Huaizhong R. answered • 05/19/25
Ph.D. Experienced & knowledgeable in Math Learning/Teaching
The matrix A has 3 distinct eigenvalues if and only if its characteristic polynomial |xE-A| has no multiple roots, where |xE-A| is the determinant of the matrix xE-A, E being the 3x3 identity matrix (with 1's on the diagonal, and 0 elsewhere). Let P(x) denote this polynomial, which is of degree 3. Then P has no multiple roots if and only if (P(x), P'(x)) =1, where (F(x), G(x)) denotes the greatest common divisor (gcd) of polynomials F(x) and G(x). By evaluating the determinant |xE-A|, we obtain the characteristic polynomial P(x)=x^3+2x^2-20x-k. Thus its derivative P'(x)=3x^2+4x-20=(3x+10)(x-2). Since P(x) and P'(x) have a non-constant common factor if and only if they have either x-2 or x+10/3 as a common factor, P(x) has 3 distinct roots if and only if P(x) has neither 2 nor -10/3 as roots. This amounts to k not equal to either -24 or 1400/27.
In conclusion, the question asked has a bit problem in that the condition is not given as a double inequality.
