
GaussJordan Elimination
Solve the following using GaussJordan elimination:
$\displaystyle 2x+2y+2z=0$
$\displaystyle 2x+5y+2z=1$
$\displaystyle 8x+y+4z=1$
I keep setting up augmented matrices but I do not know how to get it to reducedeschelon form at all.
Edit: I worked this out:
$\displaystyle x=\frac{1}{7}\frac{3}{7}r$
$\displaystyle y=\frac{3}{7}$
$\displaystyle z=r$
Is this correct?

Yes, that is correct! Well, done!
(It took me a moment to figure out where that "r" came from!)
You don't say if you did that by rowreducing the augmented matrix but here's how I did it.
The augmented matrix is
$\displaystyle \begin{bmatrix}2 & 2 & 2 & 0 \\2 & 5 & 2 & 1\\ 8& 1 & 4 & 1\end{bmatrix}$
Divide the first row by 2, add the original first row to the second row, and subtract the original first row times 4 from the third row to get
$\displaystyle \begin{bmatrix}1 & 1 & 1 & 0 \\ 0 & 7 & 4 & 1\\0 & 7 & 4 & 1\end{bmatrix}$
Divide the second row by 7, add the original second row to the third row, and subtract the new second row from the first row to get
$\displaystyle \begin{bmatrix}1 & 0 & \frac{3}{7} & \frac{1}{7} \\ 0 & 1 & \frac{4}{7} & \frac{1}{7} \\ 0 & 0 & 0 & 0\end{bmatrix}$.
The fact that the last row consists of all "0"s tells us that we will have an infinite number of solutions. From the first equation, $\displaystyle x+ \frac{3}{7}z= \frac{1}{7}$ so $\displaystyle x= \frac{1}{7} \frac{3}{7}z$ and $\displaystyle y+ \frac{4}{7}z= \frac{1}{7}$ so $\displaystyle y= \frac{1}{7} \frac{4}{7}z$. Now let r= z to get your soltutions.

Bah, forgot to write the $\displaystyle z$ for the y part on my paper. (Headbang) And yeah, I meant rowreduced since the following question asks to do it using Gaussianelimination
Thank you. (Clapping)
And to just check, for Gaussianelimination it would be:
$\displaystyle \begin{bmatrix}1 & 1 & 1 & 0 \\ 0 & 1 & \frac{4}{7} & \frac{1}{7} \\ 0 & 0 & 0 & 0\end{bmatrix}$