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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.

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3 Answers 3

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

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

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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?

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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:

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

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If you write $AB=I$ then you simply have $B = A^{-1}$ so the rest of what you wrote (not sure what you were trying to do exactly) follows from that. In this problem though $B$ is not the inverse of $A$, indeed $B$ isn't even assumed to be invertible. –  EuYu Nov 6 '12 at 6:35
    
Going to edit now. –  diimension Nov 6 '12 at 6:36
    
@EuYu is my revised answer correct? –  diimension Nov 6 '12 at 6:51
    
Yes, that looks fine. –  EuYu Nov 6 '12 at 6:53
    
@EuYu Thank you! –  diimension Nov 6 '12 at 6:54

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