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Question 63463: 2x-y-z=-3
x-2y+3z=6
x+y+z=6
use matrices to solve thank you
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! 2x-y-z=-3
x-2y+3z=6
x+y+z=6
(R1) 2 -1 -1 = -3
(R2) 1 -2 3 = 6
(R3) 1 1 1 = 6
R1 + R3 ---->R1
(R1) 3 0 0 = 3
(R2) 1 -2 3 = 6
(R3) 1 1 1 = 6
R2 - R3---->R2
(R1) 3 0 0 = 3
(R2) 0 -3 2 = 0
(R3) 1 1 1 = 6
3R3------>R3
(R1) 3 0 0 = 3
(R2) 0 -3 2 = 0
(R3) 3 3 3 = 18
R3 - R1 ------>R3
(R1) 3 0 0 = 3
(R2) 0 -3 2 = 0
(R3) 0 3 3 = 15
R3 + R2 ------->R3
(R1) 3 0 0 = 3
(R2) 0 -3 2 = 0
(R3) 0 0 5 = 15
5R2---->R2 and 2R3----->R3
(R1) 3 0 0 = 3
(R2) 0 -15 10 = 0
(R3) 0 0 10 = 30
R2-R3----R2
(R1) 3 0 0 = 3
(R2) 0 -15 0 = -30
(R3) 0 0 10 = 30
(R1)/3---->R1; (R2)/-15---->R2; (R3)/10---->R3
(R1) 1 0 0 = 1
(R2) 0 1 0 = 2
(R3) 0 0 1 = 3
x=1
y=2
z=3
Hope this helps. Have a nice holiday season---ptaylor
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