Question 1

Solve the problem.
-The columns of $\mathrm { I } _ { 3 } = \left[ \begin{array} { l l l } 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array} \right]$ are $\mathbf { e } _ { 1 } = \left[ \begin{array} { l } 1 \ 0 \ 0 \end{array} \right] , \mathbf { e } _ { 2 } = \left[ \begin{array} { l } 0 \ 1 \ 0 \end{array} \right] , \mathbf { e } _ { 3 } = \left[ \begin{array} { l } 0 \ 0 \ 1 \end{array} \right]$.
Suppose that $\mathrm { T }$ is a linear transformation from $R ^ { 3 }$ into $\mathcal { R } ^ { 2 }$ such that
$\mathrm { T } \left( \mathbf { e } _ { 1 } \right) = \left[ \begin{array} { r } 3 \ - 2 \end{array} \right] , \mathrm { T } \left( \mathbf { e } _ { 2 } \right) = \left[ \begin{array} { l } 5 \ 0 \end{array} \right]$, and $\mathrm { T } \left( \mathbf { e } _ { 3 } \right) = \left[ \begin{array} { r } - 5 \ 1 \end{array} \right]$
Find a formula for the image of an arbitrary $x = \left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} \right]$ in $R ^ { 3 }$.

Accepted Answer

A)   T[x1x2x3]=[3x1−2x25x1]T \left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] = \left[ \begin{array} { l } 3 x _ { 1 } - 2 x _ { 2 } \ 5 x _ { 1 } \end{array} ight]TH403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=[3x1​−2x2​5x1​​] 
B)   T[x1x2x3]=[3x1−2x25x15x2+x3]T \left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] = \left[ \begin{array} { l } 3 x _ { 1 } - 2 x _ { 2 } \ 5 x _ { 1 } \ 5 x _ { 2 } + x _ { 3 } \end{array} ight]TH403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​3x1​−2x2​5x1​5x2​+x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
C)   T[x1x2x3]=[3x1+5x2−5x35x1−2x1+x3]T \left[ \begin{array} { l } x _ { 1 } \x _ { 2 } \x _ { 3 }\end{array} ight] = \left[ \begin{array} { l } 3 x _ { 1 } + 5 x _ { 2 } - 5 x _ { 3 } \5 x _ { 1 } \- 2 x _ { 1 } + x _ { 3 }\end{array} ight]TH403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​3x1​+5x2​−5x3​5x1​−2x1​+x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
D)   T[x1x2x3]=[3x1+5x2−5x3−2x1+x3]T \left[ \begin{array} { l } x _ { 1 } \x _ { 2 } \x _ { 3 }\end{array} ight] = \left[ \begin{array} { l l } 3 x _ { 1 } + 5 x _ { 2 } - 5 x _ { 3 } \- 2 x _ { 1 } + x _ { 3 }\end{array} ight]TH403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=[3x1​+5x2​−5x3​−2x1​+x3​​] D

Question 2

Solve the problem.
-Let $\mathbf { a } _ { 1 } = \left[ \begin{array} { r } 3 \ 4 \ - 4 \end{array} \right] , \mathbf { a } _ { 2 } = \left[ \begin{array} { r } - 4 \ 1 \ 1 \end{array} \right]$, and $\mathbf { b } = \left[ \begin{array} { r } 2 \ - 10 \ 6 \end{array} \right]$
Determine whether $\mathbf { b }$ can be written as a linear combination of $\mathbf { a } _ { \mathbf { 1 } }$ and $\mathbf { a } _ { \mathbf { 2 } }$. In other words, determine whether weights $x _ { 1 }$ and $x _ { 2 }$ exist, such that $x _ { 1 } a _ { 1 } + x _ { 2 } a _ { 2 } = b$. Determine the weights $x _ { 1 }$ and $x _ { 2 }$ if possible.

