]$ 
B)  $\left[ \begin{array} { c c c c } 
41 & - 31 & 37 & - 73 \
- 14 & 34 & - 34 & 46 \
84 & 36 & - 15 & - 60 \
- 67 & 37 & - 48 & 107
\end{array} \right]$ 
C)  $\left[ \begin{array} { c c c c } 
- 49 & - 41 & 27 & 17 \
22 & 38 & - 30 & 10 \
- 78 & 18 & - 33 & 102 \
77 & 53 & - 32 & - 37
\end{array} \right]$ 
D) not possible 
E)  $\left[ \begin{array} { c c c c } 
41 & - 14 & 84 & - 67 \
- 31 & 34 & 36 & 37 \
37 & - 34 & - 15 & - 48 \
- 73 & 46 & - 60 & 107
\end{array} \right]$ 
A)  $\left[ \begin{array} { c } 
41 \
34 \
- 32 \
107
\end{array} \right]$ 
B)  $\left[ \begin{array} { c c c c } 
41 & - 31 & 37 & - 73 \
- 14 & 34 & - 34 & 46 \
84 & 36 & - 15 & - 60 \
- 67 & 37 & - 48 & 107
\end{array} \right]$ 
C)  $\left[ \begin{array} { c c c c } 
- 49 & - 41 & 27 & 17 \
22 & 38 & - 30 & 10 \
- 78 & 18 & - 33 & 102 \
77 & 53 & - 32 & - 37
\end{array} \right]$ 
D) not possible 
E)  $\left[ \begin{array} { c c c c } 
41 & - 14 & 84 & - 67 \
- 31 & 34 & 36 & 37 \
37 & - 34 & - 15 & - 48 \
- 73 & 46 & - 60 & 107
\end{array} \right]$ B

Question 3

Solve the system using Gaussian elimination.   $$\left\{ \begin{array} { l } 
x + y = - 2 \
x + z = - 4 \
y + z = - 4
\end{array} \right.$$

Accepted Answer

A)  $( 1,1,3 )$ 
B)  $( - 1 , - 1 , - 3 )$ 
C)  $( 1,1 , - 3 )$ 
D)  $( 1 , - 1 , - 3 )$ 
E)  $( - 1,1 , - 3 )$ 
A)  $( 1,1,3 )$ 
B)  $( - 1 , - 1 , - 3 )$ 
C)  $( 1,1 , - 3 )$ 
D)  $( 1 , - 1 , - 3 )$ 
E)  $( - 1,1 , - 3 )$ B

Question 4

Use matrices to solve the system of equations (if possible).Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$\left\{ \begin{aligned}
2 x + 9 y & = - 35 \
- 9 x + 2 y & = 30
\end{aligned} \right.$$

Accepted Answer

A) x = -3, y = -4 
B) x = 4, y = 3 
C) x = -4, y = -3 
D) x = -3, y = 4 
E) no solution 
A) x = -3, y = -4 
B) x = 4, y = 3 
C) x = -4, y = -3 
D) x = -3, y = 4 
E) no solution

Question 5

Solve the system of linear equations $\left\{ \begin{array} { l } 
2 x + 3 y = - 15 \
4 x - 3 y = 10
\end{array} \right.$ using the inverse matrix $\left[ \begin{array} { c c } 
\frac { 1 } { 6 } & \frac { 1 } { 6 } \
\frac { 2 } { 9 } & - \frac { 1 } { 9 }
\end{array} \right]$

Accepted Answer

A)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 5 } { 6 } \
\frac { 5 } { 2 }
\end{array} \right]$ 
B)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
- \frac { 5 } { 6 } \
- \frac { 40 } { 9 }
\end{array} \right]$ 
C)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 25 } { 18 } \
\frac { 5 } { 2 }
\end{array} \right]$ 
D)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 5 } { 2 } \
0
\end{array} \right]$ 
E)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
- \frac { 5 } { 3 } \
\frac { 5 } { 6 }
\end{array} \right]$ 
A)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 5 } { 6 } \
\frac { 5 } { 2 }
\end{array} \right]$ 
B)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
- \frac { 5 } { 6 } \
- \frac { 40 } { 9 }
\end{array} \right]$ 
C)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 25 } { 18 } \
\frac { 5 } { 2 }
\end{array} \right]$ 
D)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { l } 
\frac { 5 } { 2 } \
0
\end{array} \right]$ 
E)   $\left[ \begin{array} { l } 
x \
y
\end{array} \right] = \left[ \begin{array} { c } 
- \frac { 5 } { 3 } \
\frac { 5 } { 6 }
\end{array} \right]$

