Suppose I have a matrix where $x \in \mathbb{Z}$
$ A = \begin{bmatrix} x_{11} & x_{12} & x_{13} & \dots & x_{1n} \\ x_{21} & x_{22} & x_{23} & \dots & x_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{d1} & x_{d2} & x_{d3} & \dots & x_{dn} \end{bmatrix}$
what is the best way to arrive at the following matrix:
$ B = \begin{bmatrix} \frac{1}{x_{11}} & \frac{1}{x_{12}} & \frac{1}{x_{13}} & \dots & \frac{1}{x_{1n}} \\ \frac{1}{x_{21}} & \frac{1}{x_{22}} & \frac{1}{x_{23}} & \dots & \frac{1}{x_{2n}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{x_{d1}} & \frac{1}{x_{d2}} & \frac{1}{x_{d3}} & \dots & \frac{1}{x_{dn}} \end{bmatrix}$
You can define the Schur product (sometimes also called Hadamard product) as follows:
If $A, B \in \mathbb K^{n \times m}$ then the Schurproduct $A \ast B$ is defined as $(A \ast B)_{i,j} = (A)_{i,j} (B)_{i,j}$, i.e. the multiplication of the entries.
Using that you can say the following:
"Let $B$ the inverse of $A$ respecting the Schur product."