# Finding the determinant of a Cauchy matrix

Found this nice matrix:

$$\begin{pmatrix} \dfrac{1}{a_1+b_1} & \dfrac{1}{a_2+b_1} & \cdots & \dfrac{1}{a_n+b_1} \\ \dfrac{1}{a_2+b_1} & \dfrac{1}{a_2+b_2} & \cdots & \dfrac{1}{a_n+b_2}\\ \vdots & \vdots &\ddots & \vdots \\ \dfrac{1}{a_n + b_1} & \dfrac{1}{a_n + b_2} & \cdots & \dfrac{1}{a_n+b_n}\end{pmatrix}$$

What is its determinant? Is there a way to simplify it into a more compact form?

I was able to find the determinant for $$2\times 2$$ case:

$$\dfrac{(a_1-a_2)(b_1-b_2)}{(a_1+b_1)(a_2+b_2)(a_1+b_2)(a_2+b_1)}$$

Is there a nice form for the $$n\times n$$ case?

• maybe they must not be both zero.
– cgo
Mar 26, 2021 at 7:46
• If all the $a_i + b_j > 0$, then the given Cauchy matrix is a solution of a Sylvester equation, I believe. Mar 26, 2021 at 7:47

This is a Cauchy matrix. By applying the formula on Wikipedia to your specific Cauchy matrix, we get that the determinant is $$\dfrac{\prod_{i = 2}^{n}\prod_{j = 1}^{i-1}(a_i-a_j)(b_i-b_j)}{\prod_{i = 1}^{n}\prod_{j = 1}^{n}(a_i+b_j)}.$$