```Date: May 16, 2013 5:35 PM
Author: RGVickson@shaw.ca
Subject: Re: Linear System in Mathematica -- Solve returns only the zero<br> vector (though there are more solutions)

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,> > DinoThe 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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