My teacher said I have to solve it with gauss method, I tried to make for example $0y$ but I can't come to the result. If someone can help me I would appreciate that. $$2x+z=7$$ $$x+y=2$$ $$y-z=-2$$
I did like this: $$2x+0y+z=7$$ $$x+y+0z=2$$ $$ 0x+y-z=-2$$
I can solve your system of 3 equations and 3 unknowns for you. You have $$eq 1) 2x + 0y + z = 7$$ $$eq 2) x + y + 0z = 2$$ $$eq 3) 0z + y - z = -2$$ Now adding equations 1) and 3) I get $$ 2x + y = 5$$ Then subtracting equation 2) I find $$ x = 3 $$ Now solving equation 2) I see the result $$ y = -1$$ And finally using equation 1) again $$ 2*(3) + z = 7$$ and z = 1
So the answer is (x,y,z) = (3,-1, 1)
A good tutorial on the gauss method of elimination can be found at
http://www.purplemath.com/modules/systlin6.htm
This set of equations is in gaussian form when it is triangular. See link.
Here is another way of solving it, that uses gauss method of elimination. Consider eq1) - 2* eq2) $$0 -2y + z = 3$$ now add 2*eq3) $$0 + 0 - z = -1$$ So now we have a nice triangular set of equations that we can back substitute. This is half way done with a gauss method. $$x+y+0=2$$ $$0+y-z=-2$$ $$0+0+z=1$$ Now substituting z=1 in for the middle equation, we find that y=-1.
Using this fact and the top equation, we find that x = 3. Thus we have used the gauss method and the three answers are x=3, y=-1 and z=1.
Regards, Matt
second equation minus the third we obtain $x+z=4$ (1') (1)-(1'): we get $x=3$
This is a first step in finding x, y, and z.
