Finding triples (a,b,c) so that characteristic and minimal polynomials are different

Question

I got to find all complex triple $(a,b,c)$ so that the following matrix has different characteristic and minimal polynomial.\begin{bmatrix}2&0&0\\a&2&0\\b&c&-1\\\end{bmatrix}

Certainly the eigenvalues are 2,2,-1. Hence the characteristic and minimal polynomials are same if they are $(x-2)^2(x+1)$. And the JCF would be $ \begin{bmatrix}2&0&0\\1&2&0\\0&0&-1\\\end{bmatrix}$.

And the characteristic and minimal polynomials would be different if the minimal polynomial would be $(x-2)(x+1)$ and the JCF would be $ \begin{bmatrix}2&0&0\\0&2&0\\0&0&-1\\\end{bmatrix}$.

Now how to find all the complex triples which works in the second case i.e. the minimal polynomial is $(x+1)(x-2)$.

Answer 1

Since you edited the problem, so things will change.

Consider the matrix $$(A+I)(A-2I)=\begin{bmatrix}3&0&0\\a&3&0\\b&c&0\end{bmatrix}\begin{bmatrix}0&0&0\\a&0&0\\b&c&-3\end{bmatrix}=\begin{bmatrix}0&0&0\\3a&0&0\\ac&0&0\\\end{bmatrix}$$ With $a=0$, we can have a zero transformation, while $b,c$ can be any numbers.