# How do you solve using gaussian elimination or gauss-jordan elimination,  2x-3y-2z=10, 3x-2y+2z=0, 4z-y+3z=-1?

Jan 9, 2018

$x = 2$, $y = 0$ and $z = - 3$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}2 & - 3 & - 2 & | & 10 \\ 3 & - 2 & 2 & | & 0 \\ 4 & - 1 & 3 & | & - 1\end{matrix}\right)$

I have written the equations not in the sequence as in the question in order to get $1$ as pivot.

Perform the folowing operations on the rows of the matrix

$R 1 \leftarrow R 1 + R 2$

$A = \left(\begin{matrix}5 & - 5 & 0 & | & 10 \\ 3 & - 2 & 2 & | & 0 \\ 4 & - 1 & 3 & | & - 1\end{matrix}\right)$

$R 1 \leftarrow \frac{R 1}{5}$

$A = \left(\begin{matrix}1 & - 1 & 0 & | & 2 \\ 3 & - 2 & 2 & | & 0 \\ 4 & - 1 & 3 & | & - 1\end{matrix}\right)$

$R 2 \leftarrow R 2 - 3 R 1$; $R 3 \leftarrow R 3 - 4 R 1$

$A = \left(\begin{matrix}1 & - 1 & 0 & | & 2 \\ 0 & 1 & 2 & | & - 6 \\ 0 & 3 & 3 & | & - 9\end{matrix}\right)$

$R 3 \leftarrow R 3 - 3 R 2$

$A = \left(\begin{matrix}1 & - 1 & 0 & | & 2 \\ 0 & 1 & 2 & | & - 6 \\ 0 & 0 & - 3 & | & 9\end{matrix}\right)$

$R 3 \leftarrow R \frac{3}{- 3}$

$A = \left(\begin{matrix}1 & - 1 & 0 & | & 2 \\ 0 & 1 & 2 & | & - 6 \\ 0 & 0 & 1 & | & - 3\end{matrix}\right)$

$R 2 \leftarrow R 2 - 2 R 3$

$A = \left(\begin{matrix}1 & - 1 & 0 & | & 2 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & - 3\end{matrix}\right)$

$R 1 \leftarrow R 1 + R 2$

$A = \left(\begin{matrix}1 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & - 3\end{matrix}\right)$

Thus $x = 2$, $y = 0$ and $z = - 3$