Question 1

Use Determinants to Identify Inconsistent Systems and Systems with Dependent Equations
-$$\begin{array} { l } 
9 x + y = 32 \
9 x + y = 41
\end{array}$$

Accepted Answer

A)  system is inconsistent 
B)  system contains dependent equations 
A)  system is inconsistent 
B)  system contains dependent equations

Question 2

Apply Gaussian Elimination to Systems with More Variables than Equations
-$$\begin{array} { l } 
x + y + z = 9 \
2 x - 3 y + 4 z = 7
\end{array}$$

Accepted Answer

A)  $\left\{ \left( - \frac { 7 } { 5 } \mathrm { z } + \frac { 34 } { 5 } , \frac { 2 } { 5 } \mathrm { z } + \frac { 11 } { 5 } , \mathrm { z } \right) \right\}$ 
B)  $\left\{ \left( \frac { 3 } { 5 } z + \frac { 16 } { 5 } , - \frac { 8 } { 5 } z + \frac { 29 } { 5 } , z \right) \right\}$ 
C)  $\left\{ \left( \frac { 27 } { 5 } , \frac { 13 } { 5 } , 1 \right) \right\}$ 
D)  $\varnothing$ 
A)  $\left\{ \left( - \frac { 7 } { 5 } \mathrm { z } + \frac { 34 } { 5 } , \frac { 2 } { 5 } \mathrm { z } + \frac { 11 } { 5 } , \mathrm { z } \right) \right\}$ 
B)  $\left\{ \left( \frac { 3 } { 5 } z + \frac { 16 } { 5 } , - \frac { 8 } { 5 } z + \frac { 29 } { 5 } , z \right) \right\}$ 
C)  $\left\{ \left( \frac { 27 } { 5 } , \frac { 13 } { 5 } , 1 \right) \right\}$ 
D)  $\varnothing$

Question 3

Multiply Matrices

-$$A = \left[ \begin{array} { r r } 
- 1 & 3 \
2 & 2
\end{array} \right] , B = \left[ \begin{array} { l l } 
- 2 & 0 \
- 1 & 5
\end{array} \right]$$

Accepted Answer

A) 
$\left[ \begin{array} { l l } - 1 & 15 \ - 6 & 10 \end{array} \right]$ 
B) 
$\left[ \begin{array} { r r } 2 & 0 \ - 2 & 10 \end{array} \right]$ 
C) 
$\left[ \begin{array} { r r } 2 & - 6 \ - 1 & 7 \end{array} \right]$ 
D) 
$\left[ \begin{array} { l l } 15 & - 1 \ 10 & - 6 \end{array} \right]$ 
A) 
$\left[ \begin{array} { l l } - 1 & 15 \ - 6 & 10 \end{array} \right]$ 
B) 
$\left[ \begin{array} { r r } 2 & 0 \ - 2 & 10 \end{array} \right]$ 
C) 
$\left[ \begin{array} { r r } 2 & - 6 \ - 1 & 7 \end{array} \right]$ 
D) 
$\left[ \begin{array} { l l } 15 & - 1 \ 10 & - 6 \end{array} \right]$

Question 4

Use Inverses to Solve Matrix Equations
-$$\begin{array} { l } 
5 x + 2 y + 2 z = 32 \
4 x + 2 y - 2 z = 32 \
8 x + 2 y + 8 z = 38
\end{array}$$

Accepted Answer

A) 
$\left[ \begin{array} { r r r } 5 & 2 & 2 \ 4 & 2 & - 2 \ 8 & 2 & 8 \end{array} \right] \left[ \begin{array} { l } x \ y \ z \end{array} \right] = \left[ \begin{array} { l } 32 \ 32 \ 38 \end{array} \right]$ 
B) 
$\left[ \begin{array} { r r r } 5 & 4 & 8 \ 2 & 2 & 2 \ 2 & - 2 & 8 \end{array} \right] \left[ \begin{array} { l } \mathrm { x } \ \mathrm { y } \ \mathrm { z } \end{array} \right] = \left[ \begin{array} { l } 32 \ 32 \ 38 \end{array} \right]$ 
C) 
$$\left[ \begin{array} { r r r } 
32 & 2 & 2 \
32 & - 2 & 2 \
38 & 8 & 2
\end{array} \right] \left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { l } 
5 \
4 \
8
\end{array} \right]$$ 
D) 
$$\left[ \begin{array} { l l } 
5 & 2 \
4 & 2 \
8 & 2
\end{array} \right] \left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { l } 
2 \
- 2 \
8
\end{array} \right]$$ 
A) 
$\left[ \begin{array} { r r r } 5 & 2 & 2 \ 4 & 2 & - 2 \ 8 & 2 & 8 \end{array} \right] \left[ \begin{array} { l } x \ y \ z \end{array} \right] = \left[ \begin{array} { l } 32 \ 32 \ 38 \end{array} \right]$ 
B) 
$\left[ \begin{array} { r r r } 5 & 4 & 8 \ 2 & 2 & 2 \ 2 & - 2 & 8 \end{array} \right] \left[ \begin{array} { l } \mathrm { x } \ \mathrm { y } \ \mathrm { z } \end{array} \right] = \left[ \begin{array} { l } 32 \ 32 \ 38 \end{array} \right]$ 
C) 
$$\left[ \begin{array} { r r r } 
32 & 2 & 2 \
32 & - 2 & 2 \
38 & 8 & 2
\end{array} \right] \left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { l } 
5 \
4 \
8
\end{array} \right]$$ 
D) 
$$\left[ \begin{array} { l l } 
5 & 2 \
4 & 2 \
8 & 2
\end{array} \right] \left[ \begin{array} { l } 
x \
y \
z
\end{array} \right] = \left[ \begin{array} { l } 
2 \
- 2 \
8
\end{array} \right]$$

