\begin{vmatrix} n & 0 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 2 & \ldots& 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 0 &n-1 \\ 0 &1 & 0 &\ldots & 0 & 0 \end{vmatrix}
I know that we nedd to solve by (-1)^(n,n-1...2,1)a1n a2,n-1...an,1
N{n,n-1..2,1) = 1+2+...+(n-2)+(n-1)= n(n-1)/2
But I don't understand what is the total answer, like howwe are calculating (-1)^n(n-1)/2 * a1n a2,n-1...an,1
I would really appreciate your help!
Suppose each of your column has a single non-zero element which takes value $n, 1, \ldots, n-1$ respectively.
Factorize each of the number out from the determinant, we get a factor of $n!$.
The remaining term is a determinant of a permutation matrix.
\begin{align}\begin{vmatrix} n & 0 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 2 & \ldots& 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 &\ldots &n-2 & 0 \\ 0 & 0 & 0 & \ldots & 0 &n-1 \\ 0 &1 & 0 &\ldots & 0 & 0 \end{vmatrix} &= n!\begin{vmatrix} 1 & 0 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 1 & \ldots& 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 &\ldots &1 & 0 \\ 0 & 0 & 0 & \ldots & 0 &1 \\ 0 &1 & 0 &\ldots & 0 & 0 \end{vmatrix}\\ &= n! \begin{vmatrix} 1 & 0 & 0_{1 \times (n-2)} \\ 0_{(n-2) \times 1} & 0_{(n-2) \times 1} & I_{n-2} \\ 0 & 1 & 0_{1 \times (n-2)}\end{vmatrix} \\&=n!(-1)^{n+1}\end{align}
where the last expression is obtained by expanding along the second column.