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

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To express X in the equation AX + B = DC^-1, the solution is X = A^-1*DC^-1 - A^-1*B, which involves applying the inverse of matrix A. The confusion arises because simply rearranging the equation does not account for the properties of matrix multiplication, particularly that it is not commutative. Subtracting B from both sides leads to AX = DC^-1 - B, but the multiplication by A on the left side requires the application of A^-1 to both sides to isolate X correctly. The final answer incorporates A^-1 to maintain the equality, clarifying why it cannot simply remain as -B. Understanding these steps is crucial for solving similar matrix equations correctly.
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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, AB\neq BA
 
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 AX+B=DC^{-1}, subtracting B from both sides yields AX=DC^{-1}-B 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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