Question 5

Solve Matrix Equations
-Let $\mathrm { A } = \left[ \begin{array} { r r } 9 & 4 \ 4 & 9 \ - 3 & 6 \end{array} \right]$ and $\mathrm { B } = \left[ \begin{array} { r r } - 2 & - 1 \ 9 & - 9 \ - 5 & 4 \end{array} \right] ; \quad \mathrm { X } - \mathrm { B } = \mathrm { A }$

Accepted Answer

A) 
$\left[ \begin{array} { c c } 7 & 3 \ 13 & 0 \ - 8 & 10 \end{array} \right]$ 
B) 
$\left[ \begin{array} { r r } 11 & 5 \ - 5 & 18 \ 2 & 14 \end{array} \right]$ 
C) 
$\left[ \begin{array} { r r } 7 & 3 \ - 13 & 9 \ - 8 & - 10 \end{array} \right]$ 
D) 
 $\left[ \begin{array} { r r } 7 & 9 \ 13 & 0 \ - 8 & 10 \end{array} \right]$ 
A) 
$\left[ \begin{array} { c c } 7 & 3 \ 13 & 0 \ - 8 & 10 \end{array} \right]$ 
B) 
$\left[ \begin{array} { r r } 11 & 5 \ - 5 & 18 \ 2 & 14 \end{array} \right]$ 
C) 
$\left[ \begin{array} { r r } 7 & 3 \ - 13 & 9 \ - 8 & - 10 \end{array} \right]$ 
D) 
 $\left[ \begin{array} { r r } 7 & 9 \ 13 & 0 \ - 8 & 10 \end{array} \right]$

Question 6

Multiplicative Inverses of Matrices and Matrix Equations
1 Find the Multiplicative Inverse of a Square Matrix
-$$\mathrm { A } = \left[ \begin{array} { r r r r } 
1 & 6 & 0 & 0 \
0 & 1 & - 1 & 0 \
0 & 0 & 1 & - 8 \
0 & 0 & 0 & 1
\end{array} \right]$$

Accepted Answer

A) 
$$\left[ \begin{array} { r r r c } 
1 & - 6 & - 6 & - 48 \
0 & 1 & 1 & 8 \
0 & 0 & 1 & 8 \
0 & 0 & 0 & 1
\end{array} \right]$$ 
B) $$\left[ \begin{array} { r r r r } 
1 & 8 & - 8 & - 48 \
0 & 1 & 1 & 6 \
0 & 0 & 1 & - 6 \
0 & 0 & 0 & 1
\end{array} \right]$$ 
C) 
$$\left[ \begin{array} { r r r r } 
1 & 0 & 0 & 0 \
8 & 1 & 0 & 0 \
- 8 & 1 & 1 & 0 \
- 48 & 6 & - 6 & 1
\end{array} \right]$$ 
D) 
$$\left[ \begin{array} { r r r r } 
1 & 0 & 0 & 0 \
- 6 & 1 & 0 & 0 \
- 6 & 1 & 1 & 0 \
- 48 & 8 & 8 & 1
\end{array} \right]$$ 
A) 
$$\left[ \begin{array} { r r r c } 
1 & - 6 & - 6 & - 48 \
0 & 1 & 1 & 8 \
0 & 0 & 1 & 8 \
0 & 0 & 0 & 1
\end{array} \right]$$ 
B) $$\left[ \begin{array} { r r r r } 
1 & 8 & - 8 & - 48 \
0 & 1 & 1 & 6 \
0 & 0 & 1 & - 6 \
0 & 0 & 0 & 1
\end{array} \right]$$ 
C) 
$$\left[ \begin{array} { r r r r } 
1 & 0 & 0 & 0 \
8 & 1 & 0 & 0 \
- 8 & 1 & 1 & 0 \
- 48 & 6 & - 6 & 1
\end{array} \right]$$ 
D) 
$$\left[ \begin{array} { r r r r } 
1 & 0 & 0 & 0 \
- 6 & 1 & 0 & 0 \
- 6 & 1 & 1 & 0 \
- 48 & 8 & 8 & 1
\end{array} \right]$$

Question 7

Model Applied Situations with Matrix Operations
-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.