Accepted Answer

A)   No solution
B)    x1=−1,x2=−3x _ { 1 } = - 1 , x _ { 2 } = - 3x1​=−1,x2​=−3 
C)    x1=−2,x2=−2x _ { 1 } = - 2 , x _ { 2 } = - 2x1​=−2,x2​=−2 
D)    x1=−2,x2=−1x _ { 1 } = - 2 , x _ { 2 } = - 1x1​=−2,x2​=−1 C

Question 3

Determine whether the matrix is in echelon form, reduced echelon form, or neither.
-$$\left[ \begin{array} { r r r r } 
1 & - 5 & 3 & 4 \
0 & 0 & - 5 & - 4 \
0 & 0 & 0 & - 2 \
0 & 0 & 0 & 0
\end{array} \right]$$

Accepted Answer

A)  Neither
B)  Echelon form
C)  Reduced echelon form  B

Question 4

Solve the problem.
-$$2 x _ { 1 } - 16 x _ { 2 } + 10 x _ { 3 } = 0$$

Accepted Answer

A)    [x1x2x3]=8[x1x2x3]+5[x1x2x3]\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] = 8 \left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] + 5 \left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=8H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​+5H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​  (with  x2,x3x _ { 2 } , x _ { 3 }x2​,x3​  free) 
B)    [x1x2x3]=x2[801]+x3[−510]\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] = x _ { 2 } \left[ \begin{array} { l } 8 \ 0 \ 1 \end{array} ight] + x _ { 3 } \left[ \begin{array} { r } - 5 \ 1 \ 0 \end{array} ight] \quadH403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=x2​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​801​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​+x3​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​−510​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​  (with  x2,x3x _ { 2 } , x _ { 3 }x2​,x3​  free) 
C)    [x1x2x3]=x2[810]+x3[−501](\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] = x _ { 2 } \left[ \begin{array} { l } 8 \ 1 \ 0 \end{array} ight] + x _ { 3 } \left[ \begin{array} { r } - 5 \ 0 \ 1 \end{array} ight] \quad \left( ight.H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=x2​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​810​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​+x3​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​−501​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​(  with  x2,x3x _ { 2 } , x _ { 3 }x2​,x3​  free) 
D)    [x1x2x3]=x2[−810]+x3[501]\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] = x _ { 2 } \left[ \begin{array} { r } - 8 \ 1 \ 0 \end{array} ight] + x _ { 3 } \left[ \begin{array} { l } 5 \ 0 \ 1 \end{array} ight] \quadH403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=x2​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​−810​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​+x3​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​501​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​  (with  x2,x3x _ { 2 } , x _ { 3 }x2​,x3​  free)  Find the general solution of the simple homogeneous "system" below, which consists of a single linear equation. Give your answer as a linear combination of vectors. Let  x2x _ { 2 }x2​  and  x3x _ { 3 }x3​  be free variables.

Question 5

Solve the problem.
-Let $T : R ^ { 2 } \rightarrow \mathscr { R } ^ { 2 }$ be a linear transformation that maps $\mathbf { u } = \left[ \begin{array} { r } - 6 \ 4 \end{array} \right]$ into $\left[ \begin{array} { r } - 22 \ 12 \end{array} \right]$ and maps $\mathbf { v } = \left[ \begin{array} { r } 2 \ - 5 \end{array} \right]$ into $\left[ \begin{array} { c } 11 \ - 4 \end{array} \right]$.
Use the fact that $T$ is linear to find the image of $3 \mathbf { u } + \mathbf { v }$.

Accepted Answer

A)   [−3324]\left[ \begin{array} { r } - 33 \ 24 \end{array} ight][−3324​] 
B)   [−118]\left[ \begin{array} { r } - 11 \ 8 \end{array} ight][−118​] 
C)   [−5532]\left[ \begin{array} { r } - 55 \ 32 \end{array} ight][−5532​] 
D)   [−167]\left[ \begin{array} { r } - 16 \ 7 \end{array} ight][−167​]

Question 6

Write the system as a vector equation or matrix equation as indicated.
-Write the following system as a matrix equation involving the product of a matrix and a vector on the left side and a vector on the right side. $$\begin{array} { l l } 
2 x _ { 1 } + x _ { 2 } - 6 x _ { 3 } & = - 6 \
6 x _ { 1 } - 4 x _ { 2 } & = 2
\end{array}$$

