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Matrices question

  1. Aug 6, 2005 #1
    when you're multiplying two matrices together does it affect the answer at all if you swap the columns around in one of the matrices?
  2. jcsd
  3. Aug 6, 2005 #2
    yes it does, just look at the defnition of matrix multiplication :

    [tex](AB)_{ij} = \Sigma_k a_{ik}b_{kj}[/tex]

    try for yourself with an example

    Here is some extra info : Matrix Multiplication
  4. Aug 6, 2005 #3


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    Did you consider doing a little experimentation?

    What is [tex]\left( \begin{array}{ccc}1&2\\3&2\end{array}\right)\left(\begin{array}{ccc}2&-1\\2&1\end{array}\right)[/tex]?

    Is that the same as [tex]\left( \begin{array}{ccc}1&2\\3&2\end{array}\right)\left(\begin{array}{ccc}-1&2\\1&2\end{array}\right)[/tex]?
  5. Aug 6, 2005 #4
    Matrix multiplication is associative but not commutative.

    E.g. A(BC) equals (AB)C but AB does not equal BA.

    The Bob (2004 ©)
  6. Aug 6, 2005 #5


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    And that has what to do with the question?
  7. Aug 7, 2005 #6
    :rofl: :rofl: :rofl: :rofl: :rofl:

    answer : i dunno :rofl:
  8. Aug 7, 2005 #7
    Sometimes it does. Such as the square of a matrix.
  9. Aug 7, 2005 #8


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    When we say matrix multiplication isn't commutative, we mean in general.
    To be commutative it always has to be valid.

    As you say though, there are exceptions (multiplying with the identy-matrix or with the inverse too for example) but that doesn't change the fact the multiplication isn't commutative.
  10. Aug 7, 2005 #9


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    MY point was that the original question had to do with rearranging the columns in one of the matrices- giving a completely different matrix. It had nothing to do with the commutativity of multiplication.

    Wandering off topic is just going to confuse the original poster.
  11. Aug 7, 2005 #10
    Apologises for misunderstanding the question. Glad to feel I can make mistakes and not have my gut knotted. :cry:

    The Bob (2004 ©)
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