Answer
See below
Work Step by Step
Given: $T:M_3(R)\rightarrow R$
a) Computing $[T]^C_B$
$T(1)=1.1\\
T(x)=2.1\\
T(x^2)=2^2=4=4.1\\
T(x^3)=2^3=8 =8.1$
then we have $[T(1)]_0=1\\
[T(x)]_C=2\\
[T(x^2)]_C=4\\
[T(x^3)]_C=8$
Hence, $[T]_B^C=\begin{bmatrix}
1 & 2 & 4 & 8\end{bmatrix}$
We can notice that $[p(x)]_B=\begin{bmatrix}
0 \\ 2 \\ -3 \\ 0\end{bmatrix}$
Consequently, $[T(v)]_C=[T]^C_B[A]_B=\begin{bmatrix}
1 & 2 & 4 & 8\end{bmatrix}\begin{bmatrix}
0 \\ 2 \\ -3 \\ 0\end{bmatrix}\\
\rightarrow T(A)=-8$
b) From part (a), we obtain: $T(p(x))=T(2x-3x^2)=2.2-3.2^2=-8$
