Answer
$[(9,2)]_C=(3,-1)\\
[(4,-3]_C=(-1,-2)\\ [(4,-3)]_{C \leftarrow B}=\begin{bmatrix}
3 & -1 \\
-1 & -2
\end{bmatrix}$
Work Step by Step
We know that:
For B: $a(2,1)+b(-3,1)=(9,2)$
For C: $c(2,1)+d(-3,1)=(4,-3)$
To find $[(9,2)]_C$: $2a-3b=9\\
a+b=2\\
2a-3b+3(a+b)=15 \rightarrow 5a=15 \rightarrow a=3$
Thus $3+b=2 \rightarrow b=-1$
To find $[(4,-3)]_{C}:$
$2c-3d=4\\
c+d=-3\\
2c-3d+3(c+d)=-5 \rightarrow 5c=-5 \rightarrow c=-1$
Thus $-1+d=-3 \rightarrow d=-2$
Hence we have: $[(9,2)]_C=(3,-1)\\
[(4,-3]_C=(-1,-2)\\ [(4,-3)]_{C \leftarrow B}=\begin{bmatrix}
3 & -1 \\
-1 & -2
\end{bmatrix}$