Accepted Answer

A) 
$A = \left[ \begin{array} { l l l } 2 & 0 & 0 \ 2 & 0 & 0 \ 2 & 2 & 2 \end{array} \right] ; B = \left[ \begin{array} { r r r } - 1 & 0 & 0 \ - 1 & 0 & 0 \ - 1 & - 1 & - 1 \end{array} \right]$
$\mathrm { A } = \left[ \begin{array} { l l l } 2 & 1 & 1 \ 2 & 1 & 1 \ 2 & 2 & 2 \end{array} \right] ; \mathrm { B } = \left[ \begin{array} { l l l } 1 & 0 & 0 \ 1 & 0 & 0 \ 1 & 1 & 1 \end{array} \right]$ 
B) 
$\mathrm{A}=\left[\begin{array}{lll}
2 & 1 & 1 \
2 & 1 & 1 \
2 & 2 & 2
\end{array}\right] ; \mathrm{B}=\left[\begin{array}{lll}
1 & 0 & 0 \
1 & 0 & 0 \
1 & 1 & 1
\end{array}\right]$ 
C) 
$A = \left[ \begin{array} { l l l } 1 & 0 & 0 \ 1 & 0 & 0 \ 1 & 1 & 1 \end{array} \right] ; B = \left[ \begin{array} { r r r } - 1 & 0 & 0 \ - 1 & 0 & 0 \ - 1 & - 1 & - 1 \end{array} \right]$ 
D) $A = \left[ \begin{array} { l l l } 3 & 1 & 1 \ 3 & 1 & 1 \ 3 & 3 & 3 \end{array} \right] ; B = \left[ \begin{array} { l l l } - 2 & - 1 & - 1 \ - 2 & - 1 & - 1 \ - 2 & - 2 & - 2 \end{array} \right]$ 
A) 
$A = \left[ \begin{array} { l l l } 2 & 0 & 0 \ 2 & 0 & 0 \ 2 & 2 & 2 \end{array} \right] ; B = \left[ \begin{array} { r r r } - 1 & 0 & 0 \ - 1 & 0 & 0 \ - 1 & - 1 & - 1 \end{array} \right]$
$\mathrm { A } = \left[ \begin{array} { l l l } 2 & 1 & 1 \ 2 & 1 & 1 \ 2 & 2 & 2 \end{array} \right] ; \mathrm { B } = \left[ \begin{array} { l l l } 1 & 0 & 0 \ 1 & 0 & 0 \ 1 & 1 & 1 \end{array} \right]$ 
B) 
$\mathrm{A}=\left[\begin{array}{lll}
2 & 1 & 1 \
2 & 1 & 1 \
2 & 2 & 2
\end{array}\right] ; \mathrm{B}=\left[\begin{array}{lll}
1 & 0 & 0 \
1 & 0 & 0 \
1 & 1 & 1
\end{array}\right]$ 
C) 
$A = \left[ \begin{array} { l l l } 1 & 0 & 0 \ 1 & 0 & 0 \ 1 & 1 & 1 \end{array} \right] ; B = \left[ \begin{array} { r r r } - 1 & 0 & 0 \ - 1 & 0 & 0 \ - 1 & - 1 & - 1 \end{array} \right]$ 
D) $A = \left[ \begin{array} { l l l } 3 & 1 & 1 \ 3 & 1 & 1 \ 3 & 3 & 3 \end{array} \right] ; B = \left[ \begin{array} { l l l } - 2 & - 1 & - 1 \ - 2 & - 1 & - 1 \ - 2 & - 2 & - 2 \end{array} \right]$

Question 8

Model Applied Situations with Matrix Operations
-Adjust the contrast by changing the black to dark grey and the light grey to white. Use matrix addition to accomplish this.