Question 6

Find 8A.   $$A = \left[ \begin{array} { l l l } 
3 & 7 & 5 \
3 & 7 & 3
\end{array} \right]$$

Accepted Answer

A)  $\left[ \begin{array} { l l } 
7 & 40 \
7 & 24
\end{array} \right]$ 
B)  $\left[ \begin{array} { l l l } 
24 & 56 & 40 \
24 & 56 & 24
\end{array} \right]$ 
C)  $\left[ \begin{array} { c c c } 
24 & 7 & 5 \
3 & 7 & 3
\end{array} \right]$ 
D)  $\left[ \begin{array} { c c c } 
3 & 56 & 5 \
24 & 7 & 24
\end{array} \right]$ 
E)  $\left[ \begin{array} { c c } 
24 & 7 \
3 & 56
\end{array} \right]$ 
A)  $\left[ \begin{array} { l l } 
7 & 40 \
7 & 24
\end{array} \right]$ 
B)  $\left[ \begin{array} { l l l } 
24 & 56 & 40 \
24 & 56 & 24
\end{array} \right]$ 
C)  $\left[ \begin{array} { c c c } 
24 & 7 & 5 \
3 & 7 & 3
\end{array} \right]$ 
D)  $\left[ \begin{array} { c c c } 
3 & 56 & 5 \
24 & 7 & 24
\end{array} \right]$ 
E)  $\left[ \begin{array} { c c } 
24 & 7 \
3 & 56
\end{array} \right]$

Question 7

Find the product.   $$\left[ \begin{array} { c c c c } 
2 & 1 & 0 & 0 \
- 1 & 2 & 0 & - 4 \
0 & 0 & 2 & 0 \
0 & 4 & 0 & 2
\end{array} \right] \left[ \begin{array} { c } 
5 \
3 \
- 3 \
- 5
\end{array} \right]$$

Accepted Answer

A)  $\left[ \begin{array} { c } 
13 \
21 \
- 6 \
2
\end{array} \right]$ 
B)  $\left[ \begin{array} { c c } 
13 & 21 \
- 6 & 2
\end{array} \right]$ 
C)  $\left[ \begin{array} { c } 
2 \
- 6 \
21 \
13
\end{array} \right]$ 
D)  $\left[ \begin{array} { l l l l } 
13 & 21 & - 6 & 2
\end{array} \right]$ 
E) cannot multiply 
A)  $\left[ \begin{array} { c } 
13 \
21 \
- 6 \
2
\end{array} \right]$ 
B)  $\left[ \begin{array} { c c } 
13 & 21 \
- 6 & 2
\end{array} \right]$ 
C)  $\left[ \begin{array} { c } 
2 \
- 6 \
21 \
13
\end{array} \right]$ 
D)  $\left[ \begin{array} { l l l l } 
13 & 21 & - 6 & 2
\end{array} \right]$ 
E) cannot multiply

Question 8

Use matrices to solve the system of equations (if possible).Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.   $$\left\{ \begin{aligned}
x - 4 y + 3 z - 2 w & = 10 \
3 x - 2 y + z - 4 w & = - 22 \
- 4 x + 3 y - 2 z + w & = - 2 \
- 2 x + y - 4 z + 3 w & = - 10
\end{aligned} \right.$$

Accepted Answer

A)  (-2, -1, 8, 6) 
B) (-2, 0, 8, 6) 
C) (-2, 0, 8, -6) 
D)  (-2, -1, -8, -6) 
E) (-2, 0, -8, 6) 
A)  (-2, -1, 8, 6) 
B) (-2, 0, 8, 6) 
C) (-2, 0, 8, -6) 
D)  (-2, -1, -8, -6) 
E) (-2, 0, -8, 6)

Question 9

Use the matrix capabilities of a graphing utility to evaluate the determinant.   $\left[ \begin{array} { c c c } 
3 & 1 & - 6 \
0 & - 6 & 9 \
1 & 1 & 1
\end{array} \right]$

