Answer
$(-2,4,-3,-1)$
Work Step by Step
We know that $a\begin{bmatrix}
2 & -1\\
3 & 5
\end{bmatrix}+b\begin{bmatrix}
0 & 4\\
-1 & 1
\end{bmatrix}+c\begin{bmatrix}
1 & 1\\
1 & 1
\end{bmatrix}+d\begin{bmatrix}
3 & -1\\
2 & 5
\end{bmatrix}=\begin{bmatrix}
-10& 16\\
-15 & -14
\end{bmatrix}$.
Thus $\begin{bmatrix}
2a+c+3d & -a+4b+c-d\\
3a-b+c+d & 5a+b+c+5d
\end{bmatrix}=\begin{bmatrix}
-10 & 16\\
-15 & -14
\end{bmatrix}$
Thus $2a+c+3d=-10\\
-a+4b+c-d=16\\
3a-b+c+2d=-15\\
5a+b+c+5d=-14$
Thus $2(2a+c+3d)-(3a-b+c+2d+5a+b+c+5d)=2.(-10)-(-15-14) \rightarrow -4a-d=9\\
3(2a+c+3d)+(a+4b+c-d)-4(5a+b+c+5d)=3.(-10)+16-4.(-14) \rightarrow 5a+4d=-14\\$
Thus $4(-4a-d)-(5a+4d)= -22 \rightarrow 11a =-22 \rightarrow a=-2\\
5.(-1)+4d=-14 \rightarrow d=-1 \\
2(-2)+c+3(-1)=-10 \rightarrow c=-3\\
5(-2)+b+(-3)+5(-1)=-14 \rightarrow b=4$
Thus the vector is $(-2,4,-3,-1)$.
