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Math Help - Order of Matrix Multiplication

  1. #1
    Senior Member bugatti79's Avatar
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    Order of Matrix Multiplication

    Folks,

    What is the order of matrix multiplication for the following problem

    [K^e]=[T^e]^T[\overline K^e][T^e]

    where [\overline K^e] and [T^e] are 4*4 matrices.

    I would like to know the order before I tackle the problem..

    Thanks
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  2. #2
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    Re: Order of Matrix Multiplication

    Matrix multiplication is associative - it doesn't make a difference "when" you multiply two matricies together (although you have to keep the order straight since it's not commutative.).

    If A, B, and C are any matricies such that the dimensions make their product meaningful, then (AB)C = A(BC).
    That means:
    ABC can be calculated by first multiplying AB, then mulitplying the result on the right by C, or
    ABC can be calculated by first multiplying BC, then mulitplying the result on the left by A,
    and the result will be the same either way.
    i.e. (AB)C = A(BC).

    ("God does not care about our mathematical difficulties. He integrates empirically." - Einstein. That's a wonderful quote!)
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  3. #3
    Senior Member bugatti79's Avatar
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    Re: Order of Matrix Multiplication

    Quote Originally Posted by johnsomeone View Post
    Matrix multiplication is associative - it doesn't make a difference "when" you multiply two matricies together (although you have to keep the order straight since it's not commutative.).

    If A, B, and C are any matricies such that the dimensions make their product meaningful, then (AB)C = A(BC).
    That means:
    ABC can be calculated by first multiplying AB, then mulitplying the result on the right by C, or
    ABC can be calculated by first multiplying BC, then mulitplying the result on the left by A,
    and the result will be the same either way.
    i.e. (AB)C = A(BC).

    ("God does not care about our mathematical difficulties. He integrates empirically." - Einstein. That's a wonderful quote!)
    OK thanks.

    I am a little puzzled how they calculated

    [K^e]=\left [ \begin{matrix} \cos^2 \theta & 0.5 \sin 2\theta & - \cos^2 \theta & -0.5 \sin 2 \theta \\ 0.5sin 2 \theta & sin^2 \theta &-0.5 \sin 2 \theta & - \sin^2 \theta\\- \cos^2 \theta &-0.5 \sin 2 \theta & \cos^2 \theta & 0.5 \sin 2 \theta \\-0.5 \sin 2 \theta & - \sin^2 \theta & 0.5 \sin 2 \theta & \sin^2 \theta \end{matrix} \right] given

    [\overline K^e]=\left [ \begin{matrix} 1 & 0 & -1 & 0 \\ 0 & 0 &0 & 0\\-1&0 & 1 & 0 \\0 & 0 & 0 & 0 \end{matrix} \right] and  [T^e]=\left [ \begin{matrix} \cos \theta & \sin \theta & 0 & 0 \\ -\sin \theta & \cos \theta &0 & 0\\ 0 &0 & \cos \theta & \sin \theta \\0 & 0 & - \sin \theta & \cos \theta \end{matrix} \right]

    because if you look at the second row for  [\overline K^e] , they are all 0's so how can there be non zero terms in K^e...?
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  4. #4
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    Re: Order of Matrix Multiplication

    I just now multiplied the matricies out, and got the result you just posted. [K^e]=[T^e]^T[\overline K^e][T^e].

    (The frequent appearance there of 0.5\sin(2\theta) is just another way to write \cos(\theta)\sin(\theta).)
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  5. #5
    Senior Member bugatti79's Avatar
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    Re: Order of Matrix Multiplication

    Quote Originally Posted by bugatti79 View Post
    OK thanks.

    I am a little puzzled how they calculated

    [K^e]=\left [ \begin{matrix} \cos^2 \theta & 0.5 \sin 2\theta & - \cos^2 \theta & -0.5 \sin 2 \theta \\ 0.5sin 2 \theta & sin^2 \theta &-0.5 \sin 2 \theta & - \sin^2 \theta\\- \cos^2 \theta &-0.5 \sin 2 \theta & \cos^2 \theta & 0.5 \sin 2 \theta \\-0.5 \sin 2 \theta & - \sin^2 \theta & 0.5 \sin 2 \theta & \sin^2 \theta \end{matrix} \right] given

    [\overline K^e]=\left [ \begin{matrix} 1 & 0 & -1 & 0 \\ 0 & 0 &0 & 0\\-1&0 & 1 & 0 \\0 & 0 & 0 & 0 \end{matrix} \right] and  [T^e]=\left [ \begin{matrix} \cos \theta & \sin \theta & 0 & 0 \\ -\sin \theta & \cos \theta &0 & 0\\ 0 &0 & \cos \theta & \sin \theta \\0 & 0 & - \sin \theta & \cos \theta \end{matrix} \right]

    because if you look at the second row for  [\overline K^e] , they are all 0's so how can there be non zero terms in K^e...?
    Starting in steps I calculate  [\overline K^e] [T^e]=\left [ \begin{matrix} \cos \theta & \sin \theta & -\cos \theta & -\sin \theta \\ 0 & 0 &0 & 0\\- \cos \theta &\sin \theta & \cos \theta & -\sin \theta \\0 & 0 & 0 & 0 \end{matrix} \right]....?
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  6. #6
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    Re: Order of Matrix Multiplication

    Rather than my trying to explain anything, let me suggest that you just work it out.

    Multiply that on the left by [T^e]^t. All will become clear.
    Thanks from bugatti79
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