# Math Help - Gauss-Jordan Elimination

1. ## Gauss-Jordan Elimination

Solve the following using Gauss-Jordan elimination:

$2x+2y+2z=0$
$-2x+5y+2z=1$
$8x+y+4z=-1$

I keep setting up augmented matrices but I do not know how to get it to reduced-eschelon form at all.

Edit: I worked this out:

$x=-\frac{1}{7}-\frac{3}{7}r$

$y=-\frac{3}{7}$

$z=r$

Is this correct?

2. 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 row-reducing the augmented matrix but here's how I did it.

The augmented matrix is
$\begin{bmatrix}2 & 2 & 2 & 0 \\-2 & 5 & -2 & 1\\ 8& 1 & 4 & -1\end{bmatrix}$

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
$\begin{bmatrix}1 & 1 & 1 & 0 \\ 0 & 7 & 4 & -1\\0 & -7 & -4 & -1\end{bmatrix}$

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
$\begin{bmatrix}1 & 0 & \frac{3}{7} & -\frac{1}{7} \\ 0 & 1 & \frac{4}{7} & \frac{1}{7} \\ 0 & 0 & 0 & 0\end{bmatrix}$.

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, $x+ \frac{3}{7}z= -\frac{1}{7}$ so $x= -\frac{1}{7}- \frac{3}{7}z$ and $y+ \frac{4}{7}z= \frac{1}{7}$ so $y= \frac{1}{7}- \frac{4}{7}z$. Now let r= z to get your soltutions.

3. Bah, forgot to write the $z$ for the y part on my paper. And yeah, I meant row-reduced since the following question asks to do it using Gaussian-elimination

Thank you.

And to just check, for Gaussian-elimination it would be:

$\begin{bmatrix}1 & 1 & 1 & 0 \\ 0 & 1 & \frac{4}{7} & \frac{1}{7} \\ 0 & 0 & 0 & 0\end{bmatrix}$