Non trivial solution to a linear system

Hello,

I am trying to solve a linear system of the form A*x=B with A = K-eigenfreq1(3)*M and B = [0 0 0 0 0]'

K and M are 5x5 matrices and eigenfreq1(3) is just a scalar.

However whenever I use the A\B command to solve the system I get the trivial solution x=[0 0 0 0 0]' and I am told that this solution is not unique. Is there a way to get the other, non trivial, solutions?

[Karim]

out of curiosity, why do you not use the build in function eig ?

[ EigenModes , EigenFreq ] = eig( K , M );

[Serge El Asmar]

Because whenever I do I get this error message that I dont know how to solve.

[Serge El Asmar]

If you have encountered that before and can help it would be amazing thanks!

[Karim]

See below for an example. This way you obtain all the natural frequencies and the corresponding modes.

K = rand(5);
M = rand(5);
[ EigenModes , EigenFreq ] = eig( K , M )

EigenModes = 5×5

1.0000 -0.5408 0.3037 0.1309 0.1009 0.1474 -1.0000 -0.7636 -0.7115 -1.0000 -0.7752 -0.5870 -0.5500 -0.6974 0.0851 -0.8842 0.2302 1.0000 1.0000 0.2623 0.6343 0.0215 -0.5822 -0.2689 0.6139

EigenFreq = 5×5

-5.8058 0 0 0 0 0 1.5359 0 0 0 0 0 -0.4406 0 0 0 0 0 -0.0414 0 0 0 0 0 0.1681

EigenFreq = diag( EigenFreq )

EigenFreq = 5×1

-5.8058 1.5359 -0.4406 -0.0414 0.1681

Answer 1

[VBBV]

syms omega2
K = rand(5);
M = rand(5);
eigF = 0.1; % scalar
eqn1 = det(K-omega2*M) == 0;
eigenfreq1=vpasolve(eqn1,omega2)

eigenfreq1 = 

eigenmodes1 = (K-eigF*M)\ones(5,1)  %  

eigenmodes1 = 5×1

0.2999 1.0135 -0.1678 1.7206 -0.3961

[VBBV]

Since input B vector are all zeros, non-trivial solution dont exist. If your input vector are all ones you can get non-trivial solution

Answer 2

[Torsten]

However whenever I use the A\B command to solve the system I get the trivial solution x=[0 0 0 0 0]' and I am told that this solution is not unique. Is there a way to get the other, non trivial, solutions?

null(A) gives you a basis for the kernel of A.