How do you solve the system #x+2y+5z=-1#, #2x-y+z=2#, #3x+4y-4y=14#?

1 Answer
Feb 2, 2018

Answer:

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

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,2,5,|,-1),(2,-1,1,|,2),(3,4,-4,|,14))#

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

#R2larrR2-2R1#; #R3larrR3-3R1#

#A=((1,2,5,|,-1),(0,-5,-9,|,4),(0,-2,-19,|,17))#

#R2larr(R2)/(-5)#

#A=((1,2,5,|,-1),(0,1,9/5,|,-4/5),(0,-2,-19,|,17))#

#R1larrR1-2R2#; #R3larrR3+2R2#

#A=((1,0,7/5,|,3/5),(0,1,9/5,|,-4/5),(0,0,-77/5,|,77/5))#

#R3larr(R3)*(-5/77)#

#A=((1,0,7/5,|,3/5),(0,1,9/5,|,-4/5),(0,0,1,|,-1))#

#R1larrR1-7/5R3#; #R2larrR2-9/5R3#

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

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