For the matrix $A=\begin{bmatrix}3 & 1\\7 & 5\end{bmatrix}, 8A^{-1}=$___________.

$(8I-A)$

The correct answer is Option (3) → $(8I-A)$

$A^2=\begin{bmatrix}16 & 8\\56 & 32\end{bmatrix}=8\begin{bmatrix}2&1\\7 & 4\end{bmatrix}$

so $A=\frac{1}{8}A^2+I$

multiplying eq by $A^{-1}$

so $AA^{-1}=\frac{1}{8}A^2A^{-1}+IA^{-1}$

so $I=\frac{1}{8}A+A^{-1}$

so $A^{-1}=8I-A$