Putnam 2014, Problem A-2

Let $A$ be the $n \times n$ matrix whose entry in the $i$-th row and $j$-th column is
$$\frac{1}{\min(i, j)}$$
for $1 \leq i, j \leq n$. Compute $\det(A)$.

Solution:
Let \(v_1, \dots, v_n\) denote the rows of \(A\). The determinant is unchanged if we replace \(v_n\) by \(v_n - v_{n-1}\), then \(v_{n-1}\) by \(v_{n-1} - v_{n-2}\), and so on, eventually replacing \(v_k\) by \(v_k - v_{k-1}\) for \(k \geq 2\). Since \(v_{k-1}\) and \(v_k\) agree in their first \(k-1\) entries, and the \(k\)-th entry of \(v_k - v_{k-1}\) is \(\frac{1}{k} - \frac{1}{k-1}\), the result is an upper triangular matrix with diagonal entries \(1, \frac{1}{2} - 1, \frac{1}{3} - \frac{1}{2}, \dots, \frac{1}{n} - \frac{1}{n-1}\). The determinant is then  
\[
\prod_{k=2}^n \left( \frac{1}{k} - \frac{1}{k-1} \right) = \prod_{k=2}^n \left( \frac{-1}{k(k-1)} \right) = \frac{(-1)^{n-1}}{(n-1)!n!}.
\]  
Remark: The inverse of \(A\) for \(n \geq 2\) is the matrix \(B\) given by  
\[
B_{ij} = \begin{cases} 
-1 & i = j = 1, \\
-2i^2 & 1 < i = j < n, \\
-(n-1)n & i = j = n, \\
ij & |i - j| = 1, \\
0 & \text{otherwise}.
\end{cases}
\]