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Math Help - Proof for inverse matrices

  1. #1
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    Proof for inverse matrices

    Prove that for three non-singular matrices A, B, C that (ABC)^{-1}=C^{-1}B^{-1}A^{-1}

    I know that for two square matrices, C, D, say, (CD)^{-1}=D^{-1}C^{-1}. I tried using this result to prove (ABC)^{-1}=C^{-1}B^{-1}A^{-1}, by considering AB as one square matrix, and multiplying by C, and here is my problem.
    (AB)^{-1}=B^{-1}A^{-1}, I don't know how to continue after this. Or am I off the right track?

    Thanks!
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  2. #2
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    You're on the right track. Since we know that for all nonsingular matrices A,B, that (AB)^{-1}=B^{-1}A^{-1}
    So for three nonsingular matrices (ABC)^{-1}=((AB)C)^{-1}=C^{-1}(AB)^{-1}=C^{-1}B^{-1}A^{-1}
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  3. #3
    Senior Member roninpro's Avatar
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    You may also verify the rule directly.

    Recall that two matrices A,B are inverses if AB=BA=I. We claim that (ABC)^{-1}=C^{-1}B^{-1}A^{-1}. We can check it by computing ABC(C^{-1}B^{-1}A^{-1}) and (C^{-1}B^{-1}A^{-1})ABC. Everything should work out nicely.
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  4. #4
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    Quote Originally Posted by MattMan View Post
    You're on the right track. Since we know that for all nonsingular matrices A,B, that (AB)^{-1}=B^{-1}A^{-1}
    So for three nonsingular matrices (ABC)^{-1}=((AB)C)^{-1}=C^{-1}(AB)^{-1}=C^{-1}B^{-1}A^{-1}
    thank you very much, I wasn't sure how exactly to work it out from where I left.
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