How do you solve the system #3x+5y-2z=20#, #4x-10y-z=-25#, #x+y-z=5#?

1 Answer
Oct 30, 2017

Answer:

#x = 1, y = 3, z = -1#

Explanation:

We can solve this system of equations using an augmented matrix and Gauss Jordan Elimination. Recalling that we can use whichever order of the equations in the matrix we wish, I will opt to put the equation with the leading #x# as the top row:

#[(1,1,-1,|,5),(3,5,-2,|,20),(4,-10,-1,|,-25)]#

Now we begin with the row eliminations. I will use the notation #R_n# to refer to row n:

#[(1,1,-1,|,5),(3,5,-2,|,20),(4,-10,-1,|,-25)] {:(),(-3R_1->),(-4R_1):}#

#[(1,1,-1,|,5),(0,2,1,|,5),(0,-14,3,|,-45)] {:(),(R_2 div 2->),(+7R_2):}#

#[(1,1,-1,|,5),(0,1,1/2,|,5/2),(0,0,10,|,-10)] {:(),(->),(R_3 div 10):}#

#[(1,1,-1,|,5),(0,1,1/2,|,5/2),(0,0,1,|,-1)] {:(+R_3),(-1/2R_3->),():}#

#[(1,1,0,|,4),(0,1,0,|,3),(0,0,1,|,-1)] {:(-R_2),(->),():}#

#[(1,0,0,|,1),(0,1,0,|,3),(0,0,1,|,-1)] {:(x),(y),(z):}#

We have arrived at the solution set #{1,3,-1}#. We can verify this using all of the original equations:

#{(3(1)+5(3)-2(-1)=3+15+2=20),(4(1)-10(3)-(-1)=4-30+1=-25),((1)+(3)-(-1)=5):} #