Does polynomial keep inverse?

Question

Let $A=(a_{i,j})_{n\times n}$ be an invertible matrix with the positive rational entry. Let $p(x)$ be a rational polynomial. Consider the following matrix \begin{align*} B=\left(p(a_{i,j})\right)_{n\times n}. \end{align*} Assume that $a_{i,j}$ is not a zero of $p(x)$ for any $1\leq i,j\leq n$. Then I am wondering whether $B$ is invertible?

A more general question can be raised as follows: Let $f(x)$ be a real function. If each $a_{i,j}$ is not a zero of $f(x)$, then when is the matrix $D=\left(f(a_{i,j})\right)_{n\times n}$ invertible?

Any help will be appreciated!:)

Answer 1

No. Take $p(x)=(x-1)^2$ and let $A=\left[\begin{smallmatrix}2&0\\0&2\end{smallmatrix}\right]$. Then$$\begin{vmatrix}p(2)&p(0)\\p(0)&p(2)\end{vmatrix}=\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,$$but $\det A\neq0$.

Answer 2

I am not giving you a straight answer, but will say something that will help you see what all things can happen.

Actually we can easily see the other direction. All matrices are square and size $n\times n$, for some fixed $n>1$.

Let $A$ be a fixed matrix with all entries different. For an arbitrary matrix $X$ we can find a polynomial $p(x)$ such that $x_{i,j}= p(a_{i,j})$. This is Lagrange Interpolation Theorem (Numerical mathematics, or Chinese Remainder Theorem).

So this covers $X$ singular as well as non-singular.

Now you should be able to see.