Answer
See answer below
Work Step by Step
We know that: $a(7-4x)+b(5x)=6-4x\\
c(1-2x)+d(2+x)=6-4x$
Thus $a(7-4x)+b(5x)\\
=7a-4ax+5bx
=7a-x(4a-5b)
=6-4x$
Thus $7a=6 \\
4a-5b=-4\\
\rightarrow a=\frac{6}{7}\\
4\frac{6}{7} -5b=-4 \rightarrow b=-\frac{4}{35}$
Hence, we have $v_B=(\frac{6}{7},-\frac{4}{35})$
Thus $ c(1-2x)+d(2+x)\\
=c-2cx+2d+dx\\
=(c+2d)+x(-2c+d)\\
=6-4x$
Thus $c+2d=6 \\
-2c+d=-4\\
2(c+2d)+(-2c+d)=4 \rightarrow 5d=8 \rightarrow d=\frac{8}{5}\\
c+2\frac{8}{5}=6 \rightarrow c=\frac{14}{5}$
Hence, we have $v_C=(\frac{14}{5},\frac{8}{5})$
From problem 21 we have: $P_{C \leftarrow B}=\begin{bmatrix}
3 & -2\\
2 & 1
\end{bmatrix}$
Since $v_C=P_{C \leftarrow B}v_B \rightarrow \begin{bmatrix}
3 & -2\\
2 & 1
\end{bmatrix}\begin{bmatrix}
\frac{6}{7}\\
\frac{-4}{35}
\end{bmatrix}=\begin{bmatrix}
\frac{19}{5}\\
\frac{8}{5}
\end{bmatrix}=[P]_C$
