# Math Help - Gauss-Jordan Method of elimination

1. ## 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?

Thanks,

2. 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.

HI Guys,

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

Thanks,

3. Originally Posted by Quick
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.

HI Guys,

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

Thanks,

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

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

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

Thanks,
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 $z = 0.$

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

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.

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

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.

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

Well Adrian, did this make any sense at all to you?

5. 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