Question #4c170

1 Answer
Apr 12, 2017

Answer:

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

Explanation:

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.