How do you solve the system #-6x-2y-z=-17#, #5x+y-6z=19#, and #-4x-6y-6z=-20#?

1 Answer
Jan 13, 2018

Answer:

#x=2#, #y=3# and #z=-1#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((-6,-2,-1,|,-17),(5,1,-6,|,19),(-4,-6,-6,|,-20))#

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

#R1larrR1+R2#; #R3larrR3+R2#

#A=((-1,-1,-7,|,2),(5,1,-6,|,19),(1,-5,-12,|,-1))#

#R2larr(R2)+5R1#; #R3larrR3+R1#

#A=((-1,-1,-7,|,2),(0,-4,-41,|,29),(0,-6,-19,|,1))#

#R1larrR1*(-1)#; #R2larrR2-R3#

#A=((1,1,7,|,-2),(0,2,-22,|,28),(0,-6,-19,|,1))#

#R3larrR3+3R2#

#A=((1,1,7,|,-2),(0,2,-22,|,28),(0,0,-85,|,85))#

#R3larr(R3)/(-85)#

#A=((1,1,7,|,-2),(0,2,-22,|,28),(0,0,1,|,-1))#

#R1larrR1-7R3#; #R2larrR1+22R3#

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

#R2larr(R2)/2#

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

#R1larrR1-R2#

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

Thus #x=2#, #y=3# and #z=-1#