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.
