# How do you solve -x + 2y - 3z = 0 , 2x + z = 0 , and 3x - 4y + 4z = 2 using matrices?

Mar 17, 2018

The solution is $S = \left(\begin{matrix}x = \frac{2}{5} \\ y = - 1 \\ z = - \frac{4}{5}\end{matrix}\right)$

#### Explanation:

Perform the Gauss- Jordan elimination on the augmented matrix

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

Make the pivot in the first column by

${R}_{1} \leftarrow \left(- 1\right) \times {R}_{1}$

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

Eliminate the first column by

${R}_{2} \leftarrow {R}_{2} - 2 {R}_{1}$ and ${R}_{3} \leftarrow {R}_{3} - 3 {R}_{1}$

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

Make the pivot in the second column by

${R}_{2} \leftarrow {R}_{2} / 4$

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

Eliminate the second column by

${R}_{3} \leftarrow {R}_{3} - 2 {R}_{2}$ and ${R}_{1} \leftarrow {R}_{1} + 2 {R}_{2}$

$= \left(\begin{matrix}1 & 0 & \frac{1}{2} & | & 0 \\ 0 & 1 & - \frac{5}{4} & | & 0 \\ 0 & 0 & - \frac{5}{2} & | & 2\end{matrix}\right)$

Make the pivot in the third column by

${R}_{3} \leftarrow {R}_{3} / \left(- \frac{5}{2}\right)$

$= \left(\begin{matrix}1 & 0 & \frac{1}{2} & | & 0 \\ 0 & 1 & - \frac{5}{4} & | & 0 \\ 0 & 0 & 1 & | & - \frac{4}{5}\end{matrix}\right)$

Eliminate the third column by

${R}_{1} \leftarrow {R}_{1} - \frac{1}{2} {R}_{3}$ and ${R}_{2} \leftarrow {R}_{2} + \frac{5}{4} {R}_{3}$

$= \left(\begin{matrix}1 & 0 & 0 & | & \frac{2}{5} \\ 0 & 1 & 0 & | & - 1 \\ 0 & 0 & 1 & | & - \frac{4}{5}\end{matrix}\right)$

The solution is

$S = \left(\begin{matrix}x = \frac{2}{5} \\ y = - 1 \\ z = - \frac{4}{5}\end{matrix}\right)$