roots of 3*3 matrix determinant

Hello all,

I have a matrix as follows;

M = [k(1,1)-12*w k(1,2) k(1,3);k(2,1) k(2,2)-2*w k(2,3);k(3,1) k(3,2) k(3,3)-4*w ];

i want to solve and to find three roots of det(M)=0

how can i solve this?

thanks

[Walter Roberson]

Do you have the symbolic toolbox? if so then use solve()

[Cem Eren Aslan]

Yes, i have, but result is as follows;

root(z^3 - (131891655112916159*z^2)/26388279066624 + (692241875910773148691733542859037*z)/309485009821345068724781056 - 5152005758176027364447462631233389326948016877/42535295865117307932921825928971026432, z, 1)

[Walter Roberson]

solve(det(M), 'maxdegree', 3)

Answer 1

[David Goodmanson]

Hi Cern,

I assume you mean the three values of w that give a 0 determinant. Eig works for this.

K = rand(3,3)
c = diag([12 2 4]);  % the three coefficients of w down the diagonal
Knew = inv(c)*K      % multiply the first row of K by 1/12, second row by 1/2, etc.
w = eig(Knew)        % roots
% check, should be small; they are.
det(K-w(1)*c)
det(K-w(2)*c)
det(K-w(3)*c)

K =
0.4169    0.3829    0.3334
0.3801    0.0297    0.9758
0.2133    0.4723    0.5554

Knew =
0.0347    0.0319    0.0278
0.1901    0.0148    0.4879
0.0533    0.1181    0.1389

w =
0.3447
0.0218
-0.1781

In practice, for larger matrices one would use c\K instead of inv(c)*K but the latter expression seems clearer in these circumstances.