How can I show that and $n\times{n}$ matrix of the form in the description has a determinant of zero for $n>2$?

Question

In General, $n>2$, $a_{i,j}=a_{i,j-1}+1$ and the matrix will be of the following form:

$\begin{bmatrix}1&2&3&...&n\\n+1&n+2&n+3&...&n+n\\2n+1&2n+2&2n+3&...&2n+n\\...&...&...&...&...\\(n-1)n+1&(n-1)n+2&(n-1)n+3&...&n^2\end{bmatrix}_{n \times {n}}$

I tried to show this by row operations, and I got some nice patterns but I was getting into my third page and it seemed never ending. I'm working on uploading the pictures of my scratch work.

Answer 1

Subtract the first row from the second and third rows, and the new rows will be linearly dependent.

Answer 2

Simply note that the row space is spanned by $(1, 1, \cdots, 1)$ and $(1, 2, , \cdots, n)$, and hence has dimension 2. When $n>2$, the row space is not of full rank so the determinant is zero.

Answer 3

$$\begin{align} |A|&=\begin{vmatrix}1&2&3&\cdots & n\\ n&n&n&\cdots &n\\ 2n&2n&2n&\cdots&2n\\ \cdots & \cdots & \cdots & \cdots & \cdots\\ (n-1)n&(n-1)n&(n-1)n&\cdots&(n-1)n\end{vmatrix}\\ &=n\times 2n\times\cdots\times (n-1)n\begin{vmatrix}1&2&3&\cdots&n\\ 1&1&1&\cdots &1\\ 1&1&1&\cdots &1\\ \cdots & \cdots & \cdots & \cdots & \cdots\\ 1&1&1&\cdots &1\end{vmatrix}\\ &=n^{n-1}(n-1)!\begin{vmatrix}1&2&3&\cdots & n\\ 1&1&1&\cdots &1\\ 0&0&0&\cdots &0\\ \cdots & \cdots & \cdots & \cdots & \cdots\\ 0&0&0&\cdots &0\end{vmatrix} \end{align}$$

so $rk(A)=2$ and for $n\gt 2$ we have $|A|=0$