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
Since TPH closed the other thread, there is no place else to answer this question.
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.Originally Posted by adrian1116
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 $\displaystyle x_3 = 4$) and $\displaystyle z = 0.$
$\displaystyle
\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 $\displaystyle x_1$ into the solution replacing variable $\displaystyle x_3 .$
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.
$\displaystyle
\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 $\displaystyle x_2$ into the solution replacing variable $\displaystyle x_4 .$
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: $\displaystyle x_1 = 4,\ x_2 = 6, x_5 = -6, z = 42.$ It's a solution because variables $\displaystyle x_1\text{ and }x_2$ are in as desired. There were no constraints on the slack variables so $\displaystyle x_5$ turns out negative.
$\displaystyle
\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?