How do you solve using gaussian elimination or gauss-jordan elimination, x+2y+2z=9, x+y+z=9, 3x-y+3z=10?

Apr 6, 2018

$x = 9$, $y = \frac{17}{4}$ and $z = - \frac{17}{4}$

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

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

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

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

$A = \left(\begin{matrix}1 & 2 & 2 & | & 9 \\ 0 & - 1 & - 1 & | & 0 \\ 0 & - 7 & - 3 & | & - 17\end{matrix}\right)$

$R 1 \leftarrow R 1 + 2 R 2$; $R 3 \leftarrow R 3 - 7 R 2$

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

$R 2 \leftarrow R 2 + \frac{1}{4} R 3$

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

$R 2 \leftarrow \frac{R 2}{- 1}$; $R 3 \leftarrow \frac{R 3}{4}$

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

Thus $x = 9$, $y = \frac{17}{4}$ and $z = - \frac{17}{4}$