# How do you solve the following system?: 7x+2y-z=3, 8x+3y+5z=9, 8x-8y-2z=4

Jan 15, 2018

The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}\frac{131}{219} \\ - \frac{29}{219} \\ \frac{202}{219}\end{matrix}\right)$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix,

The last equation is first

$A = \left(\begin{matrix}8 & - 8 & - 2 & | & 4 \\ 8 & 3 & 5 & | & 9 \\ 7 & 2 & - 1 & | & 3\end{matrix}\right)$

Perform the row operations

Make the first column the pivot , $R 1 \leftarrow \frac{R 1}{8}$

$\left(\begin{matrix}1 & - 1 & - \frac{1}{4} & | & \frac{1}{2} \\ 8 & 3 & 5 & | & 9 \\ 7 & 2 & - 1 & | & 3\end{matrix}\right)$

Eliminate the first column, $R 2 \leftarrow \left(R 2 - 8 R 1\right)$ and $R 3 \leftarrow \left(R 3 - 7 R 1\right)$

$\left(\begin{matrix}1 & - 1 & - \frac{1}{4} & | & \frac{1}{2} \\ 0 & 11 & 7 & | & 5 \\ 0 & 9 & \frac{3}{4} & | & - \frac{1}{2}\end{matrix}\right)$

Make the pivot in the second column, $R 2 \leftarrow \frac{R 2}{11}$

$\left(\begin{matrix}1 & - 1 & - \frac{1}{4} & | & \frac{1}{2} \\ 0 & 1 & \frac{7}{11} & | & \frac{5}{11} \\ 0 & 9 & \frac{3}{4} & | & - \frac{1}{2}\end{matrix}\right)$

Eliminate the second column, $R 3 \leftarrow \left(R 3 - 9 R 2\right)$

$\left(\begin{matrix}1 & - 1 & - \frac{1}{4} & | & \frac{1}{2} \\ 0 & 1 & \frac{7}{11} & | & \frac{5}{11} \\ 0 & 0 & - \frac{219}{44} & | & - \frac{101}{22}\end{matrix}\right)$

Make the pivot in the third column, $R 3 \leftarrow \left(R 3 \times - \frac{44}{219}\right)$

$\left(\begin{matrix}1 & - 1 & - \frac{1}{4} & | & \frac{1}{2} \\ 0 & 1 & \frac{7}{11} & | & \frac{5}{11} \\ 0 & 0 & 1 & | & \frac{202}{219}\end{matrix}\right)$

Eliminate the third column, $R 2 \leftarrow R 2 - \frac{7}{11} R 3$

$\left(\begin{matrix}1 & - 1 & - \frac{1}{4} & | & \frac{1}{2} \\ 0 & 1 & 0 & | & - \frac{29}{219} \\ 0 & 0 & 1 & | & \frac{202}{219}\end{matrix}\right)$

$R 1 \leftarrow \left(R 1 + R 2\right)$ and $R 1 \leftarrow R 1 + \frac{1}{4} R 3$

$\left(\begin{matrix}1 & 0 & 0 & | & \frac{131}{219} \\ 0 & 1 & 0 & | & - \frac{29}{219} \\ 0 & 0 & 1 & | & \frac{202}{219}\end{matrix}\right)$