Answer
See below
Work Step by Step
a) Obtain:
$\begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}=(1-0)+3.0x^2+(0-1)x^3=1-x^3\\
\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}=(0-0)+3.x^2+(0-0)x^3=3x^2\\
\begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix}=(0-0)+3.0x^2+(1-0)x^3=x^3\\
\begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix}=(0-1)+3.0x^2+(0-0)x^3=-1$
then $E_{11}=1-x^3=1.1+0.x+0.x^2+(-1).x^3 \\
E_{12}=3x^2=0.1+0.x+3.x^2+0.x^3\\
E_{21}=x^3=0.1+0.x+0.x^2+1.x^3\\
E_{22}=-1=-1.1+0.x+0.x^2+0.x^3$
We have:
$[T(E_{11})]_C=\begin{bmatrix}
1\\
0 \\
0\\-1
\end{bmatrix}\\
[T(E_{12})]_C=\begin{bmatrix}
0\\
0 \\
3\\0
\end{bmatrix}\\
[T(E_{21})]_C=\begin{bmatrix}
0\\
0 \\
0\\1
\end{bmatrix}\\
[T(E_{22})]_C=\begin{bmatrix}
-1 \\
0 \\
0\\0
\end{bmatrix}$
Hence, $[T]^C_B=\begin{bmatrix}
1 & 0 & 0 & -1\\
0 & 0 & 0 & 0 \\
0 & 3 & 0 & 0\\-1 & 0 & 1 & 0
\end{bmatrix}$
b) $E_{11}=1-x^3=0.1+1.1+(-1)x^3+0.x^2 \\
E_{12}=3x^2=0.x+0.1+0.x^3+3.x^2\\
E_{21}=x^3=0.x+0.1+1.x^3+0.x^2\\
E_{22}=-1=0.x+0.1+0.x^3+3.x^2$
We have:
$[T(E_{11})]_C=\begin{bmatrix}
0\\
1 \\
-1 \\0
\end{bmatrix}\\
[T(E_{12})]_C=\begin{bmatrix}
0\\
0 \\
0\\3
\end{bmatrix}\\
[T(E_{21})]_C=\begin{bmatrix}
0\\
0 \\
1\\0
\end{bmatrix}\\
[T(E_{22})]_C=\begin{bmatrix}
0 \\
-1 \\
0\\0
\end{bmatrix}$
Hence, $[T]^C_B=\begin{bmatrix}
0 & 0 & 0& 0\\
0 & 1 & -1 & 0 \\
1 & -1 & 0 & 0\\0 & 0 & 0 & 3
\end{bmatrix}$
