# How do you solve using gaussian elimination or gauss-jordan elimination, x_1 + 3x_2 +x_3 + x_4= 3, 2x_1- 2x_2 + x_3 + 2x_4 =8 and 3x_1 + x_2 + 2x_3 - x_4 =-1?

Jan 7, 2018

#### Answer:

$P = \left\{\left(\frac{- 6 - 5 p}{8} , \frac{2 - p}{8} , p , 3\right) , p \in \mathbb{R}\right\}$

#### Explanation:

$\left(\begin{matrix}1 & 3 & 1 & 1 & | & 3 \\ 2 & - 2 & 1 & 2 & | & 8 \\ 3 & 1 & 2 & - 1 & | & - 1\end{matrix}\right) \approx \left(\begin{matrix}1 & 3 & 1 & 1 & | & 3 \\ 0 & - 8 & - 1 & 0 & | & 2 \\ 0 & - 8 & - 1 & - 4 & | & - 10\end{matrix}\right)$
${R}_{2} = {R}_{2} - 2 \times {R}_{1}$
${R}_{3} = {R}_{3} - 3 \times {R}_{1}$
${R}_{3} = {R}_{3} - {R}_{2}$

$\left(\begin{matrix}1 & 3 & 1 & 1 & | & 3 \\ 0 & - 8 & - 1 & 0 & | & 2 \\ 0 & 0 & 0 & - 4 & | & - 12\end{matrix}\right) \approx \left(\begin{matrix}1 & 3 & 1 & 1 & | & 3 \\ 0 & - 8 & - 1 & 0 & | & 2 \\ 0 & 0 & 0 & 1 & | & 3\end{matrix}\right)$
${R}_{3} = {R}_{3} \times - \frac{1}{4}$
${R}_{1} = {R}_{1} - {R}_{3}$

$\left(\begin{matrix}1 & 3 & 1 & 0 & | & 0 \\ 0 & - 8 & - 1 & 0 & | & 2 \\ 0 & 0 & 0 & 1 & | & 3\end{matrix}\right) \approx \left(\begin{matrix}1 & 3 & 1 & 0 & | & 0 \\ 0 & 1 & \frac{1}{8} & 0 & | & \frac{1}{4} \\ 0 & 0 & 0 & 1 & | & 3\end{matrix}\right)$
${R}_{2} = {R}_{2} \times - \frac{1}{8}$
${R}_{1} = {R}_{1} - 3 \times {R}_{2}$

$\left(\begin{matrix}1 & 0 & \frac{5}{8} & 0 & | & - \frac{3}{4} \\ 0 & 1 & \frac{1}{8} & 0 & | & \frac{1}{4} \\ 0 & 0 & 0 & 1 & | & 3\end{matrix}\right)$

${x}_{3} = p \textcolor{w h i t e}{\mathbb{Q} \mathbb{Q} \mathbb{Q} Q} p \in \mathbb{R}$ (as a parameter)

$\implies {x}_{1} = - \frac{3}{4} - \frac{5}{8} p = \frac{- 6 - 5 p}{8}$

$\implies {x}_{2} = \frac{1}{4} - \frac{1}{8} p = \frac{2 - p}{8}$

$\implies {x}_{4} = 3$

$P = \left\{\left(\frac{- 6 - 5 p}{8} , \frac{2 - p}{8} , p , 3\right) , p \in \mathbb{R}\right\}$