How do you solve the system #x+2y-4z=13#, #3x+4y-2z=19#, and #3x+2z=3#?

Question

1948 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-16T13:00:28

#x=2#, #y=5/2# and #z=-3/2#

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,2,-4,|,13),(3,4,-2,|,19),(3,0,2,|,3))#

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# ; #R3larrR3-3R1#

#A=((1,2,-4,|,13),(0,-2,10,|,-20),(0,-6,14,|,-36))#

#R3larrR3-3R2#

#A=((1,2,-4,|,13),(0,-2,10,|,-20),(0,0,-16,|,24))#

#R3larr(R3)/(-16)#

#A=((1,2,-4,|,13),(0,-2,10,|,-20),(0,0,1,|,-3/2))#

#R1larrR1+4R3# ; #R2larrR2-10R3#

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

#R1larrR1+R2#

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

#R2larr(R2)/(-2)#

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

Thus, solution of equation system is #x=2#, #y=5/2# and #z=-3/2#