How do you solve #x + 2y + z = 6#, #2x - y - z = 0#, and #3x + 2y +z = 10# using matrices?

Question

5432 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-17T05:03:18

#x=2#
#y=0#
#z=4#

Given -

#x+2y+z=6#
#2x-y-z=0#
#3x+2y+z=10#
Answer is developed from the template I created in Excel

enter image source here

Answer 2 · 2017-12-17T21:06:59

#x=2#, #y=0# and #z=4#

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,2,1,|,6),(2,-1,-1,|,0),(3,2,1,|,10))#

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

#R2larrR2-2R1# ; #R3larrR3-3R1#

#A=((1,2,1,|,6),(0,-5,-3,|,-12),(0,-4,-2,|,-8))#

#R1larrR1-(R3)/2# ; #R2larrR2-R3#

#A=((1,0,0,|,2),(0,-1,-1,|,-4),(0,-4,-2,|,-8))#

#R3larrR3-4R2#

#A=((1,0,0,|,2),(0,-1,-1,|,-4),(0,0,2,|,8))#

#R2larr(R2)/(-1)# ; #R3larr(R3)/2#

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

#R2larrR2-R3#

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

Thus #x=2#, #y=0# and #z=4#