How Is X Expressed Using Inverse Matrices in the Equation AX + B = DC^-1?

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Homework Help Overview

The discussion revolves around expressing the variable X in the equation AX + B = DC^-1, where A, B, C, and D are square matrices. The context is linear algebra, specifically focusing on the manipulation of matrix equations and the properties of inverse matrices.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial attempt to rearrange the equation and question why this approach did not yield the expected result. There is a focus on the non-commutative property of matrix multiplication and its implications for the manipulation of the equation.

Discussion Status

The discussion is active, with participants seeking clarification on the steps involved in manipulating the equation. Some have expressed confusion regarding the introduction of the inverse matrix A^-1 in the solution, while others are exploring the implications of matrix multiplication rules. There is no explicit consensus, but participants are engaging with the problem and providing insights.

Contextual Notes

Participants are working under the assumption that A and C are invertible matrices, and there is a focus on the sizes of the matrices involved. The discussion also reflects a common challenge in understanding matrix operations and their properties.

cmcc3119
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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!
 
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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]
 
I don't see anything wrong with that, assuming X is the same size as the other four matrices.
 
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?
 
cmcc3119 said:
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?
 
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 :)
 
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
 
You're welcome and, yes, I'll take a look.
 

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