Answer
$[(2,-5,0)]_C=(4,-1,-1)\\
[(3,0,5]_C=(1,1,1)\\
[(8,-2,-9)]_C=(-2,5,-4)\\
P_{C \leftarrow B}=\begin{bmatrix}
4 &1 & -2 \\
-1 & 1 & 5\\
-1 & 1 & -4
\end{bmatrix}$
Work Step by Step
We know that:
For B: $a_1(1,-1,1)+a_2(2,0,1)+a_3(0,1,3)=(2,-5,0)$
For C: $b_1(1,-1,1)+b_2(2,0,1)+b_3(0,1,3)=(8,-2,-9)$
To find $[(-5,-3)]_C$: $a_1+2a_2=2\\
-a_1+a_3=-5\\
a_1+a_2+3a_3=0\\
a_1+2a_2+6(-a_1+a_3) -2(a_1+a_2+3a_3)=-28 \rightarrow -7a_1=-28 \rightarrow a_1=4$
Thus $4+2a_2=2 \rightarrow a_2=-1\\
-4+a_3=-5 \rightarrow a_3=-1$
To find $[(4,28)]_{C}:$
$b_1+2b_2=3\\
-b_1+b_3=-5\\
b_1+b_2+3b_3=0\\
b_1+2b_2+6(-b_1+b_3)-2(b_1+b_1+3b_3)=-7 \rightarrow -7b_1=-7 \rightarrow b_1=1$
Thus $1+2b_2=3 \rightarrow b_2=1\\
-1+b_3=0 \rightarrow b_3=1$
Hence we have: $[(2,-5,0)]_C=(4,-1,-1)\\
[(3,0,5]_C=(1,1,1)\\
[(8,-2,-9)]_C=(-2,5,-4)\\
P_{C \leftarrow B}=\begin{bmatrix}
4 &1 & -2 \\
-1 & 1 & 5\\
-1 & 1 & -4
\end{bmatrix}$