Accepted Answer

A)  $\left[ \begin{array} { l l l } 1 & 3 & 1 \ 1 & 3 & 1 \ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { l l l } - 1 & - 1 & - 1 \ - 1 & - 1 & - 1 \ - 1 & - 1 & - 1 \end{array} \right] = \left[ \begin{array} { l l l } 0 & 2 & 0 \ 0 & 2 & 0 \ 2 & 2 & 2 \end{array} \right]$ 
B) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { r r r } 
0 & - 1 & 0 \
0 & - 1 & 0 \
- 1 & - 1 & - 1
\end{array} \right] = \left[ \begin{array} { l l l } 
1 & 2 & 1 \
1 & 2 & 1 \
2 & 2 & 2
\end{array} \right]$$ 
C) 
$\left[ \begin{array} { l l l } 1 & 3 & 1 \ 1 & 3 & 1 \ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { l l l } 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{array} \right] = \left[ \begin{array} { l l l } 2 & 4 & 2 \ 2 & 4 & 2 \ 4 & 4 & 4 \end{array} \right]$ 
D) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { r r r } 
0 & - 1 & 0 \
0 & - 1 & 0 \
- 1 & - 1 & - 1
\end{array} \right] = \left[ \begin{array} { l l l } 
0 & 2 & 0 \
0 & 2 & 0 \
2 & 2 & 2
\end{array} \right]$$ 
A)  $\left[ \begin{array} { l l l } 1 & 3 & 1 \ 1 & 3 & 1 \ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { l l l } - 1 & - 1 & - 1 \ - 1 & - 1 & - 1 \ - 1 & - 1 & - 1 \end{array} \right] = \left[ \begin{array} { l l l } 0 & 2 & 0 \ 0 & 2 & 0 \ 2 & 2 & 2 \end{array} \right]$ 
B) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { r r r } 
0 & - 1 & 0 \
0 & - 1 & 0 \
- 1 & - 1 & - 1
\end{array} \right] = \left[ \begin{array} { l l l } 
1 & 2 & 1 \
1 & 2 & 1 \
2 & 2 & 2
\end{array} \right]$$ 
C) 
$\left[ \begin{array} { l l l } 1 & 3 & 1 \ 1 & 3 & 1 \ 3 & 3 & 3 \end{array} \right] + \left[ \begin{array} { l l l } 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{array} \right] = \left[ \begin{array} { l l l } 2 & 4 & 2 \ 2 & 4 & 2 \ 4 & 4 & 4 \end{array} \right]$ 
D) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { r r r } 
0 & - 1 & 0 \
0 & - 1 & 0 \
- 1 & - 1 & - 1
\end{array} \right] = \left[ \begin{array} { l l l } 
0 & 2 & 0 \
0 & 2 & 0 \
2 & 2 & 2
\end{array} \right]$$

Question 9

Add and Subtract Matrices
-Let $A = \left[ \begin{array} { r } 3 \ - 5 \ - 4 \end{array} \right]$ and $B = \left[ \begin{array} { r } - 5 \ 3 \ 9 \end{array} \right]$. Find $A + B$.

Accepted Answer

A) 
$\left[ \begin{array} { r } - 2 \ - 2 \ 5 \end{array} \right]$ 
B) 
$\left[ \begin{array} { l l l } - 2 & - 2 & 5 \end{array} \right]$ 
C) 
$\left[ \begin{array} { r r } 3 & - 5 \ - 5 & 3 \ - 4 & 9 \end{array} \right]$ 
D) 
$\left[ \begin{array} { l } 2 \ 5 \ 6 \end{array} \right]$ 
A) 
$\left[ \begin{array} { r } - 2 \ - 2 \ 5 \end{array} \right]$ 
B) 
$\left[ \begin{array} { l l l } - 2 & - 2 & 5 \end{array} \right]$ 
C) 
$\left[ \begin{array} { r r } 3 & - 5 \ - 5 & 3 \ - 4 & 9 \end{array} \right]$ 
D) 
$\left[ \begin{array} { l } 2 \ 5 \ 6 \end{array} \right]$

Question 10

Use Matrices and Gaussian Elimination to Solve Systems
-$$\begin{array} { l r } 
x + y + z - w = & 6 \
2 x - y + 3 z + 4 w = & - 4 \
4 x + 2 y - z - w = & - 13 \
- x - 2 y + 4 z + 3 w = & 12
\end{array}$$

Accepted Answer

A)  $\{ ( - 4,3,5 , - 2 ) \}$ 
B)  $\{ ( 4 , - 3 , - 5,2 ) \}$ 
C)  $\left\{ \left( - \frac { 1 } { 4 } , \frac { 1 } { 3 } , \frac { 1 } { 5 } , - \frac { 1 } { 2 } \right) \right\}$ 
D)  $\left\{ \left( \frac { 1 } { 4 } , - \frac { 1 } { 3 } , - \frac { 1 } { 5 } , \frac { 1 } { 2 } \right) \right\}$ 
A)  $\{ ( - 4,3,5 , - 2 ) \}$ 
B)  $\{ ( 4 , - 3 , - 5,2 ) \}$ 
C)  $\left\{ \left( - \frac { 1 } { 4 } , \frac { 1 } { 3 } , \frac { 1 } { 5 } , - \frac { 1 } { 2 } \right) \right\}$ 
D)  $\left\{ \left( \frac { 1 } { 4 } , - \frac { 1 } { 3 } , - \frac { 1 } { 5 } , \frac { 1 } { 2 } \right) \right\}$

Question 11

Model Applied Situations with Matrix Operations
-Adjust the contrast by changing the black to light grey and the light grey to black. Use matrix addition to accomplish this.

