# Let $A$ and $B$ be arbitrary $n \times n$ matrices. If $B$ is invertible, prove that $AB^{-1} = B^{-1}A$ if and only if $AB=BA$.

Let $$A$$ and $$B$$ be arbitrary $$n \times n$$ matrices. If $$B$$ is invertible, prove that $$AB^{−1} = B^{−1}A$$ if and only if $$AB=BA$$. Would appreciate a solution to this proof.

I'm confused whether my proof should be showing how $$AB^{-1} = B^{−1}A$$ or I should prove $$AB = BA$$ from $$AB^{−1} =B^{−1}A$$ or prove that $$AB^{−1} =B^{−1}A$$ from $$AB=BA$$.

And if so, should I use $$\Longrightarrow$$ sign for each new line or $$\Longleftrightarrow$$ (given that this sign means if and only if)?

• I remember seeing the same question recently, but I might be mistaken.. Aug 15, 2021 at 4:31

To prove ,$$AB^{-1}=B^{-1}A$$ iff $$AB=BA$$
$$AB^{-1}=B^{-1}A$$
$$\Leftrightarrow$$ $$BAB^{-1}B=BB^{-1}AB$$
$$\Leftrightarrow$$ $$BAI=IAB$$
$$\Leftrightarrow$$ $$BA=AB$$
$$\Leftrightarrow$$ $$AB=BA$$