Accepted Answer

A) -72 
B) -71 
C) -70 
D) -74 
E) -73 
A) -72 
B) -71 
C) -70 
D) -74 
E) -73

Question 10

Find all minors of the matrix.   $\left[ \begin{array} { c c c } 
- 7 & 0 & 4 \
2 & 4 & 3 \
3 & - 3 & 5
\end{array} \right]$

Accepted Answer

A)  $M _ { 11 } = 1 , M _ { 12 } = - 18 , M _ { 13 } = 12 , M _ { 21 } = - 47 , M _ { 22 } = 21 , M _ { 23 } = - 16 , M _ { 31 } = - 29 , M _ { 32 } = - 28 , M _ { 33 } = 29$ 
B)  $M _ { 11 } = 12 , M _ { 12 } = - 47 , M _ { 13 } = 21 , M _ { 21 } = - 16 , M _ { 22 } = - 29 , M _ { 23 } = - 28 , M _ { 31 } = 29 , M _ { 32 } = 1 , M _ { 33 } = - 18$ 
C)  $M _ { 11 } = 29 , M _ { 12 } = 1 , M _ { 13 } = - 18 , M _ { 21 } = 12 , M _ { 22 } = - 47 , M _ { 23 } = 21 , M _ { 31 } = - 16 , M _ { 32 } = - 29 , M _ { 33 } = - 28$ 
D)  $M _ { 11 } = - 18 , M _ { 12 } = 12 , M _ { 13 } = - 47 , M _ { 21 } = 21 , M _ { 22 } = - 16 , M _ { 23 } = - 29 , M _ { 31 } = - 28 , M _ { 32 } = 29 , M _ { 33 } = 1$ 
E)  $M _ { 11 } = - 16 , M _ { 12 } = - 29 , M _ { 13 } = - 28 , M _ { 21 } = 29 , M _ { 22 } = 1 , M _ { 23 } = - 18 , M _ { 31 } = 12 , M _ { 32 } = - 47 , M _ { 33 } = 21$ 
A)  $M _ { 11 } = 1 , M _ { 12 } = - 18 , M _ { 13 } = 12 , M _ { 21 } = - 47 , M _ { 22 } = 21 , M _ { 23 } = - 16 , M _ { 31 } = - 29 , M _ { 32 } = - 28 , M _ { 33 } = 29$ 
B)  $M _ { 11 } = 12 , M _ { 12 } = - 47 , M _ { 13 } = 21 , M _ { 21 } = - 16 , M _ { 22 } = - 29 , M _ { 23 } = - 28 , M _ { 31 } = 29 , M _ { 32 } = 1 , M _ { 33 } = - 18$ 
C)  $M _ { 11 } = 29 , M _ { 12 } = 1 , M _ { 13 } = - 18 , M _ { 21 } = 12 , M _ { 22 } = - 47 , M _ { 23 } = 21 , M _ { 31 } = - 16 , M _ { 32 } = - 29 , M _ { 33 } = - 28$ 
D)  $M _ { 11 } = - 18 , M _ { 12 } = 12 , M _ { 13 } = - 47 , M _ { 21 } = 21 , M _ { 22 } = - 16 , M _ { 23 } = - 29 , M _ { 31 } = - 28 , M _ { 32 } = 29 , M _ { 33 } = 1$ 
E)  $M _ { 11 } = - 16 , M _ { 12 } = - 29 , M _ { 13 } = - 28 , M _ { 21 } = 29 , M _ { 22 } = 1 , M _ { 23 } = - 18 , M _ { 31 } = 12 , M _ { 32 } = - 47 , M _ { 33 } = 21$

Question 11

Write a cryptogram by assigning a number to each letter in the alphabet such as 1 = A, 2 = B, 3 = C and so on (with 0 assigned to a blank space) for the message "TWO IF BY LAND" using the following matrix.   $$\left[ \begin{array} { c c c } 
0 & 1 & - 1 \
- 1 & - 1 & 1 \
- 1 & 1 & 0
\end{array} \right]$$

