Determin what sign does the product of Antidiagonal elements in $n\times n$ matrix have.
How does one go about solving this ? I know that ht sign depends on wheater the permutation is odd or even, but I dont know how to determine it for $n \times n$ matrix.
The product of antidiagonal elements of the $n\times n$ matrix $A=[a_{ij}]$ is given by
$$ a_{1,n}a_{2,n-1}\ldots a_{n-1,2}a_{n,1} $$
and to determine the sign from which to precede it for calculating the determinant of $A$, it is necessary to examine the permutation
$$ \left(\begin{array}{ccccccc} 1 & 2 & 3 & \ldots & n-2 & n-1 & n \\ n & n-1 & n-2 & \ldots & 3 & 2 & 1\end{array}\right), $$
which we simply write
$$ (\begin{array}{ccccccc} n & n-1 & n-2 & \ldots & 3 & 2 & 1 \end{array}), $$
and calculate how many transpositions it is necessary to operate on it to obtain the natural sequence. To do this, we note that to bring n to its natural position, n-1 exchanges are necessary with the elements on its right, thus obtaining the permutation
$$ (\begin{array}{ccccccc} n-1 & n-2 & \ldots & 2 & 1 & n \end{array}). $$
In the same manner, $n-2$ exchanges are necessary to operate similarly on $n-1$ to obtain
$$ (\begin{array}{ccccccc} n-2 & n-3 & \ldots & 2 & 1 & n-1 & n\end{array}) $$
etc. Ultimately it is clear that
$$ (n-1)+(n-2)+\ldots+2+1 = \frac{n(n-1)}{2} $$
transpositions are needed in total to get the natural position of the indexes.
It follows that the sign we are looking for is given by $\;{(-1)^{\displaystyle n(n-1)/2}},\;$ thus getting the following table for all $n\geq 2$:
$$ \begin{array}{c} 2 && 3 && 4 && 5 && 6 && 7 && 8 && 9\\[1ex] - && - && + && + && - && - && + && +\end{array} \;\;\;\cdots $$