I have $3$ equations and I want to find $x,y,z$. I don't want to use gauss elimination so I find another method to do it ,but I am not sure if it will work in all cases. Example I have first equation:
$2x-3y+z=13\tag1$
second equation
$4x+y-5z=-9\tag2$
third equation
$$-2x+4y+3z=-8$$
My system is first $(1) + (2) =$ new equation $A$ .When i will do that i will eliminate on of this $x,y,z$ i mean in first equation and second equation i will have multiple with a number so one variable $(x,y,z)$ will be opposite.In this case i will multiple the second equation by $3$, so my $y$ will be $+3y$.When i add $(1)$ + $(2)$ (but i will have multiple by $3$) the $y$ will gone. As a result it will be,
2x-3y+z=13 + 3*(4x+y-5z=-9) ->
2x-3y+z=13 + 12x+3y-5z=9 -> 14x-4z=22 this is A the new equation.
I will do the same with second equation and the third equation.I will add those but first i will eliminate a variable $(x,y,z)$. In this case it will be $-4$ so that $y$ will gone again.
4x+y-5z=-9 + -2x+4y+3z=-8 -> -4*(4x+y-5z=-9) + -2x+4y+3z=-8 ->
-16x-4y-20z=36 + -2x+4y+3z=-8 -> -18x-17z=+28 this is new B equation.
I have $A$ equation and $B$ equation i have eliminated $y$ and i have now $2\times 2$ system.What i am gona do next is eliminate another variable $(x,y,x)$ in this case i have already eliminated $y$,so i will choose between $x$ or $z$.I choose $x$. I will add in equation $A+4$ and it will be $18x-8z=26$ and i will add the $B$ equation $-18x-17z=28$. The result of $A+B$ is $-25z=54\Rightarrow z=-2.16$. Where did I go wrong? Why doesn’t this method work? After I will replace $z$ on equation $A$ or $B$ and I will find $x$.
Your mistake is in
$ 3(4x+y-5z=-9) = 12x +3y -15 z = -27 $
IMHO, Gauss elimination is much cleaner and is essentially a more organized representation of what you are trying to do.