# How do you solve using gaussian elimination or gauss-jordan elimination, 2x + y - z = -2, x + 3y + 2z = 4, 3x + 3y - 3z = -10?

Jul 20, 2017

$\left(x , y , z\right) = \left(\frac{4}{3} , - \frac{4}{3} , \frac{10}{3}\right)$

#### Explanation:

So let's make the matrix of the system, which is :

$\left[\begin{matrix}2 & 1 & - 1 & - 2 \\ 1 & 3 & 2 & 4 \\ 3 & 3 & - 3 & - 10\end{matrix}\right]$

Now let's do our calculations :

$\left[\begin{matrix}2 & 1 & - 1 & - 2 \\ 1 & 3 & 2 & 4 \\ 3 & 3 & - 3 & - 10\end{matrix}\right] \rightarrow \left[\begin{matrix}1 & - 2 & - 3 & - 6 \\ 1 & 3 & 2 & 4 \\ 0 & - 6 & - 9 & - 22\end{matrix}\right] \rightarrow$

$\left[\begin{matrix}1 & - 2 & - 3 & - 6 \\ 0 & 5 & 5 & 10 \\ 0 & - 6 & - 9 & - 22\end{matrix}\right] \rightarrow \left[\begin{matrix}1 & - 2 & - 3 & - 6 \\ 0 & 1 & 1 & 2 \\ 0 & - 6 & - 9 & - 22\end{matrix}\right] \rightarrow$

$\left[\begin{matrix}1 & - 2 & - 3 & - 6 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & - 3 & - 10\end{matrix}\right] \rightarrow \left[\begin{matrix}1 & 0 & - 1 & - 2 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & \frac{10}{3}\end{matrix}\right] \rightarrow$

$\left[\begin{matrix}1 & 0 & 0 & \frac{4}{3} \\ 0 & 1 & 0 & - \frac{4}{3} \\ 0 & 0 & 1 & \frac{10}{3}\end{matrix}\right]$

So we get : $\left(x , y , z\right) = \left(\frac{4}{3} , - \frac{4}{3} , \frac{10}{3}\right)$

(Just check my aritmetic, I tend to make mistakes)