Accepted Answer

A)    [21−6641][x1x2x3]=[−62]\left[ \begin{array} { r r r } 2 & 1 & - 6 \ 6 & 4 & 1 \end{array} ight] \left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] = \left[ \begin{array} { r } - 6 \ 2 \end{array} ight][26​14​−61​]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=[−62​] 
B)    [21−66−40][x1x2x3]=[−62]\left[ \begin{array} { r r r } 2 & 1 & - 6 \ 6 & - 4 & 0 \end{array} ight] \left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} ight] = \left[ \begin{array} { r } - 6 \ 2 \end{array} ight][26​1−4​−60​]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=[−62​] 
C)    [261−4−60][x1x2]=[−62]\left[ \begin{array} { r r } 2 & 6 \ 1 & - 4 \ - 6 & 0 \end{array} ight] \left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \end{array} ight] = \left[ \begin{array} { r } - 6 \ 2 \end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​21−6​6−40​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​[x1​x2​​]=[−62​] 
D)    [x1x2x36−40][21−6]=[−62]\left[ \begin{array} { c c r } x _ { 1 } & x _ { 2 } & x _ { 3 } \ 6 & - 4 & 0 \end{array} ight] \left[ \begin{array} { r } 2 \ 1 \ - 6 \end{array} ight] = \left[ \begin{array} { r } - 6 \ 2 \end{array} ight][x1​6​x2​−4​x3​0​]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​21−6​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=[−62​]

Question 7

Find the standard matrix of the linear transformation T.
-T: $R ^ { 2 } \rightarrow R ^ { 2 }$ first performs a vertical shear that maps $\mathbf { e } _ { 1 }$ into $\mathbf { e } _ { 1 } + 3 \mathbf { e } _ { 2 }$, but leaves the vector $\mathbf { e } _ { 2 }$ unchanged, then reflects the result through the horizontal $x _ { 1 }$-axis.

Accepted Answer

A)   [−1−301]\left[ \begin{array} { r r } - 1 & - 3 \ 0 & 1 \end{array} ight][−10​−31​] 
B)   [130−1]\left[ \begin{array} { r r } 1 & 3 \ 0 & - 1 \end{array} ight][10​3−1​] 
C)   [10−3−1]\left[ \begin{array} { r r } 1 & 0 \ - 3 & - 1 \end{array} ight][1−3​0−1​] 
D)   [−103−1]\left[ \begin{array} { r r } - 1 & 0 \ 3 & - 1 \end{array} ight][−13​0−1​]

Question 8

The augmented matrix is given for a system of equations. If the system is consistent, find the general solution.
Otherwise state that there is no solution.
-$$\left[ \begin{array} { r r r r } 
1 & 2 & - 3 & - 6 \
0 & 1 & 4 & 7 \
0 & 0 & 0 & 2
\end{array} \right]$$

Accepted Answer

A)    x1=−20+11x3x _ { 1 } = - 20 + 11 x _ { 3 }x1​=−20+11x3​  x2=7−4x3x3=2\begin{array} { l } x _ { 2 } = 7 - 4 x _ { 3 } \x _ { 3 } = 2\end{array}x2​=7−4x3​x3​=2​ 
B)   No solution
C)    x1=−6−2x2+3x3x _ { 1 } = - 6 - 2 x _ { 2 } + 3 x _ { 3 }x1​=−6−2x2​+3x3​  x2x _ { 2 }x2​  is free x3x _ { 3 }x3​  is free
D)    x1=−20+11x3\mathrm { x } _ { 1 } = - 20 + 11 \mathrm { x } _ { 3 }x1​=−20+11x3​  x2=7−4x3x _ { 2 } = 7 - 4 x _ { 3 }x2​=7−4x3​  x3x _ { 3 }x3​  is free

Question 9

Use the row reduction algorithm to transform the matrix into echelon form or reduced echelon form as indicated.
-Find the reduced echelon form of the given matrix. $$\left[ \begin{array} { c r r r r } 
1 & 4 & - 5 & 1 & 2 \
2 & 5 & - 4 & - 1 & 4 \
- 3 & - 9 & 9 & 2 & 10
\end{array} \right]$$

Accepted Answer

A)   [1000260100−800018]\left[ \begin{array} { r r r r r } 1 & 0 & 0 & 0 & 26 \0 & 1 & 0 & 0 & - 8 \0 & 0 & 0 & 1 & 8\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​100​010​000​001​26−88​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
B)    [14−51201−21000018]\left[ \begin{array} { r r r r r } 1 & 4 & - 5 & 1 & 2 \0 & 1 & - 2 & 1 & 0 \0 & 0 & 0 & 1 & 8\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​100​410​−5−20​111​208​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
C)   [14−50−601−20−800018]\left[ \begin{array} { r r r r r } 1 & 4 & - 5 & 0 & - 6 \0 & 1 & - 2 & 0 & - 8 \0 & 0 & 0 & 1 & 8\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​100​410​−5−20​001​−6−88​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
D)   [10302601−20−800018]\left[ \begin{array} { r r r r r } 1 & 0 & 3 & 0 & 26 \0 & 1 & - 2 & 0 & - 8 \0 & 0 & 0 & 1 & 8\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​100​010​3−20​001​26−88​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​

