How do you solve using gaussian elimination or gauss-jordan elimination, #x+y+z=1#, #3x+y-3z=5# and #x-2y-5z=10#?

1 Answer
Feb 13, 2016

Answer:

#((x),(y),(z)) = ((6),(-7),(2)) #

Explanation:

The linear equation:
#x+y+z =1 #
#3x+y-3z=5#
#x-2y-5z=10 #
can be written in Matrix form as:
#(x, y, z)((1,1,1),(3,1,-3),(1,-2,-5)) = ((1), (5), (10))#
Let's now use gauss elimination:
#((1,1,1, |, 1),(3,1,-3, |, 5),(1,-2,-5, |, 10))#

#color(red)(R_2=-3R_1+R_2)# #((1,1,1,|, 1),(0,-2,-6,|, 2),(1,-2,-5, |, 10))#

#color(red)(R_3 =-R_1+R_3)# #((1,1,1,|, 1),(0,-2,-6,|, 2),(0,-3,-6, |, 9))#

#color(red)(R_3 =-3/2R_2+R_3)# #((1,1,1,|, 1),(0,-2,-6,|, 2),(0,0,3, |, 6))#
Now what we have is:
#x+y+z= 1# --------------------------------------------------------- (1)
#" " -2y-6z = 2# ---------------------------------------------------------(2)
#" " 3z = 6 #; #z= 2# substitute this into (2) above solve for y
#y = -7# sub for y and z into (1) #x= 6 #