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Math Help - Clarifification on matricies

  1. #1
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    Clarifification on matricies

    Hey guys, i just need clarification on how to start this problem.



    \begin{bmatrix}<br />
4 & -1\\ <br />
3 & 1<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x\\ <br />
y<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
3\\<br />
4<br />
\end{bmatrix}

    Basically how do i get this to look like a regular system of equations?
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Mp5xm8 View Post
    Hey guys, i just need clarification on how to start this problem.



    \begin{bmatrix}<br />
4 & -1\\ <br />
3 & 1<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x\\ <br />
y<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
3\\<br />
4<br />
\end{bmatrix}

    Basically how do i get this to look like a regular system of equations?
    to get "regular" system of equations from that

     4x -y = 3

     3x+y= 4

    it's just matter of multiplying those matrices do you know how to do that ?
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  3. #3
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    this hasn't come up in a while, i kinda forgot how to do it
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Mp5xm8 View Post
    this hasn't come up in a while, i kinda forgot how to do it

    to multiply two matrices AB=C


    A= \begin{bmatrix}<br />
a_{11} & a_{12} & a_{13}\\ <br />
a_{21} & a_{22} & a_{23} \\<br />
a_{31} & a_{32} & a_{33}<br />
\end{bmatrix}

     B = \begin{bmatrix}<br />
b_{11} & b_{12} & b_{13}\\ <br />
b_{21} & b_{22} & b_{23} \\<br />
b_{31} & b_{32} & b_{33}<br />
\end{bmatrix}


    \begin{bmatrix}<br />
a_{11} & a_{12} & a_{13}\\ <br />
a_{21} & a_{22} & a_{23} \\<br />
a_{31} & a_{32} & a_{33}<br />
\end{bmatrix} \cdot \begin{bmatrix}<br />
b_{11} & b_{12} & b_{13}\\ <br />
b_{21} & b_{22} & b_{23} \\<br />
b_{31} & b_{32} & b_{33}<br />
\end{bmatrix} = \begin{bmatrix}<br />
a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31} & a_{11}b_{12}+a_{12}b_{22}+a_{13}b_{32} & a_{11}b_{13}+a_{12}b_{23}+a_{13}b_{33}\\<br /> <br />
a_{21}b_{11}+a_{22}b_{21}+a_{23}b_{31} & a_{21}b_{12}+a_{22}b_{22}+a_{23}b_{32} & a_{21}b_{13}+a_{22}b_{23}+a_{23}b_{33}\\<br /> <br />
a_{31}b_{11}+a_{32}b_{21}+a_{33}b_{31} & a_{31}b_{12}+a_{32}b_{22}+a_{33}b_{32} & a_{31}b_{13}+a_{32}b_{23}+a_{13}b_{33}\\<br />
\end{bmatrix}<br />


    than


     C= \begin{bmatrix}<br />
a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31} & a_{11}b_{12}+a_{12}b_{22}+a_{13}b_{32} & a_{11}b_{13}+a_{12}b_{23}+a_{13}b_{33}\\<br /> <br />
a_{21}b_{11}+a_{22}b_{21}+a_{23}b_{31} & a_{21}b_{12}+a_{22}b_{22}+a_{23}b_{32} & a_{21}b_{13}+a_{22}b_{23}+a_{23}b_{33}\\<br /> <br />
a_{31}b_{11}+a_{32}b_{21}+a_{33}b_{31} & a_{31}b_{12}+a_{32}b_{22}+a_{33}b_{32} & a_{31}b_{13}+a_{32}b_{23}+a_{13}b_{33}\\<br />
\end{bmatrix}<br />


    two matrices are equal if their elements are equal :

    A= \begin{bmatrix}<br />
a_{11} & a_{12} & a_{13}\\ <br />
a_{21} & a_{22} & a_{23} \\<br />
a_{31} & a_{32} & a_{33}<br />
\end{bmatrix}

     B = \begin{bmatrix}<br />
b_{11} & b_{12} & b_{13}\\ <br />
b_{21} & b_{22} & b_{23} \\<br />
b_{31} & b_{32} & b_{33}<br />
\end{bmatrix}

     a_{ij} = b_{ij} where i,j = 1,2,3... depending on the size of the matrix

    if that is true than A=B
    Last edited by yeKciM; August 27th 2010 at 03:17 PM.
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  5. #5
    Senior Member yeKciM's Avatar
    Joined
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    Quote Originally Posted by Mp5xm8 View Post
    Hey guys, i just need clarification on how to start this problem.



    \begin{bmatrix}<br />
4 & -1\\ <br />
3 & 1<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x\\ <br />
y<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
3\\<br />
4<br />
\end{bmatrix}

    Basically how do i get this to look like a regular system of equations?

    although when given system like that, perhaps it's point do solve it using matrix not write them as....
    but if you need to solve it like that ...

     A= \begin{bmatrix}<br />
4 & -1\\ <br />
3 & 1<br />
\end{bmatrix}

     C= \begin{bmatrix}<br />
3\\<br />
4<br />
\end{bmatrix}

     B = \begin{bmatrix}<br />
x\\ <br />
y<br />
\end{bmatrix}

    to solve it you need inverse matrix A ... so when multiply A from the left you get

     AB=C
     A^{-1}AB=A^{-1}C
     IB =A^{-1}C
     B =A^{-1}C

    P.S. you shouldn't write so many lines... i did so you could remember what's that all about (i hope you know to find inverse)
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