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

• Apr 18th 2012, 01:21 AM
E77
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

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.
• Apr 18th 2012, 06:50 AM
HallsofIvy
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 \$\displaystyle A^{-1}BCD^{-1}A= I\$
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
\$\displaystyle AA^{-1}BCD^{-1}A= AI\$
\$\displaystyle BCD^{-1}A= A\$

Now eliminate the "A" on the right of the left side by multiplying both sides by \$\displaystyle A^{-1}\$ on the right
\$\displaystyle BCD^{-1}AA^{-1}= AA^{-1}\$
\$\displaystyle BCD^{-1}= I\$

Can you continue from here?
• Apr 18th 2012, 02:55 PM
E77
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 \$\displaystyle A^{-1}BCD^{-1}A= I\$
Okay, just as you did before, eliminate the \$\displaystyle A^{-1}\$ on the left of the left side by multiplying both sides by A on the left
\$\displaystyle AA^{-1}BCD^{-1}A= AI\$
\$\displaystyle BCD^{-1}A= A\$

Now eliminate the "A" on the right of the left side by multiplying both sides by \$\displaystyle A^{-1}\$ on the right
\$\displaystyle BCD^{-1}AA^{-1}= AA^{-1}\$
\$\displaystyle BCD^{-1}= I\$

Can you continue from here?

Thank you for your reply. I'll have a go at it now and get back to you.
• Apr 18th 2012, 03:14 PM
E77
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?