Answer
See below
Work Step by Step
a) Obtain $T(1)=0\\T(x)=1\\T(x^2)=2x\\T(x^3)=3x^2$
then we have:
$T(1)=0=0.1+0.x+0.x^2\\
T(x)=1=1.1+0.x+0.x^2\\
T(x^2)=2x=0.1+2.x+0.x^2\\
T(x^3)=3x^2=0.1+0.x+3.x^2$
as $[T(1)]_C=\begin{bmatrix}
0\\
0 \\0
\end{bmatrix}\\
[T(2)]_C=\begin{bmatrix}
1\\0 \\
0
\end{bmatrix}\\
[T(3)]_C=\begin{bmatrix}
0\\2\\
0
\end{bmatrix}\\
[T(4)]_C=\begin{bmatrix}
0 \\
0 \\ 3
\end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 3
\end{bmatrix}$
b) Obtain $T(x^3)=3x^2\\T(x^3+1)=3x^2\\T(x^3+x)=3x^2+1\\T(x^3+x^2)=3x^2+2x$
then we have:
$T(x^3)=3x^2=0.1-3(1+x)+3(1+x+x^2\\
T(x^3+1)=1=0.1-3(1+x)+3(1+x+x^2)\\
T(x^3+x)=2x=1.1-3(1+x)+3(1+x+x^2)\\
T(x^3+x^2)=3x^2=0.1-1(1+x)+3(1+x+x^2)$
as $[T(1)]_C=\begin{bmatrix}
0\\
-3 \\-3
\end{bmatrix}\\
[T(2)]_C=\begin{bmatrix}
0 \\
-3 \\ -3
\end{bmatrix}\\
[T(3)]_C=\begin{bmatrix}
1\\-3\\
-3
\end{bmatrix}\\
[T(4)]_C=\begin{bmatrix}
-2\\
-1 \\ 3
\end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
0 & 0 & 1 & -3\\
-3 & -3 & -3 & -1\\
3& 3 & 3 & 3
\end{bmatrix}$
