# How do you solve using gaussian elimination or gauss-jordan elimination, x+y+z=2, 2x-3y+z=-11, -x+2y-z=8?

Jun 1, 2018

The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}\frac{1}{3} \\ \frac{10}{3} \\ - \frac{5}{3}\end{matrix}\right)$

#### Explanation:

The augmented matrix is

$A = \left(\begin{matrix}1 & 1 & 1 & | & 2 \\ 2 & - 3 & 1 & | & - 11 \\ - 1 & 2 & - 1 & | & 8\end{matrix}\right)$

The pivot is in the the first column of the first row

Perform the operations on the rows

$R 2 \leftarrow R 2 - 2 R 1$ and $R 3 \leftarrow R 3 + R 1$

$\implies$, $\left(\begin{matrix}1 & 1 & 1 & | & 2 \\ 0 & - 5 & - 1 & | & - 15 \\ 0 & 3 & 0 & | & 10\end{matrix}\right)$

Make the pivot in the second column

$R 2 \leftarrow \frac{R 2}{- 5}$

$\implies$, $\left(\begin{matrix}1 & 1 & 1 & | & 2 \\ 0 & 1 & \frac{1}{5} & | & 3 \\ 0 & 3 & 0 & | & 10\end{matrix}\right)$

Eliminate the second column

$R 1 \leftarrow R 1 - R 2$ and $R 3 \leftarrow R 3 - 3 R 2$

$\implies$, $\left(\begin{matrix}1 & 0 & \frac{4}{5} & | & - 1 \\ 0 & 1 & \frac{1}{5} & | & 3 \\ 0 & 0 & - \frac{3}{5} & | & 1\end{matrix}\right)$

Make the pivot in the third column

$R 3 \leftarrow \left(R 3 \times - \frac{5}{3}\right)$

$\implies$, $\left(\begin{matrix}1 & 0 & \frac{4}{5} & | & - 1 \\ 0 & 1 & \frac{1}{5} & | & 3 \\ 0 & 0 & 1 & | & - \frac{5}{3}\end{matrix}\right)$

Eliminate the third column

$R 1 \leftarrow \left(R 1 - \frac{4}{5} \times R 3\right)$ and $R 2 \leftarrow \left(R 2 - \frac{1}{5} \times R 3\right)$

$\implies$, $\left(\begin{matrix}1 & 0 & 0 & | & \frac{1}{3} \\ 0 & 1 & 0 & | & \frac{10}{3} \\ 0 & 0 & 1 & | & - \frac{5}{3}\end{matrix}\right)$

The solution is

$\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}\frac{1}{3} \\ \frac{10}{3} \\ - \frac{5}{3}\end{matrix}\right)$