# 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:

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

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

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

Now interchange the last row with the second row:

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

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 $\displaystyle x_4$. We also observe that the last two rows mean that $\displaystyle x_3=1$.

Now substituting into the equation corresponding to our second row:

$\displaystyle 2x_2+4+2x_4=4$

so:

$\displaystyle x_2=-x_4$

Now substituting into the equation corresponding to our first row:

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

so:

$\displaystyle x_1=-x_4$

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