Answer
$[(-7,4,4)]_C=(0,\frac{5}{3},\frac{-7}{3})\\
[(4,2,-1]_C=(3,\frac{1}{3},-\frac{2}{3})\\
[(8,-2,-9)]_C=(5,-4,-4)\\
P_{C \leftarrow B}=\begin{bmatrix}
0 & 3& 5 \\
\frac{5}{3} & -\frac{2}{3} & -4\\
-\frac{7}{3} & \frac{1}{3} & -4
\end{bmatrix}$
Work Step by Step
We know that:
$a_1(1,1,0)+a_2(0,1,1)+a_3(3,-1,-1)=(-7,4,4)$
$b_1(1,1,0)+b_2(0,1,1)+b_3(3,-1,-1)=(4,2,-1)$
$c_1(1,1,0)+c_2(0,1,1)+b_3(3,-1,-1)=(-7,5,0)$
To find $[(-7,4,4)]_C$: $a_1+3a_3=-7\\
a_1+a_2-a_3=4\\
a_2-a_3=4\\
\rightarrow a_1+4=4 \rightarrow a_1=0\\
0+3a_3=-7 \rightarrow a_3=-\frac{7}{3} \\
a_2-(-\frac{7}{3})=4 \rightarrow a_2=\frac{5}{3}$
To find $[(4,2,-1)]_{C}:$
$b_1+3b_3=4\\
b_1+b_2-b_3=2\\
b_2-b_3=-1\\
\rightarrow b_1-1=2 \rightarrow b_1=3\\
3+3b_3=4 \rightarrow b_3=\frac{1}{3} \\
b_2-\frac{1}{3}=-1 \rightarrow b_2=\frac{-2}{3}$
To find $[(8,-2,-9)]_{C}:$
$c_1+3c_3=-7\\
c_1+c_2-c_3=5\\
c_2-c_3=0\\
\rightarrow c_1+0=5 \rightarrow c_1=5\\
5+3c_3=-7 \rightarrow c_3=-4 \\
c_2-(-4)=0 \rightarrow c_2=-4$
Hence we have: $[(-7,4,4)]_C=(0,\frac{5}{3},\frac{-7}{3})\\
[(4,2,-1]_C=(3,\frac{1}{3},-\frac{2}{3})\\
[(8,-2,-9)]_C=(5,-4,-4)\\
P_{C \leftarrow B}=\begin{bmatrix}
0 & 3& 5 \\
\frac{5}{3} & -\frac{2}{3} & -4\\
-\frac{7}{3} & \frac{1}{3} & -4
\end{bmatrix}$
