Question 1

Accepted Answer

Each element in row 1 is multiplied by -1 and added to the corresponding element in row 3. The result is the new row 3. All other rows (including row 1)are left unchanged. (Explanations may vary.)

Question 2

Solve the problem.
-An investor has $400,000 to invest in stocks, bonds, and commodities. If he plans to put three times as much into stocks as in bonds, how can he distribute his money among the three types of Investments? (Let x denote the amount put into stocks, y the amount put into bonds, and z the Amount put into commodities. Let all amounts be in dollars, and let z be the parameter.)

Accepted Answer

A)    x=300,000−z/2,y=100,000−z/2,0≤z≤400,000 x=300,000-z / 2, y=100,000-z / 2,0 \leq z \leq 400,000 x=300,000−z/2,y=100,000−z/2,0≤z≤400,000 
B)    x=100,000−z/2,y=300,000−z/2,0≤z≤400,000 x=100,000-z / 2, y=300,000-z / 2,0 \leq z \leq 400,000 x=100,000−z/2,y=300,000−z/2,0≤z≤400,000 
C)    x=300,000−3z/4,y=100,000−z/4,0≤z≤400,000 x=300,000-3 z / 4, y=100,000-z / 4,0 \leq z \leq 400,000 x=300,000−3z/4,y=100,000−z/4,0≤z≤400,000 
D)    x=100,000−3z/4,y=300,000−z/4,0≤z≤400,000 x=100,000-3 z / 4, y=300,000-z / 4,0 \leq z \leq 400,000 x=100,000−3z/4,y=300,000−z/4,0≤z≤400,000 C

Question 3

If a variable is expressed in terms of another variable, what is the other variable called?

Accepted Answer

A parameter

Question 4

Solve the system of equations by using the inverse of the coefficient matrix.
-$$
\begin{array}{l}
x-y+3 z=5 \
2 x+z=1 \
x+4 y+z=-7
\end{array}
$$

Accepted Answer

A)    (1,−2,0)  (1,-2,0)  (1,−2,0)  
B)    (1,0,−2)  (1,0,-2)  (1,0,−2)  
C)    (0,−2,1)  (0,-2,1)  (0,−2,1)  
D)   No inverse, no solution for system

Question 5

Find the inverse, if it exists, for the matrix.
-$$
\left[\begin{array}{rr}
8 & 2 \
-7 & 1
\end{array}\right]
$$

Accepted Answer

A)    [1−278] \left[\begin{array}{rr}1 & -2 \ 7 & 8\end{array}ight] [17​−28​] 
B)   
C)   
D)   No inverse

Question 6

Use a graphing calculator to find the matrix product and/or sum.
-Find $ \mathrm{A}(\mathrm{B}+\mathrm{C}) $.
$$
A=\left[\begin{array}{rrrr}
19 & -14 & 18 & 18 \
30 & 5 & -7 & 2 \
11 & 3 & -2 & 1 \
-5 & 32 & 5 & 10
\end{array}\right] B=\left[\begin{array}{rrr}
5 & 17 & 12 \
7 & 6 & -4 \
-12 & -9 & 15 \
3 & -2 & -14
\end{array}\right] C=\left[\begin{array}{rrr}
7 & -4 & 3 \
-16 & -19 & 20 \
-6 & -15 & -8 \
5 & 16 & 17
\end{array}\right]
$$

Accepted Answer

A)    [174249241457521487149166202−358−461502] \left[\begin{array}{rrr}174 & 249 & 241 \ 457 & 521 & 487 \ 149 & 166 & 202 \ -358 & -461 & 502\end{array}ight] \quad H403z M403 1759 V0 H319 V1759 v1200 v1759 h84z"/>​174457149−358​249521166−461​241487202502​M347 1759 V0 H263 V1759 v1200 v1759 h84z"/>​  
B)    [−15837305259580297972068817458−28] \left[\begin{array}{rrr}-158 & 37 & 305 \ 259 & 580 & 297 \ 97 & 206 & 88 \ 174 & 58 & -28\end{array}ight] H403z M403 1759 V0 H319 V1759 v1200 v1759 h84z"/>​−15825997174​3758020658​30529788−28​M347 1759 V0 H263 V1759 v1200 v1759 h84z"/>​ 
C)   [174373054575214879720688174−461502] \left[\begin{array}{rrr}174 & 37 & 305 \ 457 & 521 & 487 \ 97 & 206 & 88 \ 174 & -461 & 502\end{array}ight] H403z M403 1759 V0 H319 V1759 v1200 v1759 h84z"/>​17445797174​37521206−461​30548788502​M347 1759 V0 H263 V1759 v1200 v1759 h84z"/>​ 
D)   Does not exist

