How do you solve the system #x-4y+4z=-1#, #y-3z=5#, and #3x-4y+6z=1#?

1 Answer
Mar 12, 2018

Answer:

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

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

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

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

#R3larrR3-3R1#

#A=((1,-4,4,|,-1),(0,1,-3,|,5),(0,8,-6,|,4))#

#R1larrR1+4R2#; #R3larrR3-8R2#

#A=((1,0,-8,|,19),(0,1,-3,|,5),(0,0,18,|,-36))#

#R3larr(R3)/18#

#A=((1,0,-8,|,19),(0,1,-3,|,5),(0,0,1,|,-2))#

#R1larrR1+8R3#; #R2larrR2+3R3#

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

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