How do you solve the system #3x+2y-3z=-2#, #7x-2y+5z=-14#, #2x+4y+z=6#?

1 Answer
Mar 13, 2018

Answer:

#x=-22/13#, #y=29/13# and #z=6/13#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((3,2,-3,|,-2),(7,-2,5,|,-14),(2,4,1,|,6))#

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

#R1larrR1-R3#; #R2larrR2-3R3#

#A=((1,-2,-4,|,-8),(1,-14,2,|,-32),(2,4,1,|,6))#

#R2larrR2-R1#; #R3larrR3-2R1#

#A=((1,-2,-4,|,-8),(0,-12,6,|,-24),(0,8,9,|,22))#

#R3larrR3+2/3R2#

#A=((1,-2,-4,|,-8),(0,-12,6,|,-24),(0,0,13,|,6))#

#R3larr(R3)/13#

#A=((1,-2,-4,|,-8),(0,-12,6,|,-24),(0,0,1,|,6/13))#

#R1larrR1+4R3#; #R2larrR2-6R3#

#A=((1,-2,0,|,-80/13),(0,-12,0,|,-348/13),(0,0,1,|,6/13))#

#R2larr(R2)/(-12)#

#A=((1,-2,0,|,-80/13),(0,1,0,|,29/13),(0,0,1,|,6/13))#

#R1larrR1+2R2#

#A=((1,0,0,|,-22/13),(0,1,0,|,29/13),(0,0,1,|,6/13))#

Thus #x=-22/13#, #y=29/13# and #z=6/13#