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Math Help - echelon method problem...

  1. #1
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    echelon method problem...

    Can someone please help me solve this problem using the Gauss-Jordan Method:

    3x-5y-2z=-9
    -4x+3y+z=11
    8x-5y+4z=6
    Last edited by halloweengrl24; December 2nd 2010 at 05:22 AM. Reason: changed problem.
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  2. #2
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    Hello, halloweengrl24!

    \text{Solve using the Gauss-Jordan Method:}

    . . . \begin{array}{cccc}3x-5y-2z&=& -9 & [1] \\<br />
-4x+3y+z &=& 11 & [2] \\8x-5y+4z &=& 6 & [3] \end{array}

    I worked ahead a few steps and ran into some difficulty.
    Looking back, I found these preliminary steps to be helpful.

    Multiply [2] by -1: . 4x - 3y - z \:=\:\text{-}11

    And reorder the equations: . \begin{array}{ccc}4x-3y - z &=& \text{-}11 \\ 8x - 5y + 4z &=& 6 \\ 3x - 5y - 2z &=& \text{-}9 \end{array}


    \text{We have: } \;\left|\begin{array}{ccc|c}4 & \text{-}3 & \text{-}1 & \text{-}11 \\ 8 & \text{-}5 & 9 & 6 \\ 3 & \text{-}5 & \text{-}2 & \text{-}9 \end{array}\right|


    \begin{array}{c}R_1-R_3 \\ \\ \\ \end{array} \left|\begin{array}{ccc|c}<br />
1 & 2 & 1 & \text{-}2 \\<br />
8 & \text{-}5 & 4 & 6 \\ 3 & \text{-}5 & \text{-}2 & \text{-}9 \end{array}\right|


    \begin{array}{c} \\ R_2 - 8R_1 \\ R_3 - 3R_1\end{array} \left|\begin{array}{ccc|c} 1 & 2 & 1 & \text{-}2 \\ 0 & \text{-}21 & \text{-}4 & 22 \\ 0 & \text{-}11 & \text{-}5 & \text{-}3 \end{array} \right|


    \begin{array}{c} \\ R_2-2R_3 \\ (\text{-}1)R_3 \end{array} \left|\begin{array}{ccc|c} 1 & 2 & 1 & \text{-}2 \\ 0 & 1 & 6 & 28 \\ 0 & 11 & 5 & 3 \end{array}\right|


    \begin{array}{c}R_1 - 2R_2 \\ \\ R_3 - 11R_2 \end{array} \left| \begin{array}{ccc|c} 1 & 0 & \text{-}11 & \text{-}58 \\ 0 & 1 & 6 & 28 \\ 0 & 0 & \text{-}61 & \text{-}305 \end{array}\right|


    . . \begin{array}{c} \\ \\ (\text{-}\frac{1}{61})R_3 \end{array} \left|<br />
\begin{array}{ccc|c} 1 & 0 & \text{-}11 & \text{-}58 \\ 0 & 1 & 6 & 28 \\ 0 & 0 & 1 & 5 \end{array} \right|


    \begin{array}{c} R_1 + 11R_3 \\ R_2 - 6R_3 \\ \\ \end{array} \left| \begin{array}{ccc|c} 1 & 0 & 0 & \text{-}3 \\ 0 & 1 & 0 & \text{-}2 \\ 0 & 0 & 1 & 5 \end{array}\right|


    \text{Therefore: }\;\begin{Bmatrix}x &=& \text{-}3 \\ y &=& \text{-}2 \\ z &=& 5 \end{Bmatrix}

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