Accepted Answer

A) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { r r r } 
2 & - 2 & 2 \
2 & - 2 & 2 \
- 2 & - 2 & - 2
\end{array} \right] = \left[ \begin{array} { l l l } 
3 & 1 & 3 \
3 & 1 & 3 \
1 & 1 & 1
\end{array} \right]$$ 
B) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { l l l } 
- 2 & - 2 & - 2 \
- 2 & - 2 & - 2 \
- 2 & - 2 & - 2
\end{array} \right] = \left[ \begin{array} { l l l } 
3 & 1 & 3 \
3 & 1 & 3 \
1 & 1 & 1
\end{array} \right]$$ 
C) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { r r r } 
1 & - 1 & 1 \
1 & - 1 & 1 \
- 1 & - 1 & - 1
\end{array} \right] = \left[ \begin{array} { l l l } 
3 & 1 & 3 \
3 & 1 & 3 \
1 & 1 & 1
\end{array} \right]$$ 
D)

$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { l l l } 
2 & 2 & 2 \
2 & 2 & 2 \
2 & 2 & 2
\end{array} \right] = \left[ \begin{array} { l l l } 
3 & 1 & 3 \
3 & 1 & 3 \
1 & 1 & 1
\end{array} \right]$$ 
A) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { r r r } 
2 & - 2 & 2 \
2 & - 2 & 2 \
- 2 & - 2 & - 2
\end{array} \right] = \left[ \begin{array} { l l l } 
3 & 1 & 3 \
3 & 1 & 3 \
1 & 1 & 1
\end{array} \right]$$ 
B) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { l l l } 
- 2 & - 2 & - 2 \
- 2 & - 2 & - 2 \
- 2 & - 2 & - 2
\end{array} \right] = \left[ \begin{array} { l l l } 
3 & 1 & 3 \
3 & 1 & 3 \
1 & 1 & 1
\end{array} \right]$$ 
C) 
$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { r r r } 
1 & - 1 & 1 \
1 & - 1 & 1 \
- 1 & - 1 & - 1
\end{array} \right] = \left[ \begin{array} { l l l } 
3 & 1 & 3 \
3 & 1 & 3 \
1 & 1 & 1
\end{array} \right]$$ 
D)

$$\left[ \begin{array} { l l l } 
1 & 3 & 1 \
1 & 3 & 1 \
3 & 3 & 3
\end{array} \right] + \left[ \begin{array} { l l l } 
2 & 2 & 2 \
2 & 2 & 2 \
2 & 2 & 2
\end{array} \right] = \left[ \begin{array} { l l l } 
3 & 1 & 3 \
3 & 1 & 3 \
1 & 1 & 1
\end{array} \right]$$

Question 12

Evaluate a Third-Order Determinant
-The area of a triangle with vertices $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$, and $\left( x _ { 3 } , y _ { 3 } \right)$ is
$$\text { Area } = \pm \frac { 1 } { 2 } \left| \begin{array} { l l l } 
x _ { 1 } & y _ { 1 } & 1 \
x _ { 2 } & y _ { 2 } & 1 \
x _ { 3 } & y _ { 3 } & 1
\end{array} \right| \text {, }$$
where the symbol $\pm$ 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 $( 6,10 ) , ( 8 , - 8 )$, and $( - 9 , - 4 )$.

Accepted Answer

A)  149 
B)  298 
C)  3 
D)  6 
A)  149 
B)  298 
C)  3 
D)  6

Question 13

Perform Scalar Multiplication
-Let $\mathrm { A } = \left[ \begin{array} { r } 1 \ - 3 \ 2 \end{array} \right]$ and $\mathrm { B } = \left[ \begin{array} { r } - 1 \ 3 \ - 2 \end{array} \right]$. Find $\mathrm { A } - 4 \mathrm {~B}$.

