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The exchange matrix

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Multiply the following matrices

    Multiply A * B * C

    A =
    0 0 1
    0 1 0
    1 0 0

    B =
    1 2 3
    4 5 6
    7 8 9

    C =
    0 0 1
    0 1 0
    1 0 0


    3. The attempt at a solution

    Why not just exchange row 1 with row 3 and then you get an identity matrix and the answer will be

    7 8 9
    4 5 6
    1 2 3?

    The book says the answer is

    9 8 7
    6 5 4
    3 2 1

    Essentially it looks like they're exchanging column 3 with 1.
     
  2. jcsd
  3. May 7, 2012 #2
    Wait a second, maybe my answer and the book's answer is the same. If so, I'd like to know.
     
  4. May 7, 2012 #3

    tiny-tim

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    hi robertjford80! :smile:
    do you mean A*C = I ?

    yes, but that doesn't help because A*B*C ≠ A*C*B​

    put A (or C) on the left, it exchanges row 1 with row 3

    put A (or C) on the right, it exchanges column 1 with column 3 :wink:
     
  5. May 7, 2012 #4
    I don't understand, TT, are you saying my strategy was illegal? And if so, why?
     
  6. May 7, 2012 #5

    tiny-tim

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    i'm not sure what your strategy was

    if i'm understanding it correctly, you've only calculated A*B :confused:
     
  7. May 7, 2012 #6
    Well in that case, why not exchange rows 1 and 3 for A and B, multiply, then exchange rows 1 and 3 again for B and C and multiply. In that case B would be the same in the beginning as in the end.
     
  8. May 7, 2012 #7

    tiny-tim

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    multiply what?

    (exchanging the rows or the columns was the multiplication)

    (and i said columns for C !)
     
  9. May 7, 2012 #8
    never mind, I give up.
     
  10. May 7, 2012 #9

    Mark44

    Staff: Mentor

    Multiplying B on the left (premultiplying) by A causes the 1st and 3rd rows of B to be swapped.

    Multiplying AB on the right (postmultiplying) by C (= A) causes the 1st and 3rd rows of AB to be swapped, taking you right back to B.

    The net result is that ABC = B.
     
  11. May 7, 2012 #10

    tiny-tim

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    no, columns :wink:
     
  12. May 7, 2012 #11

    Ray Vickson

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    No. Multiplying by A on the left swaps rows; multiplying by C on the right swaps columns.

    RGV
     
  13. May 7, 2012 #12

    Mark44

    Staff: Mentor

    I stand corrected.
     
  14. May 7, 2012 #13

    tiny-tim

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    here's an interesting way of doing it …

    number the rows and columns -1, 0, 1 (instead of 1, 2, 3) …

    then Ai,j = Ci,j = δi,-j, so (ABC)i,j = … :smile:
     
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