Finding the determinant of anti-diagonal matrix

Question

How would one find the determinant of an anti-diagonal matrix ($n \times n$), without using eigenvalues and/or traces (those I haven't learned yet):

My initial idea was to swap the first and n-th row, then the second and $n-1$-th row and so on, until I get a diagonal determinant, however how many swaps do I have to perform for that to happen? (Note: I do know the sign changes so I'll have $-1$ to some power times the now diagonal determinant, the problem is to find to what power).

For example: $\left| \begin{matrix} 0 & 0 & 0 & \dots & 0 & n-1 \\ 0 & 0 & 0 & \dots & n-1 & 0 \\ \vdots \\ 0 & n-1 & 0 & \dots & 0 & 0 \\ n-1 & 0 & 0 & \dots & 0 & 0 \end{matrix} \right|$.

Answer 1

Your idea is good. If $n$ is even, you will need $\frac{n}{2}$ swap operations

$$ n \leftrightarrow 1, (n-1) \leftrightarrow 2, \ldots, \frac{n}{2}+1 \leftrightarrow \frac{n}{2}. $$

If $n$ is odd, you will need $\frac{n-1}{2}$ swap operations. Try to check that this works for small values of $n$.

Answer 2

The number of swaps is clearly $\;\biggl\lfloor\dfrac n2\biggr\rfloor$.