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 February 20th, 2012, 11:28 PM #1 Newbie   Joined: Feb 2012 Posts: 2 Thanks: 0 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.
 February 21st, 2012, 12:51 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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?
 February 21st, 2012, 01:24 PM #3 Newbie   Joined: Feb 2012 Posts: 2 Thanks: 0 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!
February 21st, 2012, 03:00 PM   #4
Math Team

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From: North America, 42nd parallel

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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.

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