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Math Help - Matrix again

  1. #1
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    Matrix again

    1.) If X = P^{-1}AP and A^3 = I, prove that X^3 = I

    Since order of multiplication don't matter when you're multiplying three matrices and more ( I deduced this from the multiplication of matrices (AB)C=A(BC) ),
    X = IA
    X = A
    Multiplying A^2 on both sides,
    A^2X = A^3
    A^2X = I
    This is where I'm stuck..


    2.) For A = \begin{bmatrix} 2 & 1 & -1 \\-1 & 2 & 1 \\0 & 6 & 1 \end{bmatrix} and B = \begin{bmatrix} 4 & 7 & -3 \\-1 & -2 & 1 \\6 & 12 & -5 \end{bmatrix}
    calculate AB and hence solve the system of equations
    4a+7b-3c = -8
    -a-2b+c = 3
    6a+12b-5c = -15

    I got the identity matrix for AB.But I do not know how to use it to solve the system of equations, I had applied the identity matric to the system of equations so I got 4a=-8, -2b = 3 and -5c = -15, but this does not seem to be the way..

    Thank you for your time!
    Last edited by Tutu; March 18th 2013 at 02:32 AM.
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  2. #2
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    Re: Matrix again

     \displaystyle \begin{align*} X^3 &= \left( P^{-1}AP\right)^3 \\ &= P^{-1}APP^{-1}APP^{-1}AP \\ &= P^{-1}AIAIAP \\ &= P^{-1}AAAP \\ &= P^{-1}A^3P \\ &= P^{-1}IP \\ &= P^{-1}P \\ &= I \end{align*}
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  3. #3
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    Re: Matrix again

    Certainly didn't think of that, Thank you again!
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