Accepted Answer

A) $\begin{array} { l l l l l l l l l } 
- 13 & - 11 & 12 & - 4 & 10 & - 10 & - 38 & 12 & 3 \
- 15 & - 3 & 9 & - 27 & 23 & 2 & & &
\end{array}$ 
B) $\begin{array} { l l l l l l l l l } 
- 4 & 10 & - 10 & - 38 & 12 & 3 & - 15 & - 3 & 9 \
- 27 & 23 & 2 & - 13 & - 11 & 12 & & &
\end{array}$ 
C) 
$\begin{array} { l l l l l l l l l } 
- 27 & 23 & 2 & - 13 & - 11 & 12 & - 4 & 10 & - 10 \
- 38 & 12 & 3 & - 15 & - 3 & 9 & & &
\end{array}$ 
D) $\begin{array} { l l l l l l l l l } 
- 15 & - 3 & 9 & - 27 & 23 & 2 & - 13 & - 11 & 12 \
- 4 & 10 & - 10 & - 38 & 12 & 3 & & &
\end{array}$ 
E) $\begin{array} { c c c c c c c c c } 
- 38 & 12 & 3 & - 15 & - 3 & 9 & - 27 & 23 & 2 \
- 13 & - 11 & 12 & - 4 & 10 & - 10 & & &
\end{array}$ 
A) $\begin{array} { l l l l l l l l l } 
- 13 & - 11 & 12 & - 4 & 10 & - 10 & - 38 & 12 & 3 \
- 15 & - 3 & 9 & - 27 & 23 & 2 & & &
\end{array}$ 
B) $\begin{array} { l l l l l l l l l } 
- 4 & 10 & - 10 & - 38 & 12 & 3 & - 15 & - 3 & 9 \
- 27 & 23 & 2 & - 13 & - 11 & 12 & & &
\end{array}$ 
C) 
$\begin{array} { l l l l l l l l l } 
- 27 & 23 & 2 & - 13 & - 11 & 12 & - 4 & 10 & - 10 \
- 38 & 12 & 3 & - 15 & - 3 & 9 & & &
\end{array}$ 
D) $\begin{array} { l l l l l l l l l } 
- 15 & - 3 & 9 & - 27 & 23 & 2 & - 13 & - 11 & 12 \
- 4 & 10 & - 10 & - 38 & 12 & 3 & & &
\end{array}$ 
E) $\begin{array} { c c c c c c c c c } 
- 38 & 12 & 3 & - 15 & - 3 & 9 & - 27 & 23 & 2 \
- 13 & - 11 & 12 & - 4 & 10 & - 10 & & &
\end{array}$

Question 12

Use a determinant to find an equation of the line passing through the points. ​
(0, 0), (-8, 2)
​

Accepted Answer

A) 2x + 8y = 0 
B) 8x - 2y = 2 
C) 2x + 8y = 8 
D) 8x - 2y = 0 
E) 2x - 8y = 0 
A) 2x + 8y = 0 
B) 8x - 2y = 2 
C) 2x + 8y = 8 
D) 8x - 2y = 0 
E) 2x - 8y = 0

Question 13

Use Cramer's Rule to solve (if possible) the system of equations.   $\left\{ \begin{array} { r } 
- 0.4 x + 0.8 y = 1.6 \
0.2 x + 0.3 y = 2.3
\end{array} \right.$

Accepted Answer

A)  $\left( - \frac { 34 } { 7 } , - \frac { 31 } { 7 } \right)$ 
B)  $\left( \frac { 34 } { 7 } , \frac { 31 } { 7 } \right)$ 
C)  $\left( \frac { 34 } { 7 } , - \frac { 31 } { 7 } \right)$ 
D)  $\left( \frac { 31 } { 7 } , \frac { 34 } { 7 } \right)$ 
E)  $\left( - \frac { 34 } { 7 } , \frac { 31 } { 7 } \right)$ 
A)  $\left( - \frac { 34 } { 7 } , - \frac { 31 } { 7 } \right)$ 
B)  $\left( \frac { 34 } { 7 } , \frac { 31 } { 7 } \right)$ 
C)  $\left( \frac { 34 } { 7 } , - \frac { 31 } { 7 } \right)$ 
D)  $\left( \frac { 31 } { 7 } , \frac { 34 } { 7 } \right)$ 
E)  $\left( - \frac { 34 } { 7 } , \frac { 31 } { 7 } \right)$

