Let $M(n)$ be an $n\times n$ matrix in the variables $x\_1,\dots,x\_n$ with entries
$$M\_{i,j}(n)=\frac{x\_{\max(i,j)}}{x\_{\min(i,j)}}, \qquad 1\leq i,j\leq n.$$
I'm interested in the following:



> 
> **Questions.**
> 
> 
> (1) Is there a neat or "closed form" evaluation for the determinant $\det M(n)$?
> 
> 
> (2) Is there an explicit formula for the inverse of $M(n)$?
> 
> 
> 


Thank you.



minimum-maximum entries matrix
Let us write $$a\_r=\frac{x\_{r+1}}{x\_r}$$ for $r=1\cdots n$. 


We can then write the matrix $M(n)$ in the form 
$$\begin{pmatrix} 1 & a\_1 & a\_1a\_2& \cdots & a\_1a\_2\cdots a\_{n-1} \\ a\_1 & 1 & a\_2& \cdots & a\_2\cdots a\_{n-1}\\ \vdots & \vdots & \vdots &\ddots & \vdots\\ a\_1a\_2\cdots a\_{n-1}& a\_2\cdots a\_{n-1} &\cdots & a\_{n-1} & 1\end{pmatrix} $$ 
We do now Gauss elimination, and reduce $a\_{n-1}$ times the $(n-1)$-th row from the $n$-th row. We then get in the last row $0, 0, \ldots (1-a\_{n-1}^2)$.
But this means that $det(M(n))=det(M(n-1))(1-a\_{n-1}^2)$ and by induction
$$det(M(n)) = \prod\_{r=1}^{n-1} (1-a\_r^2)=\prod\_{r=1}^{n-1}(1-\frac{x\_{r+1}^2}{x\_r^2}).$$
By using inductively the Gauss elimination mentioned above, one can also get the inverse of $M(n)$.


