How do you solve the system #x-y-2z=-6#, #3x+2y=-25#, and #-4x+y-z=12#?

Question

2623 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-28T19:17:51

#x=-5#, #y=-5# and #z=3#

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,-1,-2,|,-6),(3,2,0,|,-25),(-4,1,-1,|,12))#

I have written the equations not in the sequence as in the question in order to get #1# as pivot.

Perform the folowing operations on the rows of the matrix

#R2larrR2-3R1#, #R3larrR3+4R1#;

#A=((1,-1,-2,|,-6),(0,5,6,|,-7),(0,-3,-9,|,-12))#

#R3larr(R3)/(-3)#;

#A=((1,-1,-2,|,-6),(0,5,6,|,-7),(0,1,3,|,4))#

#R1larrR1+R3#, #R2larrR2-5R3#;

#A=((1,0,1,|,-2),(0,0,-9,|,-27),(0,1,3,|,4))#

#R2larr(R2)/(-9)#;

#A=((1,0,1,|,-2),(0,0,1,|,3),(0,1,3,|,4))#

#R1larrR1-R2#, #R3larrR3-3R2#;

#A=((1,0,0,|,-5),(0,0,1,|,3),(0,1,0,|,-5))#

Thus, #x=-5#, #y=-5# and #z=3#