Matrix with repeated eigenvalues

[Shambles]

The question is asking for what values of x will the matrix have at least one repeated eigenvalue (algebraic multiplicity of 2 or greater). The matrix is
| 3 0 0 |
| 0 x 2 | So naturally a normal attempt to find the eigenvalue in a question with only intergers
| 0 2 x | I would continue with:

| $$\lambda$$-3 0 0 |
| 0 $$\lambda$$-x -2|
| 0 -2 $$\lambda$$-x|

And then finding the determinant would continue with ($$\lambda$$-3)($$\lambda$$-x)($$\lambda$$-x) - (-2)(-2)($$\lambda$$-3) etc...

Except with 2 unknown variable it inevitably becomes a problem that I run into with more theoretical questions vs questions dealing only with numbers. As far as I know an eigenvalue with an algebraic multiplicity of >1 doesn't even have a geometric significance. Clearly I am approaching this entire question from the wrong angle and could use a push in the right direction if anyone is so able. Muchly appreciated.

 

[HallsofIvy]

I don't see why you need to be that theoretical!

The characteristic equation for the eigenvalues is obviously [itex](\lambda- 3)((\lambda- x)^2- 4)= 0. 3 is an obvious solution to that so in order to have a double root, either 3 is a root of $(\lambda- x)^2- 4= 0$ or that equation has a double root. If 3 is a root then we must have $(3- x)^2- 4= x^2- 6x+ 5= (x- 3)(x- 2)= 0$ so that x= 2 or 3. If $(\lambda- x)^2- 4= \lambda^2- 2x\lambda+ x^2- 4= 0$ has a double root, then its discriminant must be 0: $4x^2- 4(x^2- 4)= 4$ is never 0 so that can't happen.[/itex]