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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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 Dino Posts: 2 Registered: 5/16/13
Linear System in Mathematica -- Solve returns only the zero vector
(though there are more solutions)

Posted: May 16, 2013 5:46 AM

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.

Dino

Date Subject Author
5/16/13 Dino
5/16/13 RGVickson@shaw.ca
5/17/13 Roland Franzius
5/17/13 Dino