# How do you solve the system -6x-2y-z=-17, 5x+y-6z=19, and -4x-6y-6z=-20?

Jan 13, 2018

$x = 2$, $y = 3$ and $z = - 1$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}- 6 & - 2 & - 1 & | & - 17 \\ 5 & 1 & - 6 & | & 19 \\ - 4 & - 6 & - 6 & | & - 20\end{matrix}\right)$

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

$R 1 \leftarrow R 1 + R 2$; $R 3 \leftarrow R 3 + R 2$

$A = \left(\begin{matrix}- 1 & - 1 & - 7 & | & 2 \\ 5 & 1 & - 6 & | & 19 \\ 1 & - 5 & - 12 & | & - 1\end{matrix}\right)$

$R 2 \leftarrow \left(R 2\right) + 5 R 1$; $R 3 \leftarrow R 3 + R 1$

$A = \left(\begin{matrix}- 1 & - 1 & - 7 & | & 2 \\ 0 & - 4 & - 41 & | & 29 \\ 0 & - 6 & - 19 & | & 1\end{matrix}\right)$

$R 1 \leftarrow R 1 \cdot \left(- 1\right)$; $R 2 \leftarrow R 2 - R 3$

$A = \left(\begin{matrix}1 & 1 & 7 & | & - 2 \\ 0 & 2 & - 22 & | & 28 \\ 0 & - 6 & - 19 & | & 1\end{matrix}\right)$

$R 3 \leftarrow R 3 + 3 R 2$

$A = \left(\begin{matrix}1 & 1 & 7 & | & - 2 \\ 0 & 2 & - 22 & | & 28 \\ 0 & 0 & - 85 & | & 85\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{- 85}$

$A = \left(\begin{matrix}1 & 1 & 7 & | & - 2 \\ 0 & 2 & - 22 & | & 28 \\ 0 & 0 & 1 & | & - 1\end{matrix}\right)$

$R 1 \leftarrow R 1 - 7 R 3$; $R 2 \leftarrow R 1 + 22 R 3$

$A = \left(\begin{matrix}1 & 1 & 0 & | & 5 \\ 0 & 2 & 0 & | & 6 \\ 0 & 0 & 1 & | & - 1\end{matrix}\right)$

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

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

$R 1 \leftarrow R 1 - R 2$

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

Thus $x = 2$, $y = 3$ and $z = - 1$