Question #4c170

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Answer 1 · 2017-04-12T19:45:59

$x = 3, y = -1, and z = -4$

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

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

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

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

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

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

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

Perform Elementary Row Operations until an identity matrix is obtained.

$R_2+2R_1toR_2$

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

$R_3+5R_1toR_3$

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

$13R_3-37R_2toR_3$

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

$R_3/32toR_3$

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

$R_2- 3R_3toR_2$

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

$R_2/13toR_2$

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

$R_1 - 2R_3toR_1$

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

$R_1 - 6R_2toR_1$

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

$-1R_1toR_1$

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

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

Check:

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

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

This checks.