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Math Help - matrix multiplication,

  1. #1
    Junior Member
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    matrix multiplication,

    Hi,

    I have a matrix that looks like the following.
    <br />
A = \left[\begin{matrix} 3 & 9 \\ 5 & 2 \\ 1 & 4 \end{matrix}\right]<br />

    by doing the following, I can add the two columns.

    <br />
\left[\begin{matrix} 3 & 9 \\ 5 & 2 \\ 1 & 4 \end{matrix}\right] * \left[\begin{matrix} 1 \\ 1 \end{matrix}\right] = \left[\begin{matrix} 12 \\ 7 \\ 5 \end{matrix}\right]<br />

    My question is:

    Is there any clever way to multiply columns in the matrix.

    <br />
\left[\begin{matrix} 3 & 9 \\ 5 & 2 \\ 1 & 4 \end{matrix}\right] * X = \left[\begin{matrix} 18 \\ 10 \\ 4 \end{matrix}\right]<br />

    Can X be a matrix of any kind?

    Regards
    Craig.
    Last edited by craigmain; September 7th 2010 at 02:29 AM.
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  2. #2
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    X must be a 2 x 1 matrix. You could let it have elements a and b, and try solving simultaneously.
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  3. #3
    MHF Contributor
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    You need to take into account the dimensions of the matrix.

    You know that for matrices A and B, the number of columns in A has to equal the number of rows in B to be able to multiply them. And if A has dimensions m \times n and B has dimensions n \times p, then AB has dimensions  m \times p.


    So you have a 3 \times 2 multiplied by X to give 3 \times 1.

    That means X has dimensions 2 \times 1.

    There are two ways to solve this.

    The first is to write X as \left[\begin{matrix}x\\y\end{matrix}\right] so that your equation is

    \left[\begin{matrix}3 & 9\\ 5 & 1\\ 1 & 4\end{matrix}\right] \left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}18\\10\\4\end{matrix}\right].

    Perform the multiplication on the LHS to get

    3x + 9y = 18
    5x + y = 10
    x + 4y = 4.

    Solve for x and y.


    The other way is to create a "pseudo-inverse" matrix.

    If you have an equation

    A\mathbf{x} = B

    only square matrices have inverses, so to create a square matrix on the LHS, premultiply both sides by A^T.


    A^TA\mathbf{x} = A^TB.

    Since A^TA is square, you can multiply both sides by its inverse (as long as it's invertible).


    A^TA\mathbf{x} = A^TB

    (A^TA)^{-1}A^TA\mathbf{x} = (A^TA)^{-1}A^TAB

    I\mathbf{x} = (A^TA)^{-1}A^TB

    \mathbf{x} = (A^TA)^{-1}A^TB.
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by craigmain View Post
    Hi,

    I have a matrix that looks like the following.
    <br />
A = \left[\begin{matrix} 3 & 9 \\ 5 & 1 \\ 1 & 4 \end{matrix}\right]<br />

    by doing the following, I can add the two columns.

    <br />
\left[\begin{matrix} 3 & 9 \\ 5 & 1 \\ 1 & 4 \end{matrix}\right] * \left[\begin{matrix} 1 \\ 1 \end{matrix}\right] = \left[\begin{matrix} 12 \\ 7 \\ 5 \end{matrix}\right]<br />

    My question is:

    Is there any clever way to multiply columns in the matrix.

    <br />
\left[\begin{matrix} 3 & 9 \\ 5 & 1 \\ 1 & 4 \end{matrix}\right] * X = \left[\begin{matrix} 18 \\ 10 \\ 4 \end{matrix}\right]<br />

    Can X be a matrix of any kind?

    Regards
    Craig.
    this can be done

     \displaystyle \begin{bmatrix}<br />
a_{11} & a_{12}\\ <br />
a_{21} & a_{22} \\<br />
a_{31} & a_{32} <br />
\end{bmatrix} \cdot \begin{bmatrix}<br />
b_{11} \\<br />
b_{12} <br /> <br />
\end{bmatrix}

    meaning that matrix A(m x n) can be multiplied with another matrix B (n x m) only if matrix B have same number of the columns as the matrix A have rows

    important is that (mostly)

     A\cdot B \ne B\cdot A

    only in some cases can be  AB = BA


    as for the your problem there

     \displaystyle \left[\begin{matrix} 3 & 9 \\ 5 & 1 \\ 1 & 4 \end{matrix}\right] * X = \left[\begin{matrix} 18 \\ 10 \\ 4 \end{matrix}\right]

    your X can be  X= \begin{bmatrix}<br />
x_1 \\<br />
 x_2 <br />
\end{bmatrix}

    so when multiplied you get system of equations like :

     3x_1+9x_2=18
     5x_1+x_2=10
     x_1+4x_2=4



    Edit : sorry Prove It I have no idea why i didn't see your post .... early hm... very very sorry (P.S. when i refreshed this thread it wasn't there, and i waited to my post be posted at least 5 min. don't know what just happen .... )
    Last edited by yeKciM; September 7th 2010 at 02:45 AM.
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