How do you solve this system of equations: #-x + y - z = 2, 2x - 2y - z = - 1 , and 3x + 2y + z = 6#?

1 Answer
Dec 14, 2017

The solution is #((x),(y),(z))=((1),(2),(-1))#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

The augmented matrix is

#A=((-1,1,-1,|,2),(2,-2,-1,|,-1),(3,2,1,|,6))#

Perform the row operations

#R1larrR1xx(-1)#

#A=((1,-1,1,|,-2),(2,-2,-1,|,-1),(3,2,1,|,6))#

#R2larr(R2-2R1)#, and #R3larr(R3-3R1)#

#A=((1,-1,1,|,-2),(0,0,-3,|,3),(0,5,-2,|,12))#

#R2harrR3#

#A=((1,-1,1,|,-2),(0,5,-2,|,12),(0,0,-3,|,3))#

#R3larr(R3/-3)#

#A=((1,-1,1,|,-2),(0,5,-2,|,12),(0,0,1,|,-1))#

#R2larr(R2+2R3)#

#A=((1,-1,1,|,-2),(0,5,0,|,10),(0,0,1,|,-1))#

#R2larr(R2/2)#

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

#R1larr(R1+R2)#

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

#R1larr(R1-R3)#

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