Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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

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

share|cite|improve this answer

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?

share|cite|improve this answer

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:

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.