# How do you solve using gaussian elimination or gauss-jordan elimination, 2x-3y-z=2, -x+2y-5z=-13, 5x-y-z=-5?

Dec 17, 2017

Taken you almost to the end. I will let you finish it off.
Check my working!!!

#### Explanation:

$\textcolor{g r e e n}{\left[\begin{matrix}2 & - 3 & - 1 & | & 2 \\ - 1 & 2 & - 5 & | & - 13 \\ 5 & - 1 & - 1 & | & - 50\end{matrix}\right]}$

${R}_{1} \div 2 \textcolor{w h i t e}{\text{ddd}} {R}_{3} \div 5$
$\textcolor{w h i t e}{\text{dddddd}} \downarrow$

$\textcolor{g r e e n}{\left[\begin{matrix}1 & - \frac{3}{2} & - \frac{1}{2} & | & 1 \\ - 1 & 2 & - 5 & | & - 13 \\ 1 & - \frac{1}{5} & - \frac{1}{5} & | & - 10\end{matrix}\right]}$

${R}_{1} + {R}_{2} \textcolor{w h i t e}{\text{ddd}} {R}_{3} + {R}_{2}$
$\textcolor{w h i t e}{\text{dddddd}} \downarrow$

$\textcolor{g r e e n}{\left[\begin{matrix}0 & \frac{1}{2} & - \frac{11}{2} & | & 1 \\ - 1 & 2 & - 5 & | & - 13 \\ 0 & \frac{9}{5} & - \frac{26}{5} & | & - 10\end{matrix}\right]}$

${R}_{1} \times 2 \textcolor{w h i t e}{\text{ddd}} {R}_{3} \times 5$
$\textcolor{w h i t e}{\text{dddddd}} \downarrow$

$\textcolor{g r e e n}{\left[\begin{matrix}0 & 1 & - 11 & | & 2 \\ - 1 & 2 & - 5 & | & - 13 \\ 0 & 9 & - 26 & | & - 50\end{matrix}\right]}$

${R}_{2} - 2 {R}_{1} \textcolor{w h i t e}{\text{ddd}} {R}_{3} - 9 {R}_{1}$
$\textcolor{w h i t e}{\text{dddddddd}} \downarrow$

$\textcolor{g r e e n}{\left[\begin{matrix}0 & 1 & - 11 & | & 2 \\ - 1 & 0 & - 27 & | & - 17 \\ 0 & 0 & - 125 & | & - 68\end{matrix}\right]}$

${R}_{2} \div \left(- 1\right) \textcolor{w h i t e}{\text{ddd}} {R}_{3} \div \left(- 125\right)$
$\textcolor{w h i t e}{\text{dddddddd}} \downarrow$

$\textcolor{g r e e n}{\left[\begin{matrix}0 & 1 & - 11 & | & 2 \\ 1 & 0 & 27 & | & 17 \\ 0 & 0 & 1 & | & + \frac{68}{125}\end{matrix}\right]}$

You should be able to finish this now.

PLEASE MAKE SURE THERE ARE NO MISTAKES. Not checked!