Any linear equation people here? Need some help.

[Ace]

So I'm doing an a bit of a pre-assessment for something, and I feel like I am missing something on this question:

Solve the following and explain your answer:

x₁ - 2x₂ + 3x₃ = 0

-2x₁ - 3x₂ - 4x₃ = 0

2x₁ - 4x₂ + 4x₃ = 0

Now I know how to solve a normal 3 variable system, but with this, since they are all set to 0, how exactly would I solve it? Or would I just solve it by saying that there are an infinite number of solutions since any variable == 0 would mean the other two could be anything. Maybe I'm just brain farting, but some help would be sweet.

[FusSioN]

I remembered I studied this months ago but forgot it 3 months later.

[Cameron]

Should only be the trivial solution of x1=x2=x3=0.  You would get more than one solution if you had a free variable, but that isn't the case here.  Also just for the sake of it in mathematica:

Code: [Select]

Solve[x - 2 y + 3 z == 0 && -2 x - 3 y - 4 z == 0 && 2 x - 4 y + 4 z == 0, {x, y, z}]
{{x -> 0, y -> 0, z -> 0}}


[prozajik]

Isn't just the exact solution 0? What method did you use for solving it? There are quite a few methods, simplest being discriminant method and maybe gaussian elimination.

I tried solving it with gaussian and came up with sth like this

1 -2  3 0
-2 -3 -4 0
 2 -4  4 0

1. line * 2 -> 2 -4 3 0
switch 1. line and 2. line

-2 -3 -4 0
 0 -7  6 0
 0 -7  7 0

-2 -3 -4 0
 0 -7  6 0
 0  0  1 0

z = 0
y = 0
x = 0

Might be wrong, did it during a lecture lol, i went through some online equation solvers and they all came up with 0 too.

The only question i have is if it isn't some kind of 'exception' as in 1=0, but i doubt it

EDIT:Agreed with Cameron

[Ace]

Thanks guys, I was using Gaussian and got the same answers, but I just thought I was missing something, seemed too simple.