# How do you solve using gaussian elimination or gauss-jordan elimination, 2x + 4y−6z = 42, x + 2y+ 3z = 3, 3x−4y+ 4z = −16?

Nov 20, 2017

The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}4 \\ 4 \\ - 3\end{matrix}\right)$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & 2 & - 3 & | & 21 \\ 1 & 2 & 3 & | & 3 \\ 3 & - 4 & 4 & | & - 16\end{matrix}\right)$

Perform the row operations

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

$A = \left(\begin{matrix}1 & 2 & - 3 & | & 21 \\ 0 & 0 & 6 & | & - 18 \\ 0 & - 10 & 13 & | & - 79\end{matrix}\right)$

$R 2 \leftrightarrow R 3$ and $R 3 \leftarrow \frac{R 3}{6}$

$A = \left(\begin{matrix}1 & 2 & - 3 & | & 21 \\ 0 & - 10 & 13 & | & - 79 \\ 0 & 0 & 1 & | & - 3\end{matrix}\right)$

$R 2 \leftarrow \left(R 2 - 13 R 3\right)$ and $R 2 \leftarrow \frac{R 2}{- 10}$

$A = \left(\begin{matrix}1 & 2 & - 3 & | & 21 \\ 0 & 1 & 0 & | & 4 \\ 0 & 0 & 1 & | & - 3\end{matrix}\right)$

$R 1 \leftarrow \left(R 1 - 2 R 2\right)$ and $R 1 \leftarrow R 1 + 3 R 3$

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