How do you solve the system x+2y+5z=-1, 2x-y+z=2, 3x+4y-4y=14?

Feb 2, 2018

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

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

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

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

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

$A = \left(\begin{matrix}1 & 2 & 5 & | & - 1 \\ 0 & - 5 & - 9 & | & 4 \\ 0 & - 2 & - 19 & | & 17\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 5}$

$A = \left(\begin{matrix}1 & 2 & 5 & | & - 1 \\ 0 & 1 & \frac{9}{5} & | & - \frac{4}{5} \\ 0 & - 2 & - 19 & | & 17\end{matrix}\right)$

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

$A = \left(\begin{matrix}1 & 0 & \frac{7}{5} & | & \frac{3}{5} \\ 0 & 1 & \frac{9}{5} & | & - \frac{4}{5} \\ 0 & 0 & - \frac{77}{5} & | & \frac{77}{5}\end{matrix}\right)$

$R 3 \leftarrow \left(R 3\right) \cdot \left(- \frac{5}{77}\right)$

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

$R 1 \leftarrow R 1 - \frac{7}{5} R 3$; $R 2 \leftarrow R 2 - \frac{9}{5} R 3$

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

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