Question 7

Solve the problem.
-An appliance dealer has two stores in the town of Washingwell. During a given week, they have a beginning inventory of 
Washing Dish
Machines Washers
$ \mathrm{B}=\left[\begin{array}{ll}44 & 25 \ 51 & 38\end{array}\right] \begin{array}{l}\text { Store 1 } \ \text { Store 2 }\end{array} $
a purchase matrix of
Washing Dish
Machines Washers
$ P=\left[\begin{array}{ll}5 & 3 \ 7 & 0\end{array}\right] \begin{array}{l}\text { Store 1 } \ \text { Store 2 }\end{array} $
and an ending inventory of
Washing Dish
Machines Washers
$ E=\left[\begin{array}{ll}8 & 1 \ 3 & 2\end{array}\right] \begin{array}{l}\text { Store 1 } \ \text { Store 2 }\end{array} $
Find the sales matrix.

Accepted Answer

A)  
B)   
C)   
D)

Question 8

Find the matrix product, if possible.
-

Accepted Answer

A)    [−22−7−13] \left[\begin{array}{rr}-2 & 2 \ -7 & -13\end{array}ight] [−2−7​2−13​] 
B)    [04−53] \left[\begin{array}{rr}0 & 4 \ -5 & 3\end{array}ight] [0−5​43​] 
C)    [2−2−133] \left[\begin{array}{rr}2 & -2 \ -13 & 3\end{array}ight] [2−13​−23​] 
D)    [−10−655] \left[\begin{array}{rr}-10 & -6 \ 5 & 5\end{array}ight] [−105​−65​]

Question 9

Solve the system of equations by using the inverse of the coefficient matrix if it exists and by the echelon method if the
inverse doesn't exist.
-$ -2 x-6 y=-2 $
 $ 2 x-y=-5 $

Accepted Answer

A)    (−1,2)  (-1,2)  (−1,2)  
B)    (1,−2)  (1,-2)  (1,−2)  
C)    (−2,1)  (-2,1)  (−2,1)  
D)    (2,−1)  (2,-1)  (2,−1)

Question 10

Use a graphing calculator to find the matrix product and/or sum.
-Find $ A B+A C $.
$$
A=\left[\begin{array}{rrrr}
17 & -14 & 18 & 18 \
30 & 5 & -7 & 2 \
11 & 3 & -2 & 1 \
-5 & 32 & 5 & 10
\end{array}\right] B=\left[\begin{array}{rrr}
5 & 17 & 12 \
7 & 6 & -4 \
-12 & -9 & 15 \
3 & -2 & -14
\end{array}\right] C=\left[\begin{array}{rrr}
7 & -4 & 3 \
-16 & -19 & 20 \
-6 & -15 & -8 \
5 & 16 & 17
\end{array}\right]
$$

Accepted Answer

A)    [15032814575214879720688174−461502] \left[\begin{array}{rrr}150 & 3 & 281 \ 457 & 521 & 487 \ 97 & 206 & 88 \ 174 & -461 & 502\end{array}ight] H403z M403 1759 V0 H319 V1759 v1200 v1759 h84z"/>​15045797174​3521206−461​28148788502​M347 1759 V0 H263 V1759 v1200 v1759 h84z"/>​ 
B)    [−1683281259580297972068817458−28] \left[\begin{array}{rrr}-168 & 3 & 281 \ 259 & 580 & 297 \ 97 & 206 & 88 \ 174 & 58 & -28\end{array}ight] H403z M403 1759 V0 H319 V1759 v1200 v1759 h84z"/>​−16825997174​358020658​28129788−28​M347 1759 V0 H263 V1759 v1200 v1759 h84z"/>​ 
C)   [150223211457521487149166202−358−461502] \left[\begin{array}{rrr}150 & 223 & 211 \ 457 & 521 & 487 \ 149 & 166 & 202 \ -358 & -461 & 502\end{array}ight] H403z M403 1759 V0 H319 V1759 v1200 v1759 h84z"/>​150457149−358​223521166−461​211487202502​M347 1759 V0 H263 V1759 v1200 v1759 h84z"/>​ 
D)   Does not exist

