Answer
See below
Work Step by Step
Obtain $T(x)=Ax=\begin{bmatrix}
1 & 3\\
-1 & 1\\
-2 & 0\\
5 & 2
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}=\begin{bmatrix}
x+3y\\
-x+y\\
-2x\\
5x+2y
\end{bmatrix}$
then we have:
$T(1,1)=(4,0,-2,7)=7(0,0,0,1)-2(0,0,1,0)+0(0,1,0,0)+4(1,0,0,0)\\ T(1,0)=(1,-1,-2,5)=5(0,0,0,1)-2(0,0,1,0)-1(0,1,0,0)+1(1,0,0,0)$
Thus,
$T[(1,1)]_C=\begin{bmatrix}
7\\
-2 \\
0\\4
\end{bmatrix}\\
T[(1,0)]_C=\begin{bmatrix}
5\\
-2 \\
-1 \\ 1
\end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
7 & 5 \\
-2 & -2\\
0 & -1\\4 & 1 \end{bmatrix}$
We can notice that $[v]_B=\begin{bmatrix}
4 & -6 \end{bmatrix}$
From Theorem 6.4.5,
$$[T(v)]_C=[T]^C_B[v]_B\\
\rightarrow [T(v)]_C=\begin{bmatrix}
7 & 5 \\
-2 & -2\\
0 & -1\\4 & 1 \end{bmatrix}\begin{bmatrix}
4 & -6\end{bmatrix}=\begin{bmatrix}
-2\\ 4 \\ 6 \\ 10\end{bmatrix}\\
\rightarrow T(v)=-2(0,0,0,1)+4(0,0,1,0)+6(0,1,0,0)+10(1,0,0,0)=(10,6,4,-2)$$
