Suppose that matric $AB+A+B=0$, how to prove that $AB=BA$?

Question

Assume that $A$ and $B$ and $n\times n$ matrices, and $$ AB+A+B=0. $$
How can we prove that $AB=BA$?

Thank you in advance. Any help is much appreciated.

Answer 1

If $$ A+B+AB=0, \tag{1} $$ then $$ (I+A)(I+B)=I+A+B+AB=I, $$ and hence $I+A$ is the inverse of $I+B$, which implies that $$ I=(I+B)(I+A)=I+B+A+BA, $$ or $$ A+B+BA=0. \tag{2} $$ Combination of $(1)$ and $(2)$ provides that $$ AB=BA. $$

Answer 2

Hint:

$$0=AB+A+B=(A+I)(B+I)-I \implies (B+I)(A+I)=I\;\;\text{(why?)}$$

and we're almost done since

$$(B+I)(A+I)=I\implies BA+B+A=0\ldots$$