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Regarding transpose of matrix products

 
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Feb6-13, 02:15 PM   #1
 

Regarding transpose of matrix products


Starting out a Lin Alg class - my prof wrote this on the board.

(ABC-1Dt)t = DC-1BtAt

On the right hand side, I get why D is D, why A and B are now both transpose, but why is C still inverse? I know the rule (D-1)t = (Dt)-1, but I do not see how the heck it applies here or what would make the original equality true.
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Feb6-13, 03:01 PM   #2
 
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It's incorrect, it should be

[tex](ABC^{-1}D^t)^t = D (C^t)^{-1} B^t A^t[/tex]

You should ask your prof for clarification.
Feb6-13, 03:59 PM   #3
 
Ugh, I KNEW it was incorrect. He wrote it on the board right before we took a test - he was answering questions - so despite knowing the write answer as you put it, I wrote the wrong thing on the exam. I would have asked, but out the tests came.

This might have something to do with his being an octogenarian. Half hour late to his office hours too and forgets who the heck you are. Son of a...

Thanks for your help. Also, how do you make the script look so nice for writing math?
Feb6-13, 04:06 PM   #4
 
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Regarding transpose of matrix products


Quote by MoreDrinks View Post
Also, how do you make the script look so nice for writing math?
http://www.physicsforums.com/showpos...17&postcount=3
Feb6-13, 04:15 PM   #5

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FWIW, some people write ##C^{-T}## instead of ##(C^{-1})^T## or ##(C^T)^{-1}##

Maybe you misread ##C^{-T}## as ##C^{-1}##.
Feb6-13, 10:07 PM   #6
 
Quote by AlephZero View Post
FWIW, some people write ##C^{-T}## instead of ##(C^{-1})^T## or ##(C^T)^{-1}##

Maybe you misread ##C^{-T}## as ##C^{-1}##.
Could be - wouldn't be in keeping with his writing style though. And it wouldn't be the first mistake we've dealt with.

Thanks a bunch!
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