# How do you solve using gaussian elimination or gauss-jordan elimination, 2x + y - 3z = - 3, 3x + 2y + 4z = 5, -4x - y + 2z = 4?

Jul 3, 2018

The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- 1 \\ 2 \\ 1\end{matrix}\right)$

#### Explanation:

Perform the Gauss-Jordan Elimination on the augmented matrix

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

Make the pivot in the first column $R 1 \leftarrow \frac{R 1}{2}$

$\left(\begin{matrix}1 & \frac{1}{2} & - \frac{3}{2} & | & - \frac{3}{2} \\ 3 & 2 & 4 & | & 5 \\ - 4 & - 1 & 2 & | & 4\end{matrix}\right)$

Eliminate the first column $R 2 \leftarrow \left(R 2 - 3 R 1\right)$ and $R 3 \leftarrow \left(R 3 + 4 R 1\right)$

$\left(\begin{matrix}1 & \frac{1}{2} & - \frac{3}{2} & | & - \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{17}{2} & | & \frac{19}{2} \\ 0 & 1 & - 4 & | & - 2\end{matrix}\right)$

Make the pivot in the $2$nd column swap $R 2 \leftrightarrow R 3$

$\left(\begin{matrix}1 & \frac{1}{2} & - \frac{3}{2} & | & - \frac{3}{2} \\ 0 & 1 & - 4 & | & - 2 \\ 0 & \frac{1}{2} & \frac{17}{2} & | & \frac{19}{2}\end{matrix}\right)$

Eliminate the second column $R 3 \leftarrow \left(R 3 - \frac{1}{2} \left(R 2\right)\right)$, $R 1 \leftarrow \left(R 1 - \frac{R 2}{2}\right)$

$\left(\begin{matrix}1 & 0 & \frac{1}{2} & | & - \frac{1}{2} \\ 0 & 1 & - 4 & | & - 2 \\ 0 & 0 & \frac{21}{2} & | & \frac{21}{2}\end{matrix}\right)$

Make the pivot in the $3 r d$ column $R 3 \leftarrow \left(R 3 \cdot \frac{2}{21}\right)$

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

Eliminate the $3 r d$ column $R 1 \leftarrow \left(R 1 - \frac{R 3}{2}\right)$ and $R 2 \leftarrow \left(R 2 + 4 R 3\right)$

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

The solution is

$\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- 1 \\ 2 \\ 1\end{matrix}\right)$