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February 20th, 2012, 11:28 PM   #1
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Invertible matrices equation problem

I have this one question I do not understand, please explain it.

Suppose A and B are both 2x2 invertible matrices. Is there a 2x2 matrix X that satisfies the equation AB(3A + 2BX^T) B^-1 A^-1 = I(subscript 2)? If so, find X.


Thank You.
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February 21st, 2012, 12:51 AM   #2
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Re: Invertible matrices equation problem

yes, since A,B are invertible multiply on the left by B^-1A^-1

B^-1A^-1AB(3A + 2BX^T)B^-1A^-1 = B^-1A^-1(IDEN)

now multiply on the right by AB

(IDEN)(3A + 2BX^T)B^-1A^-1AB = B^-1A^-1AB

so you have

(3A + 2BX^T) = (IDEN)

lets use I for identity

2BX^T = I - 3A

multiply on the left by B^-1 noting the scaler 2 commutes...

2B^-1BX^T = B^-1(I - 3A)

X = (1/2)[B^-1(I - 3A)]^T

now, if you substitute this for X in the original, everything reduces to the identity, can you do that and let me know how it goes?
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February 21st, 2012, 01:24 PM   #3
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Re: Invertible matrices equation problem

Yes, I followed your steps.

Still, a couple questions arise from that. In the first step, we multiplied on left by B^-1 A^-1, how come after you completed the second step (multiplying by AB) the B^-1 A^-1 were no longer there?


Also, once we solve for X, you are saying that 1/2(B^-1 (I-3A))^T should be substituted into the original and then solved and that will be the final solution?


Thank you for your help!
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February 21st, 2012, 03:00 PM   #4
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Re: Invertible matrices equation problem

Quote:
Originally Posted by Jascha
Yes, I followed your steps.

Still, a couple questions arise from that. In the first step, we multiplied on left by B^-1 A^-1, how come after you completed the second step (multiplying by AB) the B^-1 A^-1 were no longer there?


Also, once we solve for X, you are saying that 1/2(B^-1 (I-3A))^T should be substituted into the original and then solved and that will be the final solution?


Thank you for your help!
You're welcome.

B^-1A^-1AB cancels from the inside out to give ( I ) by using the associative property like this...

B^-1A^-1AB = B^-1(A^-1A)B = B^-1( I )B = B^-1B = I

I used the associative property after the second line of my post and after the fourth line of my post and after the tenth line.

When you solve for X you can stop, that is the solution. However, back substituting and getting the left hand side of the equation to equal the right hand side is a good idea because it helps you get practice learning techniques necessary to solve the equation in the first place and it confirms if you got the right answer.

Try the back substituting, i'll help you if you get stuck.
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