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Math Help - Matrices proof

  1. #1
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    Matrices proof

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

    I did this
    X^{3} = (P^{-1}AP)(P^{-1}AP)(P^{-1}AP) = (P^{-3}A^{3}P^{3}) = (P^{-3+3}A^{3}) = A^{3} = I

    I used the law of indices, is it right to apply this for matrices?

    Please help me, I am not sure if I did it right here!

    Thank you!
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  2. #2
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    Re: Matrices proof

    Hey Tutu.

    Hint: Group the P*P^(-1) = I (I is identity matrix) together to get a lot of cancellation and then use A^3 = I to get your final result.
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  3. #3
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    Re: Matrices proof

    Thank you!
    I thought I could not change the order within the multiplication, but so is it that I can do it in thie case? Why?
    Is it
    X^3 = (P^(-1)PA)(P^(-1)PA)(P^(-1)PA)
    = P^(-3) (PA)^3
    = I
    Wait, I did this using laws of indices again, can you please explain how to do it, I'm afraid I do not understand..

    Thank you ever so much!
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  4. #4
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    Re: Matrices proof

    your proof is incorrect.

    X^3 = (P^{-1}AP)^3 = (P^{-1}AP)(P^{-1}AP)(P^{-1}AP)

    = P^{-1}A(PP^{-1})A(PP^{-1})AP = P^{-1}AIAIAP = P^{-1}AAAP = P^{-1}A^3P

    NOT P^{-3}A^3P^3.

    and since A^3 = I,

    P^{-1}A^3P = P^{-1}IP = P^{-1}P = I.

    no "switching around of order" is required.
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  5. #5
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    Re: Matrices proof

    Thank you!
    Does that mean that in matrices, PP^(-1) = P? I thought it equalled to I
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  6. #6
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    Re: Matrices proof

    Sorry, ignore the previous post. I get it now!
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