Question 14

Evaluate the expression. $$\left[ \begin{array} { l } 
- 2 \
- 9 \
- 4
\end{array} \right] \left( \left[ \begin{array} { l l } 
- 3 & 5
\end{array} \right] + \left[ \begin{array} { l l } 
7 & - 5
\end{array} \right] \right)$$

Accepted Answer

A)  $\left[ \begin{array} { c c c c } 
6 & - 10 & - 14 & 10 \
27 & - 45 & - 63 & 45 \
12 & - 20 & - 28 & 20
\end{array} \right]$ 
B)  $\left[ \begin{array} { c } 
- 8 \
- 36 \
- 16
\end{array} \right]$ 
C)  $\left[ \begin{array} { c c c } 
- 8 & - 36 & - 16 \
0 & 0 & 0
\end{array} \right]$ 
D) not possible 
E)  $\left[ \begin{array} { c c } 
- 8 & 0 \
- 36 & 0 \
- 16 & 0
\end{array} \right]$ 
A)  $\left[ \begin{array} { c c c c } 
6 & - 10 & - 14 & 10 \
27 & - 45 & - 63 & 45 \
12 & - 20 & - 28 & 20
\end{array} \right]$ 
B)  $\left[ \begin{array} { c } 
- 8 \
- 36 \
- 16
\end{array} \right]$ 
C)  $\left[ \begin{array} { c c c } 
- 8 & - 36 & - 16 \
0 & 0 & 0
\end{array} \right]$ 
D) not possible 
E)  $\left[ \begin{array} { c c } 
- 8 & 0 \
- 36 & 0 \
- 16 & 0
\end{array} \right]$

Question 15

Use the inverse formula $A ^ { - 1 } = \frac { 1 } { a d - b c } \left[ \begin{array} { c c } 
d & - b \
- c & a
\end{array} \right]$ to find the inverse of the 2×2 matrix (if it exists).  
  $\left[ \begin{array} { c c } 
1 & - 2 \
- 3 & 2
\end{array} \right]$

Accepted Answer

A)   $- \frac { 1 } { 4 } \left[ \begin{array} { c c } 
- 2 & - 2 \
3 & 1
\end{array} \right]$ 
B)   $- \frac { 1 } { 4 } \left[ \begin{array} { l l } 
- 2 & - 2 \
- 3 & - 1
\end{array} \right]$ 
C)   $- \frac { 1 } { 4 } \left[ \begin{array} { l l } 
2 & 2 \
3 & 1
\end{array} \right]$ 
D)   $- \frac { 1 } { 4 } \left[ \begin{array} { c c } 
- 2 & 2 \
3 & - 1
\end{array} \right]$ 
E)   $- \frac { 1 } { 4 } \left[ \begin{array} { c c } 
2 & - 2 \
3 & 1
\end{array} \right]$ 
A)   $- \frac { 1 } { 4 } \left[ \begin{array} { c c } 
- 2 & - 2 \
3 & 1
\end{array} \right]$ 
B)   $- \frac { 1 } { 4 } \left[ \begin{array} { l l } 
- 2 & - 2 \
- 3 & - 1
\end{array} \right]$ 
C)   $- \frac { 1 } { 4 } \left[ \begin{array} { l l } 
2 & 2 \
3 & 1
\end{array} \right]$ 
D)   $- \frac { 1 } { 4 } \left[ \begin{array} { c c } 
- 2 & 2 \
3 & - 1
\end{array} \right]$ 
E)   $- \frac { 1 } { 4 } \left[ \begin{array} { c c } 
2 & - 2 \
3 & 1
\end{array} \right]$

Question 16

Solve the system by Gauss - Jordan elimination.   $$\left\{ \begin{aligned}
\frac { 1 } { 3 } x + \frac { 3 } { 4 } y - \frac { 2 } { 3 } z & = - 4 \
x + \frac { 1 } { 2 } y + \frac { 1 } { 3 } z & = 30 \
\frac { 1 } { 6 } x - \frac { 1 } { 8 } y - z & = - 30
\end{aligned} \right.$$

Accepted Answer

A)  $( - 12 , - 16 , - 30 )$ 
B)  $( - 12,16 , - 30 )$ 
C)  $( 12,16,30 )$ 
D)  $( 0,16 , - 30 )$ 
E)  $( 12 , - 16,30 )$ 
A)  $( - 12 , - 16 , - 30 )$ 
B)  $( - 12,16 , - 30 )$ 
C)  $( 12,16,30 )$ 
D)  $( 0,16 , - 30 )$ 
E)  $( 12 , - 16,30 )$

Question 17

Use the matrix capabilities of a graphing utility to find the determinant of the matrix $\left[ \begin{array} { c c c } 
0.2 & - 0.6 & 0.6 \
0.8 & 0.4 & 0 \
- 0.4 & - 1.4 & 1.0
\end{array} \right]$ .