Accepted Answer

A) 
$\left[ \begin{array} { r } 5 \ - 15 \ 10 \end{array} \right]$ 
B) 
$\left[ \begin{array} { r } - 3 \ 9 \ - 6 \end{array} \right]$ 
C) 
$\left[ \begin{array} { r } - 5 \ 15 \ - 10 \end{array} \right]$ 
D) 
$\left[ \begin{array} { r } 5 \ - 6 \ 4 \end{array} \right]$ 
A) 
$\left[ \begin{array} { r } 5 \ - 15 \ 10 \end{array} \right]$ 
B) 
$\left[ \begin{array} { r } - 3 \ 9 \ - 6 \end{array} \right]$ 
C) 
$\left[ \begin{array} { r } - 5 \ 15 \ - 10 \end{array} \right]$ 
D) 
$\left[ \begin{array} { r } 5 \ - 6 \ 4 \end{array} \right]$

Question 14

Multiply Matrices

-$$A = \left[ \begin{array} { r r r } 
1 & - 6 & 1 \
- 3 & 4 & - 2
\end{array} \right] , B = \left[ \begin{array} { l } 
- 1 \
- 4 \
- 7
\end{array} \right]$$

Accepted Answer

A) $\left[ \begin{array} { r } 16 \ 1 \end{array} \right]$ 
B)  $\mathrm { AB }$ is not defined. 
C)  [ $\left. \begin{array} { l l } 16 & 1 \end{array} \right]$ 
D) 
$$\left[ \begin{array} { r c r } 
1 & - 6 & 1 \
- 3 & 4 & - 2 \
- 1 & - 4 & - 7
\end{array} \right]$$ 
A) $\left[ \begin{array} { r } 16 \ 1 \end{array} \right]$ 
B)  $\mathrm { AB }$ is not defined. 
C)  [ $\left. \begin{array} { l l } 16 & 1 \end{array} \right]$ 
D) 
$$\left[ \begin{array} { r c r } 
1 & - 6 & 1 \
- 3 & 4 & - 2 \
- 1 & - 4 & - 7
\end{array} \right]$$

Question 15

Apply Gaussian Elimination to Systems with More Variables than Equations
-$$\begin{array} { l } 
x + y + z = 7 \
x - y + 2 z = 7
\end{array}$$

Accepted Answer

A)  $\left\{ \left( - \frac { 3 } { 2 } z + 7 , \frac { 1 } { 2 } z , z \right) \right\}$ 
B)  $\{ ( - 3 z + 14,2 z - 7 , z ) \}$ 
C)  $\{ ( 4,1,2 ) \}$ 
D)  $\{ ( 8 , - 3,2 ) \}$ 
A)  $\left\{ \left( - \frac { 3 } { 2 } z + 7 , \frac { 1 } { 2 } z , z \right) \right\}$ 
B)  $\{ ( - 3 z + 14,2 z - 7 , z ) \}$ 
C)  $\{ ( 4,1,2 ) \}$ 
D)  $\{ ( 8 , - 3,2 ) \}$

Question 16

Solve a System of Linear Equations in Two Variables Using Cramer's Rule
-$$\begin{array} { l } 
4 x - 2 y = 26 \
4 x + 3 y = 51
\end{array}$$

Accepted Answer

A)  $\{ ( 9,5 ) \}$ 
B)  $\{ ( 5,9 ) \}$ 
C)  $\{ ( - 5,9 ) \}$ 
D)  $\{ ( 9 , - 5 ) \}$ 
A)  $\{ ( 9,5 ) \}$ 
B)  $\{ ( 5,9 ) \}$ 
C)  $\{ ( - 5,9 ) \}$ 
D)  $\{ ( 9 , - 5 ) \}$

Question 17

Inconsistent and Dependent Systems and Their Applications
1 Apply Gaussian Elimination to Systems Without Unique Solutions
-$$\begin{aligned}
x + y + z + w & = 7 \
3 x - 2 z + 5 w & = 11 \
- 4 x + 3 y + w & = 4 \
- x - y - z - w & = 6
\end{aligned}$$

Accepted Answer

A)  $\varnothing$ 
B)  $\left\{ \left( \frac { 3 } { 2 } , 1 , \frac { 1 } { 3 } , - 2 \right) \right\}$ 
C)  $\left\{ \left( \frac { 7 } { 4 } , - \frac { 1 } { 2 } , 5 , - \frac { 1 } { 6 } \right) \right\}$ 
D)  $\left\{ \left( - 11 , \frac { 7 } { 19 } , \frac { 6 } { 19 } , - 4 \right) \right\}$ 
A)  $\varnothing$ 
B)  $\left\{ \left( \frac { 3 } { 2 } , 1 , \frac { 1 } { 3 } , - 2 \right) \right\}$ 
C)  $\left\{ \left( \frac { 7 } { 4 } , - \frac { 1 } { 2 } , 5 , - \frac { 1 } { 6 } \right) \right\}$ 
D)  $\left\{ \left( - 11 , \frac { 7 } { 19 } , \frac { 6 } { 19 } , - 4 \right) \right\}$

