4
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.