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Thread: I simply cannot do this Gauss elimination question

  1. #1
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    I simply cannot do this Gauss elimination question

    $\displaystyle
    2x - y + 3z + t = 1
    $
    $\displaystyle
    -2x + y - 2z = 1
    $
    $\displaystyle
    4x - 2y + 7z + 3t = 4
    $


    I got it to

    (2 -1 3 1 1)
    (0 0 1 1 2)

    and I don't know what to do from there : /
    Last edited by thomasashcroft; Aug 15th 2009 at 06:27 PM. Reason: MATH Tags
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  2. #2
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    Quote Originally Posted by thomasashcroft View Post
    $\displaystyle
    2x - y + 3z + t = 1
    $
    $\displaystyle
    -2x + y - 2z = 1
    $
    $\displaystyle
    4x - 2y + 7z + 3t = 4
    $


    I got it to

    (2 -1 3 1 1)
    (0 0 1 1 2)

    and I don't know what to do from there : /
    Get the reduced row echelon form. Then it should be clear that you have to introduce two parameters .... Perhaps $\displaystyle t = \lambda$ and $\displaystyle x = \mu$ where each parameter can be any real number. Now get y and z in terms of these parameters and you have a solution.
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    Quote Originally Posted by mr fantastic View Post
    Get the reduced row echelon form. Then it should be clear that you have to introduce two parameters .... Perhaps $\displaystyle t = \lambda$ and $\displaystyle x = \mu$ where each parameter can be any real number. Now get y and z in terms of these parameters and you have a solution.
    It's not possible to get the reduced row echelon form is it?
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  4. #4
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    Quote Originally Posted by thomasashcroft View Post
    It's not possible to get the reduced row echelon form is it?
    Why do you think that? Of course it is.

    PlanetMath: reduced row echelon form
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  5. #5
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    Well all I would have to do is multiply the first row by 1/2 which would leave me with 1 - 0.5 1.5 0.5 0.5 in the first row.. and then it'd be in r.r.e.f. But how does that help me see that y should be arbitrary?
    Last edited by thomasashcroft; Aug 15th 2009 at 07:09 PM. Reason: Misspelling
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  6. #6
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    Quote Originally Posted by thomasashcroft View Post
    Well all I would have to do is multiply the first row by 1/2 which would leave me with 1 - 0.5 1.5 0.5 0.5 in the first row.. and then it'd be in r.r.e.f. But how does that help me see that y should be arbitrary?
    $\displaystyle R_1 \longrightarrow R_1 - 3 R_2$:

    $\displaystyle \left( \begin{array}{cccc}
    2 & -1 & 0 & -2 \\
    0 & 0 & 1 & 1
    \end{array} \right|$ $\displaystyle \left. \begin{array}{c}
    -5 \\
    2
    \end{array} \right)$

    which is equivalent to the linear system:

    $\displaystyle 2x - y - 2t = -5$

    $\displaystyle z + t = 2$

    Now there is choice. I chose $\displaystyle t = \lambda$ and $\displaystyle x = \mu$ where both parameters can be any real number.

    Then:

    $\displaystyle z + \lambda = 2 \Rightarrow z = 2 - \lambda$.

    $\displaystyle 2 \mu - y - 2 \lambda = -5 \Rightarrow y = 2 \mu + 3$.

    Other choices leading to equivalent solutions can obviously be made.
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  7. #7
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    Quote Originally Posted by thomasashcroft View Post
    $\displaystyle
    2x - y + 3z + t = 1
    $
    $\displaystyle
    -2x + y - 2z = 1
    $
    $\displaystyle
    4x - 2y + 7z + 3t = 4
    $


    I got it to

    (2 -1 3 1 1)
    (0 0 1 1 2)

    and I don't know what to do from there : /

    The initial tablau is:

    $\displaystyle
    \begin{array}{rrrr|r}
    2&-1&3&1&1\\
    -2&1&-2&0&1\\
    4&-2&7&3&4
    \end{array}
    $

    add the first row to the second:

    $\displaystyle
    \begin{array}{rrrr|r}
    2&-1&3&1&1\\
    0&0&1&1&2\\
    4&-2&7&3&4
    \end{array}
    $

    Now subtract twice the first row from the third:

    $\displaystyle
    \begin{array}{rrrr|r}
    2&-1&3&1&1\\
    0&0&1&1&2\\
    0&0&1&1&2
    \end{array}
    $

    Removing the duplicate leaves you with:

    $\displaystyle
    \begin{array}{rrrr|r}
    2&-1&3&1&1\\
    0&0&1&1&2
    \end{array}
    $

    NOw let $\displaystyle t=\alpha$ and $\displaystyle y=\beta$ and solve for $\displaystyle x$ and $\displaystyle z$ in terms of $\displaystyle \alpha$ and $\displaystyle \beta$.

    CB
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