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3x3 Matrix Proof Help watch

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    (Original post by SimonM)
    I'm fairly sure that powers of some matrix are commutative
    I'm 70% sure it doesn't generally though.

    Oh and unless I'm missing something here couldn't you just do the first one by multiplying pre- and post- on both sides by the inverse of M?
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    (Original post by ukdragon37)
    Don't think so. A=M^2 B=M^3 the matrices are different so I think it just falls into the rule above as post- and pre- multiplying would yield different answers. Caution is the better part of prudence (something like that :shifty: )
    Matrix multiplication is associative, so if we use your examples for A and B, we get:
    M^2 * M^3 = (M*M) * (M*M*M) = (MM)MMM = MMMMM = MMM(MM) = M^3 * M^2
    Lo and behold, AB = BA.
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    (Original post by Brook Taylor)
    Matrix multiplication is associative, so if we use your examples for A and B, we get:
    M^2 * M^3 = (M*M) * (M*M*M) = (MM)MMM = MMMMM = MMM(MM) = M^3 * M^2
    Oh well that's me told :p:
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    (Original post by ukdragon37)
    I'm 70% sure it doesn't generally though.

    Oh and unless I'm missing something here couldn't you just do the first one by multiplying pre- and post- on both sides by the inverse of M?
    Generally \mathbf{A B} \not= \mathbf{B A}

    But \mathbf{A}^n \mathbf{A}^m = \mathbf{A}^{m+n} = \mathbf{A}^k \mathbf{A}^l (assuming m+n=k+l and m,n,l,k are positive integers)
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    what's this pre- and post- multiplying business?
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    (Original post by Guaglio)
    what's this pre- and post- multiplying business?
    Let A=B
    Pre-multiplying by M means multiply by M in front on both sides.
    MA=MB
    Post-multiplying by M means multiply by M at the end on both sides.
    AM=BM

    The order matters since matrices are not generally commutative, i.e. AB =/= BA.
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    (Original post by Guaglio)
    what's this pre- and post- multiplying business?
    OK.

    Since

    \mathbf{A B} \not = \mathbf{B A}

    It matters which side of an matrix which you multiply on. Therefore we say whether we are "pre-" or "post-" multiplying based on which side we are multiplying on
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    AHA!! I didn't know that was allowed...
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    btw the solution was pre- and post-multiplying both sides of the initial property given in the Q by  M^{-1} . If you were interested. I don't know if that was what you'd done.

    Thanks for the help anyway
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    (Original post by Guaglio)
    btw the solution was pre- and post-multiplying both sides of the initial property given in the Q by  M^{-1} . If you were interested. I don't know if that was what you'd done.

    Thanks for the help anyway
    (Original post by ukdragon37)
    Oh and unless I'm missing something here couldn't you just do the first one by multiplying pre- and post- on both sides by the inverse of M?
    At least I know I can do something right :p:
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    HAHAHA sorry I completely missed that! Thanks mate!
 
 
 
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