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Math Help - I simply cannot do this Gauss elimination question

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

    <br />
2x - y + 3z + t = 1<br />
    <br />
-2x + y - 2z = 1<br />
    <br />
4x - 2y + 7z + 3t = 4<br />


    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; August 15th 2009 at 07:27 PM. Reason: MATH Tags
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  2. #2
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    Quote Originally Posted by thomasashcroft View Post
    <br />
2x - y + 3z + t = 1<br />
    <br />
-2x + y - 2z = 1<br />
    <br />
4x - 2y + 7z + 3t = 4<br />


    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 t = \lambda and 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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  3. #3
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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 t = \lambda and 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; August 15th 2009 at 08:09 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?
    R_1 \longrightarrow R_1 - 3 R_2:

    \left( \begin{array}{cccc}<br />
2 & -1 & 0 & -2 \\<br />
0 & 0 & 1 & 1<br />
\end{array} \right| \left. \begin{array}{c}<br />
-5 \\<br />
2<br />
\end{array} \right)

    which is equivalent to the linear system:

    2x - y - 2t = -5

    z + t = 2

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

    Then:

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

    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
    <br />
2x - y + 3z + t = 1<br />
    <br />
-2x + y - 2z = 1<br />
    <br />
4x - 2y + 7z + 3t = 4<br />


    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:

     <br />
\begin{array}{rrrr|r}<br />
2&-1&3&1&1\\<br />
-2&1&-2&0&1\\<br />
4&-2&7&3&4<br />
\end{array}<br />

    add the first row to the second:

     <br />
\begin{array}{rrrr|r}<br />
2&-1&3&1&1\\<br />
0&0&1&1&2\\<br />
4&-2&7&3&4<br />
\end{array}<br />

    Now subtract twice the first row from the third:

     <br />
\begin{array}{rrrr|r}<br />
2&-1&3&1&1\\<br />
0&0&1&1&2\\<br />
0&0&1&1&2<br />
\end{array}<br />

    Removing the duplicate leaves you with:

     <br />
\begin{array}{rrrr|r}<br />
2&-1&3&1&1\\<br />
0&0&1&1&2<br />
\end{array}<br />

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

    CB
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