# How do you solve the system x-2y+8z=-4, x-2y+6z=-2, 2x-4y+19z=-11?

$\left(x , y , z\right) = t V + W , \forall t \in \mathbb{R}$

#### Explanation:

$\left(\begin{matrix}1 & - 2 & 8 \\ 1 & - 2 & 6 \\ 2 & - 4 & 19\end{matrix}\right) \cdot \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- 4 \\ - 2 \\ - 11\end{matrix}\right)$

$\left(\begin{matrix}1 & - 2 & 8 \\ 1 - 1 & - 2 + 2 & 6 - 8 \\ 2 - 2 & - 4 + 4 & 19 - 16\end{matrix}\right) \cdot \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- 4 \\ - 2 + 4 \\ - 11 + 8\end{matrix}\right)$

$\left(\begin{matrix}1 & - 2 & 8 \\ 0 & 0 & - 2 \\ 0 & 0 & 3\end{matrix}\right) \cdot \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- 4 \\ 2 \\ - 3\end{matrix}\right)$

$z = - 1$ and $x - 2 y - 8 = - 4 R i g h t a r r o w x = 2 y + 4$

$\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}2 y + 4 \\ y \\ - 1\end{matrix}\right) = y \cdot \left(\begin{matrix}2 \\ 1 \\ 0\end{matrix}\right) + \left(\begin{matrix}4 \\ 0 \\ - 1\end{matrix}\right)$