Answer
$(\frac{-34}{3},12,\frac{-55}{3},\frac{56}{3})$.
Work Step by Step
We know that $a\begin{bmatrix}
-1 & 1\\
0 & 1
\end{bmatrix}+b\begin{bmatrix}
1 & 3\\
-1 & 0
\end{bmatrix}+c\begin{bmatrix}
1 & 0\\
1 & 2
\end{bmatrix}+d\begin{bmatrix}
0 & -1\\
2 & 3
\end{bmatrix}=\begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}$.
Thus $\begin{bmatrix}
-a+b+c & a+3b-d\\
-b+c+2d & a+2c+3d
\end{bmatrix}=\begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}$
Thus $-a+b+c=5\\
a+3b-d=6\\
-b+c+2d=7\\
a+2c+3d=8$
Thus $-a+b+c+(a+3b-d)+4(-b+c+2d)=39 \rightarrow 5c+7d=39\\
-a+b+c+(-b+c+2d)+(a+3b-d)=20 \rightarrow 4c+5d=20\\$
Thus $4(5c+7d)-5(4c+5d)=56 \rightarrow 3d =56 \rightarrow d=\frac{56}{3}\\
4c+5\frac{56}{3}=20 \rightarrow c=\frac{-55}{3}\\
-b-\frac{55}{3}+2\frac{56}{3}=7 \rightarrow b=12\\
a+3.12-\frac{56}{3}=6 \rightarrow a=\frac{-34}{3}$
Thus the vector is $(\frac{-34}{3},12,\frac{-55}{3},\frac{56}{3})$.
