Finding the eigenvalues for each value of $a \in \mathbb R$

Question

Let $$A = \begin{bmatrix} a^2 & 0 & a \\ 0 & 2a & 0 \\ 0 & 0 & a+2 \end{bmatrix}$$ For each value of $a \in \mathbb R$, find the eigenvalues of $A$.

I thought it should be whenever $\ a$ makes the determinant of $A$ zero so it should be $\ 0, -2, $ but apparently the answers are $\ -1,0,2 $ which I don't understand why?

$\ a = -1 , 2, \ \det (A) \not = 0 $

Answer 1

If the question is

find for each value of $\ a $, the eigen values of $\ A $

then the answer is $a^2, 2a, a+2$ since $A$ is triangular.

If the answer is

$-1,0,2$

then the question is "find for which value of $a$ the matrix $A$ has repeated eigenvalues."

Answer 2

We have $\det(A- \lambda I)=(a^2-\lambda)(2a-\lambda)(a+2- \lambda)$.

Hence the eigenvalues of $A$ are: $a^2,2a$ and $a+2$.

Answer 3

Eigenvalues of a matrix are usually computed by finding the values of $\lambda$ such that $\det(A-\lambda I)=0$. The determinant of $A$ is not relevant.

The answer will be three expressions that would in general contain $a$, so I do not see how they can be -1, 0 and 2.

Answer 4

$A^{2}-a^{2}I,A^{2}-(a+2)I$ adn $A^{2}-2aI$ all have one row of zeros so these are singular and the eigen values are $a^{2},a+2$ and $2a$.