How do you solve using gaussian elimination or gauss-jordan elimination, # 2x - y + 3z = 24#, #2y - z = 14#, #7x - 5y = 6#?

1 Answer
Sep 1, 2017

Answer:

The solution is #((x),(y),(z))=((8),(10),(6))#

Explanation:

The augmented matrix is

#((2,-1,3,:,24),(0,2,-1,:,14),(7,-5,0,:,6))#

We can perform the Gauss-Jordan elimination

#R3 harr R2#, #=>#, #((2,-1,3,:,24),(7,-5,0,:,6),(0,2,-1,:,14))#

#R2larrR2-3R1#, #=>#, #((2,-1,3,:,24),(1, -2,-9,:,-66),(0,2,-1,:,14))#

#R2larr2R2-R1#, #=>#, #((2,-1,3,:,24),(0, -3,-21,:,-156),(0,2,-1,:,14))#

#R3larr3R3+2R2#, #=>#, #((2,-1,3,:,24),(0, -3,-21,:,-156),(0,0,-45,:,-270))#

#R3larr(R3)/(-45)#, #=>#, #((2,-1,3,:,24),(0, -3,-21,:,-156),(0,0,1,:,6))#

#R2larrR2+21R3#, #=>#, #((2,-1,3,:,24),(0, -3,0,:,-30),(0,0,1,:,6))#

#R2larr(R2)/(-3)#, #=>#, #((2,-1,3,:,24),(0, 1,0,:,10),(0,0,1,:,6))#

#R1larrR1+R2#, #=>#, #((2,0,3,:,34),(0, 1,0,:,10),(0,0,1,:,6))#

#R1larrR1-3R3#, #=>#, #((2,0,0,:,16),(0, 1,0,:,10),(0,0,1,:,6))#

#R1larr(R1)/2#, #=>#, #((1,0,0,:,8),(0, 1,0,:,10),(0,0,1,:,6))#