You're right in that it is nonsingular. Otherwise, it would have a zero eigenvalue, so it would not be annihilated by a polynomial with a constant term.
$A$'s minimal polynomial divides $x^3-1$, so its eigenvalues are all third roots of unity.
If they were both $1$, $A$ would be identity (it can't have nontrivial Jordan blocks, because $x^3-1$ is squarefree), so at least one of them is nonreal.
$A$'s entries are real, so nonreal complex roots occur in conjugate pairs, and the sum of the two conjugate nonreal third roots of unity is $-1$.
In general, for an $n\times n$ matrix, the trace of such a matrix will be an integer of the form $n-3k$ with $0< k\leq \lfloor n/2\rfloor$ (and any such integer can be found as a trace of such a matrix, consider block matrices with diagonal blocks equal to $A$ or a $1\times 1$ matrix $(1)$ and $0$ elsewhere).