Question 10

Determine whether the matrix is in echelon form, reduced echelon form, or neither.
-$$\left[ \begin{array} { r r r r } 
1 & 0 & - 3 & - 5 \
0 & 1 & - 3 & 4 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
\end{array} \right]$$

Accepted Answer

A)  Echelon form
B)  Reduced echelon form
C)  Neither

Question 11

Determine whether the matrix is in echelon form, reduced echelon form, or neither.
-$$\left[ \begin{array} { r r r r } 
1 & 4 & 5 & - 7 \
0 & 1 & - 4 & - 6 \
0 & 2 & 1 & 6
\end{array} \right]$$

Accepted Answer

A)  Echelon form
B)  Reduced echelon form
C)  Neither

Question 12

Solve the problem.
-Suppose an economy consists of three sectors: Energy (E), Manufacturing (M), and Agriculture (A) Sector E sells $70 \%$ of its output to $\mathrm { M }$ and $30 \%$ to $\mathrm { A }$.
Sector M sells $30 \%$ of its output to E, $50 \%$ to A, and retains the rest.
Sector A sells $15 \%$ of its output to E, $30 \%$ to $M$, and retains the rest.
Denote the prices (dollar values) of the total annual outputs of the Energy, Manufacturing, and Agriculture sectors by $\mathrm { p } _ { \mathrm { e } } , \mathrm { p } _ { \mathrm { m } }$, and $\mathrm { p } _ { \mathrm { a } }$, respectively. If possible, find equilibrium prices that make each sector's income match its expenditures.
Find the general solution as a vector, with $\mathrm { p } _ { \mathrm { a } }$ free.

Accepted Answer

A)   [pepmpa]=[0.465pa0.593papa]\left[ \begin{array} { l } p _ { e } \ p _ { m } \ p _ { a } \end{array} ight] = \left[ \begin{array} { c } 0.465 \mathrm { pa } ^ { 0.593 } \mathrm { pa } \ p _ { a } \end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​pe​pm​pa​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=[0.465pa0.593papa​​] 
B)   [pepmpa]=[0.356pa0.686papaa]\left[ \begin{array} { l } \mathrm { p } _ { \mathrm { e } } \\mathrm { p } _ { \mathrm { m } } \\mathrm { p } _ { \mathrm { a } }\end{array} ight] = \left[ \begin{array} { c } 0.356 \mathrm { pa } \0.686 \mathrm { pa } \\mathrm { pa } _ { \mathrm { a } }\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​pe​pm​pa​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​0.356pa0.686papaa​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
C)   [pepmpa]=[0.607paa0.481papa]\left[ \begin{array} { l } p _ { e } \p _ { m } \p _ { a }\end{array} ight] = \left[ \begin{array} { c } 0.607 \mathrm { pa } _ { a } \0.481 \mathrm { pa } ^ { p _ { a } }\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​pe​pm​pa​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=[0.607paa​0.481papa​​] 
D)   [pepmpa]=[0.308pa0.716papa]\left[ \begin{array} { l } \mathrm { p } _ { \mathrm { e } } \\mathrm { p } _ { \mathrm { m } } \\mathrm { p } _ { \mathrm { a } }\end{array} ight] = \left[ \begin{array} { c } 0.308 \mathrm { p } _ { \mathrm { a } } \0.716 \mathrm { pa } \\mathrm { p } _ { \mathrm { a } }\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​pe​pm​pa​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​0.308pa​0.716papa​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​

Question 13

The augmented matrix is given for a system of equations. If the system is consistent, find the general solution.
Otherwise state that there is no solution.
-$$\left[ \begin{array} { r r r r } 
1 & 2 & - 3 & 5 \
0 & 1 & 4 & - 5 \
0 & 0 & 0 & 0
\end{array} \right]$$

