How do you solve #x + y + z = 3#, #-x + 3y + 2z = -8#, and #5y + z = 2#?

1 Answer
Jan 13, 2018

Answer:

#x=5#, #y=1# and #z=-3#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

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

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+R1#;

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

#R2larrR2*5#; #R3larrR3*4#

#A=((1,1,1,|,3),(0,20,15,|,-25),(0,20,4,|,8))#

#R3larrR3-R2#

#A=((1,1,1,|,3),(0,20,15,|,-25),(0,0,-11,|,33))#

#R3larr(R3)/(-11)#

#A=((1,1,1,|,3),(0,20,15,|,-25),(0,0,1,|,-3))#

#R1larrR1-R3#; #R2larrR2-15R3#

#A=((1,1,0,|,6),(0,20,0,|,20),(0,0,1,|,-3))#

#R2larr(R2)/20#

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

#R1larrR1-R2#

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

Thus #x=5#, #y=1# and #z=-3#