# Question 4c170

Apr 12, 2017

$x = 3 , y = - 1 , \mathmr{and} z = - 4$

#### Explanation:

Given:
$2 x + y - z = 9$
$- x + 6 y + 2 z = - 17$
$5 x + 7 y + z = 4$

Write the equation $- x + 6 y + 2 z = - 17$ as the first row of an Augmented Matrix :

[ (-1,6,2,|,-17) ]

Add a row for the equation $2 x + y - z = 9$:

[ (-1,6,2,|,-17), (2,1,-1,|,9) ]

Add a row for the equation $5 x + 7 y + z = 4$:

[ (-1,6,2,|,-17), (2,1,-1,|,9), (5,7,1,|,4) ]

Perform Elementary Row Operations until an identity matrix is obtained.

${R}_{2} + 2 {R}_{1} \to {R}_{2}$

[ (-1,6,2,|,-17), (0,13,3,|,-25), (5,7,1,|,4) ]

${R}_{3} + 5 {R}_{1} \to {R}_{3}$

[ (-1,6,2,|,-17), (0,13,3,|,-25), (0,37,11,|,-81) ]

$13 {R}_{3} - 37 {R}_{2} \to {R}_{3}$

[ (-1,6,2,|,-17), (0,13,3,|,-25), (0,0,32,|,-128) ]

${R}_{3} / 32 \to {R}_{3}$

[ (-1,6,2,|,-17), (0,13,3,|,-25), (0,0,1,|,-4) ]

${R}_{2} - 3 {R}_{3} \to {R}_{2}$

[ (-1,6,2,|,-17), (0,13,0,|,-13), (0,0,1,|,-4) ]

${R}_{2} / 13 \to {R}_{2}$

[ (-1,6,2,|,-17), (0,1,0,|,-1), (0,0,1,|,-4) ]

${R}_{1} - 2 {R}_{3} \to {R}_{1}$

[ (-1,6,0,|,-9), (0,1,0,|,-1), (0,0,1,|,-4) ]

${R}_{1} - 6 {R}_{2} \to {R}_{1}$

[ (-1,0,0,|,-3), (0,1,0,|,-1), (0,0,1,|,-4) ]

$- 1 {R}_{1} \to {R}_{1}$

[ (1,0,0,|,3), (0,1,0,|,-1), (0,0,1,|,-4) ]#

We have an identity matrix on the left, therefore, the solution set is on the right, $x = 3 , y = - 1 , \mathmr{and} z = - 4$.

Check:

$2 \left(3\right) + \left(- 1\right) - \left(- 4\right) = 9$
$- \left(3\right) + 6 \left(- 1\right) + 2 \left(- 4\right) = - 17$
$5 \left(3\right) + 7 \left(- 1\right) + \left(- 4\right) = 4$

$9 = 9$
$- 17 = - 17$
$4 = 4$

This checks.