# How do you solve using gaussian elimination or gauss-jordan elimination, 2x + 5y - 2z = 14, 5x -6y + 2z = 0, 4x - y + 3z = -7?

Jun 14, 2017

The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}2 \\ 0 \\ - 5\end{matrix}\right)$

#### Explanation:

We perform the Gauss-Jordan elimination

$\left(\begin{matrix}2 & 5 & - 2 & 14 \\ 5 & - 6 & 2 & 0 \\ 4 & - 1 & 3 & - 7\end{matrix}\right)$

Exchange $R 2 \leftrightarrow R 1$

$\left(\begin{matrix}5 & - 6 & 2 & 0 \\ 2 & 5 & - 2 & 14 \\ 4 & - 1 & 3 & - 7\end{matrix}\right)$

Divide $R 1$ by $5$

$\left(\begin{matrix}1 & - 1.2 & 0.4 & 0 \\ 2 & 5 & - 2 & 14 \\ 4 & - 1 & 3 & - 7\end{matrix}\right)$

$R 2 \leftrightarrow R 2 - 2 R 1$

$\left(\begin{matrix}1 & - 1.2 & 0.4 & 0 \\ 0 & 7.4 & - 2.8 & 14 \\ 4 & - 1 & 3 & - 7\end{matrix}\right)$

$R 3 \leftrightarrow R 3 - 4 R 1$

$\left(\begin{matrix}1 & - 1.2 & 0.4 & 0 \\ 0 & 7.4 & - 2.8 & 14 \\ 0 & 3.8 & 1.4 & - 7\end{matrix}\right)$

$R 2 \leftrightarrow \frac{R 2}{7.4}$

$\left(\begin{matrix}1 & - 1.2 & 0.4 & 0 \\ 0 & 1 & - 0.378 & 1.892 \\ 0 & 3.8 & 1.4 & - 7\end{matrix}\right)$

$R 3 \leftrightarrow R 3 - 3.8 R 2$

$\left(\begin{matrix}1 & - 1.2 & 0.4 & 0 \\ 0 & 1 & - 0.378 & 1.892 \\ 0 & 0 & 2.838 & - 14.189\end{matrix}\right)$

$R 3 \leftrightarrow \frac{R 3}{2.838}$

$\left(\begin{matrix}1 & - 1.2 & 0.4 & 0 \\ 0 & 1 & - 0.378 & 1.892 \\ 0 & 0 & 1 & - 5\end{matrix}\right)$

$R 1 \leftrightarrow R 1 - 0.4 R 3$

$\left(\begin{matrix}1 & - 1.2 & 0 & 2 \\ 0 & 1 & - 0.378 & 1.892 \\ 0 & 0 & 1 & - 5\end{matrix}\right)$

$R 2 \leftrightarrow R 2 + 0.378 R 3$

$\left(\begin{matrix}1 & - 1.2 & 0 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - 5\end{matrix}\right)$

$R 1 \leftrightarrow R 1 + 1.2 R 2$

$\left(\begin{matrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - 5\end{matrix}\right)$