# Solve for X: 2 x 2 matrices.

• May 6th 2010, 10:47 PM
learner123
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.

http://img293.imageshack.us/i/examre...questions.jpg/ 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,

http://img293.imageshack.us/i/examre...questions.jpg/
• May 7th 2010, 05:02 AM
Opalg
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" $\displaystyle \frac CA$ should mean the product of C and the inverse of A. But that could be either $\displaystyle A^{-1}C$ or $\displaystyle CA^{-1}$. 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 $\displaystyle A(X+B) - C = 0$, leading to the solution $\displaystyle X = A^{-1}(C-AB) = A^{-1}C - B$. But if you multiply the original equation on the right by A then it becomes $\displaystyle (X+B)A - C = 0$, leading to the solution $\displaystyle X = (C-BA)A^{-1} = CA^{-1} - B$. Those two solutions are different, and they are both equally valid.
• May 7th 2010, 06:50 AM
learner123
Thank you
well thank you for your respond Opalg.