find all values of k for which A is invertible $\begin{bmatrix}
k &k  &0 \\ 
k^2 &2  &k \\ 
 0& k & k
\end{bmatrix}$
what I did is find the det first: 
$$\det= k(2k-k^2)-k(k^3-0)-0(k^3 -0)=2k^2-k^3-k^4$$
when $det = 0$  the matrix isn't invertible
$$2k^2-k^3-k^4=0$$
$$k^2(k^2 +k-2)=0$$
$$k^2+k-2=0$$
$$(k+2)(k-1)=0$$
$k = -2$ or $k = 1$.
I am lost here how to find the value for k when the matrix is invertible.

Almost all square matrices are invertible. It is very special, i.e. singular, for a square matrix to be non-invertible. As you say, $\det = 2k^2-k^3-k^4$. This factorises to give $k^2(2+k)(1-k)$.
Your matrix is invertible for all values of $k$ except $k=0$, $k=-2$ or $k=1$. The matrix is invertible for all of the other values of $k$. There are three isolated values that make the matrix non-invertible. We can let $k$ be any complex number, i.e. $k$ can belong to an infinite, two-dimensional plane. out of all of the infinite points in the complex plane, only three isolated points give a singular matrix.
