# Given that $A$ is a square matrix such that $A^2 -4A -3I =0$ how do I find $(A+2I)^{-1}$

The way I tried to solve this is to find out how much $(A+2I)$ equals then find it's inverse so:

$$A^2 -4A -3I == A^2 -3A-A -2I-I=0$$

$$A^2 -3A-I = (A+2I)$$ Do I simply just inverte the left equation to get the inverse?

Try long division: $$0=A^2-4A-3I=(A+2I)(A-6I)+9I.$$ That implies $$(A+2I)(6I-A)=9I.$$
The question is really about $A+2I$ rather than about $A$. So you may write $B=A+2I$, and rewrite the relation $A^2-4A-3I=0$ in terms of $B$ by substituting $A=B-2I$. That gives after expansion and simplification $B^2-8B+9I=0$. Since the constant term is nonzero, you can find the inverse of $B$ from this after writing it as $B(B-8I)=-9I$ and dividing by $-9$. Finally rewrite $B=A+2I$.
In the end this is a bit more work than directly doing long division of the polynomial $X^2-4X-3I$ by $X+2I$, giving a nonzero remainder, and then setting $X=A$ for an expression that allows directly solving $(A+2I)^{-1}$ in terms of $A$. But simplifying the question when possible is usually a fairly sure first step to more easily find a solution.