# How do you solve using gaussian elimination or gauss-jordan elimination, 2x₁+4x₂-6x₃= 10, 3x₁+3x₂-3x₃= 6?

Jan 6, 2018

$P = \left(\left\{- 1 - p , 3 + 2 p , p\right\} , p \in \mathbb{R}\right)$

#### Explanation:

$\left(\begin{matrix}2 & 4 & - 6 & | & 10 \\ 3 & 3 & - 3 & | & 6\end{matrix}\right) \approx \left(\begin{matrix}6 & 12 & - 18 & | & 30 \\ 6 & 6 & - 6 & | & 12\end{matrix}\right)$
${R}_{1} = {R}_{1} \times 3$
${R}_{2} = {R}_{2} \times 2$
${R}_{2} = {R}_{2} - {R}_{1}$

$\left(\begin{matrix}6 & 12 & - 18 & | & 30 \\ 0 & - 6 & 12 & | & - 18\end{matrix}\right) \approx \left(\begin{matrix}1 & 2 & - 3 & | & 5 \\ 0 & 1 & - 2 & | & 3\end{matrix}\right)$
${R}_{1} = {R}_{1} \times \frac{1}{6}$
${R}_{2} = {R}_{2} \times - \frac{1}{6}$
${R}_{1} = {R}_{1} - 2 \times {R}_{2}$

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

We have to set ${x}_{3}$ as parameter $p$:
${x}_{1} = - 1 - p$
${x}_{2} = 3 + 2 p$
${x}_{3} = p \textcolor{w h i t e}{\mathbb{Q} \mathbb{Q} \mathbb{Q} \mathbb{Q} \mathbb{Q} Q} p \in \mathbb{R}$

$P = \left(\left\{- 1 - p , 3 + 2 p , p\right\} , p \in \mathbb{R}\right)$