Answer
See answer below
Work Step by Step
We know that: $a_1(-4+x-6x^2)+a_2(6+2x^2)+a_3(-6-2x+4x^2)=5-x+3x^2\\
b_1(1-x+3x^2)+b_2(2)+b_2(3+x^2)=5-x+3x^2$
We have $a_1(-4+x-6x^2)+a_2(6+2x^2)+a_3(-6-2x+4x^2)\\
=-4a_1+a_1x-6a_1x^2+6a_2+2a_2x^2-6a_3-2a_3x+4a_3x^2\\
=(-4a_1+6a_2-6a_3)+x(a_1-2a_3)+x^2(-6a_1+2a_2+4a_3)\\
=5-x+3x^2$
We obtain $ \begin{bmatrix}
-4 & 6 & -6 | 5\\
1 & 0 & -2|-1\\
-6 & 2 & 4 | 3
\end{bmatrix} \approx \begin{bmatrix}
-4 & 6 & -6 | 5\\
0 & 6 & -14|1\\
0 & 2 & -8 | -3
\end{bmatrix} \approx \begin{bmatrix}
-4 & 6 & -6 | 5\\
0 & 6 & -14|1\\
0 & 0 & 10 | 10
\end{bmatrix} \\
\rightarrow 10a_3=10 \rightarrow a_3=1\\
6a_2-14.1=1 \rightarrow a_2=\frac{5}{2}\\
-4a_1+6\frac{5}{2}-6.1=5 \rightarrow a_1=1$
Hence, we have $v_B=(1,\frac{5}{2},1)$
Thus $ b_1(1-x+3x^2)+b_2(2)+b_3(3+x^2)\\
=b_1-b_1x+3b_1x^2+2b_2+3b_3+b_3x^2 \\
=(b_1+2b_2+3b_3)+x(-b_1)+x^2(3b_1+b_3)
=5-x+3x^2$
Thus $b_1+2b_2+3b_3=5 \\
-b_1=-1 \rightarrow b_1=1\\
3b_1+b_3=3 \rightarrow 3.1+b_3=3 \rightarrow b_3=0\\
1+2b_2+3.0=5 \rightarrow b_2=2$
Hence, we have $v_C=(1,2,0)$
From problem 22 we have: $P_{C \leftarrow B}=\begin{bmatrix}
-1 & 0 & 2 \\
3 & 0 & -1\\
-3 & 2 & -2
\end{bmatrix}$
Since $v_C=P_{C \leftarrow B}v_B \rightarrow \begin{bmatrix}
-1 & 0 & 2 \\
3 & 0 & -1\\
-3 & 2 & -2
\end{bmatrix}\begin{bmatrix}
1\\
\frac{5}{2}\\
1
\end{bmatrix}=\begin{bmatrix}
1\\
2\\
0
\end{bmatrix}=[P]_C$
