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

1 Answer
Feb 2, 2018

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