Accepted Answer

A) -0.008 
B) -0.080 
C) -0.040 
D) 0.024 
E) -0.016 
A) -0.008 
B) -0.080 
C) -0.040 
D) 0.024 
E) -0.016

Question 18

Use Cramer's Rule to solve (if possible) the system of equations.   $$\left\{ \begin{array} { r } 
4 x - y + z = - 20 \
2 x + 2 y + 3 z = 10 \
5 x - 2 y + 6 z = 2
\end{array} \right.$$

Accepted Answer

A) (6, 2, 6) 
B) (6, -2, 6) 
C) (-6, 2, 6) 
D) (7, -2, 6) 
E) (-6, -2, 6) 
A) (6, 2, 6) 
B) (6, -2, 6) 
C) (-6, 2, 6) 
D) (7, -2, 6) 
E) (-6, -2, 6)

Question 19

Find x and y. $$\left[ \begin{array} { c c c } 
x - 7 & 3 & - 8 y \
4 & - 8 x & - 7 \
- 9 & y - 3 & - 3
\end{array} \right] = \left[ \begin{array} { c c c } 
7 x - 19 & 3 & 24 \
4 & - 16 & - 7 \
- 9 & - 6 & - 3
\end{array} \right]$$

Accepted Answer

A) x = -3, y = 3 
B) x = 3, y = -4 
C) no solution 
D) x = 2, y = -3 
E) x = 2, y = -4 
A) x = -3, y = 3 
B) x = 3, y = -4 
C) no solution 
D) x = 2, y = -3 
E) x = 2, y = -4

Question 20

Use the matrix capabilities of a graphing utility to solve the following system of linear equations:   $$\left\{ \begin{array} { l l } 
- 14 x + 14 y + 21 z & = 10 \
7 x - 7 y & = - 5 \
7 x + 28 z & = 20
\end{array} \right.$$

Accepted Answer

A)   $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 20 } { 7 } \
\frac { 25 } { 7 } \
0
\end{array} \right]$ 
B)   $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 10 } { 7 } \
0 \
\frac { - 15 } { 7 }
\end{array} \right]$ 
C)   $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 15 } { 7 } \
\frac { 10 } { 7 } \
\frac { 5 } { 7 }
\end{array} \right]$ 
D)   $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 5 } { 7 } \
\frac { - 20 } { 7 } \
\frac { 10 } { 7 }
\end{array} \right]$ 
E)  $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 15 } { 7 } \
\frac { - 10 } { 7 } \
\frac { 10 } { 7 }
\end{array} \right]$ 
A)   $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 20 } { 7 } \
\frac { 25 } { 7 } \
0
\end{array} \right]$ 
B)   $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 10 } { 7 } \
0 \
\frac { - 15 } { 7 }
\end{array} \right]$ 
C)   $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 15 } { 7 } \
\frac { 10 } { 7 } \
\frac { 5 } { 7 }
\end{array} \right]$ 
D)   $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { 5 } { 7 } \
\frac { - 20 } { 7 } \
\frac { 10 } { 7 }
\end{array} \right]$ 
E)  $\left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { c } 
\frac { - 15 } { 7 } \
\frac { - 10 } { 7 } \
\frac { 10 } { 7 }
\end{array} \right]$

Find $A B$ . $A = \left[ \begin{array} { c c } - 2 & 0 \\0 & 4\end{array} \right] , B = \left[ \begin{array} { c c } 3 & 0 \\0 & - 2\end{array} \right]$

Use the matrix capabilities of a graphing utility to find AB, if possible. $A = \left[ \begin{array} { c c } 4 & - 5 \\- 4 & 2 \\- 3 & - 9 \\- 5 & 8\end{array} \right] , B = \left[ \begin{array} { c c c c } - 1 & - 9 & 8 & - 7 \\- 9 & - 1 & - 1 & 9\end{array} \right]$

