# How do you solve using gaussian elimination or gauss-jordan elimination, 3y + 2z = 4, 2x − y − 3z = 3, 2x+ 2y − z = 7?

Dec 23, 2017

$\left\{\left(\frac{13 + 7 p}{6} , \frac{4 - 2 p}{3} , p\right) , p \in \mathbb{R}\right\}$

#### Explanation:

$0 x + 3 y + 2 z = 4$
2x−y−3z=3
2x+2y−z=7

$\left(\begin{matrix}0 & 3 & 2 \quad \quad | 4 \\ 2 & - 1 & - 3 | 3 \\ 2 & 2 & - 1 | 7\end{matrix}\right) \approx \left(\begin{matrix}2 & - 1 & - 3 | 3 \\ 2 & 2 & - 1 | 7 \\ 0 & 3 & 2 \quad \quad | 4\end{matrix}\right) \approx$
${R}_{3} \Leftrightarrow {R}_{1}$
${R}_{2} = {R}_{2} - {R}_{1}$

$\approx \left(\begin{matrix}2 & - 1 & - 3 | 3 \\ 0 & 3 & 2 \quad | 4 \\ 0 & 3 & 2 \quad | 4\end{matrix}\right) \approx \left(\begin{matrix}2 & - 1 & - 3 | 3 \\ 0 & 3 & 2 \quad | 4 \\ 0 & 0 & 0 \quad | 0\end{matrix}\right) \approx$
${R}_{3} = {R}_{3} - {R}_{2}$
${R}_{2} = \frac{1}{3} \cdot {R}_{2}$

$\approx \left(\begin{matrix}2 & - 1 & - 3 | 3 \\ 0 & 1 & \frac{2}{3} \quad | \frac{4}{3} \\ 0 & 0 & 0 \quad | 0\end{matrix}\right) \approx \left(\begin{matrix}2 & 0 & - \frac{7}{3} | \frac{13}{3} \\ 0 & 1 & \frac{2}{3} \quad | \frac{4}{3} \\ 0 & 0 & 0 \quad | 0\end{matrix}\right) \approx \left(\begin{matrix}1 & 0 & - \frac{7}{6} | \frac{13}{6} \\ 0 & 1 & \frac{2}{3} \quad | \frac{4}{3} \\ 0 & 0 & 0 \quad | 0\end{matrix}\right)$
${R}_{1} = {R}_{1} + {R}_{2}$
${R}_{1} = \frac{1}{2} \cdot {R}_{1}$

$x - \frac{7}{6} z = \frac{13}{6}$

$y + \frac{2}{3} z = \frac{4}{3}$

Last row has all zeros which means we have to substitute for parameter. z seems to be the best way to go.

$z = p$ (p as parameter)

$x - \frac{7}{6} p = \frac{13}{6} \quad \implies \quad x = \frac{13}{6} + \frac{7}{6} p = \frac{13 + 7 p}{6}$

$y + \frac{2}{3} p = \frac{4}{3} \quad \implies \quad y = \frac{4}{3} - \frac{2}{3} p = \frac{4 - 2 p}{3}$

$\left\{\left(\frac{13 + 7 p}{6} , \frac{4 - 2 p}{3} , p\right) , p \in \mathbb{R}\right\}$