Question 18

Write the system of linear equations represented by the augmented matrix. Use x, y, z, and, if necessary, w for the variables.
-$$\left[ \begin{array} { r r r r | r } 
5 & 1 & 0 & 2 & 12 \
- 1 & 8 & 1 & 0 & - 6 \
9 & 0 & 0 & 6 & - 11 \
0 & 3 & 0 & - 3 & - 4
\end{array} \right]$$

Accepted Answer

A) 
$5 x + y + 2 w = 12$
$- x + 8 y + z = - 6$
$9 x + 6 w = - 11$
$3 y - 3 w = - 4$ 
B) 
$5 x + y + z + 2 w = 12$
$- x + 8 y + z + w = - 6$
$9 x + y + z + 6 w = - 11$
$x + 3 y + z - 3 w = - 4$ 
C) 
$5 x + y + 2 w = 12$
$x + 8 y + z = - 6$
$9 x + 6 w = - 11$
$3 y + 3 w = - 4$ 
D) 
$5 x + y + 2 z = 12$
$- x + 8 y + z = - 6$
$9 x + 6 y = - 11$
$3 x - 3 y = - 4$ 
A) 
$5 x + y + 2 w = 12$
$- x + 8 y + z = - 6$
$9 x + 6 w = - 11$
$3 y - 3 w = - 4$ 
B) 
$5 x + y + z + 2 w = 12$
$- x + 8 y + z + w = - 6$
$9 x + y + z + 6 w = - 11$
$x + 3 y + z - 3 w = - 4$ 
C) 
$5 x + y + 2 w = 12$
$x + 8 y + z = - 6$
$9 x + 6 w = - 11$
$3 y + 3 w = - 4$ 
D) 
$5 x + y + 2 z = 12$
$- x + 8 y + z = - 6$
$9 x + 6 y = - 11$
$3 x - 3 y = - 4$

Question 19

Solve Matrix Equations
-Let $\mathrm { A } = \left[ \begin{array} { r r r } 2 & 1 & - 2 \ 3 & 0 & 0 \ 1 & - 2 & 2 \end{array} \right]$ and $\mathrm { B } = \left[ \begin{array} { r r r } - 1 & - 2 & - 2 \ 0 & 1 & 1 \ 3 & 0 & 2 \end{array} \right] ; \quad 2 \mathrm {~B} - 2 \mathrm {~A} = \mathrm { X }$

Accepted Answer

A) 
$X = \left[ \begin{array} { r r r } - 6 & - 6 & 0 \ - 6 & 2 & 2 \ 4 & 4 & 0 \end{array} \right]$ 
B) 
$X = \left[ \begin{array} { r r r } - 6 & 2 & 2 \ 4 & 4 & 0 \ - 6 & - 6 & 0 \end{array} \right]$ 
C) 
$X = \left[ \begin{array} { r r r } - 6 & - 6 & 0 \ - 6 & 1 & 1 \ 4 & 4 & 0 \end{array} \right]$ 
D) 
$X = \left[ \begin{array} { r r r } - 6 & 1 & 1 \ 4 & 4 & 0 \ - 6 & - 6 & 0 \end{array} \right]$ 
A) 
$X = \left[ \begin{array} { r r r } - 6 & - 6 & 0 \ - 6 & 2 & 2 \ 4 & 4 & 0 \end{array} \right]$ 
B) 
$X = \left[ \begin{array} { r r r } - 6 & 2 & 2 \ 4 & 4 & 0 \ - 6 & - 6 & 0 \end{array} \right]$ 
C) 
$X = \left[ \begin{array} { r r r } - 6 & - 6 & 0 \ - 6 & 1 & 1 \ 4 & 4 & 0 \end{array} \right]$ 
D) 
$X = \left[ \begin{array} { r r r } - 6 & 1 & 1 \ 4 & 4 & 0 \ - 6 & - 6 & 0 \end{array} \right]$

Question 20

Inconsistent and Dependent Systems and Their Applications
1 Apply Gaussian Elimination to Systems Without Unique Solutions
-$$\begin{array} { r } 
x + y + z = 7 \
x - y + 2 z = 7 \
2 x + 3 z = 14
\end{array}$$

Accepted Answer

A)  $\left\{ \left( - \frac { 3 z } { 2 } + 7 , \frac { z } { 2 } , z \right) \right\}$ 
B)  $\left\{ \left( - \frac { 3 z } { 2 } - 7 , \frac { z } { 2 } , z \right) \right\}$ 
C)  $\left\{ \left( - \frac { 3 z } { 2 } + 7,2 z , z \right) \right\}$ 
D)  $\left\{ \left( - \frac { 3 z } { 2 } - 7,2 z , z \right) \right\}$ 
A)  $\left\{ \left( - \frac { 3 z } { 2 } + 7 , \frac { z } { 2 } , z \right) \right\}$ 
B)  $\left\{ \left( - \frac { 3 z } { 2 } - 7 , \frac { z } { 2 } , z \right) \right\}$ 
C)  $\left\{ \left( - \frac { 3 z } { 2 } + 7,2 z , z \right) \right\}$ 
D)  $\left\{ \left( - \frac { 3 z } { 2 } - 7,2 z , z \right) \right\}$

