Suppose $$A = \begin{pmatrix}0&0&-1&&&\\ 1&0&0&-1&\ddots\\ &\ddots&\ddots&\ddots&\ddots\\ &&1&0&0&-1\\ &&&1&0&0&-1\end{pmatrix}$$
For an $n \times n$ matrix, I see that the inverse of $A$ does not always exist. I am not sure but it looks like a Heisenberg matrix.
If $n$ is divisible by $3$, then I see that $\det(A)=1$. For any other $n$, $\det(A)=0$ and thus $A$ is non-invertible. Not sure if there is a general formula that can show this?
For example a $9 \times 9$ matrix is like the following
If your image is correct then your $\LaTeX$ has two extra columns at the end that you should delete. Also, you mean Hessenberg matrix.
Anyway, because the matrix is so sparse this is actually quite straightforward to do by cofactor expansion. Write $M_n$ for this matrix. If you cofactor expand along the first two columns you end up with a matrix with first row $-1, 0, 0, \dots$ and if you then cofactor expand along this row you find that
$$\det M_n = - \det M_{n-3}.$$
Now it suffices to compute $\det M_1 = 0, \det M_2 = 0, \det M_3 = -1$; we get
$$\det M_n = \begin{cases} 0 & \text{ if } & 3 \nmid n \\ (-1)^{\frac{n}{3}} & \text{ if } & 3 \mid n \end{cases}$$