# How do you solve the following linear system: 2x-y+3z= 7 , x-2y-z=2, 4x+y-2z=-4 ?

Mar 7, 2017

$\left(x , y , z\right) = \left(\frac{3}{13} , - \frac{22}{13} , \frac{21}{13}\right)$

#### Explanation:

Taking:

$2 x - y + 3 z = 7 , x - 2 y - z = 2 , 4 x + y - 2 z = - 4$

We can write it in matrix form first:

$\left(\begin{matrix}2 & - 1 & 3 \\ 1 & - 2 & - 1 \\ 4 & 1 & - 2\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}7 \\ 2 \\ - 4\end{matrix}\right)$

Good thing about that presentation is that you can start looking for column and row vector linear dependence, in which case there is no solution.....or an infinite number of solutions.

But it looks OK, so we write the augmented matrix [can't get the type-setting right for that on Socratic so it will look a bit odd] and go to row reduction:

$\left(\begin{matrix}2 & - 1 & 3 \\ 1 & - 2 & - 1 \\ 4 & 1 & - 2\end{matrix}\right) \left(\begin{matrix}7 \\ 2 \\ - 4\end{matrix}\right)$

We'll switch R1 and R2 to get a $1$ at ${a}_{1 , 1}$ [pointless move, in hindsight]:

$\left(\begin{matrix}1 & - 2 & - 1 \\ 2 & - 1 & 3 \\ 4 & 1 & - 2\end{matrix}\right) \left(\begin{matrix}2 \\ 7 \\ - 4\end{matrix}\right)$

Then $R 2 \to R 2 - 2 R 1 , R 3 \to R 3 - 4 R 1$

$\left(\begin{matrix}1 & - 2 & - 1 \\ 0 & 3 & 5 \\ 0 & 9 & 2\end{matrix}\right) \left(\begin{matrix}2 \\ 3 \\ - 12\end{matrix}\right)$

$R 3 \to R 3 - 3 R 2$

$\left(\begin{matrix}1 & - 2 & - 1 \\ 0 & 3 & 5 \\ 0 & 0 & - 13\end{matrix}\right) \left(\begin{matrix}2 \\ 3 \\ - 21\end{matrix}\right)$

And now we can back substitute because we actually have:

$\left(\begin{matrix}1 & - 2 & - 1 \\ 0 & 3 & 5 \\ 0 & 0 & - 13\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}2 \\ 3 \\ - 21\end{matrix}\right)$

$- 13 z = - 21 \implies z = \frac{21}{13}$

Next:

$3 y + 5 z = 3 \implies y = \frac{3 - 5 z}{3} = - \frac{22}{13}$

And finally:

$x - 2 y - z = 2 \implies x = 2 + 2 y + z$

$\implies x = 2 - 2 \cdot \frac{22}{13} + \frac{21}{13} = \frac{3}{13}$

So:

$\left(x , y , z\right) = \left(\frac{3}{13} , - \frac{22}{13} , \frac{21}{13}\right)$