# How do you solve using gaussian elimination or gauss-jordan elimination, x + y + z - 3t = 1, 2x + y + z - 5t = 0, y + z - t = 2,  3x - 2z + 2t = -7?

Jan 4, 2018

$P = \left(\left\{- 1 + 2 t , \quad - 3 t , \quad 2 + 4 t\right\} , t \in \mathbb{R}\right)$

#### Explanation:

$\left(\begin{matrix}1 & 1 & 1 & - 3 & | & 1 \\ 2 & 1 & 1 & - 5 & | & 0 \\ 0 & 1 & 1 & - 1 & | & 2 \\ 3 & 0 & - 2 & 2 & | & - 7\end{matrix}\right)$
${R}_{2} = {R}_{2} - 2 \times {R}_{1}$
${R}_{4} = {R}_{4} - 3 \times {R}_{1}$

$\left(\begin{matrix}1 & 1 & 1 & - 3 & | & 1 \\ 2 - 2 & 1 - 2 & 1 - 2 & - 5 + 6 & | & 0 - 2 \\ 0 & 1 & 1 & - 1 & | & 2 \\ 3 - 3 & 0 - 3 & - 2 - 3 & 2 + 9 & | & - 7 - 3\end{matrix}\right) \approx \left(\begin{matrix}1 & 1 & 1 & - 3 & | & 1 \\ 0 & - 1 & - 1 & 1 & | & - 2 \\ 0 & 1 & 1 & - 1 & | & 2 \\ 0 & - 3 & - 5 & 11 & | & - 10\end{matrix}\right)$
${R}_{3} = {R}_{3} + {R}_{2}$
${R}_{4} = {R}_{4} - 3 \times {R}_{2}$

( [1,1,1,-3,|,1],[0,-1,-1,1,|,-2],[0,1-1,1-1,-1+1,|,2-2],[0,-3+3,-5+3,11-3,|,-10+6] )~~( [1,1,1,-3,|,1],[0,-1,-1,1,|,-2],[0,0,0,0,|,0],[0,0,-2,8,|,-4] )~#
${R}_{4} = - \frac{1}{2} \times {R}_{4}$
${R}_{4} \Leftrightarrow {R}_{3}$
${R}_{2} = - 1 \times {R}_{2}$

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

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

$\left(\begin{matrix}1 & 1 - 1 & 1 - 1 & - 3 - 3 + 4 & | & 1 - 2 \\ 0 & 1 & 0 & 3 & | & 0 \\ 0 & 0 & 1 & - 4 & | & 2 \\ 0 & 0 & 0 & 0 & | & 0\end{matrix}\right) \approx \left(\begin{matrix}1 & 0 & 0 & - 2 & | & - 1 \\ 0 & 1 & 0 & 3 & | & 0 \\ 0 & 0 & 1 & - 4 & | & 2 \\ 0 & 0 & 0 & 0 & | & 0\end{matrix}\right)$

$x - 2 t = - 1$
$y + 3 t = 0$
$z - 4 t = 2$

Let $t$ be a parameter, $t \in \mathbb{R}$

$x = - 1 + 2 t$
$y = - 3 t$
$z = 2 + 4 t$

$P = \left(\left\{- 1 + 2 t , \quad - 3 t , \quad 2 + 4 t\right\} , t \in \mathbb{R}\right)$