Is this proof for invertible matrices correct?

[cmajor47]

Homework Statement

Prove: If A and B are both nxn matrices, A is invertible and AB=BA, then A^-1B=BA^-1

Homework Equations

(AB)^-1=B^-1A^-1

The Attempt at a Solution

A^-1B=BA^-1
A^-1BB^-1=BA^-1B^-1
A^-1BB^-1=B(A^-1B^-1)
A^-1(BB^-1)=B(BA)^-1
B^-1A^-1(I)=B^-1B(BA)^-1
(AB)^-1=I(BA)^-1
(AB)^-1$$\neq$$A^-1B^-1

Therefore A^-1B does not equal BA^-1

I've been struggling with proofs and was wondering if this was correct?

 

[gabbagabbahey]

First, you aren't told that $B^{-1}$ exists, so any proof that relied on the use of $B^{-1}$ is out of the question.

Second, even if $B^{-1}$ did exist then $(AB)^{-1}=(BA)^{-1} \Rightarrow AB=BA$ which was already given in the question and so your attempt to find a contradiction has failed.

Third try starting with one of the conditions you're given: $AB=BA$ you're also told that $A^{-1}$ exists, so why not try multiplying both sides of the equation $AB=BA$ by $A^{-1}$ from either side and see what you get...

 

[cmajor47]

I took your advice and came up with this:

AB=BA
A^-1AB=A^-1BA
IB=A^-1BA
B=A^-1BA
BA^-1=A^-1BAA^-1
BA^-1=A^-1B(AA^-1)
BA^-1=A^-1BI
BA^-1=A^-1B

Read from right to left, this proves the statement. Is this right?

 

[Dick]

It sure does.