Quick Matrices Help, if A⁻¹ B C D⁻¹ A = I, then C = ?

Hi guys,

My first post!

Just revising these matrices for an exam tomorrow. I can solve easy ones, like:

If A⁻¹ B C D = I, then C = ?

A A⁻¹ B C D = AI

BCD = A

B⁻¹BCD = B⁻¹A

CD=B⁻¹A

CDD⁻¹=B⁻¹AD⁻¹

C=B⁻¹ A D⁻¹

But, I don't think I am correctly solving the more complicated ones like:

A⁻¹ B C D⁻¹ A = I, then C = ?

I solved it as C = B⁻¹ A D A⁻¹

I don't think this is correct.

My textbook neglects these and my lecture notes have no answers.

I appreciate all help and will monitor the thread over the coming hours and post working etc if asked :)..

Thank you.

Re: Quick Matrices Help, if A⁻¹ B C D⁻¹ A = I, then C = ?

I don't understand why you can do one and not the other (which is not really "more complicated").

First, do you understand that matrix multiplication is NOT commutative? That is, you need to be careful about on which side you multiply.

You have

Okay, just as you did before, eliminate the "[tex]A^{-1}[/itex] on the left of the left side by multiplying both sides by A **on the left**

Now eliminate the "A" on the right of the left side by multiplying both sides by on the right

Can you continue from here?

Re: Quick Matrices Help, if A⁻¹ B C D⁻¹ A = I, then C = ?

Quote:

Originally Posted by

**HallsofIvy** I don't understand why you can do one and not the other (which is not really "more complicated").

First, do you understand that matrix multiplication is NOT commutative? That is, you need to be careful about on which side you multiply.

You have

Okay, just as you did before, eliminate the

on the left of the left side by multiplying both sides by A

**on the left**
Now eliminate the "A" on the right of the left side by multiplying both sides by

on the right

Can you continue from here?

Thank you for your reply. I'll have a go at it now and get back to you.

Re: Quick Matrices Help, if A⁻¹ B C D⁻¹ A = I, then C = ?

Quote:

Originally Posted by

**HallsofIvy**

First, do you understand that matrix multiplication is NOT commutative? That is, you need to be careful about on which side you multiply.

I did know this, but I wasn't applying it to the working... hopefully I have done it right below:

A⁻¹BCD⁻¹A=I

AA⁻¹BCD⁻¹A=AI

BCD⁻¹A=A

BCD⁻¹AA⁻¹=AA⁻¹

BCD⁻¹ = I

B⁻¹BCD⁻¹ = B⁻¹I

CD⁻¹D = B⁻¹ID

C = B⁻¹ID

I might have gone wrong, but I left the identity I there because B⁻¹I =/= B (unlike BI = B). Is this correct?