Let $A,B$ be $2$ square matrices such that $A+AB-2B=0$. Prove $2$ isn’t an eigenvalue of $A$.

Question

Let $A,B$ be $2$ square matrices such that $A+AB-2B=0$. Prove $2$ isn't an eigenvalue of $A$.
I tried assuming the contrary, then there exists $v$ such that $Av=2v$. I multiply both sides of the given equation by $v$ and we get $Av+ABv-2Bv=0$ which implies $2v+ABv-2Bv=0$ and $ABv-2Bv=-2v$, $(A-2I)Bv=-2v$ which means $-2$ is an eigenvalue of $(A-2I)B$. Not sure how to continue.

Answer 1

$A+AB-2B=0$

$A+AB-2B-2I=-2I$

$(A-2I)(B+I)=-2I$

$\therefore A-2I$ is non-singular.

$\therefore$ for any $v \neq 0,$

$(A-2I)v \neq 0$

$Av \neq 2v$

$\therefore 2$ cannot be an eigenvalue.