# Question 18dba

Mar 21, 2017

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

#### Explanation:

Personally, I'd write it as a matrix and row reduce for workflow simplicity:

So :

$\left(\begin{matrix}2 & 3 & - 2 \\ 4 & - 3 & 1 \\ 1 & 2 & - 4\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- 4 \\ 25 \\ - 12\end{matrix}\right)$

Becomes in augmented form:

$\left(\begin{matrix}2 & 3 & - 2 \\ 4 & - 3 & 1 \\ 1 & 2 & - 4\end{matrix}\right) \left(\begin{matrix}- 4 \\ 25 \\ - 12\end{matrix}\right)$

R2 to R2 - 2 R1; R3 to R3 - 1/2 R1#

$\left(\begin{matrix}2 & 3 & - 2 \\ 0 & - 9 & 5 \\ 0 & \frac{1}{2} & - 3\end{matrix}\right) \left(\begin{matrix}- 4 \\ 33 \\ - 10\end{matrix}\right)$

$R 3 \to R 3 + \frac{1}{18} R 2$

$\left(\begin{matrix}2 & 3 & - 2 \\ 0 & - 9 & 5 \\ 0 & 0 & - \frac{49}{18}\end{matrix}\right) \left(\begin{matrix}- 4 \\ 33 \\ - \frac{49}{6}\end{matrix}\right)$

So we have:

$\left(\begin{matrix}2 & 3 & - 2 \\ 0 & - 9 & 5 \\ 0 & 0 & - \frac{49}{18}\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) \left(\begin{matrix}- 4 \\ 33 \\ - \frac{49}{6}\end{matrix}\right)$

We back substitute starting at the bottom:

From row 3: $- \frac{49}{18} z = - \frac{49}{6} \implies z = 3$

From row 2: $- 9 y + 5 \left(3\right) = 33 \implies y = - 2$

From row 1: $2 x + 3 \left(- 2\right) - 2 \left(3\right) = - 4 \implies x = 4$

So:

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