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Math Help - Inverses

  1. #1
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    Inverses

    Find b such that A inverse exists.

    BA + I = 2A

    BA = 2A-I
    BAA^-1 = A^-1(2A-I)
    B = (2A)A^-1 - A^-1(I)
    B = 2 - A^-1

    Is this correct?
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  2. #2
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    Quote Originally Posted by millerst View Post
    BA = 2A-I
    BAA^-1 = A^-1(2A-I)
    Multiplication is not commutative. You used "from the right" on the left and "from the left" on the right. No good.

    BAA^-1 = A^-1(2A-I)
    B = (2A)A^-1 - A^-1(I)
    You did it again! How did that A^(-1) get on the right of that '2A'?

    B = (2A)A^-1 - A^-1(I)
    B = 2 - A^-1
    Ack! I'm pretty sure you mean 2I, not just 2.

    Okay, that was some lovely algebra. Did you answer the question?
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  3. #3
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    So the solution is:

    B = 2I - A^-1
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  4. #4
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    No, the solution is a learning process. Did you learn anything? Did you correct your algebra errors along the way?

    No, the solution is an answer to the question that was asked. Did you answer it? You solved for 'B' by assuming that A^(-1) exists. Is that the same thing?
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  5. #5
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    Quote Originally Posted by millerst View Post
    So the solution is:

    B = 2I - A^-1
    The problem was to "Find B such that A inverse exists".

    Does an answer in terms of A inverse even make sense?

    In both attempts here you have started by assuming that A^{-1} exists.

    Instead, from BA + I = 2A, you have I= 2A- BA= (2I- B)A. Now what is the definition of A^{-1}?
    Last edited by HallsofIvy; February 11th 2010 at 02:58 AM.
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