1. May 14, 2008

### cmcc3119

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!

2. May 14, 2008

### cristo

Staff Emeritus
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, $AB\neq BA$

3. May 14, 2008

### Defennder

I don't see anything wrong with that, assuming X is the same size as the other four matrices.

4. May 14, 2008

### cmcc3119

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. May 14, 2008

### cristo

Staff Emeritus
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?

6. May 14, 2008

### cmcc3119

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. May 14, 2008

### cmcc3119

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. May 14, 2008

### cristo

Staff Emeritus
You're welcome and, yes, I'll take a look.