Accepted Answer

A)    x1=5−2x2+3x3x _ { 1 } = 5 - 2 x _ { 2 } + 3 x _ { 3 }x1​=5−2x2​+3x3​  x2=−5−4x3x _ { 2 } = - 5 - 4 x _ { 3 }x2​=−5−4x3​  x3x _ { 3 }x3​  is free
B)    x1=5−2x2+3x3x _ { 1 } = 5 - 2 x _ { 2 } + 3 x _ { 3 }x1​=5−2x2​+3x3​  x2x _ { 2 }x2​  is free x3x _ { 3 }x3​  is free
C)    x1=15+11x3x _ { 1 } = 15 + 11 x _ { 3 }x1​=15+11x3​  x2=−5−4x3x _ { 2 } = - 5 - 4 x _ { 3 }x2​=−5−4x3​  x3=0x _ { 3 } = 0x3​=0 
D)    x1=15+11x3x _ { 1 } = 15 + 11 x _ { 3 }x1​=15+11x3​  x2=−5−4x3x _ { 2 } = - 5 - 4 x _ { 3 }x2​=−5−4x3​  x3x _ { 3 }x3​  is free

Question 14

Solve the problem.
-The table shows the amount (in g)of protein, carbohydrate, and fat supplied by one unit (100 g)of three different foods. Let $\mathrm { T }$ be the linear transformation whose standard matrix is
$$A = \left[ \begin{array} { r r r } 
1 & - 2 & 3 \
- 1 & 3 & - 4 \
- 2 & - 9 & 5
\end{array} \right]$$
Determine whether the linear transformation $\mathrm { T }$ is one-to-one and whether it maps $R ^ { 3 }$ onto $R ^ { 3 }$.

Accepted Answer

A)   Not one-to-one; onto  R3R ^ { 3 }R3 
B)   One-to-one; onto  R3R ^ { 3 }R3 
C)   Not one-to-one; not onto  R3R ^ { 3 }R3 
D)   One-to-one; not onto  R3\mathfrak { R } ^ { 3 }R3

Question 15

Find the indicated vector.
-Let $\mathbf { u } = \left[ \begin{array} { r } 4 \ - 7 \end{array} \right]$. Find $7 \mathbf { u }$.

Accepted Answer

A)   [2849]\left[ \begin{array} { l } 28 \ 49 \end{array} ight][2849​] 
B)   [28−49]\left[ \begin{array} { r } 28 \ - 49 \end{array} ight][28−49​] 
C)   [−28−49]\left[ \begin{array} { l } - 28 \ - 49 \end{array} ight][−28−49​] 
D)   [−2849]\left[ \begin{array} { r } - 28 \ 49 \end{array} ight][−2849​]

Question 16

Determine whether the matrix is in echelon form, reduced echelon form, or neither.
-$$\left[ \begin{array} { r r r r } 
1 & 6 & 2 & - 7 \
0 & 1 & - 4 & - 6 \
0 & 0 & 0 & 0
\end{array} \right]$$

Accepted Answer

A)  Echelon form
B)  Neither
C)  Reduced echelon form

Question 17

Solve the system of equations.
-$$\begin{array} { l } 
7 x _ { 1 } + 7 x _ { 2 } + x _ { 3 } = 1 \
x _ { 1 } + 8 x _ { 2 } + 8 x _ { 3 } = 8 \
9 x _ { 1 } + x _ { 2 } + 9 x _ { 3 } = 9
\end{array}$$

Accepted Answer

A)    (−1,1,1) ( - 1,1,1 ) (−1,1,1)  
B)    (0,0,1) ( 0,0,1 ) (0,0,1)  
C)    (1,−1,1) ( 1 , - 1,1 ) (1,−1,1)  
D)    (0,1,0) ( 0,1,0 ) (0,1,0)

Question 18

Solve the problem.
-Let $A = \left[ \begin{array} { r r r } 1 & - 3 & 2 \ - 3 & 4 & - 1 \ 2 & - 5 & 3 \end{array} \right]$ and $\mathbf { b } = \left[ \begin{array} { r } 2 \ 4 \ - 4 \end{array} \right]$.
Define a transformation $\mathrm { T } : \mathfrak { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ by $\mathrm { T } ( \mathrm { x } ) = \mathrm { Ax }$.
If possible, find a vector $\mathbf { x }$ whose image under $\mathrm { T }$ is $\mathbf { b }$. Otherwise, state that $\mathbf { b }$ is not in the range of the transformation $\mathrm { T }$.

Accepted Answer

A)   [422]\left[ \begin{array} { l } 4 \ 2 \ 2 \end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​422​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
B)   [440]\left[ \begin{array} { l } 4 \ 4 \ 0 \end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​440​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
C)   [40−4]\left[ \begin{array} { r } 4 \0 \- 4\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​40−4​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
D)    b\mathbf { b }b  is not in the range of the transformation T.

