Answer
See below
Work Step by Step
Given: $T:R^3 \rightarrow P^3$
and $T(a,b,c)=2a-(a+b-c)x+(2c-a)x^3$
and the standard bases $B,C: B=\{(1,0,0);(0,1,0);(0,0,1)\}\\
C=\{1,x,x^2,x^3\}$
a) Computing $[T]^C_B$
$T(1,0,0)=2.1-(1+0-0)x+(2.0-1)x^3=2-x-x^3\\
T(0,1,0)=2.0-(0+1-0)x+(2.0-0)x^3=-x\\
T(0,0,1)=2.0-(0+0-1)x+(2.1-0)x^3=x+2x^3$
then we have $T(1,0,0)=2-x-x^3=2.1-1.x+0.x^2-1.x^3\\
T(0,1,0)=-x=0.1-1.x+0.x^2+0.x^3\\
T(0,0,1)=x+2x^3=0.1+1.x+0.x^2+2.x^3$
$[T(1,0,0)]_C=\begin{bmatrix}
2\\
-1 \\ 0 \\ -1
\end{bmatrix}\\
[T(0,1,0)]_C=\begin{bmatrix}
0 \\ -1 \\ 0 \\ 0
\end{bmatrix} \\
[T(0,0,1)]_C=\begin{bmatrix}
0 \\ 1\\ 0 \\ 2
\end{bmatrix} $
Hence, $[T]_B^C=\begin{bmatrix}
2 & 0 & 0 \\
-1 & -1 & 1 \\ -1 & 0 & 2\end{bmatrix}$
We can notice that $[v]_B=\begin{bmatrix}
2 \\
-1\\5
\end{bmatrix}$
Consequently, $[T(v)]_C=[T]^C_B[v]_B=\begin{bmatrix}
2 &0 & 0 \\
-1 & -1 & 1\\ -1 & 0 & 2
\end{bmatrix}\begin{bmatrix}
2 \\
-1\\5
\end{bmatrix}=\begin{bmatrix}
4 \\
4 \\ 0 \\ 8
\end{bmatrix}\\
\rightarrow T(v)= 4+4x+8x^3$
b) From part (a), we obtain: $T(v)=T(2,-1,5)=2.2-[2+(-1)-5)x+(2.5-2)x^3=4+4x+8x^3$
