
Re: Linear System in Mathematica  Solve returns only the zero vector (though there are more solutions)
Posted:
May 16, 2013 5:35 PM


On Thursday, May 16, 2013 2:46:02 AM UTC7, 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.

