# Gauss method

• Nov 3rd 2008, 06:46 AM
Aiviekste92
Gauss method
I have a problem (Headbang) to solve with the Gauss metohd.

{ X1 + X2 + 3X3 + 2X4 + 2X5 = 3
{ X1 + X2 + X3 + 2X4 = 1
{ X1 + X2 - X3 + 2X4 - 2X5 = -1
{ -X1 + X2 + X3 = 1

Thanks!
You are THE BEST :)
• Nov 3rd 2008, 11:33 PM
CaptainBlack
Quote:

Originally Posted by Aiviekste92
I have a problem (Headbang) to solve with the Gauss metohd.

{ X1 + X2 + 3X3 + 2X4 + 2X5 = 3
{ X1 + X2 + X3 + 2X4 = 1
{ X1 + X2 - X3 + 2X4 - 2X5 = -1
{ -X1 + X2 + X3 = 1

Thanks!
You are THE BEST :)

The augmented matrix is:

$
\left[
\begin{array}{ccccc}
1&1&3&2&3\\
1&1&1&2&1\\
1&1&-1&2&-1\\
-1&1&1&0&1
\end{array}
\right]
$

Now subtract the first row from the second and third rows and add it to the third row to get:

$
\left[
\begin{array}{ccccc}
1&1&3&2&3\\
0&0&-2&0&-2\\
0&0&-4&0&-4\\
0&2&4&2&4
\end{array}
\right]
$

Now interchange the last row with the second row:

$
\left[
\begin{array}{ccccc}
1&1&3&2&3\\
0&2&4&2&4\\
0&0&-4&0&-4\\
0&0&-2&0&-2
\end{array}
\right]
$

Now because the last two rows are identical it means that one of the variables will be indeterminate, and we might as well let this be $x_4$. We also observe that the last two rows mean that $x_3=1$.

Now substituting into the equation corresponding to our second row:

$2x_2+4+2x_4=4$

so:

$
x_2=-x_4
$

Now substituting into the equation corresponding to our first row:

$x_1-x_4+3+2x_4=3$

so:

$
x_1=-x_4
$

CB
• Nov 4th 2008, 12:18 AM
Aiviekste92
thank you ever so much (Clapping)