$$\left[ \begin{array}{cc} 2 & -6&-4 \\ -2 & 7& 3 \\ 3&-8&-6 \end{array} \right]^{-1} +2 \cdot A= \left[ \begin{array}{cc} 7 & 2&5 \\ 6 & -10& -7 \\ 5&-8&-5 \end{array} \right]$$
I can seem to solve this question. I move the inverse matrix to the right of the equation and add both those matrices. Finally I multiply 1/2 to the final matrix for A but that doesn't seem to work
You have something like $B^{-1}+2A=C\Rightarrow A=1/2(C-B^{-1})$. Now the only problem is to compute $B^{-1}=1/2 \left(\begin{array}{ccc}18& -4 & 10\\ -3 & 0 & 2\\ -5 & -2 & 2\end{array}\right)$ You should look up inverting a matrix.