Is it possible to avoid symbolic math for below query

I have two matrices a and b. I need to find the value of x such that the determinant of (a + bx) is 0. The size of the matrices is 4x4. So in effect I need the roots of 4th order polynomial in variable x.

I did it by using the symbolic math tool box and below code :

syms l; char_matrix=a + l*b; determinant=det(char_matrix); R=solve(determinant);

This code is working but its taking too long for solving . Is there any way I can avoid symbolic math in such a situation as I think symbolic math takes longer than numerical math. Thank you for your time.

Answer 1

[Andrew Newell]

As long as B has a nonzero determinant, you could recast it as an eigenvalue problem:

det(A+Bx) = det(B)*det(inv(B)*A+Ix) = 0,

where I is the identity, and you could use the following code:

x = -eig(B\A)

[Sakshi]

Thank you so much. If I understood correctly you have eliminated the need to use symbolic math toolbox for this case. It has saved several hours of my programming. Profusely thank you !!

[Andrew Newell]

Glad to help.

Answer 2

[Jan]

You can do it numerically:

R = fzero(@(x) det(a + x * b), x0)

with a suiting initial value x0.

[Sakshi]

Thank you sir. However, I would not be able to give an initial value of x0. I will keep both the solutions in mind for future. Thank you once again for your time.

Answer 3

[John D'Errico]

If A and B are known, then this is a simple problem using roots. I'll use my sympoly toolbox to show what is happening, and a way to solve it. Pick two arbitrary matrices.

>> A = magic(4)
A =
    16     2     3    13
     5    11    10     8
     9     7     6    12
     4    14    15     1

>> B = round(rand(4)*5)
B =
     1     2     2     3
     4     3     3     3
     5     2     2     1
     0     2     4     5

See that the determinant is a polynomial of 4th degree in x.

>> det(A+B*x)
ans =
    1125*x^2 + 406*x^3 + 4*x^4

>> roots(det(A+B*x))
ans =
            0
            0
      -98.649
       -2.851

There are 4 solutions here as you would expect. They need not all be real.