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

Answer:

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))#