Solve the system using Gaussian elimination. $\left\{ \begin{array} { l } x + y = - 2 \\x + z = - 4 \\y + z = - 4\end{array} \right.$

Use matrices to solve the system of equations (if possible).Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $\left\{ \begin{aligned}2 x + 9 y & = - 35 \\- 9 x + 2 y & = 30\end{aligned} \right.$

Solve the system of linear equations $\left\{ \begin{array} { l } 2 x + 3 y = - 15 \\4 x - 3 y = 10\end{array} \right.$ using the inverse matrix $\left[ \begin{array} { c c } \frac { 1 } { 6 } & \frac { 1 } { 6 } \\\frac { 2 } { 9 } & - \frac { 1 } { 9 }\end{array} \right]$

Find 8A. $A = \left[ \begin{array} { l l l } 3 & 7 & 5 \\3 & 7 & 3\end{array} \right]$

Find the product. $\left[ \begin{array} { c c c c } 2 & 1 & 0 & 0 \\- 1 & 2 & 0 & - 4 \\0 & 0 & 2 & 0 \\0 & 4 & 0 & 2\end{array} \right] \left[ \begin{array} { c } 5 \\3 \\- 3 \\- 5\end{array} \right]$

Use the matrix capabilities of a graphing utility to evaluate the determinant. $\left[ \begin{array} { c c c } 3 & 1 & - 6 \\0 & - 6 & 9 \\1 & 1 & 1\end{array} \right]$

Find all minors of the matrix. $\left[ \begin{array} { c c c } - 7 & 0 & 4 \\2 & 4 & 3 \\3 & - 3 & 5\end{array} \right]$

Write a cryptogram by assigning a number to each letter in the alphabet such as 1 = A, 2 = B, 3 = C and so on (with 0 assigned to a blank space) for the message "TWO IF BY LAND" using the following matrix. $\left[ \begin{array} { c c c } 0 & 1 & - 1 \\- 1 & - 1 & 1 \\- 1 & 1 & 0\end{array} \right]$

Use a determinant to find an equation of the line passing through the points. (0, 0), (-8, 2)

Use Cramer's Rule to solve (if possible) the system of equations. $\left\{ \begin{array} { r } - 0.4 x + 0.8 y = 1.6 \\0.2 x + 0.3 y = 2.3\end{array} \right.$

Evaluate the expression. $\left[ \begin{array} { l } - 2 \\- 9 \\- 4\end{array} \right] \left( \left[ \begin{array} { l l } - 3 & 5\end{array} \right] + \left[ \begin{array} { l l } 7 & - 5\end{array} \right] \right)$

Use the inverse formula $A ^ { - 1 } = \frac { 1 } { a d - b c } \left[ \begin{array} { c c } d & - b \\- c & a\end{array} \right]$ to find the inverse of the 2×2 matrix (if it exists). $\left[ \begin{array} { c c } 1 & - 2 \\- 3 & 2\end{array} \right]$

Solve the system by Gauss - Jordan elimination. $\left\{ \begin{aligned}\frac { 1 } { 3 } x + \frac { 3 } { 4 } y - \frac { 2 } { 3 } z & = - 4 \\x + \frac { 1 } { 2 } y + \frac { 1 } { 3 } z & = 30 \\\frac { 1 } { 6 } x - \frac { 1 } { 8 } y - z & = - 30\end{aligned} \right.$

Use the matrix capabilities of a graphing utility to find the determinant of the matrix $\left[ \begin{array} { c c c } 0.2 & - 0.6 & 0.6 \\0.8 & 0.4 & 0 \\- 0.4 & - 1.4 & 1.0\end{array} \right]$ .

