# Thread: Solve for X: 2 x 2 matrices.

1. ## Solve for X: 2 x 2 matrices.

Hello everyone,

I am here asking for maths help, the question I have difficult with is question 2 , in my exam practice sheet. Here is the following:

Exam Revision Questions.jpg

Requirement:
Solve for X:

We know we need to use this method , which are two types:

Type 1: AX = B and Type 2: XA = B

But the issue is we must first do operations that will set it up to these two forms.

We know that we must remove the fraction because Matrices cant divided. I would assume we would times both sides by A to remove the fraction. But the problem is getting those two forms. With these operations.

Here is my working out: But i don't believe they're correct.

Workings out.jpg

So my problem is I cant get AX = B or XA = B. So I can solve for X, If you want anymore details please tell me and I hope posted this in the correct section as well , I apologize if I didn't.

Kindest Regards,

2. This is a very badly worded question because it is ambiguous. As you correctly say, matrices can't be divided. The reason for that is that the "quotient" $\frac CA$ should mean the product of C and the inverse of A. But that could be either $A^{-1}C$ or $CA^{-1}$. Those products are not the same, and so there are two possible solutions to the question.

If you take the equation (X+B)-\frac CA = 0 and multiply it on the left by A then you get $A(X+B) - C = 0$, leading to the solution $X = A^{-1}(C-AB) = A^{-1}C - B$. But if you multiply the original equation on the right by A then it becomes $(X+B)A - C = 0$, leading to the solution $X = (C-BA)A^{-1} = CA^{-1} - B$. Those two solutions are different, and they are both equally valid.

3. ## Thank you

well thank you for your respond Opalg.