# How do you solve the system x+3y-2z=8, 3x+2y-3z=15, and 4x+2y+3z=-1?

Jan 3, 2018

$x = \frac{134}{47}$, $y = \frac{24}{47}$ and $z = - \frac{85}{47}$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & 3 & - 2 & | & 8 \\ 3 & 2 & - 3 & | & 15 \\ 4 & 2 & 3 & | & - 1\end{matrix}\right)$

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

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

$A = \left(\begin{matrix}1 & 3 & - 2 & | & 8 \\ 0 & - 7 & 3 & | & - 9 \\ 0 & - 10 & 11 & | & - 25\end{matrix}\right)$

$R 2 \leftarrow R 2 - R 3$

$A = \left(\begin{matrix}1 & 3 & - 2 & | & 8 \\ 0 & 3 & - 8 & | & 16 \\ 0 & - 10 & 11 & | & - 25\end{matrix}\right)$

$R 1 \leftarrow R 1 - R 2$

$A = \left(\begin{matrix}1 & 0 & 6 & | & - 8 \\ 0 & 3 & - 8 & | & 16 \\ 0 & - 10 & 11 & | & - 25\end{matrix}\right)$

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

$A = \left(\begin{matrix}1 & 0 & 6 & | & - 8 \\ 0 & 1 & - \frac{8}{3} & | & \frac{16}{3} \\ 0 & - 10 & 11 & | & - 25\end{matrix}\right)$

$R 3 \leftarrow R 3 + 10 R 2$

$A = \left(\begin{matrix}1 & 0 & 6 & | & - 8 \\ 0 & 1 & - \frac{8}{3} & | & \frac{16}{3} \\ 0 & 0 & - \frac{47}{3} & | & \frac{85}{3}\end{matrix}\right)$

$R 3 \leftarrow R 3 \cdot \frac{- 3}{47}$

$A = \left(\begin{matrix}1 & 0 & 6 & | & - 8 \\ 0 & 1 & - \frac{8}{3} & | & \frac{16}{3} \\ 0 & 0 & 1 & | & - \frac{85}{47}\end{matrix}\right)$

$R 1 \leftarrow R 1 - 6 R 3$ ; $R 2 \leftarrow R 2 + \frac{8}{3} R 3$

$A = \left(\begin{matrix}1 & 0 & 0 & | & \frac{134}{47} \\ 0 & 1 & 0 & | & \frac{24}{47} \\ 0 & 0 & 1 & | & - \frac{85}{47}\end{matrix}\right)$

Thus $x = \frac{134}{47}$, $y = \frac{24}{47}$ and $z = - \frac{85}{47}$