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

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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 $n$.

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  • $\begingroup$ That I kinda figured out for small values of n, but then what would be the solution for that example given above, since I don't know if n is even or odd. $\endgroup$ – gvidoje Dec 8 '15 at 17:48
  • $\begingroup$ You can divide it into two cases and write what happens if $n$ is even and what happens if $n$ is odd. This is ok - after all, the answer does depend on whether $n$ is even or odd. $\endgroup$ – levap Dec 8 '15 at 18:01
  • $\begingroup$ My teacher wrote it like $ \left(-1\right)^{\frac{n\left(n-1\right)}{2}} $ my guess was because 1/2 of the elements are even and 1/2 are odd.. yet still not 100% sure how she got that. $\endgroup$ – gvidoje Dec 8 '15 at 18:02
  • $\begingroup$ It's just a fancy way of writing the determinant as $(n-1)^n (-1)^{\frac{n(n-1)}{2}}$ instead of $\begin{cases} (n-1)^n(-1)^{\frac{n}{2}} & n \text{ is even} \\ (n-1)^n(-1)^{\frac{n-1}{2}} & n \text { is odd}. \end{cases}$. $\endgroup$ – levap Dec 8 '15 at 18:09
  • $\begingroup$ But both ways are correct. To see why it is the same, note that if $n$ is even then $n - 1$ is odd and so $(-1)^{\frac{n}{2}} = \left( (-1)^{\frac{n}{2}} \right)^{n-1} = (-1)^{\frac{n(n-1)}{2}}$ and that if $n$ is odd then $n - 1$ is even and so $(-1)^{\frac{n-1}{2}} = \left( (-1)^{\frac{n-1}{2}} \right)^{n} = (-1)^{\frac{n(n-1)}{2}}$ so in both cases, the sign is equal to $(-1)^{\frac{n(n-1)}{2}}$. $\endgroup$ – levap Dec 8 '15 at 18:11
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The number of swaps is clearly $\;\biggl\lfloor\dfrac n2\biggr\rfloor$.

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