Answer
See below
Work Step by Step
Given: $T:R^3 \rightarrow P^3$
and $T(a+bx)=2a-(a+b-c)x+(2c-a)x^3$
a) Computing $[T]^C_B$
$T(1)=T(1+0x)=\begin{bmatrix}
1- 0& 0\\
-2.0 & -1+0
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\\
T(x)=T(0+1x)=\begin{bmatrix}
0-1 & 0\\
-2.1 & -0+1
\end{bmatrix}=\begin{bmatrix}
-1 & 0\\
-2 & 1
\end{bmatrix}$
then we have $T(1)=\begin{bmatrix}
-1 & 0\\
-2 & 1
\end{bmatrix}=1.E_{11}+0.E_{12}+0.E_{21}-1.E_{22}\\
T(x)=\begin{bmatrix}
-1 & 0\\
-2 & 1
\end{bmatrix}=-1.E_{11}+0.E_{12}-2.E_{21}+1.E_{22}$
$[T(1,0,0)]_C=\begin{bmatrix}
1\\
0 \\ 0 \\ -1
\end{bmatrix}\\
[T(x)]_C=\begin{bmatrix}
-1 \\ 0 \\ -2 \\ 1
\end{bmatrix} $
Hence, $[T]_B^C=\begin{bmatrix}
1 & -1 \\
0 & 0 \\ 0 & -2 \\ -1 & 1\end{bmatrix}$
We can notice that $[p(x)]_B=\begin{bmatrix}
-2 \\
3 \end{bmatrix}$
Consequently, $[T(v)]_C=[T]^C_B[p(x)]_B=\begin{bmatrix}
1 & -1\\
0 & 0\\ 0 & -2 \\ -1 & 1
\end{bmatrix}\begin{bmatrix}
-2 \\
3
\end{bmatrix}=\begin{bmatrix}
-5 \\
0\\ -6 \\ 5
\end{bmatrix}\\
\rightarrow T(p(x))= \begin{bmatrix}
-5 & 0\\
-6 & 5
\end{bmatrix}$
b) From part (a), we obtain: $T(p(x))=T(-2+3x)=\begin{bmatrix}
-2-3 & 0\\
-2.3 & 2+3
\end{bmatrix}=\begin{bmatrix} -5& 0\\
-6 & 5
\end{bmatrix}$