Question 19

Solve the problem.
-Find the general solution of the homogeneous system below. Give your answer as a vector. $$\begin{array} { r } 
x _ { 1 } + 2 x _ { 2 } - 3 x _ { 3 } = 0 \
4 x _ { 1 } + 7 x _ { 2 } - 9 x _ { 3 } = 0 \
- x _ { 1 } - 4 x _ { 2 } + 9 x _ { 3 } = 0
\end{array}$$

Accepted Answer

A)   [x1x2x3]=x3[3−31]\left[ \begin{array} { l } x _ { 1 } \x _ { 2 } \x _ { 3 }\end{array} ight] = x _ { 3 } \left[ \begin{array} { r } 3 \- 3 \1\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=x3​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​3−31​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
B)   [x1x2x3]=[−331]\left[ \begin{array} { l } x _ { 1 } \x _ { 2 } \x _ { 3 }\end{array} ight] = \left[ \begin{array} { r } - 3 \3 \1\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​−331​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
C)   [x1x2x3]=x3[−331]\left[ \begin{array} { l } x _ { 1 } \x _ { 2 } \x _ { 3 }\end{array} ight] = x _ { 3 } \left[ \begin{array} { r } - 3 \3 \1\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=x3​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​−331​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​ 
D)   [x1x2x3]=x3[−330]\left[ \begin{array} { l } x _ { 1 } \x _ { 2 } \x _ { 3 }\end{array} ight] = x _ { 3 } \left[ \begin{array} { r } - 3 \3 \0\end{array} ight]H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​x1​x2​x3​​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​=x3​H403z M403 1759 V0 H319 V1759 v0 v1759 h84z"/>​−330​M347 1759 V0 H263 V1759 v0 v1759 h84z"/>​

Question 20

Find the indicated vector.
-Let $\mathbf { u } = \left[ \begin{array} { r } - 1 \ 1 \end{array} \right] , \mathbf { v } = \left[ \begin{array} { r } - 4 \ 3 \end{array} \right]$. Find $\mathbf { u } - \mathbf { v }$.

Accepted Answer

A)   [3−2]\left[ \begin{array} { r } 3 \ - 2 \end{array} ight][3−2​] 
B)   [−2−7]\left[\begin{array}{l}-2 \-7\end{array}ight][−2−7​] 
C)   [−54]\left[ \begin{array} { r } - 5 \ 4 \end{array} ight][−54​] 
D)   [−45]\left[ \begin{array} { r } - 4 \ 5 \end{array} ight][−45​]

Exam 1: Linear Equations in Linear Algebra

Determine whether the matrix is in echelon form, reduced echelon form, or neither. - $\left[ \begin{array} { r r r r } 1 & - 5 & 3 & 4 \\0 & 0 & - 5 & - 4 \\0 & 0 & 0 & - 2 \\0 & 0 & 0 & 0\end{array} \right]$

Solve the problem. - $2 x _ { 1 } - 16 x _ { 2 } + 10 x _ { 3 } = 0$

The augmented matrix is given for a system of equations. If the system is consistent, find the general solution. Otherwise state that there is no solution. - $\left[ \begin{array} { r r r r } 1 & 2 & - 3 & - 6 \\0 & 1 & 4 & 7 \\0 & 0 & 0 & 2\end{array} \right]$

Use the row reduction algorithm to transform the matrix into echelon form or reduced echelon form as indicated. -Find the reduced echelon form of the given matrix. $\left[ \begin{array} { c r r r r } 1 & 4 & - 5 & 1 & 2 \\2 & 5 & - 4 & - 1 & 4 \\- 3 & - 9 & 9 & 2 & 10\end{array} \right]$

Determine whether the matrix is in echelon form, reduced echelon form, or neither. - $\left[ \begin{array} { r r r r } 1 & 0 & - 3 & - 5 \\0 & 1 & - 3 & 4 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{array} \right]$

Determine whether the matrix is in echelon form, reduced echelon form, or neither. - $\left[ \begin{array} { r r r r } 1 & 4 & 5 & - 7 \\0 & 1 & - 4 & - 6 \\0 & 2 & 1 & 6\end{array} \right]$

The augmented matrix is given for a system of equations. If the system is consistent, find the general solution. Otherwise state that there is no solution. - $\left[ \begin{array} { r r r r } 1 & 2 & - 3 & 5 \\0 & 1 & 4 & - 5 \\0 & 0 & 0 & 0\end{array} \right]$

Find the indicated vector. -Let $\mathbf { u } = \left[ \begin{array} { r } 4 \\ - 7 \end{array} \right]$ . Find $7 \mathbf { u }$ .

