Solution space to homogenous equations.

[CalculusSandwich]

Ok so my question was:

The directions state:

Find the solution space of the following systems of linear homogeneous equations:

x-y+z-w=0
2x+y-z+2w=0
2y+3z+w=0

Is the same as finding the solutions, because I did that and got (-z, -3z, 3z, z). So i said
the solution space is actually the subspace you get. For instance, this problem yields a one dimensional subspace.
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And someone responded:

I don't think that solution set is right.

The vector -1,-3,3,1 does not give you 0 when used as weights in the original equation. It is a system of 4 unknowns in 3 rows there has to be a free variable somewhere. That means when you write your free variable in terms of the other variables, there should be a 0 in that solution set.

The solution (-1,-3,3,1) does not satisfy ax=0.

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Am i right or am i wrong here?

 

[Defennder]

I think you made an error somewhere. Substituting your answer into the first equation doesn't give 0. How did you get (-z, 3z, 3z, z) ?

 

[HallsofIvy]

Ok so my question was:

The directions state:

Find the solution space of the following systems of linear homogeneous equations:

x-y+z-w=0
2x+y-z+2w=0
2y+3z+w=0

Is the same as finding the solutions, because I did that and got (-z, -3z, 3z, z). So i said
the solution space is actually the subspace you get. For instance, this problem yields a one dimensional subspace.

I suggest you try that again. That is not at all what I get.

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And someone responded:

I don't think that solution set is right.

The vector -1,-3,3,1 does not give you 0 when used as weights in the original equation. It is a system of 4 unknowns in 3 rows there has to be a free variable somewhere. That means when you write your free variable in terms of the other variables, there should be a 0 in that solution set.

The solution (-1,-3,3,1) does not satisfy ax=0.

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Am i right or am i wrong here?