Question 11

Use graphing calculator to find the inverse of the matrix. Give 5 decimal places.
-$$
A=\left[\begin{array}{ccc}
1 / 3 & 2 / 7 & 4 / 5 \
17 / 11 & 5 & 0 \
18 / 13 & -5 / 6 & 8 / 7
\end{array}\right]
$$

Accepted Answer

A)    [001010100] \left[\begin{array}{lll}0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0\end{array}ight] H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​001​010​100​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
B)    [2.659520.15830−1.86167−9.04238−0.338246.329663.371290.05484−1.48490] \left[\begin{array}{rrr}2.65952 & 0.15830 & -1.86167 \ -9.04238 & -0.33824 & 6.32966 \ 3.37129 & 0.05484 & -1.48490\end{array}ight] H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​2.65952−9.042383.37129​0.15830−0.338240.05484​−1.861676.32966−1.48490​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
C)    [−1.105570.192160.773900.341720.14061−0.239201.58861−0.13028−0.23703] \left[\begin{array}{rrr}-1.10557 & 0.19216 & 0.77390 \ 0.34172 & 0.14061 & -0.23920 \ 1.58861 & -0.13028 & -0.23703\end{array}ight] H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​−1.105570.341721.58861​0.192160.14061−0.13028​0.77390−0.23920−0.23703​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
D)   [33.51.250.64706500.72222−1.21.14286] \left[\begin{array}{lll}3 & 3.5 & 1.25 \ 0.64706 & 5 & 0 \ 0.72222 & -1.2 & 1.14286\end{array}ight] H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​30.647060.72222​3.55−1.2​1.2501.14286​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​

Question 12

The sizes of two matrices A and B are given. Find the sizes of the product AB and the product BA, whenever these products exist.
-$ A $ is $ 3 \times 2 $, and B is $ 2 \times 3 $.

Accepted Answer

A)    2×2;3×3 2 	imes 2 ; 3 	imes 3 2×2;3×3 
B)    3×3 3 	imes 3 3×3 ; BA does not exist.
C)    AB \mathrm{AB} AB  does not exist:  2×2 2 	imes 2 2×2 
D)    3×3;2×2 3 	imes 3 ; 2 	imes 2 3×3;2×2

Question 13

Find the value.
-Let $ \mathrm{A}=\left[\begin{array}{ll}-5 & 2\end{array}\right] $ and $ \mathrm{B}=\left[\begin{array}{ll}1 & 0\end{array}\right] ; 2 \mathrm{~A}+3 \mathrm{~B} $

Accepted Answer

A)    [−22] \left[\begin{array}{ll}-2 & 2\end{array}\right] [−2​2​] 
B)    [−94] \left[\begin{array}{ll}-9 & 4\end{array}\right] [−9​4​] 
C)    [−74] [-74] [−74] 
D)    [−104] [-104] [−104]

Question 14

An m × n zero matrix serves as an m × n .

Accepted Answer

The answer of An m × n zero matrix serves...

Question 15

Solve the problem.
-A company makes three chocolate candies: cherry, almond, and raisin. Matrix A gives the amount of ingredients in one batch. Matrix B gives the costs of ingredients from suppliers J and K. What is
The cost of 100 batches of each candy using ingredients from supplier J? 
$$
A=\left[\begin{array}{ccc}
\text { sugar choc milk } \
6 & 8 & 1 \
6 & 4 & 1 \
5 & 7 & 1
\end{array}\right] \text { cherry }
$$
$$
B=\left[\begin{array}{lll}
\mathrm{J} & \mathrm{K} \
4 & 3 \
4 & 5 \
2 & 2
\end{array}\right] \text { sugar }
$$

Accepted Answer

A)    $12,000 \$ 12,000 $12,000 
B)    $6000 \$ 6000 $6000 
C)    $15,000 \$ 15,000 $15,000 
D)    $15,200 \$ 15,200 $15,200

Question 16

Use the Gauss-Jordan method to solve the system of equations.
-$$
\begin{array}{l}
5 x-y+z=8 \
7 x+y+z=6 \
12 x+2 z=14
\end{array}
$$

Accepted Answer

A)    (−z+76,z−136,z)  \left(\frac{-z+7}{6}, \frac{z-13}{6}, z\right)  (6−z+7​,6z−13​,z)  
B)    (z+76,−z−136,z)  \left(\frac{z+7}{6}, \frac{-z-13}{6}, z\right)  (6z+7​,6−z−13​,z)  
C)    (−z+76,z+136,z)  \left(\frac{-z+7}{6}, \frac{z+13}{6}, z\right)  (6−z+7​,6z+13​,z)  
D)    (z−76,−z−136,z)  \left(\frac{z-7}{6}, \frac{-z-13}{6}, z\right)  (6z−7​,6−z−13​,z)

Question 17

How many solutions are there to a dependent system?