Determine whether the matrix is in echelon form, reduced echelon form, or neither. - $\left[ \begin{array} { r r r r } 1 & 6 & 2 & - 7 \\0 & 1 & - 4 & - 6 \\0 & 0 & 0 & 0\end{array} \right]$

Solve the system of equations. - $\begin{array} { l } 7 x _ { 1 } + 7 x _ { 2 } + x _ { 3 } = 1 \\x _ { 1 } + 8 x _ { 2 } + 8 x _ { 3 } = 8 \\9 x _ { 1 } + x _ { 2 } + 9 x _ { 3 } = 9\end{array}$

Solve the problem. -Find the general solution of the homogeneous system below. Give your answer as a vector. $\begin{array} { r } x _ { 1 } + 2 x _ { 2 } - 3 x _ { 3 } = 0 \\4 x _ { 1 } + 7 x _ { 2 } - 9 x _ { 3 } = 0 \\- x _ { 1 } - 4 x _ { 2 } + 9 x _ { 3 } = 0\end{array}$

Find the indicated vector. -Let $\mathbf { u } = \left[ \begin{array} { r } - 1 \\ 1 \end{array} \right] , \mathbf { v } = \left[ \begin{array} { r } - 4 \\ 3 \end{array} \right]$ . Find $\mathbf { u } - \mathbf { v }$ .

Exam 1: Linear Equations in Linear Algebra

Determine whether the matrix is in echelon form, reduced echelon form, or neither. - [1−53400−5−4000−20000]\left[ \begin{array} { r r r r } 1 & - 5 & 3 & 4 \\0 & 0 & - 5 & - 4 \\0 & 0 & 0 & - 2 \\0 & 0 & 0 & 0\end{array} \right]​1000​−5000​3−500​4−4−20​​

Solve the problem. - 2x1−16x2+10x3=02 x _ { 1 } - 16 x _ { 2 } + 10 x _ { 3 } = 02x1​−16x2​+10x3​=0

Determine whether the matrix is in echelon form, reduced echelon form, or neither. - [10−3−501−3400000000]\left[ \begin{array} { r r r r } 1 & 0 & - 3 & - 5 \\0 & 1 & - 3 & 4 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{array} \right]​1000​0100​−3−300​−5400​​

Determine whether the matrix is in echelon form, reduced echelon form, or neither. - [145−701−4−60216]\left[ \begin{array} { r r r r } 1 & 4 & 5 & - 7 \\0 & 1 & - 4 & - 6 \\0 & 2 & 1 & 6\end{array} \right]​100​412​5−41​−7−66​​

Find the indicated vector. -Let u=[4−7]\mathbf { u } = \left[ \begin{array} { r } 4 \\ - 7 \end{array} \right]u=[4−7​] . Find 7u7 \mathbf { u }7u .

Determine whether the matrix is in echelon form, reduced echelon form, or neither. - [162−701−4−60000]\left[ \begin{array} { r r r r } 1 & 6 & 2 & - 7 \\0 & 1 & - 4 & - 6 \\0 & 0 & 0 & 0\end{array} \right]​100​610​2−40​−7−60​​

Solve the system of equations. - 7x1+7x2+x3=1x1+8x2+8x3=89x1+x2+9x3=9\begin{array} { l } 7 x _ { 1 } + 7 x _ { 2 } + x _ { 3 } = 1 \\x _ { 1 } + 8 x _ { 2 } + 8 x _ { 3 } = 8 \\9 x _ { 1 } + x _ { 2 } + 9 x _ { 3 } = 9\end{array}7x1​+7x2​+x3​=1x1​+8x2​+8x3​=89x1​+x2​+9x3​=9​

Find the indicated vector. -Let u=[−11],v=[−43]\mathbf { u } = \left[ \begin{array} { r } - 1 \\ 1 \end{array} \right] , \mathbf { v } = \left[ \begin{array} { r } - 4 \\ 3 \end{array} \right]u=[−11​],v=[−43​] . Find u−v\mathbf { u } - \mathbf { v }u−v .