# Invertible matrix proof

1. Nov 3, 2008

### cmajor47

1. The problem statement, all variables and given/known data
Prove: If A and B are both nxn matrices, A is invertible and AB=BA, then A-1B=BA-1

2. Relevant equations
(AB)-1=B-1A-1

3. 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?

2. Nov 3, 2008

### 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...

3. Nov 3, 2008

### 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?

4. Nov 3, 2008

### Dick

It sure does.

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