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Thread: 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 $\displaystyle (ABC)^{-1}=C^{-1}B^{-1}A^{-1}$

    I know that for two square matrices, C, D, say, $\displaystyle (CD)^{-1}=D^{-1}C^{-1}$. I tried using this result to prove $\displaystyle (ABC)^{-1}=C^{-1}B^{-1}A^{-1}$, by considering AB as one square matrix, and multiplying by C, and here is my problem.
    $\displaystyle (AB)^{-1}=B^{-1}A^{-1}$, 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 $\displaystyle A,B$, that $\displaystyle (AB)^{-1}=B^{-1}A^{-1}$
    So for three nonsingular matrices $\displaystyle (ABC)^{-1}=((AB)C)^{-1}=C^{-1}(AB)^{-1}=C^{-1}B^{-1}A^{-1}$
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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 $\displaystyle A,B$ are inverses if $\displaystyle AB=BA=I$. We claim that $\displaystyle (ABC)^{-1}=C^{-1}B^{-1}A^{-1}$. We can check it by computing $\displaystyle ABC(C^{-1}B^{-1}A^{-1})$ and $\displaystyle (C^{-1}B^{-1}A^{-1})ABC$. 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 $\displaystyle A,B$, that $\displaystyle (AB)^{-1}=B^{-1}A^{-1}$
    So for three nonsingular matrices $\displaystyle (ABC)^{-1}=((AB)C)^{-1}=C^{-1}(AB)^{-1}=C^{-1}B^{-1}A^{-1}$
    thank you very much, I wasn't sure how exactly to work it out from where I left.
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