# Nontrivial solutions of equation

Here, I have one problem in finding nontrivial solutions of a system of equations. I want to choose one variable, for example X1 and to get solutions of X2(X1) and X3(X1). It is not difficult when I have three unknowns, it can be solved by hand quickly, but if I have system with many unknowns, how can I find nontrivial solutions of a system with chosen unknown for example first X1 and to find others in function of X1. Here is a an example. I need X2(X1) and X3(X1)

 A={{A11, A12, A13}, {A21, A22, A23}, {A31, A32, A33}};
X={{X1}, {X2}, {X3}};

Solve[(A-λ*IdentityMatrix).X==0,{X2,X3}]


or with four unknowns I am looking for nontrivial solutions X2(X1),X3(X1),X4(X1)

 A={{A11, A12, A13,A14}, {A21, A22, A23, A24}, {A31, A32, A33, A34}};
X={{X1}, {X2}, {X3}, {X4}};

Solve[(A-λ*IdentityMatrix).X==0,{X2,X3,X4}]

• Solve[(A - \[Lambda]*IdentityMatrix).X == 0, {X2, X3}, MaxExtraConditions -> All] ? – Dr. belisarius May 18 '13 at 1:22
• How about using Eigensystem? Eigensystem[A] – Jens May 18 '13 at 1:31
• ...if $\lambda$ is in fact an eigenvalue of your system, then you can use @Jens's suggestion and pick out an appropriate eigenvector that you can then normalize. If not, then you have a nonsingular system of linear equations, easily handled by Solve[]. – J. M. is in limbo May 18 '13 at 1:43
• @Jens Eigensystem gave a large output. If I already have a lambda eigenvalues, how to get X2(lambda,X1) and X3(lambda,X1) directly? Solve doesn't work, Eigensystem gave a mess with root, I already have roots and.. – Pipe May 18 '13 at 10:20
• Try to do it by hand and see if you can get a smaller answer, without knowing anything about the matrix coefficients and $\lambda$. – Jens May 18 '13 at 15:38