Clearly the $1 \times 1$ and the $2 \times 2$ matrices $(1)$ and $\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}$ are invertible. I am wondering about the following: if I take any matrix that has in the first line $(1 \ 2 \ \dots \ n)$ and in each other line a different permutation of the numbers $\{1, \dots,n\}$, will it always be invertible?
I think this is wrong for higher $n$; it might have to do with the fact that we get more permutations than we can fit in the rows and with the Leibniz formula. But still, it works with all (or most) $3 \times 3$ cases, so I am really getting curious.
If it turns out that this does not work, I would still be interested in knowing if there is at least one way of doing it that works (for every $n$) and if it is relevant that we have chosen $\{1,\dots,n\}$ and not any other (real, different, positive) entries.
Edit:
As pointed out in the comments and in user1551's answer, this is not always the case, not even when in each column and in each row each element appears exactly once. I would still like to know why all the $3 \times 3$ matrices (of this kind) are always invertible and, above all, why the "shift matrix" mentioned in Hw Chu's comment is indeed always invertible (and whether there are other procedures that always work).