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Math Help - Matrices 2

  1. #1
    Member
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    Matrices 2

    Help! No matter how many times I do still can't get the right answer for C! Someone show the formula please?

    A, B and C are square matrices such that BA=B^{-1} and ABC=(AB)^{-1}. Show that A^{-1}=B^2=C. If B=\left(\begin{array}{ccc}1&2&0\\0&-1&0\\1&0&1\end{array}\right), find C and A.

    Answer:
    A=\left(\begin{array}{ccc}1&0&0\\0&1&0\\-2&-2&1\end{array}\right) C=\left(\begin{array}{ccc}1&0&0\\0&1&0\\2&2&1\end{  array}\right)
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  2. #2
    MHF Contributor
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    Hi

    Starting from BA=B^{-1}, multiply both sides by B on the left : B^2A=BB^{-1}=I_2 and then by A^{-1} on the right : B^2AA^{-1}=I_2A^{-1} \implies B^2=A^{-1}

    For the second equality, start from ABC=(AB)^{-1} = B^{-1}A^{-1} and do the same
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