# Thread: Linear Algebra Identity Understanding

1. ## Linear Algebra Identity Understanding

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!

2. ## Re: Linear Algebra Identity Understanding

Originally Posted by PodoTheGreat
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 $\displaystyle -2A$.
Do you agree that that $A(-A-I)=-A^2-A~?$
If so, then can you explain you difficulty?

3. ## Re: Linear Algebra Identity Understanding

Matrices obey the "distributive law": A(B+ C)= AB+ AC. That is what you are using when you factor.