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.
Best Answer
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 $D_n$.