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Thread: Solve for X: 2 x 2 matrices.

  1. #1
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    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,



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  2. #2
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    Opalg's Avatar
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    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.
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  3. #3
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    Thank you

    well thank you for your respond Opalg.
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