Determinant of the $n\times n$ matrix whose $jk$-th entry is $1/\min\{j,k\}$

Question

So on my exam I got a True/False question that asked the following:

For any $n\in\mathbb N$, compute the determinant of the matrix

\begin{bmatrix} 1&1&1&\cdots&1 \\1&1/2&1/2&\cdots&1/2 \\1&1/2&1/3&\cdots&1/3 \\\vdots&\vdots&\vdots&\ddots&\vdots \\1&1/2&1/3&\cdots&1/n \end{bmatrix}

This is the $n\times n$ matrix whose $jk$-th entry is $1/\min\{j,k\}$.

I recognize there has to be some formula that makes calculating the determinant easier, but I'm confused as to what it is.

Answer 1

For any given $n\in\Bbb{N}$ denote the matrix by $A_n$. For $n\geq 2$, applying Laplace expansion along the last row and starting at the bottom right (the entry $\frac{1}{n}$) an working right to left yields $$\det A_n=\frac{1}{n}\det A_{n-1}-\frac{1}{n-1}\det A_{n-1}=-\frac{1}{n(n-1)}\det A_{n-1},$$ because the subdeterminants at the coefficients $\frac{1}{1}$ through $\frac{1}{n-2}$ are all zero; removing the $k$-th column makes the rows $k-1$, $k$ and $k+1$ linearly dependent. Of course $\det A_1=1$ so $$\det A_n=(-1)^{n+1}\frac{1}{n!(n-1)!}=(-1)^{n+1}\frac{n}{(n!)^2}.$$

Answer 2

I'll take $n=4$ for simplicity of exposition. We develop against the last line: $$-\begin{vmatrix} 1 & 1 & 1 \\ 1/2 & 1/2 & 1/2 \\ 1/2 & 1/3 & 1/3 \end{vmatrix} +\dfrac{1}{2}\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1/2 & 1/2 \\ 1 & 1/3 & 1/3 \end{vmatrix} -\dfrac{1}{3}\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1/2 & 1/2 \\ 1 & 1/2 & 1/3 \end{vmatrix} +\dfrac{1}{4}\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1/2 & 1/2 \\ 1 & 1/2 & 1/3 \end{vmatrix}$$ The first determinant is $0$ (two proportional lines), the second is $0$ (two identical columns) the last two are the same and equal the original determinant in size $n-1=3$. This shows that $$D_4=-\dfrac{1}{12}D_3$$ if $D_n$ is the determinant of your $n\times n$ matrix. You can generalize these calculations to prove that $$D_{n+1}=-\dfrac{1}{n(n+1)}D_n$$ Together with $D_1=1$ this is enough to find a general formula for $D_n$.