Answer
See answer below
Work Step by Step
We know that: $a_1\begin{bmatrix}
1 & 0\\
-1 & -2
\end{bmatrix}+a_2\begin{bmatrix}
0& -1\\
3& 0
\end{bmatrix}+a_3\begin{bmatrix}
3 & 5\\
0 & 0
\end{bmatrix}+a_4\begin{bmatrix}
-2 & -4\\
0 & 0
\end{bmatrix}=\begin{bmatrix}
-1 & -1\\
-4 & 5
\end{bmatrix}\\ b_1\begin{bmatrix}
1 & 1\\
1 & 1
\end{bmatrix}+b_2\begin{bmatrix}
1& 1\\
1& 0
\end{bmatrix}+a_3\begin{bmatrix}
1& 1\\
0 & 0
\end{bmatrix}+a_4\begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}=\begin{bmatrix}
-1 & -1\\
-4 & 5
\end{bmatrix}$
We obtain $ \begin{bmatrix}
1 & 0 & 3 & -2 | -1\\
0 & -1 & 5 &-4|-1\\
-1 & 3 & 0 & 0 | -4\\
-2 & 0 & 0& 0 | 5
\end{bmatrix} \approx \begin{bmatrix}
2 & 2 & 1 & 0 | -1\\
0 & -1 & 5 &-4|-1\\
-1 & 3 & 0 & 0 | -4\\
-2 & 0 & 0& 0 | 5
\end{bmatrix} \\
-2a_1=5 \rightarrow a_1=-\frac{5}{2}\\
-a_1+3a_2=-4 \rightarrow a_2=-\frac{13}{6}\\
2a_1+2a_2+a_3=-1 \rightarrow a_3=\frac{25}{3}\\
\rightarrow a_4=-\frac{17}{2}$
Hence, we have $v_B=(-\frac{5}{2},-\frac{13}{6},-\frac{25}{3},-\frac{17}{2})$
For C: $\begin{bmatrix}
1 & 1 & 1 & 1 |-1\\
1 & 1 & 1 & 0 | -1\\
1 & 1 & 0 & 0 | -4\\
1 & 0 & 0 & 0 | 5
\end{bmatrix}\\
b_1=5\\
b_1+b_2=-4 \rightarrow b_2=-9\\
b_1+b_2+b_3=-1 \rightarrow b_3=3\\
b1+b_2+b_3+b_4=-1 \rightarrow b_4=0$
Hence, we have $v_C=(5,-9,3,0)$
From problem 25 we have: $P_{C \leftarrow B}=\begin{bmatrix}
-2 & 0 & 0 & 0 \\
1 & 3 & 0 & 0\\
1 & -4& 5 & -4 \\
1 & 1 & -2 & 2
\end{bmatrix}$
Since $v_C=P_{C \leftarrow B}v_B \rightarrow \begin{bmatrix}
-2 & 0 & 0 & 0 \\
1 & 3 & 0 & 0\\
1 & -4& 5 & -4 \\
1 & 1 & -2 & 2
\end{bmatrix} \begin{bmatrix}
-\frac{5}{2}\\
\frac{-13}{6}\\
\frac{25}{3}\\
-\frac{17}{2}
\end{bmatrix}=\begin{bmatrix}
5\\
-9\\
3 \\
0
\end{bmatrix}=[P]_C$
