Answer
See below
Work Step by Step
From Theorem 6.4.5:
$$[T(A)]_C=[T]^C_B[A)]_B$$
a) Computing $[T]^C_B$
$T(E_{11})=\begin{bmatrix}
2.1-0+0 & -1+3.0\\
0 & -1-0+3.0
\end{bmatrix}=\begin{bmatrix}
2 & -1 \\0 & -1
\end{bmatrix} \\
T(E_{12})=\begin{bmatrix}
2.0-1+0 & -0+3.0\\
0 & -1-0+3.0
\end{bmatrix}=\begin{bmatrix}
-1 & 0\\0 & -1
\end{bmatrix}\\
T(E_{21})=\begin{bmatrix}
2.0-0+0 & -0+3.0\\
0 & -0-0+3.1
\end{bmatrix}=\begin{bmatrix}
0 & 0 \\0 & 3
\end{bmatrix}\\
T(E_{22})=\begin{bmatrix}
2.0-0+1 & 0+3.1\\
0 & -0-0+3.0
\end{bmatrix}=\begin{bmatrix}
1 & 3 \\0 & 0
\end{bmatrix}$
then we have $T(E_{11})=\begin{bmatrix}
2 & -1 \\0 & -1
\end{bmatrix}=2.E_{11}-1.E_{12}+0.E_{21}-1.E_{22}\\
T(E_{12})=\begin{bmatrix}
-1 & 0 \\0 & -1
\end{bmatrix}=-1.E_{11}+0.E_{12}+0.E_{21}-1.E_{22}\\
T(E_{21})=\begin{bmatrix}
0 & 0 \\0 & 3
\end{bmatrix}=0.E_{11}+0.E_{12}+0.E_{21}+3.E_{22}\\
T(E_{22})=\begin{bmatrix}
1 & 3\\0 & 0
\end{bmatrix}=1.E_{11}+3.E_{12}+0.E_{21}+0.E_{22}$
Therefore, $[T(E_{11})]_C=\begin{bmatrix}
2 \\ -1 \\ 0 \\ -1
\end{bmatrix}\\
[T(E_{12})]_C=\begin{bmatrix}-1 \\ 0 \\ 0 \\ -1
\end{bmatrix}\\
[T(E_{21})]_C=\begin{bmatrix}
0 \\ 0 \\ 0 \\ 3
\end{bmatrix}\\
[T(E_{22})]_C=\begin{bmatrix}
1 \\ 3 \\ 0 \\ 0
\end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
2 & -1 & 0 & 1 \\
-1 & 0 & 0 & 3\\ 0 & 0 & 0 & 0 \\ -1 & -1 & 3 & 0 \end{bmatrix}$
We can notice that $[A]_B=\begin{bmatrix}
-7 \\
2 \\ 1 \\ 3 \end{bmatrix}$
Consequently, $[T(v)]_C=[T]^C_B[p(x)]_B=\begin{bmatrix}
2 & -1 & 0 & 1 \\
-1 & 0 & 0 & 3\\ 0 & 0 & 0 & 0 \\ -1 & -1 & 3 & 0 \end{bmatrix}\begin{bmatrix}
-7 \\
2 \\ 1 \\ 3 \end{bmatrix}=\begin{bmatrix}
-19 \\
-2 \\ 0 \\ 8
\end{bmatrix}\\
\rightarrow T(A)=\begin{bmatrix}
-19 & -2 \\ 0 & 8\end{bmatrix}$
b) From part (a), we obtain: $T(p(x))=T(\begin{bmatrix}
-7 & 2 \\ 1 & -3\end{bmatrix})=\begin{bmatrix}
2.(-7)-2+(-3) & 7+3.(-3) \\ 0 & 7-2+3.1\end{bmatrix}=\begin{bmatrix}
-19 & -2 \\ 0 & 8\end{bmatrix}$
