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

Jan 6, 2018

$P = \left[\left(- 2 , 1 , 0\right)\right]$

#### Explanation:

I'm assuming that in the second row is missing y. Solving this matrix:

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

$\left(\begin{matrix}6 & 6 & - 9 & | & - 6 \\ - 6 & 2 & 4 & | & - 2 \\ - 6 & - 9 & 15 & | & 9\end{matrix}\right)$
${R}_{2} = {R}_{2} + {R}_{1}$
${R}_{3} = {R}_{3} + {R}_{1}$

$\left(\begin{matrix}6 & 6 & - 9 & | & - 6 \\ 0 & 8 & - 5 & | & - 8 \\ 0 & - 3 & 6 & | & 3\end{matrix}\right)$
${R}_{2} = {R}_{2} \times 3$
${R}_{3} = {R}_{3} \times 8$

$\left(\begin{matrix}6 & 6 & - 9 & | & - 6 \\ 0 & 24 & - 15 & | & - 24 \\ 0 & - 24 & 48 & | & 24\end{matrix}\right)$
${R}_{3} = {R}_{3} + {R}_{2}$
${R}_{1} = {R}_{1} \times \frac{1}{3}$

$\left(\begin{matrix}2 & 2 & - 3 & | & - 2 \\ 0 & 24 & - 15 & | & - 24 \\ 0 & 0 & 33 & | & 0\end{matrix}\right)$
${R}_{2} = {R}_{2} \times \frac{1}{3}$
${R}_{3} = {R}_{3} \times \frac{5}{33}$

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

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

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

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

$P = \left[\left(- 2 , 1 , 0\right)\right]$

(too long)