Answer
See below
Work Step by Step
a) Obtain:
$T(1)=(1+0x+0x^2)=(1,2)\\
T(x)=(0+x+0x^2)=(0,1)\\
T(x^2)=(0+0x+x^2)=(-3,-2)$
then $T(1)=(1,2)=1.(1,0)+2.(0,1)\\
T(x)=(0,1)=0.(1,0)+1.(0,1)\\
T(x^2)=(-3,-2)=-3.(1,0)+(-2).(0,1)$
We have:
$[T(1)]_C=\begin{bmatrix}
1\\
2\end{bmatrix}\\
[T(x)]_C=\begin{bmatrix}
0\\
1
\end{bmatrix}\\
[T(x^2)]_C=\begin{bmatrix}
-3\\
-2
\end{bmatrix}$
Hence, $[T]^C_B=\begin{bmatrix}
1 & 0 & -3\\2 & 1 & -2
\end{bmatrix}$
b) $T(1)=(1+0x+0x^2)=(-1,1)\\
T(1+x)=(1+x+0x^2)=(1,3)\\
T(1+x+x^2)=(1+x+x^2)=(-2,1)$
then $T(1)=(1,2)=-1.(1,-1)+1.(2,1)\\
T(1+x)=(1,3)=-\frac{5}{3}.(1,-1)+\frac{3}{4}.(2,1)\\
T(1+x+x^2)=(-2,-1)=-\frac{4}{3}.(1,-1)+(-\frac{1}{3}).(2,1)$
We have:
$[T(1)]_C=\begin{bmatrix}
-1\\
1\end{bmatrix}\\
[T(1+x)]_C=\begin{bmatrix}
-\frac{5}{3}\\
\frac{4}{3}
\end{bmatrix}\\
[T(1+x+x^2)]_C=\begin{bmatrix}
-\frac{4}{3}\\
-\frac{1}{3}
\end{bmatrix}$
Hence, $[T]^C_B=\begin{bmatrix}
-1 & \frac{-5}{3} & -\frac{4}{3}\\1 & \frac{4}{3} & -\frac{1}{3}
\end{bmatrix}$