Accepted Answer

The answer of How many solutions are there to a...

Question 18

Solve the system of equations by using the inverse of the coefficient matrix.
-$
\begin{array}{l}
2 x+4 y+z=-10 \
4 x-4 y-z=-8 \
5 x+y+4 z=-1
\end{array}
$

Accepted Answer

A)    (−3,4,−2)  (-3,4,-2)  (−3,4,−2)  
B)    (−3,−2,4)  (-3,-2,4)  (−3,−2,4)  
C)    (4,−2,−3)  (4,-2,-3)  (4,−2,−3)  
D)   No inverse, no solution for system

Question 19

Solve the problem.
-A simplified economy has only two industries, the electric company and the gas company. Each dollar's worth of the electric company's output requires 0.20 of its own output and 0.4 of the gas Company's output. Each dollar's worth of the gas company's output requires 0.50 of its own output
And 0.7 of the electric company's output. Construct the input-output matrix.

Accepted Answer

A)   
B)   
C)   
D)

Question 20

Solve the system of equations. Let z be the parameter.
-$$
\begin{array}{l}
-3 x+y+6 z=-7 \
7 x+3 y+4 z=-14
\end{array}
$$

Accepted Answer

A)    (7+14z,−91+7z,z)  (7+14 z,-91+7 z, z)  (7+14z,−91+7z,z)  
B)    (716+78z,−9116−278z,z)  \left(\frac{7}{16}+\frac{7}{8} z,-\frac{91}{16}-\frac{27}{8} z, z\right)  (167​+87​z,−1691​−827​z,z)  
C)    (716+78z,−7+6z,z)  \left(\frac{7}{16}+\frac{7}{8} z,-7+6 z, z\right)  (167​+87​z,−7+6z,z)  
D)    (716+78z,−7−6z,z} \left(\frac{7}{16}+\frac{7}{8} z,-7-6 z, z\right\} (167​+87​z,−7−6z,z}

Exam 2: Systems of Linear Equations and Matrices

If a variable is expressed in terms of another variable, what is the other variable called?

Solve the system of equations by using the inverse of the coefficient matrix. - $\begin{array}{l}x-y+3 z=5 \\2 x+z=1 \\x+4 y+z=-7\end{array}$

Find the inverse, if it exists, for the matrix. - $\left[\begin{array}{rr}8 & 2 \\-7 & 1\end{array}\right]$

Solve the system of equations by using the inverse of the coefficient matrix if it exists and by the echelon method if the inverse doesn't exist. - $-2 x-6 y=-2$ $2 x-y=-5$

Use graphing calculator to find the inverse of the matrix. Give 5 decimal places. - $A=\left[\begin{array}{ccc}1 / 3 & 2 / 7 & 4 / 5 \\17 / 11 & 5 & 0 \\18 / 13 & -5 / 6 & 8 / 7\end{array}\right]$

The sizes of two matrices A and B are given. Find the sizes of the product AB and the product BA, whenever these products exist. - $A$ is $3 \times 2$ , and B is $2 \times 3$ .

Find the value. -Let $\mathrm{A}=\left[\begin{array}{ll}-5 & 2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 0\end{array}\right] ; 2 \mathrm{~A}+3 \mathrm{~B}$

An m × n zero matrix serves as an m × n .

Use the Gauss-Jordan method to solve the system of equations. - $\begin{array}{l}5 x-y+z=8 \\7 x+y+z=6 \\12 x+2 z=14\end{array}$

How many solutions are there to a dependent system?

Solve the system of equations by using the inverse of the coefficient matrix. - $\begin{array}{l}2 x+4 y+z=-10 \\4 x-4 y-z=-8 \\5 x+y+4 z=-1\end{array}$

Solve the system of equations. Let z be the parameter. - $\begin{array}{l}-3 x+y+6 z=-7 \\7 x+3 y+4 z=-14\end{array}$

Exam 2: Systems of Linear Equations and Matrices

If a variable is expressed in terms of another variable, what is the other variable called?

