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Math Help - Matrix Notation and Matrix Multiplication

  1. #1
    Newbie Erich's Avatar
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    Matrix Notation and Matrix Multiplication

    If A and B are n by n matrices with all entries equal to 1, find \ (ab)_{ij}. Summation notation turns the product AB, and the law (AB)C=A(BC), into
    \\ (ab)_{ij}= \displaystyle \sum_{k}\ a_{ik}b_{kj}\         \displaystyle \sum_{j}\left(\sum_{k}\ a_{ik}b_{kj}\right)\ c_{jl}\ =\displaystyle \sum_{k}\ a_{ik}\left(\sum_j\ b_{kj}c_{il}\right)
    Compute both sides if C is also n by n with every \ c_{jl}=2
    I haven't even done any work on this yet, because first I don't even understand what the question wants, and second I don't even know where to begin.
    Last edited by Erich; February 8th 2011 at 08:18 PM. Reason: Trying to make my problem clearer
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  2. #2
    A Plied Mathematician
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    Could you please insert some punctuation into your big equation there? It's impossible to parse as is.
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  3. #3
    Newbie Erich's Avatar
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    That's how it's written in my book. I am not really after a solution, more like am more interested in how I could start working this out. What I think that this question is basically asking, is for a proof of the associative property using the above formulas, but I don't know where to start.
    Last edited by Erich; February 8th 2011 at 10:05 PM.
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  4. #4
    A Plied Mathematician
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    It looks like you're asked to assume that A_{ij}=B_{ij}=1 for all i,j, and C_{ij}=2 for all i,j. Then you're asked to compute (AB)_{ij}, and show that ((AB)C)_{ij}=(A(BC))_{ij}.

    For the first, you have that

    \displaystyle (AB)_{ij}=\sum_{k=1}^{n}A_{ik}B_{kj}=\sum_{k=1}^{n  }(1\cdot 1)=n.

    For the second, plug into your formula:

    \displaystyle((AB)C)_{il}=\sum_{j}\left(\sum_{k}\ a_{ik}b_{kj}\right)\ c_{jl}\ \overset{?}{=}\sum_{k}\ a_{ik}\left(\sum_j\ b_{kj}c_{jl}\right)=(A(BC))_{il}, yielding the test

    \displaystyle((AB)C)_{il}=\sum_{j}\left(\sum_{k}1\  right)2\overset{?}{=}\sum_{k}1\left(\sum_j 2\right)=(A(BC))_{il}, which yields

    \displaystyle((AB)C)_{il}=\sum_{j}2n\overset{?}{=}  \sum_{k}1\cdot 2n=(A(BC))_{il}.

    Can you finish?
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  5. #5
    Newbie Erich's Avatar
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    If all I needed to do was compute both sides I believe you've already finished. I could probably even build a 2x2 that supports this. Thank you!!
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  6. #6
    A Plied Mathematician
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    You're welcome. Have a good one!
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  7. #7
    Newbie Erich's Avatar
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    Thought I would let you know Ackbeet that I found the solution to be 2n^2, which made so much more sense after looking at your work, and learned something new about summations and matrices. I would like to thank you once again for giving me a new perspective between proofs, associative law, and computations with summations and matrices.
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    A Plied Mathematician
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    You're welcome!
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