Answer
See below
Work Step by Step
a) Obtain:
$T(1,0,0)=(1-2.0)\cos x+(3.0+0)\sin x=\cos x\\
T(0,1,0)=(0-2.0)\cos x+(3.1+0)\sin x=3\sin x\\
T(0,0,1)=(0-2.1)\cos x+(3.0+1)\sin x=-2\cos x+\sin x$
then $T(1,0,0)=\cos x=1.\cos x+0.\sin x\\
T(0,1,0)=3\sin x=0.\sin x+3.\sin x\\
T(0,0,1)=-2\cos x+\sin x=-2.\cos x+1.\sin x$
We have:
$[T(1,0,0)]_C=\begin{bmatrix}
1\\0
\end{bmatrix}\\
[T(0,1,0)]_C=\begin{bmatrix}
0\\ 3
\end{bmatrix}\\
[T(0,0,1)]_C=\begin{bmatrix}
-2 \\ 1
\end{bmatrix}$
Hence, $[T]^C_B=\begin{bmatrix}
1 & 0 & -2\\0 & 3 & 1
\end{bmatrix}$
b)Obtain:
$T(2,-1,-1)=[2-2.(-1)]\cos x+[3.(-1)+(-1)]\sin x=4\cos x-4\sin x\\
T(1,3,5)=(1-2.5)\cos x+[3.3+5]\sin x=-9\cos x+14\sin x\\
T(0,4,-1)=[0-2.(-1)]\cos x+[3.4+(-1)]\sin x=2\cos x+11\sin x$
then $T(2,-1,-1)=4\cos x-4\sin x=4.(\cos x-\sin x)+0.(\cos x+\sin x)\\
T(0,1,0)=-9\sin x+14\sin x=-\frac{23}{2}.(\cos x-\sin x)+\frac{5}{2}.(\cos x+\sin x)\\
T(0,4,1)=2\cos x=-\frac{9}{2}.(\cos x-\sin x)+\frac{13}{2}.(\cos x+\sin x)$
We have:
$[T(2,-1,-1)]_C=\begin{bmatrix}
4\\0
\end{bmatrix}\\
[T(1,3,5)]_C=\begin{bmatrix}
-\frac{23}{2}\\ \frac{5}{2}
\end{bmatrix}\\
[T(0,4,-1)]_C=\begin{bmatrix}
-\frac{9}{2} \\ \frac{13}{2}
\end{bmatrix}$
Hence, $[T]^C_B=\begin{bmatrix}
4& -\frac{23}{2} & \frac{-9}{2}\\ 0& \frac{5}{2} & \frac{13}{2}
\end{bmatrix}$
