Prove that A is invertible if $A^2 - 4A -7I = 0$. The $2 \times 2$ matrix $A$ satisfies  

$$A^2 - 4A -7I = 0,$$ 

where $I$ is the identity matrix. Prove that $A$ is invertible.

I'm not sure how to do this. Help would be appreciated.
Observe that
$$
\left(\frac17(A-4I)\right)A=I
$$
you can rewrite $A^2 - 4A - 7I = 0$ as $$A\left(\frac17A-\frac47I\right) = I.$$  therefore $A$ is invertible and $$A^{-1} = \frac17A-\frac47I.  $$