Answer
See answer below
Work Step by Step
We know that: $a(9,2)+b(4,-3)=(5,-3)\\
c(2,1)+d(-3,1)=(-5,3)$
Thus $a(9,2)+b(4,-3)=(5,-3) \\
\rightarrow (9a+4b,2a-3b)=(5,-3)$
Thus $9a+4b=5 \\
2a-3b=-3\\
3(9a+4b)+4(2a-3b)=3.5+4.(-3) \rightarrow 35a=-3 \rightarrow a=-\frac{3}{35}\\
2(-\frac{3}{35})-3b=-3 \rightarrow b=-\frac{37}{35}$
Hence, we have $v_B=(-\frac{3}{35},-\frac{37}{35})$
Thus $c(2,1)+d(-3,1)=(-5,3) \\
\rightarrow (2c-3d,c+d)=(-5,3)$
Thus $2c-3d=-5 \\
c+d=3\\
2c-3d+3(c+d)=-5 +3.3 \rightarrow 5c=4 \rightarrow c=\frac{4}{5}\\
\frac{4}{5}+d=3 \rightarrow d=\frac{11}{5}$
Hence, we have $v_C=(\frac{4}{5},\frac{11}{5})$
From problem 17 we have: $P_{C \leftarrow B}=\begin{bmatrix}
3 & -1\\
-1 & -2
\end{bmatrix}$
Since $v_C=P_{C \leftarrow B}v_B \rightarrow \begin{bmatrix}
3 & -1\\
-1 & -2
\end{bmatrix}\begin{bmatrix}
-\frac{3}{35}\\
-\frac{37}{35}
\end{bmatrix}=\begin{bmatrix}
\frac{4}{5}\\
\frac{11}{5}
\end{bmatrix}$
