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

• Soupy11
In summary, the homework statement is that a and b are both nonsingular matrices and A^2B^2 = I, then (AB)^-1=BA. The Attempt at a Solution is (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? Yes, you need to use (AB)^{-1} = B^{-1} A^{-1} to solve for BA. Remember, you can multiply with new matrices on the left and right side as
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?

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.

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

AB(BA) = I which can be rewritten

A^2B^2 = I

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)

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

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

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?

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.

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?

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

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

## 1. How do you prove (AB)^-1 = BA when A^2B^2 = I?

To prove this equation, we can use the definition of matrix inverse, which states that (AB)^-1 = B^-1A^-1. We can rewrite the given equation as (AB)(AB) = I, and then take the inverse of both sides to get (AB)^-1 = (AB)^-1. From here, we can substitute B^-1A^-1 for (AB)^-1 to get B^-1A^-1 = BA. This shows that (AB)^-1 = BA when A^2B^2 = I.

## 2. What is the importance of proving (AB)^-1 = BA when A^2B^2 = I?

Proving this equation is important because it shows that the inverse of a product of matrices is equal to the product of the inverses in a specific case. This can be helpful in solving various problems in linear algebra and other areas of mathematics.

## 3. Can this equation be proven using any other method?

Yes, there are other methods that can be used to prove this equation. One way is to use the fact that (AB)^-1 = B^-1A^-1 to rewrite the equation as (AB)^-1 = (BA)^-1. Then, we can multiply both sides by AB to get AB(AB)^-1 = AB(BA)^-1. By the associative property of matrix multiplication, this becomes (ABA)B^-1 = (BAB)A^-1. Since A^2B^2 = I, we can substitute I for A^2B^2 in this equation to get AB = BA.

## 4. Are there any restrictions on the matrices A and B for this equation to hold true?

Yes, for this equation to hold true, A and B must be square matrices of the same size. Additionally, the product of A^2B^2 must equal the identity matrix, I.

## 5. How can this equation be applied in real-world situations?

This equation has applications in various fields, including engineering, physics, and computer science. For example, it can be used to solve systems of linear equations, perform transformations in 3D graphics, and in the design of control systems for robotics. It is also used in cryptography and coding theory.

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