How do you solve the system #x+3y-2z=8#, #3x+2y-3z=15#, and #4x+2y+3z=-1#?

1 Answer
Jan 3, 2018

Answer:

#x=134/47#, #y=24/47# and #z=-85/47#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,3,-2,|,8),(3,2,-3,|,15),(4,2,3,|,-1))#

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

#R2larrR2-3R1# ; #R3larrR3-4R1#

#A=((1,3,-2,|,8),(0,-7,3,|,-9),(0,-10,11,|,-25))#

#R2larrR2-R3#

#A=((1,3,-2,|,8),(0,3,-8,|,16),(0,-10,11,|,-25))#

#R1larrR1-R2#

#A=((1,0,6,|,-8),(0,3,-8,|,16),(0,-10,11,|,-25))#

#R2larr(R2)/3#

#A=((1,0,6,|,-8),(0,1,-8/3,|,16/3),(0,-10,11,|,-25))#

#R3larrR3+10R2#

#A=((1,0,6,|,-8),(0,1,-8/3,|,16/3),(0,0,-47/3,|,85/3))#

#R3larrR3*(-3)/47#

#A=((1,0,6,|,-8),(0,1,-8/3,|,16/3),(0,0,1,|,-85/47))#

#R1larrR1-6R3# ; #R2larrR2+8/3R3#

#A=((1,0,0,|,134/47),(0,1,0,|,24/47),(0,0,1,|,-85/47))#

Thus #x=134/47#, #y=24/47# and #z=-85/47#