Proving that matrix in equation is invertible
$A^2-4A-7I=0$ $\Longrightarrow$ $A^2-4A=7I$ $\Longrightarrow$ $A\cdot \frac17(A-4I)=I$. So $A$ is invertible. Its inverse is $\frac17(A-4I)$.
Proving that matrix in equation is invertible
$A^2-4A-7I=0$ $\Longrightarrow$ $A^2-4A=7I$ $\Longrightarrow$ $A\cdot \frac17(A-4I)=I$. So $A$ is invertible. Its inverse is $\frac17(A-4I)$.