Is it possible to avoid symbolic math for below query

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Sakshi
Sakshi am 8 Jun. 2011
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.

Akzeptierte Antwort

Andrew Newell
Andrew Newell am 8 Jun. 2011
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)
  2 Kommentare
Sakshi
Sakshi am 9 Jun. 2011
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
Andrew Newell am 9 Jun. 2011
Glad to help.

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Weitere Antworten (2)

Jan
Jan am 8 Jun. 2011
You can do it numerically:
R = fzero(@(x) det(a + x * b), x0)
with a suiting initial value x0.
  1 Kommentar
Sakshi
Sakshi am 9 Jun. 2011
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.

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John D'Errico
John D'Errico am 9 Jun. 2011
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.

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