# How do you solve the system x-4y+4z=-1, y-3z=5, and 3x-4y+6z=1?

Mar 12, 2018

$x = 3$, $y = - 1$ and $z = - 2$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & - 4 & 4 & | & - 1 \\ 0 & 1 & - 3 & | & 5 \\ 3 & - 4 & 6 & | & 1\end{matrix}\right)$

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

$R 3 \leftarrow R 3 - 3 R 1$

$A = \left(\begin{matrix}1 & - 4 & 4 & | & - 1 \\ 0 & 1 & - 3 & | & 5 \\ 0 & 8 & - 6 & | & 4\end{matrix}\right)$

$R 1 \leftarrow R 1 + 4 R 2$; $R 3 \leftarrow R 3 - 8 R 2$

$A = \left(\begin{matrix}1 & 0 & - 8 & | & 19 \\ 0 & 1 & - 3 & | & 5 \\ 0 & 0 & 18 & | & - 36\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{18}$

$A = \left(\begin{matrix}1 & 0 & - 8 & | & 19 \\ 0 & 1 & - 3 & | & 5 \\ 0 & 0 & 1 & | & - 2\end{matrix}\right)$

$R 1 \leftarrow R 1 + 8 R 3$; $R 2 \leftarrow R 2 + 3 R 3$

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

Thus $x = 3$, $y = - 1$ and $z = - 2$