Answer
See below
Work Step by Step
Obtain $T(0,1,0)=\begin{bmatrix}
0 & 1\\
0 & -3
\end{bmatrix}\\ T(0,0,1)=\begin{bmatrix}
0 & 3\\
-1 & 0
\end{bmatrix} \\T(1,0,0)=\begin{bmatrix}
0 & -1\\
5 & 0
\end{bmatrix}$
then we have:
$T(0,1,0)=\begin{bmatrix}
0 & 1\\
0 & -3
\end{bmatrix}=0.E_{21}-3.E_{22}+0.E_{11}+1.E_{12}\\ T(0,0,1)=\begin{bmatrix}
0 & 3\\
-1 & 0
\end{bmatrix}=-1.E_{21}+0.E_{22}+0.E_{11}+3.E_{12} \\T(1,0,0)=\begin{bmatrix}
0 & -1\\
5 & 0
\end{bmatrix}=5.E_{21}+0.E_{22}+0.E_{11}-1.E_{12}$
as $[T(1)]_C=\begin{bmatrix}
1\\
1
\end{bmatrix}\\
[T(0,1,0)]_C=\begin{bmatrix}
0 \\ -3 \\ 0 \\ 1
\end{bmatrix} \\
[T(0,0,1)]_C=\begin{bmatrix}
-1 \\0 \\ 0 \\ 3
\end{bmatrix} \\ [T(1,0,0)]_C=\begin{bmatrix}
5 \\ 0 \\ 0 \\ -1
\end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
0 & -1 & 5 \\
3 & 0 & 0 \\
0 & 0 & 0 \\ 1 & 3 & -1\end{bmatrix}$
We can notice that $[v]_B=\begin{bmatrix}
1 & -2 & -2\end{bmatrix}$
From Theorem 6.4.5,
$$[T(v)]_C=[T]^C_B[v]_B\\
\rightarrow [T(v)]_C=\begin{bmatrix}
0 & -1 & 5 \\
3 & 0 & 0 \\
0 & 0 & 0 \\ 1 & 3 & -1\end{bmatrix}\begin{bmatrix}
1 & -2 & -2\end{bmatrix}=\begin{bmatrix}
-8 & -3 & 0 & -3\end{bmatrix}\\
\rightarrow T(v)=\begin{bmatrix}
0 & -3\\8 & -3\end{bmatrix}$$
