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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)?

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    $\begingroup$ I remember seeing the same question recently, but I might be mistaken.. $\endgroup$
    – Kenta S
    Aug 15, 2021 at 4:31

1 Answer 1

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To prove ,$AB^{-1}=B^{-1}A$ iff $AB=BA$

$AB^{-1}=B^{-1}A$

$\Leftrightarrow$ $BAB^{-1}B=BB^{-1}AB$

$\Leftrightarrow$ $BAI=IAB$

$\Leftrightarrow$ $BA=AB$

$\Leftrightarrow$ $AB=BA$

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