Determinant Calculation Increasing entries

Question

The problem is to find the determinant of $$A = \left|\begin{array}{ccccccc}1 & 2 & 3 & \cdots & n-2 & n-1 & n \\ 2 & 3 & 4 & \cdots & n-1 & n & n \\ 3 & 4 & 5 & \cdots & n & n & n \\ 4 & 5 & 6 & \cdots & n & n & n \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots \\ n & n & n & \cdots & n & n & n\end{array}\right|$$ Not sure how to approach this problem; I've tried row-reduction to put this in triangular form, but doing so requires several row switches which doesn't seem to be a clean solution. Would appreciate some help!

Answer 1

$$A = \left|\begin{array}{ccccccc}1 & 2 & 3 & \cdots & n-2 & n-1 & n \\ 2 & 3 & 4 & \cdots & n-1 & n & n \\ 3 & 4 & 5 & \cdots & n & n & n \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots \\ n & n & n & \cdots & n & n & n\end{array}\right|$$

$$ \downarrow (R_i \rightarrow R_i-R_{i-1})$$

$$A = \left|\begin{array}{ccccccc}1 & 1 & 1 & \cdots & 1 & 1 & 1 \\ 2 & 1 & 1 & \cdots & 1 & 1 & 0 \\ 3 & 1 & 1 & \cdots & 1 & 0 & 0\\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 1 & 0 & 0 & \cdots & 0 & 0 & 0\end{array}\right|$$

$$ \downarrow (C_i \rightarrow R_i-R_{i+1})$$

$$A = \left|\begin{array}{ccccccc}1 & 0 & 0 & \cdots & 0 & 0 & 1 \\ -1 & 0 & 0 & \cdots & 0 & 1 & 0 \\ -1 & 0 & 0 & \cdots & 1 & 0 & 0\\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots \\ n & 0 & 0 & \cdots & 0 & 0 & 0\end{array}\right|$$

$=(-1)^{(n+1)}{n}{\left|\begin{array}{ccccccc} 0 & 0 & \cdots & 0 & 0 & 1 \\ 0 & 0 & \cdots & 0 & 1 & 0 \\ 0 & 0 & \cdots & 1 & 0 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 1 & 0 & 0 & \cdots & 0 & 0 & \end{array}\right|}=(-1)^{n+1}(n)\sigma(n)$

Where, $\sigma(n)=1$ if $n=4k+${$1,0$} and $\sigma=-1$ if $n=4k+${$-1,2$}.

Answer 2

Here is a relatively quick way always subtracting the second last column from the last one and then expanding:

$$\left|\begin{array}{ccccccc} 1 & 2 & 3 & \cdots & n-2 & n-1 & n \\ 2 & 3 & 4 & \cdots & n-1 & n & n \\ 3 & 4 & 5 & \cdots & n & n & n \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots \\ n & n & n & \cdots & n & n & n\end{array}\right| = \left|\begin{array}{ccccccc}1 & 2 & 3 & \cdots & n-2 & n-1 & 1 \\ 2 & 3 & 4 & \cdots & n-1 & n & 0 \\ 3 & 4 & 5 & \cdots & n & n & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots \\ n & n & n & \cdots & n & n & 0\end{array}\right|$$ $$=(-1)^n\left|\begin{array}{cccccc} 2 & 3 & 4 & \cdots & n-1 & n \\ 3 & 4 & 5 & \cdots & n & n \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ n & n & n & \cdots & n & n \end{array}\right| =\ldots$$

$$\ldots = (-1)^{n+(n-1)+\cdots +2}\cdot n = (-1)^{\frac{(n+2)(n-1)}{2}}\cdot n$$