- #1

Karate Chop

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- Thread starter Karate Chop
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- #1

Karate Chop

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- #2

marlon

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Karate Chop said:

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

marlon

- #3

HallsofIvy

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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]?

- #4

The Bob

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

The Bob (2004 ©)

- #5

HallsofIvy

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And that has what to do with the question?

- #6

marlon

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:rofl: :rofl: :rofl: :rofl: :rofl:HallsofIvy said:And that has what to do with the question?

answer : i dunno :rofl:

marlon

- #7

primarygun

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Sometimes it does. Such as the square of a matrix.E.g. A(BC) equals (AB)C but AB does not equal BA.

- #8

TD

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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.

- #9

HallsofIvy

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Wandering off topic is just going to confuse the original poster.

- #10

The Bob

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Apologises for misunderstanding the question. Glad to feel I can make mistakes and not have my gut knotted.HallsofIvy said:And that has what to do with the question?

The Bob (2004 ©)

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