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Math Help - Gauss Jordan Elimination and Augmented Matrices

  1. #1
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    Gauss Jordan Elimination and Augmented Matrices

    Not a specially question but I am really bad at them so I would want to know if there is any quick ways or methods to do them except technology. Or like any ways to check the answers as I seem to get various answers

    Just for example, how to do

    1 2 1
    2 -3 4

    Those different colors are the ones equal to!

    Like for that example how do you know that you cannot go any further(no solutions))

    Please help, i need quick ways to do and check
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  2. #2
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    Hello, someone21!

    I'm not sure what your question is,
    . . but I can solve the system for you . . .


    \left|\begin{array}{cc|c}1 & 2 & 1 \\ 2 & \text{-}3 & 4\end{array}\right|

    We have: . \left|\begin{array}{cc|c}1 & 2 & 1 \\ 2 & \text{-}3 & 4 \end{array}\right|

    . \begin{array}{c} \\ R_2-2R_1\end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & \text{-}7 & 2 \end{array}\right|

    . . . . \begin{array}{c} \\  \text{-}\frac{1}{7}R_2 \end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & 1 & \text{-}\frac{2}{7} \end{array}\right|

    . \begin{array}{c} R_1-2R_2 \\ \\ \end{array}\left|\begin{array}{cc|c}1 & 0 & \frac{11}{7} \\ \\[-4mm] 0 & 1 & \text{-}\frac{2}{7}\end{array}\right|


    Solution: . \boxed{x \:=\:\frac{11}{7}\quad y \:=\:\text{-}\frac{2}{7}}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, someone21!

    I'm not sure what your question is,
    . . but I can solve the system for you . . .



    We have: . \left|\begin{array}{cc|c}1 & 2 & 1 \\ 2 & \text{-}3 & 4 \end{array}\right|

    . \begin{array}{c} \\ R_2-2R_1\end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & \text{-}7 & 2 \end{array}\right|

    . . . . \begin{array}{c} \\  \text{-}\frac{1}{7}R_2 \end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & 1 & \text{-}\frac{2}{7} \end{array}\right|

    . \begin{array}{c} R_1-2R_2 \\ \\ \end{array}\left|\begin{array}{cc|c}1 & 0 & \frac{11}{7} \\ \\[-4mm] 0 & 1 & \text{-}\frac{2}{7}\end{array}\right|


    Solution: . \boxed{x \:=\:\frac{11}{7}\quad y \:=\:\text{-}\frac{2}{7}}

    Thanks but the thing is there are so many ways to do them so how we do it the fastest

    i mean like in that example do we try to make the top left one =1 then make the exact lower row 0 or make the number next to it 0
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