# How do you solve the system 9x + 9y + z = -112, 8x + 5y - 9z = -137, 7x + 4y + 3z = -64?

Mar 4, 2018

$x = - 9$, $y = - 4$ and $z = 5$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}9 & 9 & 1 & | & - 112 \\ 8 & 5 & - 9 & | & - 137 \\ 7 & 4 & 3 & | & - 64\end{matrix}\right)$

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

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

$A = \left(\begin{matrix}1 & 4 & 10 & | & 25 \\ 8 & 5 & - 9 & | & - 137 \\ - 1 & - 1 & 12 & | & 73\end{matrix}\right)$

$R 2 \leftarrow R 2 - 8 R 1$; $R 3 \leftarrow R 3 + R 1$

$A = \left(\begin{matrix}1 & 4 & 10 & | & 25 \\ 0 & - 27 & - 89 & | & - 337 \\ 0 & 3 & 22 & | & 98\end{matrix}\right)$

$R 2 \leftarrow R 2 + 9 R 3$

$A = \left(\begin{matrix}1 & 4 & 10 & | & 25 \\ 0 & 0 & 109 & | & 545 \\ 0 & 3 & 22 & | & 98\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{109}$

$A = \left(\begin{matrix}1 & 4 & 10 & | & 25 \\ 0 & 0 & 1 & | & 5 \\ 0 & 3 & 22 & | & 98\end{matrix}\right)$

$R 1 \leftarrow R 1 - 10 R 2$; $R 3 \leftarrow R 3 - 22 R 2$

$A = \left(\begin{matrix}1 & 4 & 0 & | & - 25 \\ 0 & 0 & 1 & | & 5 \\ 0 & 3 & 0 & | & - 12\end{matrix}\right)$

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

$A = \left(\begin{matrix}1 & 4 & 0 & | & - 25 \\ 0 & 0 & 1 & | & 5 \\ 0 & 1 & 0 & | & - 4\end{matrix}\right)$

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

$A = \left(\begin{matrix}1 & 0 & 0 & | & - 9 \\ 0 & 0 & 1 & | & 5 \\ 0 & 1 & 0 & | & - 4\end{matrix}\right)$

Thus $x = - 9$, $y = - 4$ and $z = 5$