# How do you solve the below equation using reduced row-echelon form?

## $2 x + 3 y - 4 z = 9$ $x + y + z = 3$ $2 x + y - 3 z = 2$

Feb 25, 2018

$x = - \frac{7}{11} , y = \frac{39}{11} , z = \frac{1}{11}$

#### Explanation:

Start with the augmented matrix $\left(A | b\right)$

$\left(\begin{matrix}2 & 3 & - 4 & | & 9 \\ 1 & 1 & 1 & | & 3 \\ 2 & 1 & - 3 & | & 2\end{matrix}\right)$

We carry out row reductions to transform this to the reduced row-echelon form. We first swap rows 1 and 2 (this is not necessary - but will cut down some of the computation)

$\left(\begin{matrix}1 & 1 & 1 & | & 3 \\ 2 & 3 & - 4 & | & 9 \\ 2 & 1 & - 3 & | & 2\end{matrix}\right)$

${R}_{2} \to {R}_{2} - 2 {R}_{1} , {R}_{3} \to {R}_{3} - 2 {R}_{1}$
$\left(\begin{matrix}1 & 1 & 1 & | & 3 \\ 0 & 1 & - 6 & | & 3 \\ 0 & - 1 & - 5 & | & - 4\end{matrix}\right)$

${R}_{1} \to {R}_{1} - {R}_{2} , {R}_{3} \to {R}_{3} + {R}_{2}$
$\left(\begin{matrix}1 & 0 & 7 & | & 0 \\ 0 & 1 & - 6 & | & 3 \\ 0 & 0 & - 11 & | & - 1\end{matrix}\right)$

${R}_{3} \to {R}_{3} / \left\{- 11\right\}$
$\left(\begin{matrix}1 & 0 & 7 & | & 0 \\ 0 & 1 & - 6 & | & 3 \\ 0 & 0 & 1 & | & \frac{1}{11}\end{matrix}\right)$

${R}_{1} \to {R}_{1} - 7 {R}_{3} , {R}_{2} \to {R}_{2} + 6 {R}_{3}$
$\left(\begin{matrix}1 & 0 & 0 & | & - \frac{7}{11} \\ 0 & 1 & 0 & | & \frac{39}{11} \\ 0 & 0 & 1 & | & \frac{1}{11}\end{matrix}\right)$

This gives
$x = - \frac{7}{11} , y = \frac{39}{11} , z = \frac{1}{11}$