# How do you solve using gaussian elimination or gauss-jordan elimination, -2x -5y +5z =4, -3x -y -z =10, 5x +3y -z =10?

Jan 6, 2018

$P = \text{[(43,-46,-28)]}$

#### Explanation:

$\left(\begin{matrix}- 2 & - 5 & 5 & | & 4 \\ - 3 & - 1 & - 1 & | & 10 \\ 5 & 3 & - 1 & | & 10\end{matrix}\right)$
${R}_{1} = {R}_{1} \times 5$
${R}_{2} = {R}_{2} \times 5$
${R}_{3} = {R}_{3} \times 3$

$\left(\begin{matrix}- 10 & - 25 & 25 & | & 20 \\ - 15 & - 5 & - 5 & | & 50 \\ 15 & 9 & - 3 & | & 30\end{matrix}\right)$
${R}_{2} = {R}_{2} + {R}_{3}$
${R}_{3} = {R}_{3} \times \frac{2}{3}$

$\left(\begin{matrix}- 10 & - 25 & 25 & | & 20 \\ 0 & 4 & - 8 & | & 80 \\ 10 & 6 & - 2 & | & 20\end{matrix}\right)$
${R}_{3} = {R}_{3} + {R}_{1}$
${R}_{1} = {R}_{1} \times \frac{1}{5}$
${R}_{2} = {R}_{2} \times \frac{1}{4}$

$\left(\begin{matrix}- 2 & - 5 & 5 & | & 4 \\ 0 & 1 & - 2 & | & 20 \\ 0 & - 19 & 23 & | & 40\end{matrix}\right)$
${R}_{3} = {R}_{3} + 19 \times {R}_{2}$

$\left(\begin{matrix}- 2 & - 5 & 5 & | & 4 \\ 0 & 1 & - 2 & | & 20 \\ 0 & 0 & - 15 & | & 40 + 20 \times 19\end{matrix}\right)$
${R}_{3} = {R}_{3} \times - \frac{1}{15}$

$\left(\begin{matrix}- 2 & - 5 & 5 & | & 4 \\ 0 & 1 & - 2 & | & 20 \\ 0 & 0 & 1 & | & - \frac{\cancel{5} \times 8 + \cancel{5} \times 4 \times 19}{\cancel{5} \times 3}\end{matrix}\right)$
${R}_{2} = {R}_{2} + 2 \times {R}_{3}$

$\left(\begin{matrix}- 2 & - 5 & 5 & | & 4 \\ 0 & 1 & 0 & | & 20 - 2 \times \frac{4 \times \left(2 + 19\right)}{3} \\ 0 & 0 & 1 & | & - \frac{4 \times \left(2 + 19\right)}{3}\end{matrix}\right)$
Simplifing right side

$\left(\begin{matrix}- 2 & - 5 & 5 & | & 4 \\ 0 & 1 & 0 & | & \frac{60 - 2 \times 84}{3} \\ 0 & 0 & 1 & | & - \frac{84}{3}\end{matrix}\right)$
${R}_{1} = {R}_{1} + 5 \times {R}_{2}$
${R}_{1} = {R}_{1} - 5 \times {R}_{3}$
${R}_{1} = {R}_{1} \times - \frac{1}{2}$

$\left(\begin{matrix}1 & 0 & 0 & | & - 2 + 5 \times \frac{60 - 198}{-} 6 - 5 \times \left(- \frac{84}{-} 6\right) \\ 0 & 1 & 0 & | & \frac{60 - 198}{3} \\ 0 & 0 & 1 & | & - 28\end{matrix}\right)$
Simplifing right side

$\left(\begin{matrix}1 & 0 & 0 & | & - 2 + 5 \times 6 \times \frac{10 - 33}{-} 6 - 5 \times 6 \times \left(\frac{14}{6}\right) \\ 0 & 1 & 0 & | & \frac{60 - 198}{3} \\ 0 & 0 & 1 & | & - 28\end{matrix}\right)$

$\textcolor{red}{x} = - 2 - 5 \times \left(10 - 33\right) - 5 \times 14 = - 2 + 115 - 70 \textcolor{red}{= 43}$

$\textcolor{red}{y} = \frac{60 - 198}{3} = 20 - 66 \textcolor{red}{= - 46}$

$\textcolor{red}{z = - 28}$

$P = \text{[(43,-46,-28)]}$