# How do you solve the system 3x+5y-2z=20, 4x-10y-z=-25, x+y-z=5?

Oct 30, 2017

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

#### Explanation:

We can solve this system of equations using an augmented matrix and Gauss Jordan Elimination. Recalling that we can use whichever order of the equations in the matrix we wish, I will opt to put the equation with the leading $x$ as the top row:

$\left[\begin{matrix}1 & 1 & - 1 & | & 5 \\ 3 & 5 & - 2 & | & 20 \\ 4 & - 10 & - 1 & | & - 25\end{matrix}\right]$

Now we begin with the row eliminations. I will use the notation ${R}_{n}$ to refer to row n:

$\left[\begin{matrix}1 & 1 & - 1 & | & 5 \\ 3 & 5 & - 2 & | & 20 \\ 4 & - 10 & - 1 & | & - 25\end{matrix}\right] \left.\begin{matrix}\null \\ - 3 {R}_{1} \to \\ - 4 {R}_{1}\end{matrix}\right.$

$\left[\begin{matrix}1 & 1 & - 1 & | & 5 \\ 0 & 2 & 1 & | & 5 \\ 0 & - 14 & 3 & | & - 45\end{matrix}\right] \left.\begin{matrix}\null \\ {R}_{2} \div 2 \to \\ + 7 {R}_{2}\end{matrix}\right.$

$\left[\begin{matrix}1 & 1 & - 1 & | & 5 \\ 0 & 1 & \frac{1}{2} & | & \frac{5}{2} \\ 0 & 0 & 10 & | & - 10\end{matrix}\right] \left.\begin{matrix}\null \\ \to \\ {R}_{3} \div 10\end{matrix}\right.$

$\left[\begin{matrix}1 & 1 & - 1 & | & 5 \\ 0 & 1 & \frac{1}{2} & | & \frac{5}{2} \\ 0 & 0 & 1 & | & - 1\end{matrix}\right] \left.\begin{matrix}+ {R}_{3} \\ - \frac{1}{2} {R}_{3} \to \\ \null\end{matrix}\right.$

$\left[\begin{matrix}1 & 1 & 0 & | & 4 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & - 1\end{matrix}\right] \left.\begin{matrix}- {R}_{2} \\ \to \\ \null\end{matrix}\right.$

$\left[\begin{matrix}1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & - 1\end{matrix}\right] \left.\begin{matrix}x \\ y \\ z\end{matrix}\right.$

We have arrived at the solution set $\left\{1 , 3 , - 1\right\}$. We can verify this using all of the original equations:

$\left\{\begin{matrix}3 \left(1\right) + 5 \left(3\right) - 2 \left(- 1\right) = 3 + 15 + 2 = 20 \\ 4 \left(1\right) - 10 \left(3\right) - \left(- 1\right) = 4 - 30 + 1 = - 25 \\ \left(1\right) + \left(3\right) - \left(- 1\right) = 5\end{matrix}\right.$