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Math Help - Solving a system using elementary row operations

  1. #1
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    Solving a system using elementary row operations

    If someone could please help me with this. its a question about Solving a system using elementary row operations which is honestly one of the easiest and msot simple things to do but for the life of me I cannt understand what I am doing wrong

    I solved the system so i know what hte answers are but when doing elementary row operations method i keep getting wrong answers

    the Matrix is as per below

    3 -2 1 2
    1 -1 -1 -4
    4 1 -2 0

    please if someone could could..just really frustrated at this point
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Solving a system using elementary row operations

    Show us the first step that you tried and we'll see if you're on the right track.
    Last edited by ebaines; April 24th 2013 at 12:27 PM.
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  3. #3
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    Re: Solving a system using elementary row operations

    here is how i started the soln

    Divide row1 by 3
    1 -2/3 1/3 2/3
    1 -1 -1 -4
    4 1 -2 0

    Add (-1 * row1) to row2
    1 -2/3 1/3 2/3
    0 -1/3 -4/3 -14/3
    4 1 -2 0

    Add (-4 * row1) to row3
    1 -2/3 1/3 2/3
    0 -1/3 -4/3 -14/3
    0 11/3 -10/3 -8/3

    right now im about to divide row 2 by -1/3 and will get back to you soon
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Solving a system using elementary row operations

    Looking good so far!
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  5. #5
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    Re: Solving a system using elementary row operations

    Hello, flametag2!

    I would avoid fractions when possible.
    My first step may surprise you.



    \text{Solve the system: }\;\left[\begin{array}{ccc|c} 3&\text{-}2&1&2 \\ 1&\text{-}1&\text{-}1&\text{-}4 \\ 4&1&\text{-}2&0 \end{array}\right]

    \begin{array}{c}\text{Switch }R_1 \\ \text{and }R_2 \\ \end{array}\left[\begin{array}{ccc|c} 1&\text{-}1&\text{-}1&\text{-}4 \\ 3&\text{-}2&1&2 \\ 4&1&\text{-}2&0 \end{array}\right]

    \begin{array}{c}\\ R_2-3R_1 \\ R_3-4R_1 \end{array}\left[\begin{array}{ccc|c}1&\text{-}1&\text{-}1&\text{-}4 \\ 0&1&4&14 \\ 0&5&2&16 \end{array}\right]

    \begin{array}{c}R_1+R_2 \\ \\ R_3-5R_2 \end{array}\left[\begin{array}{ccc|c}1&0&3&10 \\ 0&1&4&14 \\ 0&0&\text{-}18 & \text{-}54 \end{array}\right]

    . . . \begin{array}{c}\\ \\ \text{-}\frac{1}{18}R_3 \end{array}\left[\begin{array}{ccc|c}1&0&3&10 \\ 0&1&4&14 \\ 0&0&1&3 \end{array}\right]

    \begin{array}{c}R_1-3R_3 \\ R_2-4R_3 \\ \\ \end{array}\left[\begin{array}{ccc|c}1&0&0&1 \\ 0&1&0&2 \\ 0&0&1&3 \end{array}\right]


    Therefore: . \begin{Bmatrix}x &=& 1 \\ y &=& 2 \\ z &=&3 \end{Bmatrix}
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