Answer
$[-4+x-6x^2]_C=(-1,3,-3)\\
[6-2x]_C=(2,11,-6)\\
[-6-2x+4x^2]_C=(2,-1,-2) \\
P_{C \leftarrow B}=\begin{bmatrix}
-1& 11 & 2\\
3 & 5 & -1\\
-3 & -6 & -2
\end{bmatrix}$
Work Step by Step
We know that:
$a_1(1-x+3x^2)+a_2(2)+a_3(3+x^2)=-4+x-6x^2$
$b_1(1-x+3x^2)+b_2(2)+b_3(3+x^2)=-6-2x$
$c_1(1-x+3x^2)+c_2(2)+c_3(3+x^2)=-6-2x+x^2$
To find $[-4+x-6x^2]_C$: $a_1(1-x+3x^2)+a_2(2)+a_3(3+x^2)\\
=a_1-a_1x+3a_1x^2+2a_2+3a_3+a_3x^2\\
=(a_1+2a_2+3a_3)+x(-a_1)+x^2(3a_1+a_3)\\
=-4+x-6x^2$
Thus $a_1+2a_2=-4\\
-a_1=1 \rightarrow a_1=-1\\
3a_1+a_3=-6 \rightarrow -3+a_3=-6 \rightarrow a_3=-3\\
-1+2a_2+3(-3)=-4\rightarrow a_2=3$
To find $[-6-2x]_{C}:$
$b_1(1-x+3x^2)+b_2(2)+b_3(3+x^2)\\
=b_1-b_1+3b_1x^2+2b_2+3b_3+b_3x^2\\
=(b_1+2b_2+3b_3)+x(b_1)+x^2(3b_1+b_3)\\
=-6-2x$
Thus $b_1+2b_2+3b_3=-6\\
-b_1=-2 \rightarrow b_1=2\\
3b_1+b_3=0 \rightarrow b_3=-6 \\
2+2b_2+3(-6)=0 \rightarrow b_2=11$
To find $[-6-2x+4x^2]_{C}:$
$c_1(1-x+3x^2)+c_2(2)+c_3(3+x^2)\\
=c_1-c_1+3c_1x^2+2c_2+3c_3+c_3x^2\\
=(c_1+2c_2+3c_3)+x(c_1)+x^2(3c_1+c_3)\\
=-6-2x+4x^2$
Thus $c_1+2c_2+3c_3=-6\\
-c_1=-2 \rightarrow c_1=2\\
3c_1+c_3=4 \rightarrow c_3=-2 \\
2+2c_2-6=-6 \rightarrow c_2=-1$
Hence we have: $[-4+x-6x^2]_C=(-1,3,-3)\\
[6-2x]_C=(2,11,-6)\\
[-6-2x+4x^2]_C=(2,-1,-2) \\
P_{C \leftarrow B}=\begin{bmatrix}
-1& 11 & 2\\
3 & 5 & -1\\
-3 & -6 & -2
\end{bmatrix}$