linear algebra - determinant of a tricky matrix

I'm doing a research on matrix integrators and I ran into a problem in one particular case. To finish my proof the last thing remaining is to prove the nonsingularity of a specific matrix $$M_n: (m_{ij} = \frac{1}{a_i - a_j}, 1\leq i \leq n, 1\leq j \leq n,i\neq j;m_{ii} = \frac{c}{a_i - b} + \sum\limits_{k\neq i, 1\leq k \leq n}\frac{1}{a_i - a_k}),$$
where all $a_i, b$ are distinct.

To be more clear I provide $$M_2 = \begin{pmatrix} \frac{c}{a_1 -b} + \frac{1}{a_1 - a_2} && \frac{1}{a_1 - a_2}\\ \frac{1}{a_2 - a_1} && \frac{c}{a_2 -b} + \frac{1}{a_2 - a_1} \end{pmatrix}$$

$$M_3 = \begin{pmatrix} \frac{c}{a_1 -b} + \frac{1}{a_1 - a_2} + \frac{1}{a_1 - a_3} && \frac{1}{a_1 - a_2} && \frac{1}{a_1 - a_3}\\ \frac{1}{a_2 - a_1} && \frac{c}{a_2 -b} + \frac{1}{a_2 - a_1} + \frac{1}{a_2 - a_3} && \frac{1}{a_2 - a_3}\\ \frac{1}{a_3 - a_1} && \frac{1}{a_3 - a_2} && \frac{c}{a_3 - b} + \frac{1}{a_3 - a_1} + \frac{1}{a_3 - a_2} \end{pmatrix}$$

For $n \leq 7$ I calculated the $det(M_n) = \frac{c(c+1)...(c + n -1)}{\prod\limits_{1\leq i\leq n}(a_i - b)}$, but I have no idea how to prove this in general case.

I my particular case $c\in \mathbb N$, so this formula will prove the nonsingularity of $M_n$.

Any ideas and tips to prove the formula, or even to prove nonsingularity of $M_n$ in some other way - are very appreciated

Answer

The answer is a development of ideas from the comments of amsmath. The decisive step forward is derivation of the equation $(7)$ below.

Given a set of $n$ pairs of complex numbers
$$\{(x_1,y_2),(x_2,y_2),\dots,(x_n,y_n)\}$$ such that
$$\forall\; i\ne j:\; x_i\ne x_j,$$
define its interpolating polynomial as

$$y(x)=\sum_{i}y_i\prod_{k\ne i}\frac{x-x_k}{x_i-x_k}.\tag1$$

Differentiating the expression over $x$ and evaluating the result at $x_i$ one obtains:
$$y'_i\equiv y'(x_i)=\sum_{j\ne i}\frac1{x_i-x_j} \left[y_i-y_j\prod_{k\ne(i,j)}\frac{x_i-x_k}{x_j-x_k}\right].\tag2$$
or

$$\frac{y'_i}{\prod\limits_{k\ne i}x_i-x_k} =\sum_{j\ne i}\frac1{x_i-x_j}\left[\frac{y_i}{\prod\limits_{k\ne i}x_i-x_k} +\frac{y_j}{\prod\limits_{k\ne j}x_j-x_k}\right].\tag{2a}$$
Introducing $f_i=\frac{y_i}{\prod\limits_{k\ne i}x_i-x_k}$, $f'_i=\frac{y'_i}{\prod\limits_{k\ne i}x_i-x_k}$ the equation $(\text{2a})$ can be rewritten in matrix notation as:
$$\begin{pmatrix} \sum\limits_{i\ne1}\frac{1}{x_1-x_i}& \frac1{x_1-x_2}&\cdots&\frac1{x_1-x_n}\\ \frac1{x_2-x_1}& \sum\limits_{i\ne2}\frac{1}{x_2-x_i}&\cdots&\frac1{x_2-x_n}\\ \vdots& \vdots& \ddots&\vdots\\ \frac1{x_n-x_1}&\frac1{x_n-x_2}&\cdots&\sum\limits_{i\ne n}\frac{1}{x_n-x_i}\\ \end{pmatrix} \begin{pmatrix} \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f_1\\ \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f_2\\ \vdots\\ \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f_n\\ \end{pmatrix}= \begin{pmatrix} \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f'_1\\ \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f'_2\\ \vdots\\ \vphantom{\sum\limits_{i\ne1}\frac{1}{x_1-x_i}}f'_n\\ \end{pmatrix}.\tag3$$
or
$$A f=f'.\tag4$$

Assume now a special form of the interpolating polynomial:
$$
y_l(x)=(\lambda x+\beta)^l,\quad
\beta,\lambda\in\mathbb C,\,\lambda\ne 0;\; l\in\mathbb Z,\,0\le l$$
with corresponding $f$-vector components
$$f_{li}=\frac{(\lambda x_i+\beta)^l}{\prod\limits_{k\ne i}x_i-x_k},\quad f'_{li}=\frac{\lambda l(\lambda x_i+\beta)^{l-1}}{\prod\limits_{k\ne i}x_i-x_k}.\tag5$$

Substituting the vectors $f_l$ and $f'_l$ into $(4)$ and multiplying both sides of the resulting equation from the left by diagonal matrix $D$ with elements
$$D_{ii}=\lambda x_i+\beta,\tag6$$
one obtains:
$$DA f_l=\lambda l f_l.\tag7$$

From this one concludes that $f_l$ are the eigenvectors of the matrix $DA$ with corresponding eigenvalues $\epsilon_l=\lambda l$. Observe that we have found all $n$ (distinct) eigenvalues of the matrix.

Since the determinant of the matrix $DA+Iz$ is characteristic polynomial of the matrix $-DA$, one obtains:
$$\begin{align} &\det (DA+I z)=\prod_{l=0}^{n-1} z+\lambda l\tag8\\ &\implies \det (A+D^{-1}z)=\det{D^{-1}}\det (DA+Iz) =\prod_{l=0}^{n-1}\frac{z+\lambda l}{\lambda x_{l+1}+\beta}.\tag9 \end{align}$$

It remains only to observe that $A+D^{-1}z$ with $\lambda=1$, $\beta=-b$, $z=c$, $x_i=a_i$ is exactly your matrix $M$.