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Math Help - Gauss-Jordan Method of elimination

  1. #1
    adrian1116
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    Gauss-Jordan Method of elimination

    HI Guys,

    Can someone pls help me and teach the easiest way to solve linear problems using the Gauss-Jordan Method? Any new techniques?

    Appreciate your soonest reply.

    Thanks,
    Adrian
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by adrian1116 on a different post
    Hi,

    I am asking someone's help to show any easiest technique to solve linear equations using the Gauss-Jordan Method.

    Below is an example:

    Z-3X1-5X2 = 0
    X1+X3 = 4
    2X2+X4 = 12
    3X1+2X2+X5 = 18

    where X3, X4 AND X5 are the slack variables. solve for values of Z, X1 and X2 using the said method.

    Thanks and appreciate your soonest feedback.

    Adrian
    Quote Originally Posted by adrian1116
    HI Guys,

    Can someone pls help me and teach the easiest way to solve linear problems using the Gauss-Jordan Method? Any new techniques?

    Appreciate your soonest reply.

    Thanks,
    Adrian
    You should never ask the same question twice Adrian...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Quick
    Quote Originally Posted by adrian1116 on a different post
    Hi,

    I am asking someone's help to show any easiest technique to solve linear equations using the Gauss-Jordan Method.

    Below is an example:

    Z-3X1-5X2 = 0
    X1+X3 = 4
    2X2+X4 = 12
    3X1+2X2+X5 = 18

    where X3, X4 AND X5 are the slack variables. solve for values of Z, X1 and X2 using the said method.

    Thanks and appreciate your soonest feedback.

    Adrian
    Quote Originally Posted by adrian1116
    HI Guys,

    Can someone pls help me and teach the easiest way to solve linear problems using the Gauss-Jordan Method? Any new techniques?

    Appreciate your soonest reply.

    Thanks,
    Adrian
    You should never ask the same question twice Adrian...

    Nor should you tag a new question on to the end of an existing thread -
    it confuses the helpers (well it confuses me anyway).

    RonL
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  4. #4
    Senior Member
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    Since TPH closed the other thread, there is no place else to answer this question.

    Quote Originally Posted by adrian1116
    HI Guys,

    Can someone pls help me and teach the easiest way to solve linear problems using the Gauss-Jordan Method? Any new techniques?

    X1+X3 = 4
    2X2+X4 = 12
    3X1+2X2+X5 = 18
    Z-3X1-5X2 = 0

    Appreciate your soonest reply.

    Thanks,
    Adrian
    Hi, Adrian. How much do you know about this stuff? I'm experimenting here to see if it is possible to explain this in a short space.

    Here is your example worked out.

    Set up this tableau. Do you see how the tableau relates to your problem? We start off with the slack variables equal to the constant terms in the equations (for example x_3 = 4) and z = 0.

    <br />
\begin{tabular}{|r|r|rrrrr|}<br />
\hline<br />
      & $b$  & $x_1$ &  $x_2$ & $x_3$ &  $x_4$ & $x_5$ \\ \hline<br />
$x_3$ &  4 & \fbox{1} &  0 &  1 &  0 & 0 \\<br />
$x_4$ & 12 &  0 &  2 &  0 &  1 & 0 \\<br />
$x_5$ & 18 &  3 &  2 &  0 &  0 & 1 \\ \hline<br />
z     &  0 & -3 & -5 &  0 &  0 & 0 \\ \hline<br />
\end{tabular}<br />

    Do you know what it means to pivot? It means to reduce a column using row operations to zeroes in every row but one row that has a 1 in it. You apply those same row operations to all the other columns.

    To get the next tableau, pivot using the row and column with the box. This means we are bringing variable x_1 into the solution replacing variable x_3 .

    The next tableau is the result of the first pivoting. Keep track in the first column of what row and column was used to pivot.

    <br />
\begin{tabular}{|r|r|rrrrr|}<br />
\hline<br />
      & $b$  & $x_1$ &  $x_2$ & $x_3$ &  $x_4$ & $x_5$ \\ \hline<br />
$x_1$ &  4 &  1 &  0 &  1 &  0 & 0 \\<br />
$x_4$ & 12 &  0 & \fbox{2} &  0 &  1 & 0 \\<br />
$x_5$ &  6 &  0 &  2 &  -3 &  0 & 1 \\ \hline<br />
z     & 12 &  0 & -5 &   3 &  0 & 0 \\ \hline<br />
\end{tabular}<br />

    To get to the next tableau, pivot using the row and column with the box. This means we are bringing variable x_2 into the solution replacing variable x_4 .

    The next tableau is the result of the second pivoting. Again, keep track in the first column of what row and column was used to pivot.

    This is a solution to the equations: x_1 = 4,\ x_2 = 6, x_5 = -6, z = 42. It's a solution because variables x_1\text{ and }x_2 are in as desired. There were no constraints on the slack variables so x_5 turns out negative.

    <br />
\begin{tabular}{|r|r|rrrrr|}<br />
\hline<br />
      & $b$ & $x_1$ &  $x_2$ & $x_3$ &  $x_4$ & $x_5$ \\ \hline<br />
$x_1$ &  4 &  1 &  0 &  1 &  0  & 0 \\<br />
$x_2$ &  6 &  0 &  1 &  0 & 1/2 & 0 \\<br />
$x_5$ & -6 &  0 &  0 & -3 & -1  & 1 \\ \hline<br />
z     & 42 &  0 &  0 &  3 & 5/2 & 0 \\ \hline<br />
\end{tabular}<br />

    Well Adrian, did this make any sense at all to you?
    Last edited by JakeD; July 18th 2006 at 02:25 AM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by JakeD
    Since TPH closed the other thread, there is no place else to answer this question.
    And now it is a thread in its own right!

    RonL
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