I have a solution here that I don't quite understand.

Q: Show that if a square matrix A satisfies the equation $\displaystyle A^2+2A+I=0$ , then A must be invertible.

A:

$\displaystyle A^2+2A=-I$

$\displaystyle -A^2-2A=I$

I got this far on my own, but the next step confuses me...

$\displaystyle A(-A-2I)=I$

I get that if $\displaystyle A^-^1=(-A-2I)$ then $\displaystyle A*A^-^1=I$, however what I'm confused about is when they draw out the $\displaystyle A$ from $\displaystyle -2A$.

Could someone help me understand why $\displaystyle A/A=I$?

Thank you!