# How do you solve the system x+2y-4z=13, 3x+4y-2z=19, and 3x+2z=3?

Dec 16, 2017

$x = 2$, $y = \frac{5}{2}$ and $z = - \frac{3}{2}$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & 2 & - 4 & | & 13 \\ 3 & 4 & - 2 & | & 19 \\ 3 & 0 & 2 & | & 3\end{matrix}\right)$

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

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

$A = \left(\begin{matrix}1 & 2 & - 4 & | & 13 \\ 0 & - 2 & 10 & | & - 20 \\ 0 & - 6 & 14 & | & - 36\end{matrix}\right)$

$R 3 \leftarrow R 3 - 3 R 2$

$A = \left(\begin{matrix}1 & 2 & - 4 & | & 13 \\ 0 & - 2 & 10 & | & - 20 \\ 0 & 0 & - 16 & | & 24\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{- 16}$

$A = \left(\begin{matrix}1 & 2 & - 4 & | & 13 \\ 0 & - 2 & 10 & | & - 20 \\ 0 & 0 & 1 & | & - \frac{3}{2}\end{matrix}\right)$

$R 1 \leftarrow R 1 + 4 R 3$ ; $R 2 \leftarrow R 2 - 10 R 3$

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

$R 1 \leftarrow R 1 + R 2$

$A = \left(\begin{matrix}1 & 0 & 0 & | & 2 \\ 0 & - 2 & 0 & | & - 5 \\ 0 & 0 & 1 & | & - \frac{3}{2}\end{matrix}\right)$

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

$A = \left(\begin{matrix}1 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & | & \frac{5}{2} \\ 0 & 0 & 1 & | & - \frac{3}{2}\end{matrix}\right)$

Thus, solution of equation system is $x = 2$, $y = \frac{5}{2}$ and $z = - \frac{3}{2}$