Use Cramer's Rule to solve (if possible) the system of equations. $\left\{ \begin{array} { r } 4 x - y + z = - 20 \\2 x + 2 y + 3 z = 10 \\5 x - 2 y + 6 z = 2\end{array} \right.$

Find x and y. $\left[ \begin{array} { c c c } x - 7 & 3 & - 8 y \\4 & - 8 x & - 7 \\- 9 & y - 3 & - 3\end{array} \right] = \left[ \begin{array} { c c c } 7 x - 19 & 3 & 24 \\4 & - 16 & - 7 \\- 9 & - 6 & - 3\end{array} \right]$

Use the matrix capabilities of a graphing utility to solve the following system of linear equations: $\left\{ \begin{array} { l l } - 14 x + 14 y + 21 z & = 10 \\7 x - 7 y & = - 5 \\7 x + 28 z & = 20\end{array} \right.$

Functions and Their Graphs

Polynomial and Rational Functions

Exponential and Logarithmic Functions

Trigonometry

Analytic Trigonometry

Additional Topics In Trigonometery

Systems Of Equations and Inequalities

Sequences Series and Probability

Topics In Analytic Geometry

Analytic Geometry In Three Dimensions

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Filters

Exam 8: Matrices and Determinants

Find ABA BAB . A=[−2004],B=[300−2]A = \left[ \begin{array} { c c } - 2 & 0 \\0 & 4\end{array} \right] , B = \left[ \begin{array} { c c } 3 & 0 \\0 & - 2\end{array} \right]A=[−20​04​],B=[30​0−2​]

Solve the system using Gaussian elimination. {x+y=−2x+z=−4y+z=−4\left\{ \begin{array} { l } x + y = - 2 \\x + z = - 4 \\y + z = - 4\end{array} \right.⎩⎨⎧​x+y=−2x+z=−4y+z=−4​

Use matrices to solve the system of equations (if possible).Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. {2x+9y=−35−9x+2y=30\left\{ \begin{aligned}2 x + 9 y & = - 35 \\- 9 x + 2 y & = 30\end{aligned} \right.{2x+9y−9x+2y​=−35=30​

Find 8A. A=[375373]A = \left[ \begin{array} { l l l } 3 & 7 & 5 \\3 & 7 & 3\end{array} \right]A=[33​77​53​]

Find the product. [2100−120−400200402][53−3−5]\left[ \begin{array} { c c c c } 2 & 1 & 0 & 0 \\- 1 & 2 & 0 & - 4 \\0 & 0 & 2 & 0 \\0 & 4 & 0 & 2\end{array} \right] \left[ \begin{array} { c } 5 \\3 \\- 3 \\- 5\end{array} \right]​2−100​1204​0020​0−402​​​53−3−5​​

Use the matrix capabilities of a graphing utility to evaluate the determinant. [31−60−69111]\left[ \begin{array} { c c c } 3 & 1 & - 6 \\0 & - 6 & 9 \\1 & 1 & 1\end{array} \right]​301​1−61​−691​​

Find all minors of the matrix. [−7042433−35]\left[ \begin{array} { c c c } - 7 & 0 & 4 \\2 & 4 & 3 \\3 & - 3 & 5\end{array} \right]​−723​04−3​435​​

Use a determinant to find an equation of the line passing through the points. ​ (0, 0), (-8, 2) ​

Use Cramer's Rule to solve (if possible) the system of equations. {−0.4x+0.8y=1.60.2x+0.3y=2.3\left\{ \begin{array} { r } - 0.4 x + 0.8 y = 1.6 \\0.2 x + 0.3 y = 2.3\end{array} \right.{−0.4x+0.8y=1.60.2x+0.3y=2.3​

Evaluate the expression. [−2−9−4]([−35]+[7−5])\left[ \begin{array} { l } - 2 \\- 9 \\- 4\end{array} \right] \left( \left[ \begin{array} { l l } - 3 & 5\end{array} \right] + \left[ \begin{array} { l l } 7 & - 5\end{array} \right] \right)​−2−9−4​​([−3​5​]+[7​−5​])

Use the matrix capabilities of a graphing utility to find the determinant of the matrix [0.2−0.60.60.80.40−0.4−1.41.0]\left[ \begin{array} { c c c } 0.2 & - 0.6 & 0.6 \\0.8 & 0.4 & 0 \\- 0.4 & - 1.4 & 1.0\end{array} \right]​0.20.8−0.4​−0.60.4−1.4​0.601.0​​ .

Use Cramer's Rule to solve (if possible) the system of equations. {4x−y+z=−202x+2y+3z=105x−2y+6z=2\left\{ \begin{array} { r } 4 x - y + z = - 20 \\2 x + 2 y + 3 z = 10 \\5 x - 2 y + 6 z = 2\end{array} \right.⎩⎨⎧​4x−y+z=−202x+2y+3z=105x−2y+6z=2​