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Topic: Linear System in Mathematica -- Solve returns only the zero vector
(though there are more solutions)

Replies: 3   Last Post: May 17, 2013 3:57 AM

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RGVickson@shaw.ca

Posts: 1,643
Registered: 12/1/07
Re: Linear System in Mathematica -- Solve returns only the zero
vector (though there are more solutions)

Posted: May 16, 2013 5:35 PM
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On Thursday, May 16, 2013 2:46:02 AM UTC-7, Dino wrote:
> Hey guys,
>
> say I want to find the null space of this matrix (which I define as 'Tbl'):
>
> {{c, 1/2, 0}, {0, 1 + c, 1}, {0, 1/2, 3 + c}}
>
>
>
> I try this:
>
> Solve[Tbl.{x, y, z} == {0, 0, 0}, {x, y, z}]
>
> and I get this:
>
> {{x -> 0, y -> 0, z -> 0}}
>
>
>
> But the zero vector is not the only solution. For example if I do this:
>
> c = Sqrt[3/2] - 2
>
> then
>
> Solve[Tbl.{x, y, z} == {0, 0, 0}, {x, y, z}]
>
> returns
>
> {{x -> -(((-2 - Sqrt[6]) z)/(-4 + Sqrt[6])), y -> -(2 + Sqrt[6]) z}}
>
> which is closer to what I want. I actually want something like that with 'c' in it. (And then I have some other equation that I'll use to find the value of c. Btw, the matrix above is only an example. I actually care to find a solution to a whole range of matrices of larger and larger sizes.)
>
>
>
> Can anyone help? I guess I want to tell Mathematica somehow that c is some arbitrary constant and there may be more solutions, depending on the value of c.
>
>
>
> Thanx in advance,
>
> Dino


The same thing happens in Maple. The problem is that both Maple and Mathematica do not test the determinant, which is D = c*(2c^2 + 8c + 5)/2 and this is nonzero except for c = 0 and c = -2 +- sqrt(6)/2. When the determinant is nonzero there is only the one solution [0,0,0]. When the determinant is zero the matrix has rank 2.



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