# Thread: Linear algebra - Eigenvalues can't get the answer

1. I need to solve for the eigenvalues of the matrix 1 -1 -1 1 3 2 -1 -1 0 So I subtract lambda I from the matrix and find the determinant I get: (1-x)(3-x)(-x)-(-1)(2)(-1)-(-1)(1)(-1)+(-1)(3-x)(-1)+(-1)(1)(-x)+(1-x)(2)(-1) = -x^3 + 4x^2 - x - 2 So (x-1) is a factor and then I get this remaining -(x^2-3x-2) which can't be factored..

Did I calculate the determinant wrongly?

2. Originally Posted by zeion
I need to solve for the eigenvalues of the matrix 1 -1 -1 1 3 2 -1 -1 0 So I subtract lambda I from the matrix and find the determinant I get: (1-x)(3-x)(-x)-(-1)(2)(-1)-(-1)(1)(-1)+(-1)(3-x)(-1)+(-1)(1)(-x)+(1-x)(2)(-1) = -x^3 + 4x^2 - x - 2 So (x-1) is a factor and then I get this remaining -(x^2-3x-2) which can't be factored..

Did I calculate the determinant wrongly?
I haven't checked your determinant (I can't follow your calculation) but I will say that if you're studying eigenvalues then you ought to know how to factorise a quadratic function using surds (that the quadratic can be factorised using real numbers should be obvious from the fact that the discriminant is greater than zero).

3. In other words, any quadratic equation can be solved by completing the square or using the quadratic formula. You don't need to be able to "factor" it.