Prove that matrix is invertible

Hi.

I need to prove that if $\displaystyle A^2+5A+6I=0$ then A is invertibale.

so im not sure if thats valid, but i think i can write it like that:

$\displaystyle (A+2I)(A+3I)=0$

but it still doesn't prove that $\displaystyle A+2I=0$ or $\displaystyle A+3I=0$, since they're bote can be zero divisors and therefore don't necessarily equal 0.

so im kinda lost here.

is there another way?

TIA!

Re: Prove that matrix is invertible

from A^{2} + 5A + 6I = 0 we have:

6I = -A^{2} - 5A

I = (1/6)(-A - 5I)A

so evidently, A^{-1} = (-1/6)(A + 5I)

(not only is A invertible, we actually found the inverse!)