Matirx help please

  • Thread starter cmcc3119
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  • #1
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[SOLVED] Matirx help please

Hey, I have a quiz tomorrow and I was hoping someone could explain how this answer was formed from a question on the practice quiz.

Suppose A, B, C, D are square matrices of the same size and that A and C are invertible. Given that AX + B = DC^-1 ( Sorry it's meant to say D* C inverse), express X in terms of the other matrices.

Solution: X = A^-1*DC^-1 - A^-1*B.

I am confused as my initial reaction was to just rearrange the given equation but sure enough this was too simple and did not give the same answer as you can't just separate A from X right?

Please can someone explain the steps they took in getting that answer!
 

Answers and Replies

  • #2
cristo
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Why did it not give the same answer when you rearranged the equation? What did you get? Remember that matrix multiplication is not commutative (in general); that is, [itex]AB\neq BA[/itex]
 
  • #3
Defennder
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I don't see anything wrong with that, assuming X is the same size as the other four matrices.
 
  • #4
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well I don't understand how they have attached an extra A^-1 to the B for the final part of the solution. Why wouldn't it just remain as -B?
 
  • #5
cristo
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well I don't understand how they have attached an extra A^-1 to the B for the final part of the solution. Why wouldn't it just remain as -B?

So you've got this [itex]AX+B=DC^{-1}[/itex], subtracting B from both sides yields [itex]AX=DC^{-1}-B[/itex] right? But this isn't the same as what you want; you have a multiplication of A on the LHS. What would you do next?
 
  • #6
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Hahahaha derr... So sorry yes I just realized the answer is really REALLY simple I was forgetting to multiply the whole RHS not just DC^-1 by a denominator of 1/A.

Thank you muchly for your help though :)
 
  • #7
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Cristo if you have a chance could you please look at the other post i made tonight as no one has replied yet and I am more concerned about not understanding that one...

Sorry to be a pain and I understand if you can't be bothered!

ciao
 
  • #8
cristo
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You're welcome and, yes, I'll take a look.
 

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