Answer
$$
\begin{aligned} T\left(\begin{array}{c}{-4} \\ {5} \\ {1}\end{array}\right) =\left(\begin{array}{c}{4} \\ {22} \\ {-5}\end{array}\right) \ \end{aligned} \
$$
\begin{array}{l}{ \text { the preimage of }} \\ {\mathbf{w}=(4,1,-1) \text { is } \ (1,2,2)}\end{array}
Work Step by Step
Given $$T\left(\begin{array}{c}{v_{1}} \\ {v_{2}} \\ {v_{3}}\end{array}\right)=\left(\begin{array}{c}{2 v_{1}+v_{2}} \\ {2 v_{2}-3 v_{1}} \\ {v_{3}}\end{array}\right)$$
So, we get
$$
\begin{aligned} T\left(\begin{array}{c}{-4} \\ {5} \\ {1}\end{array}\right)=\left(\begin{array}{c}{2(-4)+5} \\ {2(5)-3(-4)} \\ {1}\end{array}\right)=\left(\begin{array}{c}{4} \\ {22} \\ {-5}\end{array}\right) \ \end{aligned} \
$$
$$
\begin{array}{l}{\text { The preimage of } \mathbf{w}=(4,1,-1) \text { is } \mathbf{a}=\left(a_{1}, a_{2}, a_{3}\right) . \text { Hence, }} \\ {\qquad \begin{aligned} T\left(\begin{array}{c}{a_{1}} \\ {a_{2}} \\ {a_{3}}\end{array}\right)=\left(\begin{array}{c}{4} \\ {1} \\ {-1}\end{array}\right) \\
&\left(\begin{array}{c}{2 a_{1}+a_{2}} \\ {2 a_{2}-3 a_{1}} \\ {a_{1}-a_{3}}\end{array}\right)=\left(\begin{array}{c}{4} \\ {1} \\ {-1}\end{array}\right) \end{aligned}}\end{array}$$
Thus, we have a system of equations
$$\begin{array}{l}{\qquad \begin{aligned} 2 a_{1}+a_{2} &=4 \\-3 a_{1}+2 a_{2} &=1 \\ a_{1}-a_{3} &=-1 \end{aligned}}\end{array}$$
$$
\begin{array}{l}{\text { Solving for } a_{1}, a_{2}, a_{3}, \text { we get } a_{1}=1, a_{2}=2, a_{3}=2 . \text { Hence, the preimage of }} \\ {\mathbf{w}=(4,1,-1) \text { is } \mathbf{a}=\left(a_{1}, a_{2}, a_{3}\right)=(1,2,2)}\end{array}
$$
