Given as such: $4(A^T+2I)^{-1} =$ \begin{bmatrix}1&1\\-3/2&1/2\end{bmatrix}
Now to find Matrix A, would I have to inverse the matrix on the RHS to make it equal to A on the LHS and perform the opposite operations? A little stumped on the operations necessary to get the right result.
Just need some clarification, thanks.
Hints: Divide both sides by $4$ to get $$(A^T+2I)^{-1}=\left[\begin{array}{cc} 1/4 & 1/4 \\ -3/8 & 1/8 \\ \end{array}\right]$$ Then take the inverse of both sides: $$A^T+2I=\left[\begin{array}{cc} 1/4 & 1/4 \\ -3/8 & 1/8 \\ \end{array}\right]^{-1}.$$ I'll leave it to you to find the inverse. Then subtract $2I$ from both sides: $$A^T=\left[\begin{array}{cc} 1/4 & 1/4 \\ -3/8 & 1/8 \\ \end{array}\right]^{-1}-2I$$ Lastly, take the transpose of both sides, and note that $(A^T)^T=A$.