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Prove: A(BC)=(AB)C for matrix multiplication

  1. Jul 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove the following theorem:
    A(BC)=(AB)C.


    2. Relevant equations
    http://en.wikipedia.org/wiki/Matrix_multiplication#Ordinary_matrix_product"


    3. The attempt at a solution
    Let A be of order m by n, B be of order n by p, and C be of order p by q. Then
    [tex](A(BC))_{ij}=\sum^{n}_{k=1} A_{ik}(BC)_{kj}=\sum^{n}_{k=1} (A_{ik}(\sum^{p}_{r=1} B_{kr}C_{rj}))[/tex]
    After this, I'm stuck. How do I put the summations together, and how should I change the indices of summation?

    Thanks!
     
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Jul 27, 2008 #2

    tiny-tim

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    Hi adartsesirhc! :smile:

    (nice LaTeX, btw!)

    Hint: You can move the ∑s as far to the left as you like. :smile:
     
  4. Jul 27, 2008 #3
    [tex](A(BC))_{ij}=\sum^{n}_{k=1} A_{ik}(BC)_{kj}=\sum^{n}_{k=1} (A_{ik}(\sum^{p}_{r=1} B_{kr}C_{rj}))=\sum^{n}_{k=1} \sum^{p}_{r=1} A_{ik}B_{kr}C_{rj}=\sum^{p}_{r=1} (\sum^{n}_{k=1} A_{ik}B_{kr})C_{rj}=\sum^{p}_{r=1} (AB)_{ir}C_{rj}=((AB)C)_{ij}[/tex].

    How's that?
     
  5. Jul 27, 2008 #4

    tiny-tim

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    Perfect! :smile:
    :biggrin: Woohoo! :biggrin:
     
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