Answer
$[(-5,-3)]_C=(-1,1)\\
[(4,28]_C=(4,-20)\\ P_{C \leftarrow B}=\begin{bmatrix}
-1 & 4 \\
1 & -20
\end{bmatrix}$
Work Step by Step
We know that:
For B: $a(6,2)+b(1,-1)=(-5,-3)$
For C: $c(6,2)+d(1,-1)=(4,28)$
To find $[(-5,-3)]_C$: $2a-b=-3\\
6a+b=-5\\
6a+b+2a-b=-8 \rightarrow 8a=-8 \rightarrow a=-1$
Thus $6(-1)+b=-5 \rightarrow b=1$
To find $[(4,28)]_{C}:$
$6c+d=4\\
2c-d=28\\
6c+d+2c-d=32 \rightarrow 8c=32 \rightarrow c=4$
Thus $6(4)+d=4 \rightarrow d=-20$
Hence we have: $[(-5,-3)]_C=(-1,1)\\
[(4,28]_C=(4,-20)\\ P_{C \leftarrow B}=\begin{bmatrix}
-1 & 4 \\
1 & -20
\end{bmatrix}$
