Find the value of x such that the determinant of A is zero.

I am have matrix A with x as the parameter: A=[2,8,5;2,2,8*x;6*x^2,4*x+2,x^3]; I want to find the value of x such that the determinant of A is zero.

    % 
format shortE    
syms x
A=[2,8,5;2,2,8*x;6*x^2,4*x+2,x^3];
xc = det(A);    
solve(xc == 0);
  end

However, it is outputting the polynomial: 372*x^3 - 124*x^2 + 8*x + 20 . I am not sure what I am doing that is incorrect.

Answer 1

[Stephen23]

There is no need to use symbolic operations, a simple numeric calculation can give you the answer:

>> fun = @(x)[2,8,5;2,2,8*x;6*x^2,4*x+2,x^3];
>> xval = fzero(@(x)det(fun(x)),0)
xval = -0.27922
>> fun(xval)
ans =
   2.000000   8.000000   5.000000
   2.000000   2.000000  -2.233791
   0.467796   0.883104  -0.021770
>> det(fun(xval))
ans =   5.5955e-014

[Amanda]

Thank you! That makes much more sense and it much easier.

[Stephen23]

@Amanda: I hope that it helps. You should accept the answer that best resolves your question.

[Walter Roberson]

There are three solutions for x, two of which are complex valued. The above finds the real valued solution.

The complete set of solutions can be found by

simplify(solve(xc == 0, 'MaxDegree', 3))

Answer 2

[Anuja Menokey]

[airy(2,Z) airy(0,Z);airy(3,Z) airy(1,Z)]

det of this matirix is zero.. I want to solve this matrix..please help me..

[Walter Roberson]

The Airy Ai function is complicated enough that you are not going to be able to get a closed form solution for that. You are probably going to need to use a numeric approach, or vpasolve()

There are an infinite number of solutions in the negative Z, and the only solution in positive Z is Z = infinity

The solutions get closer together as Z gets more negative.

Note: the determinent is never positive.

[John D'Errico]

Please, this is a different question. Do not ask it as an answer to another question.

Anyway, what is the determinant of a 2x2 matrix? Surely you can write that as ONE simple equation, in terms of those Airy functions?