How do you solve using gaussian elimination or gauss-jordan elimination, 2x + 5y - 2z = 14, 5x -6y + 2z = 0, 4x - y + 3z = -7?

1 Answer
Jun 14, 2017

The solution is ((x),(y),(z))=((2),(0),(-5))

Explanation:

We perform the Gauss-Jordan elimination

((2,5,-2,14),(5,-6,2,0),(4,-1,3,-7))

Exchange R2 harr R1

((5,-6,2,0),(2,5,-2,14),(4,-1,3,-7))

Divide R1 by 5

((1,-1.2,0.4,0),(2,5,-2,14),(4,-1,3,-7))

R2 harr R2-2R1

((1,-1.2,0.4,0),(0,7.4,-2.8,14),(4,-1,3,-7))

R3 harr R3-4R1

((1,-1.2,0.4,0),(0,7.4,-2.8,14),(0,3.8,1.4,-7))

R2 harr (R2)/7.4

((1,-1.2,0.4,0),(0,1,-0.378,1.892),(0,3.8,1.4,-7))

R3 harr R3-3.8R2

((1,-1.2,0.4,0),(0,1,-0.378,1.892),(0,0,2.838,-14.189))

R3 harr (R3)/2.838

((1,-1.2,0.4,0),(0,1,-0.378,1.892),(0,0,1,-5))

R1 harr R1-0.4R3

((1,-1.2,0,2),(0,1,-0.378,1.892),(0,0,1,-5))

R2 harr R2+0.378R3

((1,-1.2,0,2),(0,1,0,0),(0,0,1,-5))

R1 harr R1+1.2R2

((1,0,0,2),(0,1,0,0),(0,0,1,-5))