# How do you solve the system 3x+2y-3z=-2, 7x-2y+5z=-14, 2x+4y+z=6?

Mar 13, 2018

$x = - \frac{22}{13}$, $y = \frac{29}{13}$ and $z = \frac{6}{13}$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

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

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

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

$A = \left(\begin{matrix}1 & - 2 & - 4 & | & - 8 \\ 1 & - 14 & 2 & | & - 32 \\ 2 & 4 & 1 & | & 6\end{matrix}\right)$

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

$A = \left(\begin{matrix}1 & - 2 & - 4 & | & - 8 \\ 0 & - 12 & 6 & | & - 24 \\ 0 & 8 & 9 & | & 22\end{matrix}\right)$

$R 3 \leftarrow R 3 + \frac{2}{3} R 2$

$A = \left(\begin{matrix}1 & - 2 & - 4 & | & - 8 \\ 0 & - 12 & 6 & | & - 24 \\ 0 & 0 & 13 & | & 6\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{13}$

$A = \left(\begin{matrix}1 & - 2 & - 4 & | & - 8 \\ 0 & - 12 & 6 & | & - 24 \\ 0 & 0 & 1 & | & \frac{6}{13}\end{matrix}\right)$

$R 1 \leftarrow R 1 + 4 R 3$; $R 2 \leftarrow R 2 - 6 R 3$

$A = \left(\begin{matrix}1 & - 2 & 0 & | & - \frac{80}{13} \\ 0 & - 12 & 0 & | & - \frac{348}{13} \\ 0 & 0 & 1 & | & \frac{6}{13}\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 12}$

$A = \left(\begin{matrix}1 & - 2 & 0 & | & - \frac{80}{13} \\ 0 & 1 & 0 & | & \frac{29}{13} \\ 0 & 0 & 1 & | & \frac{6}{13}\end{matrix}\right)$

$R 1 \leftarrow R 1 + 2 R 2$

$A = \left(\begin{matrix}1 & 0 & 0 & | & - \frac{22}{13} \\ 0 & 1 & 0 & | & \frac{29}{13} \\ 0 & 0 & 1 & | & \frac{6}{13}\end{matrix}\right)$

Thus $x = - \frac{22}{13}$, $y = \frac{29}{13}$ and $z = \frac{6}{13}$