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Math Help - Gaussian elimination, linear equations

  1. #1
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    Gaussian elimination, linear equations

    Hey,

    Im having trouble on the following question but i get stuck after the first step i dont know how to work out the -R2 and so on.

    Question:

    x
    1 + x2 + 2x3 = 4
    2
    x1 + 3x2 + x3 = 1

    x1 + 3x2 + 3x3 = 2


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  2. #2
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    Hey man,

    Theres alot of ways to "think" about how to do it but this is mine

    You have
    "Row 1" x1 + x2 + 2x3 = 4
    "Row 2" 2x1 + 3x2 + x3 = 1
    "Row 3" −x1 + 3x2 + 3x3 = 2

    And you are aming for

    x1+x2+2x3=4
    Ax2 +Bx3=C
    Dx3=E
    Where A,B,C,D,E are constants that will come up along the way

    So the first step is to eliminate the x1's from the 2nd and 3rd equation. That is to say,

    "Row 4=Row1" ie, the top line is unchanged
    "Row 5=Row 2 (plus or minus) (some multiple of)Row 1"
    "Row 6=Row 3 (plus or minus) (some multiple of)Row 1"

    So to help you out, the first 'iteration' would be

    "Row 5=Row 2 -2 times Row 1"
    "Row 6=Row 3+ 1 times Row 1"
    Giving you

    "Row4" x1+x2+2x3=4
    "Row5" x2 -3x3=-7
    "Row6" 4x2+5x3=6

    So you need to do one more process like this to get to what you want, which then allows you to use a process of "backwards substitution" to find x3,x2 and x1 in turn.



    Hope thats clear and helps

    Btw mikel, im sorry if theres any little errors in there, rushing out, hope that helps mate!
    Last edited by mr fantastic; May 29th 2009 at 05:36 AM. Reason: Merged posts
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  3. #3
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    Hello, mikel03!

    If, by "Gaussian Elimination", you mean a matrix method,
    . . I don't see where there would be any difficulty.
    Every step is clearly indicated.


    \begin{array}{ccc} x_1 + x_2 + 2x_3 &=& 4 \\<br />
2x_1 + 3x_2 + x_3 &=& 1 \\ -x_1 + 3x_2 + 3x_3 &=& 2 \end{array}

    We have: . \begin{array}{|ccc|c|}<br />
1 &1&2&4 \\ 2&3&1&1 \\ \text{-}1&3&3&2 \end{array}


    \begin{array}{c} \\ R_2-2R_1 \\ R_3 + R_1\end{array} \begin{array}{|ccc|c|} 1&1&2&4 \\ 0&1&\text{-}3&\text{-}7 \\ 0&4&5&6 \end{array}


    \begin{array}{c}R_1-R_2 \\ \\ R_3-4R_2\end{array} \begin{array}{|ccc|c|} 1&0&0&11 \\ 0&1&\text{-}3 & \text{-}7 \\ 0&0&17&34 \end{array}


    . . . \begin{array}{c}\\ \\ \frac{1}{17}R_3\end{array} \begin{array}{|ccc|c|} 1&0&5&11 \\ 0&1&\text{-}3&\text{-}7 \\ 0&0&1&2 \end{array}


    \begin{array}{c}R_1-5R_3 \\ R_2 + 3R_3 \\ \\ \end{array} \begin{array}{|ccc|c|} 1&0&0&1 \\ 0&1&0&\text{-}1 \\ 0&0&1&2 \end{array}


    . . Therefore: . \begin{Bmatrix}x_1 &=& 1 \\ x_2 &=& \text{-}1 \\ x_3 &=& 2 \end{Bmatrix}

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  4. #4
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    Soroban, i like that very last step to get that identity matrix, ive never seen/thought of doing it like that!
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  5. #5
    MHF Contributor
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    Quote Originally Posted by OnkelTom View Post
    Soroban, i like that very last step to get that identity matrix, ive never seen/thought of doing it like that!
    That is reduced row echelon form (rref).
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