Counting Matrices

We are given an $n \times n$ matrix $\mathcal{M}_n = \begin{bmatrix} 1 & 2 & \cdots & n \\ n+1 & n+2 & \cdots & 2n \\ \vdots & \vdots & \vdots & \vdots \\ n^2-(n-1) & n^2-(n-2) & \cdots & n^2 \\ \end{bmatrix}.$

Row reduce the matrix by replacing each row (besides the first one) with the result when the first row is subtracted from it. In other words,

$\begin{bmatrix} 1 & 2 & \cdots & n \\ \left( n+1 \right) - 1 & \left( n+2 \right) - 2 & \cdots & \left( 2n \right) - n \\ \vdots & \vdots & \vdots & \vdots \\ \left( n^2-(n-1) \right) - 1 & \left( n^2-(n-2) \right) - 2 & \cdots & \left( n^2 \right) - n \\ \end{bmatrix} = \begin{bmatrix} 1 & 2 & \cdots & n \\ n & n & \cdots & n \\ \vdots & \vdots & \vdots & \vdots \\ n(n-1) & n(n-1) & \cdots & n(n-1) \\ \end{bmatrix}.$

So each row of the matrix (starting with the second) is a multiple of $n$ , implying that matrices of this form have determinant zero, provided they have enough rows for one row to be a multiple of another. Since this sequence starts with the second row, we see that $\text{det} \big(\mathcal{M}_n\big) = 0$ , when $n \ge 3$ .

Now, we need to find $\text{det} \big(\mathcal{M}_n\big), \text{ for } n = 1,2$ . For $n = 1$ , this is clearly $1$ . For $n = 2$ , we have $\text{det} \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} = 1 \times 4 - 2 \times 3 = 4 - 6 = -2$ .

Hence, $\displaystyle \sum_{n = 1}^{\infty} \text{det}\big(\mathcal{M}_n\big) = 1 - 2 + 0 = \boxed{-1}$ .