SOLUTION: Linear Systems 2x+3y-z=1 x-y-z=2 3x+2y-3z=3

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Question 20130: Linear Systems

2x+3y-z=1
x-y-z=2
3x+2y-3z=3

Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
2x + 3y -  z = 1
x - y - z = 2
3x + 2y - 3z = 3
`
Get it in echelon form
`
Ax + By + Cz = D
Ey + Fz = G
Hz = I
`
2x + 3y - z = 1
x - y - z = 2
3x + 2y - 3z = 3
`
Get rid of the x in the 2nd equation by multiplying it by -2,
getting -2x + 2y + 2z = -4, and adding 1 times the first equation to it,
getting 5y + z = -3. Replace the second equation with this one
`
2x + 3y - z = 1
5y + z - -3
3x + 2y - 3z = 3
`
Get rid of the 3x in the 3rd equation by multiplying it by 2,
getting 6x + 4y - 6z = 6, and adding -3 times the first equation to it,
namely -6x - 9y + 3z = -3, getting -5y - 3z = 3. Replace the third equation
with this one
`
2x + 3y - z = 1
5y + z = -3
-5y - 3z = 3
`
Get rid of the -5y in the 3rd equation by multiplying it by 1,
and adding 1 times the second equation to it, getting -2z = 0
Replace the third equation with this one
`
2x + 3y - z = 1
5y + z = -3
-2z = 0
`
Finish by:
(1) solving the third equation for z
(2) Substitute the value you got for z into the second equation, and
solve for y
(3) Substitute the values of z and y into the first equation and solve
for x. Answer (x, y, z) = (7/5, -3/5, 0)
Edwin

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