# Calculating general $n\times n$ determinant

I'm given determinant $\begin{vmatrix} 1 & 2 &3 & \cdots & n -1 & n \\ 2 & 3 &4 & \cdots & n & 1 \\ 3 & 4 &5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{vmatrix}$ and I have to calculate it's value only using Laplace formula and properties of antilinear $n$-forms.

First, I subtract $i-1$-th column from $i$-th column, which gives me

$\begin{vmatrix} 1 & 1 &1 & \cdots & 1 & 1 \\ 2 & 1 &1 & \cdots & 1 & 1-n \\ 3 & 1 &1 & \cdots & 1 - n & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ n & 1-n & 1 & \cdots & 1 & 1 \end{vmatrix}$.

Then, subtracting $i-1$-th row from $i$-th row yields us

$\begin{vmatrix} 1 & 1 &1 & \cdots & 1 & 1 \\ 1 & 0 &0 & \cdots & 0 & -n \\ 1 & 0 &0 & \cdots & - n & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & -n & 0 & \cdots & 0 & 0 \end{vmatrix}$.

Again, subtracting $i-1$-th column from $i$-th column

$\begin{vmatrix} 1 & 1 &0 & \cdots & 0 & 0 \\ 1 & 0 &0 & \cdots & 0 & -n \\ 1 & 0 &0 & \cdots & - n & n \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & -n & n & \cdots & 0 & 0 \end{vmatrix}$.

Two last steps before using Laplace formula are adding $i-1$ to $i$-th row

$\begin{vmatrix} 1 & 1 &0 & \cdots & 0 & 0 \\ 2 & 1 &0 & \cdots & 0 & -n \\ 2 & 0 &0 & \cdots & - n & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 2 & -n & 0 & \cdots & 0 & 0 \end{vmatrix}$

and using properties of antilinear $n$-form flipping determinant

$(-1)^\frac{n(n-1)}{2}\begin{vmatrix} 0 & 0 &0 & \cdots & 1 & 1 \\ -n & 0 &0 & \cdots & 1 & 2 \\ 0 & -n &0 & \cdots & 0 & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & -n & 2 \end{vmatrix}$.

Final step is using Laplace formula on 1st row : $$(-1)^\frac{n(n-1)}{2}\begin{vmatrix} 0 & 0 &0 & \cdots & 1 & 1 \\ -n & 0 &0 & \cdots & 1 & 2 \\ 0 & -n &0 & \cdots & 0 & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & -n & 2 \end{vmatrix} = (-1)^\frac{n(n-1)}{2}((-n)^{n-1} - 2(-n)^{n-2}) = (-1)^{\frac{n(n-1)^2}{2}}(n^{n-1} + 2n^{n - 2}).$$

But my textbook states that result should be $(-1)^{\frac{n(n-1)}{2}}\cdot\frac{1}{2}(n^n + n^{n-1})$. What I am doing wrong in my calculations?

In your second step, when you subtract row $i-1$ from row $i$, there should be a diagonal of $n$'s under your diagonal of $-n$'s, because $1-(1-n)=n$. For example, the third entry in the last column of your third determinant should be $n$, not $0$.