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Thread: Transpose of matraces

  1. #1
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    Transpose of matraces

    Hey guys, I understand matrices and their tranpose but I have this question which is just confusing... maybe it's because it's late but anyway it reads:

    Given the two vectors: A = [1, 2] B = [3, 4] calculate:

    A BT and B AT

    So I did this

    A = [1, 2]
    B =
    [3]
    [4]

    But I don't get what to do next... I can't multiple these can I?

    P.S I really must learn how to insert the proper mathematical code to format my questions correctly
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    Re: Transpose of matraces

    Quote Originally Posted by uperkurk View Post
    But I don't get what to do next... I can't multiple these can I?
    Why can't you? In both cases you have a 1x2 matrix times a 2x1 matrix. So your results should be 1x1 matrices.
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    Re: Transpose of matraces

    Nevermind I just worked it out. I was checking my results using a website that has a calculator for matraces. But it only went from 2x2 - 1x3 - 3x3 so I just assumed 1x2 and 1x1 did not exist...

    So I got this as my answer

    ABT I got 11
    BAT I got 11

    ATB I got 11
    BTA I got 11

    Which were the four questions I had to work out so for a conclusion I just wrote, not matter which way you switch them around you'll always get the same answer?
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    Re: Transpose of matraces

    Quote Originally Posted by uperkurk View Post
    ATB I got 11
    BTA I got 11
    Check these again. In this case you have a 2x1 times a 1x2. You should, therefore, get a 2x2 result. In general, when multiplying an $\displaystyle m\times p$ matrix with a $\displaystyle p\times n$ matrix, the result will be an $\displaystyle m\times n$ matrix.
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    Re: Transpose of matraces

    Oh... now I think I rushed through them too fast...

    For ATB I'm getting a 2x1 matrix of

    7
    14

    and for BTA I'm getting a matrix of

    9
    12

    is this now correct?
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    Re: Transpose of matraces

    $\displaystyle A^{\textrm T}B = \left[\begin{array}{c} 1 \\ 2\end{array}\right]\left[\begin{array}{cc}3 & 4\end{array}\right] = \left[\begin{array}{cc}1\times3 & 1\times4 \\ 2\times3 & 2\times4\end{array}\right] = \left[\begin{array}{cc}3 & 4\\6 & 8\end{array}\right]$
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    Re: Transpose of matraces

    What?! I thought when doing matraces you had to multiply and then add...

    1 x 3 + 1 x 4 = 7
    2 x 3 + 2 x 4 = 14

    Then what the hell have I been learning LOL
    Last edited by uperkurk; May 11th 2012 at 07:57 PM.
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    Re: Transpose of matraces

    Quote Originally Posted by uperkurk View Post
    What?! I thought when doing matraces you had to multiple and then add...

    1 x 3 + 1 x 4 = 7
    2 x 3 + 2 x 4 = 14

    Then what the hell have I been learning LOL
    If you multiply two matrices $\displaystyle \mathbf A$ and $\displaystyle \mathbf B$, each entry $\displaystyle \left[\mathbf{AB}\right]_{i,j}$ of the resulting product is found by multiplying (taking the dot product of) the $\displaystyle i^\mathrm{th}$ row of $\displaystyle \mathbf A$ by the $\displaystyle j^\mathrm{th}$ column of $\displaystyle \mathbf B$.

    We have a 2x1 matrix times a 1x2 matrix, so we should expect our answer to be 2x2:

    $\displaystyle \left[\begin{array}{cc}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$.

    So, using the above definition of matrix multiplication, $\displaystyle a_{11}$ is given by the dot product of the first row of $\displaystyle \mathbf A$ and the first column of $\displaystyle \mathbf B$: $\displaystyle a_{11} = 1\cdot 3 = 3$. And so on.

    Since the rows of $\displaystyle \mathbf A$ and the columns of $\displaystyle \mathbf B$ have only one entry each, there is no addition to be done.
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    Re: Transpose of matraces

    ok so I see what you did there, no addition in those cases

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