I have an $n\times n$ matrix where the signs of the elements (sign pattern) are all strictly negative except on the anti-diagonal where the sign is zero (i.e. the elements on the anti-diagonal are zero).
I think such a matrix always have full rank (i.e. invertible) but can't come up with a generic argument. For $n=3$ for example, one could directly argue that the second column vector is not in the span of the first and the third column vector is not in the span of the first two. To be explicit, for $n=3$ the sign pattern defined above would be
$\begin{bmatrix}-&-&0\\-&0&-\\0&-&-\end{bmatrix}.$
This is not true if $n= 4$. Here is a counterexample: $$ A=\pmatrix{-1&-1&-2&0\\-1&-1&0&-2\\-2&0&-1&-1\\0&-2&-1&-1},x = \pmatrix{1\\1\\-1\\-1}. $$
Obviously, for $n=2$ the claim is true.
The claim is also true for $n=3$: Let $A$ be a matrix with the special structure, $x\ne0$ such that $Ax=0$. Then $x$ has at least one positive and one negative entry. Since $n= 3$, $x$ has exactly one negative or exactly one positive entry in component $i$. Now if the $i$-th row is multiplied with $x$, the $0$-entry in $A$ and the entry $x_i$ cancel. Since the $i$-th component of $Ax$ is zero, this shows that all entries $x_j$ for $i\ne j$ are zero, which is a contradiction. Hence $A$ is invertible.