Use Determinants to Identify Inconsistent Systems and Systems with Dependent Equations - 9x+y=32 9x+y=41

Apply Gaussian Elimination to Systems with More Variables than Equations - x+y+z=9 2x-3y+4z=7

Multiply Matrices - $A = \left[ \begin{array} { r r } - 1 & 3 \\2 & 2\end{array} \right] , B = \left[ \begin{array} { l l } - 2 & 0 \\- 1 & 5\end{array} \right]$

Use Inverses to Solve Matrix Equations - 5x+2y+2z=32 4x+2y-2z=32 8x+2y+8z=38

Solve Matrix Equations -Let $\mathrm { A } = \left[ \begin{array} { r r } 9 & 4 \\ 4 & 9 \\ - 3 & 6 \end{array} \right]$ and $\mathrm { B } = \left[ \begin{array} { r r } - 2 & - 1 \\ 9 & - 9 \\ - 5 & 4 \end{array} \right] ; \quad \mathrm { X } - \mathrm { B } = \mathrm { A }$

Multiplicative Inverses of Matrices and Matrix Equations 1 Find the Multiplicative Inverse of a Square Matrix - $\mathrm { A } = \left[ \begin{array} { r r r r } 1 & 6 & 0 & 0 \\0 & 1 & - 1 & 0 \\0 & 0 & 1 & - 8 \\0 & 0 & 0 & 1\end{array} \right]$

Model Applied Situations with Matrix Operations -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.

Model Applied Situations with Matrix Operations -Adjust the contrast by changing the black to dark grey and the light grey to white. Use matrix addition to accomplish this.

Add and Subtract Matrices -Let $A = \left[ \begin{array} { r } 3 \\ - 5 \\ - 4 \end{array} \right]$ and $B = \left[ \begin{array} { r } - 5 \\ 3 \\ 9 \end{array} \right]$ . Find $A + B$ .

Use Matrices and Gaussian Elimination to Solve Systems - x+y+z-w= 6 2x-y+3z+4w= -4 4x+2y-z-w= -13 -x-2y+4z+3w= 12

Model Applied Situations with Matrix Operations -Adjust the contrast by changing the black to light grey and the light grey to black. Use matrix addition to accomplish this.

Perform Scalar Multiplication -Let $\mathrm { A } = \left[ \begin{array} { r } 1 \\ - 3 \\ 2 \end{array} \right]$ and $\mathrm { B } = \left[ \begin{array} { r } - 1 \\ 3 \\ - 2 \end{array} \right]$ . Find $\mathrm { A } - 4 \mathrm {~B}$ .

Multiply Matrices - $A = \left[ \begin{array} { r r r } 1 & - 6 & 1 \\- 3 & 4 & - 2\end{array} \right] , B = \left[ \begin{array} { l } - 1 \\- 4 \\- 7\end{array} \right]$

Apply Gaussian Elimination to Systems with More Variables than Equations - x+y+z=7 x-y+2z=7

Solve a System of Linear Equations in Two Variables Using Cramer's Rule - 4x-2y=26 4x+3y=51

Inconsistent and Dependent Systems and Their Applications 1 Apply Gaussian Elimination to Systems Without Unique Solutions - x+y+z+w =7 3x-2z+5w =11 -4x+3y+w =4 -x-y-z-w =6

Write the system of linear equations represented by the augmented matrix. Use x, y, z, and, if necessary, w for the variables. - $\left[ \begin{array} { r r r r | r } 5 & 1 & 0 & 2 & 12 \\- 1 & 8 & 1 & 0 & - 6 \\9 & 0 & 0 & 6 & - 11 \\0 & 3 & 0 & - 3 & - 4\end{array} \right]$

Inconsistent and Dependent Systems and Their Applications 1 Apply Gaussian Elimination to Systems Without Unique Solutions - x+y+z=7 x-y+2z=7 2x+3z=14

Equations and Inequalities

Functions and Graphs

Polynomial and Rational Functions

Exponential and Logarithmic Functions

Trigonometric Functions

Analytic Trigonometry

Additional Topics in Trigonometry

Systems of Equations and Inequalities

Conic Sections and Analytic Geometry

Sequences, Induction, and Probability

Prerequisites: Fundamental Concepts of Algebra

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Exam 9: Matrices and Determinants