How can I solve this equation system? #x + 5y + 2z = 19" "3x + 2y+ z = 13" "2x - y - z = 0# How should I think about this?

#x+5 y+2 z=19;(1), 3 x+2 y+ z=13; (2)#

#2 x- y- z=0 ; (3) :. 2 x = y+z# . Multiplying equation (1)

by #2# we get , #2 x + 10 y + 4 z = 38 ; # or

#y +z + 10 y + 4 z = 38 ; or 11 y +5 z =38; (4) #

Multiplying equation (2) by #2# we get ,

#6 x + 4 y + 2 z = 26 ; # or

#3(y +z) + 4 y + 2 z = 26 ; or 7 y +5 z =26; (5) #

Subtracting equation (5) from equation (4) we get,

#4 y = 12 or y =3 :. 5 z= 38 - 11 y # or

# 5 z = 38 - 11*3 or 5 z= 5 or z=1 #

#:. 2 x= y+ z or 2 x= 3 +1 or 2 x =4 or x=2 #

Solution:# x=2 , y=3 , z=1# [Ans]

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,5,2,|,19),(3,2,1,|,13),(2,-1,-1,|,0))#

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#; #R3larrR3-2R1#

#A=((1,5,2,|,19),(0,-13,-5,|,-44),(0,-11,-5,|,-38))#

#R2larrR2-R3#

#A=((1,5,2,|,19),(0,-2,0,|,-6),(0,-11,-5,|,-38))#

#R2larr(R2)/(-2)#

#A=((1,5,2,|,19),(0,1,0,|,3),(0,-11,-5,|,-38))#

#R1larrR1-5R2#; #R3larrR3+11R2#

#A=((1,0,2,|,4),(0,1,0,|,3),(0,0,-5,|,-5))#

#R3larr(R3)/(-5)#

#A=((1,0,2,|,4),(0,1,0,|,3),(0,0,1,|,1))#

#R1larrR1-2R3#

#A=((1,0,0,|,2),(0,1,0,|,3),(0,0,1,|,1))#

Thus #x=2#, #y=3# and #z=1#