0
$\begingroup$

I need to set up a proof for this problem:

Given that $A$ and $B$ are both $n\times n$ matrices. $A$ is invertible, and $AB=BA$.

Prove that $A^{-1}B=BA^{-1}$.

I'm just unsure how to go about this particular proof.

3 Answers 3

5

Hint: Multiply both sides by $A^{-1}$ on the left, then on right:

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

3

You could try to start with the fact that $AB=BA$ and multiply one side by $A^{-1}$. After that, do the same with the other side. You should find what you wanted to prove.

Question for you: what does $AA^{-1}$ equals?

0

Since A is invertible we have that $AA^{-1}=I$.

We are told$ AB = BA$ Then if you multiply both sides by $A^{-1}$we have:

$A^{-1}AB = A^{-1}BA$ $IB = A^{-1}BA B=A^{-1}BA, BA^{-1}=A^{-1}BAA^{-1}, BA^{-1}=A^{-1}B(AA^{-1}),BA^{-1}=A^{-1}BI$ $BA^{-1}=A^{-1}B$