Let $p\_1, p\_2,\dots, p\_n$ and $q\_1,q\_2,\dots,q\_n$ be a collection of complex polynomials. Let $A$ be a $n \times n$ matrix satisfying


$$a\_{ij} = \begin{cases} p\_i(x) & \text{ if } i = j, \\ q\_i(x) & \text{ otherwise} \end{cases} .$$


is there any connection between the roots of the polynomials $p\_i$'s and $q\_i$'s and the roots of the polynomial $\det A$? if not, is this true under at least under any special assumptions?


Kindly share some references.


Thank you.



Roots of determinant of matrix with polynomial entries
Let $r\_i := p\_i - q\_i$.


$${\bf A} (x) := \begin{bmatrix} p\_1 (x) & q\_1 (x) & \ldots & q\_1 (x)\\ q\_2 (x) & p\_2 (x) & \ldots & q\_2 (x)\\ \vdots & \vdots & \ddots & \vdots\\ q\_n (x) & q\_n (x) & \ldots & p\_n (x)\end{bmatrix} = \mbox{diag} \left( {\bf r} (x) \right) + {\bf q} (x) {\Bbb 1}\_n^\top$$


Using the [matrix determinant lemma](https://en.wikipedia.org/wiki/Matrix_determinant_lemma),


$$\det \left ( {\bf A} (x) \right) = \det \left( \mbox{diag} \left( {\bf r} (x) \right) \right) \left( 1 + {\Bbb 1}\_n^\top \mbox{diag}^{-1} \left( {\bf r} (x) \right) {\bf q} (x) \right) = \color{blue}{\left( 1 + \sum\_{i=1}^n \frac{q\_i(x)}{r\_i(x)} \right)\displaystyle\prod\_{i=1}^n r\_i (x)}$$


as mentioned by [Brendan McKay](https://mathoverflow.net/questions/365764/roots-of-determinant-of-matrix-with-polynomial-entries/365772#comment923581_365764) some 20 minutes ago.


