I was doing the problem $$ A+B=AB\implies AB=BA. $$
$AB=BA$ means they're invertible, but I can't figure out how to show that $A+B=AB$ implies invertibility.
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Consider the expression $$(A-\mathbb 1)(B-\mathbb 1)=AB-A-B+\mathbb 1=\mathbb 1$$
Thus $(A-\mathbb 1)$ and $(B-\mathbb 1)$ are inverse to each other, whence $$\mathbb 1= (B-\mathbb 1)(A-\mathbb 1)=BA - A - B + \mathbb 1$$
It follows that $$BA=A+B=AB$$ and we are done.
Note: here $\mathbb 1$ denotes the appropriate identity matrix.
Start with equation: $$A+B=AB$$ replace $B$ with $(B - I) + I$ in left side $$A+(B - I) + I =AB$$ $$I =AB - A - (B - I)$$ $$I =A(B - I) - (B - I)$$ $$I =(A - I)(B - I)$$ the inverse matrix of $(B - I)$ is $(A - I)$, so $(B - I)$ is invertible. Then the rest follows as in previous answer by @lulu