Solve the system of equations by using the inverse of the coefficient matrix. - x−y+3z=52x+z=1x+4y+z=−7\begin{array}{l}x-y+3 z=5 \\2 x+z=1 \\x+4 y+z=-7\end{array}x−y+3z=52x+z=1x+4y+z=−7​

Find the inverse, if it exists, for the matrix. - [82−71]\left[\begin{array}{rr}8 & 2 \\-7 & 1\end{array}\right][8−7​21​]

Find the matrix product, if possible. -

Solve the system of equations by using the inverse of the coefficient matrix if it exists and by the echelon method if the inverse doesn't exist. - −2x−6y=−2 -2 x-6 y=-2 −2x−6y=−2 2x−y=−5 2 x-y=-5 2x−y=−5

Use graphing calculator to find the inverse of the matrix. Give 5 decimal places. - A=[1/32/74/517/115018/13−5/68/7]A=\left[\begin{array}{ccc}1 / 3 & 2 / 7 & 4 / 5 \\17 / 11 & 5 & 0 \\18 / 13 & -5 / 6 & 8 / 7\end{array}\right]A=​1/317/1118/13​2/75−5/6​4/508/7​​

The sizes of two matrices A and B are given. Find the sizes of the product AB and the product BA, whenever these products exist. - A A A is 3×2 3 \times 2 3×2 , and B is 2×3 2 \times 3 2×3 .

Find the value. -Let A=[−52] \mathrm{A}=\left[\begin{array}{ll}-5 & 2\end{array}\right] A=[−5​2​] and B=[10];2 A+3 B \mathrm{B}=\left[\begin{array}{ll}1 & 0\end{array}\right] ; 2 \mathrm{~A}+3 \mathrm{~B} B=[1​0​];2 A+3 B

An m × n zero matrix serves as an m × n .

Use the Gauss-Jordan method to solve the system of equations. - 5x−y+z=87x+y+z=612x+2z=14\begin{array}{l}5 x-y+z=8 \\7 x+y+z=6 \\12 x+2 z=14\end{array}5x−y+z=87x+y+z=612x+2z=14​

How many solutions are there to a dependent system?

Solve the system of equations by using the inverse of the coefficient matrix. - 2x+4y+z=−104x−4y−z=−85x+y+4z=−1\begin{array}{l}2 x+4 y+z=-10 \\4 x-4 y-z=-8 \\5 x+y+4 z=-1\end{array}2x+4y+z=−104x−4y−z=−85x+y+4z=−1​

Solve the system of equations. Let z be the parameter. - −3x+y+6z=−77x+3y+4z=−14\begin{array}{l}-3 x+y+6 z=-7 \\7 x+3 y+4 z=-14\end{array}−3x+y+6z=−77x+3y+4z=−14​

Solve the system of equations by using the inverse of the coefficient matrix. - $\begin{array}{l}x-y+3 z=5 \\2 x+z=1 \\x+4 y+z=-7\end{array}$

Find the inverse, if it exists, for the matrix. - $\left[\begin{array}{rr}8 & 2 \\-7 & 1\end{array}\right]$

Solve the system of equations by using the inverse of the coefficient matrix if it exists and by the echelon method if the inverse doesn't exist. - $-2 x-6 y=-2$ $2 x-y=-5$

Use graphing calculator to find the inverse of the matrix. Give 5 decimal places. - $A=\left[\begin{array}{ccc}1 / 3 & 2 / 7 & 4 / 5 \\17 / 11 & 5 & 0 \\18 / 13 & -5 / 6 & 8 / 7\end{array}\right]$

The sizes of two matrices A and B are given. Find the sizes of the product AB and the product BA, whenever these products exist. - $A$ is $3 \times 2$ , and B is $2 \times 3$ .

Find the value. -Let $\mathrm{A}=\left[\begin{array}{ll}-5 & 2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 0\end{array}\right] ; 2 \mathrm{~A}+3 \mathrm{~B}$

Use the Gauss-Jordan method to solve the system of equations. - $\begin{array}{l}5 x-y+z=8 \\7 x+y+z=6 \\12 x+2 z=14\end{array}$

Solve the system of equations by using the inverse of the coefficient matrix. - $\begin{array}{l}2 x+4 y+z=-10 \\4 x-4 y-z=-8 \\5 x+y+4 z=-1\end{array}$

Solve the system of equations. Let z be the parameter. - $\begin{array}{l}-3 x+y+6 z=-7 \\7 x+3 y+4 z=-14\end{array}$