# How do you solve using gaussian elimination or gauss-jordan elimination, 2x+4y-6z=48, x+2y+3z=-6, 3x-4y+4z=-23?

May 19, 2017

Using Gauss elimination gives $x = 3$, $y = 3$, and $z = - 5$

#### Explanation:

Step 1. Rewrite the equations in matrix form
$\left(\begin{matrix}2 & 4 & - 6 & | & 48 \\ 1 & 2 & 3 & | & - 6 \\ 3 & - 4 & 4 & | & - 23\end{matrix}\right)$

Step 2. Switch ${R}_{2}$ with ${R}_{1}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 2 & 4 & - 6 & | & 48 \\ 3 & - 4 & 4 & | & - 23\end{matrix}\right)$

Step 3. $- 2 {R}_{1} + {R}_{2} \to {R}_{2}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & 0 & - 12 & | & 60 \\ 3 & - 4 & 4 & | & - 23\end{matrix}\right)$

Step 4. $- 3 {R}_{1} + {R}_{3} \to {R}_{3}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & 0 & - 12 & | & 60 \\ 0 & - 10 & - 5 & | & - 5\end{matrix}\right)$

Step 5. Switch ${R}_{2}$ with ${R}_{3}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & - 10 & - 5 & | & - 5 \\ 0 & 0 & - 12 & | & 60\end{matrix}\right)$

Step 6. $- \frac{1}{10} {R}_{2} \to {R}_{2}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & 1 & \frac{1}{2} & | & \frac{1}{2} \\ 0 & 0 & - 12 & | & 60\end{matrix}\right)$

Step 7. $- \frac{1}{12} {R}_{3} \to {R}_{3}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & 1 & \frac{1}{2} & | & \frac{1}{2} \\ 0 & 0 & 1 & | & - 5\end{matrix}\right)$

Translating back into the values $x$, $y$, and $z$
$z = - 5$
$y + \frac{1}{2} \left(- 5\right) = \frac{1}{2} \implies y = 3$
$x + 2 \left(3\right) + 3 \left(- 5\right) = - 6 \implies x = 3$