Proving (AB)^-1=BA When A^2B^2=I

[Soupy11]

Homework Statement

A and B are both nonsingular matrices and A^2B^2 = I, then (AB)^-1=BA

The Attempt at a Solution

(AA)(BB)=I

I also realize that (AB)^-1 = B^-1 A^-1 but am unsure on what connection to attack here.

Is the fact that (AB)^-1 = BA = B^-1 A^-1 an important connection?

 

[Inferior89]

Yes, you need to use
$$(AB)^{-1} = B^{-1} A^{-1}$$

Remember, you can multiply with new matrices on the left and right side as long as you do the same on the other side of the equal sign. Think about what matrices that would be suitable.

 

[Soupy11]

Ok so if (AB)^-1 = BA that implies that

AB(BA) = I which can be rewritten

A^2B^2 = I

 

[Inferior89]

Why would AB(BA) be A^2B^2? Matrix multiplication is not commutative.

I give you a small hint for a start.

AABB = I -->
A^(-1)AABB = A^(-1) -->
ABB = A^(-1)

 

[Soupy11]

Ok so if I do that for both I can also show that

AABB=I ->
AABB B^-1 = B^-1
AAB=B^-1

Which leads to

B^-1 A^-1 =(AB)^-1
-> AAB ABB = (AB)^-1

re arrange

(AB)^-1 = AABABB
(AB)^-1 = (A^2) (BA) (B^2)

Am i on the right track here?

 

[Inferior89]

Why do you start all over to multiply with B^(-1)?

AABB = I -->
A^(-1)AABB = A^(-1) -->
ABB = A^(-1)

Now do the same on this result with B as I did with A.
Don't start all over again with AABB.

 

[Soupy11]

ABB = A^-1

so if I multiply both sides by B^-1 won't I get A^-1B^-1

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

Which to me looks backwards on the Right Side - shouldn't it be B^-1 A^-1? Is it proper when multiplying both sides to keep the matrix left/right position the same for both operations?

 

[Inferior89]

So you get AB = (BA)^(-1)

Now if a matrix say C is the inverse of D then D is also inverse of C. Use this.

Edit: To clarify:

If
AB = (BA)^(-1) then
(AB)^(-1) = (BA)^(-1)^(-1)
and for any invertible matrix C, C^(-1)^(-1) = C. So:
(AB)^(-1) = BA

 

[Soupy11]

ok, listen, thanks so much I loved the way you got me through this problem...kudos.