If $A$ is a square matrix that satisfies $A^2-A+2I=0$, show that $A+I$ is invertible

Question

If $A$ is a square matrix that satisfies $A^2-A+2I=0$, show that $A+I$ is invertible. I understand how to find if $A$ is invertible but I don't know how to solve for the $A+I$ version.

Answer 1

$(A+I)(A-2I)=A^2-2A+A-2I=A^2-A-2I=-4I$

Answer 2

name $$ B = A + I \; , \; $$ $$ A = B - I $$

given $$ A^2 - A + 2I = 0 $$ $$ (B-I)^2 - (B-I) + 2 I = 0 $$ $$ B^2 - 2 B + I - B + I + 2I = 0 $$ $$ B^2 - 3 B + 4 I = 0 $$

Answer 3

Write $B=A+I$, then $A=B-I$ and $$0=(B-I)^2-(B-I)+2I=B^2-3B+4I.$$ From this, is it apparent that $B$ is invertible?