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    Why is that in this question (part c) http://i.imgur.com/1o1uiPl.png
    BA=C is changed to A=CA^(-1)

    While in this question http://i.imgur.com/bH4ZwNV.png
    MC =A is changed to C=M^(-1)A (from the markschemes)

    When should the inverse thats being multiplied through be on the left or right.


    Matrix multiplication is not commutative so XY =/= YX so if you are to produce an identity matrix you must multiply the matrix by the inverse matrix on the side the matrix is on.

    In your first examle BA = C > BA(A^-1) = C(A^-1) > BI = C(A^-1); the reason it's on the right is because you post multiplied A in BA and thus have to post-multiply C.

    Same rule applies for the second problem.

    Because you're trying to get the A on its own in the first part - so you want to get rid of the B by turning it into the identity matrix, I, which is like the 'one' of matricesTo turn it into I you need to multiply by the inverse - which you could either do as
    BA(B^-1) =C(B^-1) (which would not work as BAB^-1 will not be equivalent to IA and therefore I)
    (B^-1)BA = (B^-1)C(which is correct as (B^-1)B = I and so IA=(B^-1)C and so A=(B^-1)C

    As you can see you can't stick the inverse term 'inside' of the two matrices which are already multiplied - so you can't do B(B^-1)A = B(^-1). Therefore you have to pick to put the inverse on one side or the other; the correct form is the one in which the inverse is next to the matrix (letter) you want to 'cancel'. If you put it on one side, you have to put it on the same side of the other half of the equation - for example, if you are doing MC=A to find C, you'd write it as
    (M^-1)MC =(M^-1)A so that (M^-1)M = I can be substituted in to get IC=(M^-1)A, which is the same as C=(M^-1)A
    NOT AS (M^-1)MC=A(M^-1) -----> as if you put the inverse on the left side of the MC on the left hand side, you MUST put it on the left side of A on the right hand side

    I hope that makes sense
    Think of it as dividing to term you want to get rid of by itself
    Like how A x (1/A) can be written as A x (A^-1) and the A's would cancel to give 1, or I in terms of matrices
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