Lets define a matrix of 9s as: $$ N = \begin{bmatrix} 9&9&9\\9&9&9\\9&9&9 \end{bmatrix} $$
Lets define an exploding number as a number at position \$(x,y)\$ that can be decomposed into equal integers between all its adjacent neighbors (including itself) and the absolute value of each portion is greater than 0.
From the previous matrix, lets explode the number at position \$(1,1)\$ (0 indexed) $$ N = \begin{bmatrix} 9&9&9\\9&\color{red}9&9\\9&9&9 \end{bmatrix} $$ $$ N = \begin{bmatrix} 9+\color{red}1&9+\color{red}1&9+\color{red}1\\9+\color{red}1&\color{blue}0+\color{red}1&9+\color{red}1\\9+\color{red}1&9+\color{red}1&9+\color{red}1 \end{bmatrix} $$
$$ N = \begin{bmatrix} 10&10&10\\10&\color{red}1&10\\10&10&10 \end{bmatrix} $$
Sometimes, decomposing result into a rational number greater than 1. This is something we need to avoid when exploding numbers. In this cases the remainder will be assigned to the exploded number.
To demonstrate it, lets continue working with our previous matrix. This time we will explode the number at position \$(0,0)\$
$$ N = \begin{bmatrix} \color{red}{10}&10&10\\10&1&10\\10&10&10 \end{bmatrix} $$
Here we have 3 neightbors and the number itself. Here the equation is something like \$10/4\$ which give us 2 for each and 2 as remainder.
$$ N = \begin{bmatrix} \color{blue}2+\color{red}{2}&\color{red}{10+2}&10\\\color{red}{10+2}&\color{red}{1+2}&10\\10&10&10 \end{bmatrix} $$
$$ N = \begin{bmatrix} \color{red}{4}&12&10\\12&3&10\\10&10&10 \end{bmatrix} $$
As well, sometimes a number wont be big enough to be decomposed in equal parts (where the abs is greater than 0) between his neighbors (|rational number| < 1). In this cases we need to "borrow" from the exploded number in order to maintain the "greater than 0" condition. Lets continue with our previous example and explode the number at position \$(1,1)\$.
$$ N = \begin{bmatrix} 4&12&10\\12&\color{red}3&10\\10&10&10 \end{bmatrix} $$
$$ N = \begin{bmatrix} 4+\color{red}1&12+\color{red}1&10+\color{red}1\\12+\color{red}1&\color{blue}0+\color{red}{1}-\color{green}6&10+\color{red}1\\10+\color{red}1&10+\color{red}1&10+\color{red}1 \end{bmatrix} $$ $$ N = \begin{bmatrix} 5&13&11\\13&\color{red}{-5}&11\\11&11&11 \end{bmatrix} $$
The challenge is, given a list of \$(x,y)\$ positions and an finite non-empty array of natural numbers, return the exploded form after each number from the positions list has been exploded.
Test cases
Input: initial matrix: [[3, 3, 3], [3, 3, 3], [3, 3, 3]], numbers: [[0,0],[0,1],[0,2]]
Output: [[1, 0, 1], [5, 6, 5], [3, 3, 3]]
Input: Initial matrix: [[9, 8, 7], [8, 9, 7], [8, 7, 9]], numbers: [[0,0],[1,1],[2,2]]
Output: [[4, 11, 8],[11, 5, 10],[9, 10, 4]]
Input: Initial matrix: [[0, 0], [0, 0]], numbers: [[0,0],[0,0],[0,0]]
Output: [[-9, 3],[3, 3]]
Input: Initial Matrix: [[10, 20, 30],[30, 20, 10],[40, 50, 60]], numbers: [[0,2],[2,0],[1,1],[1,0]]
Output: [[21, 38, 13], [9, 12, 21], [21, 71, 64]]
Input: Initial Matrix: [[1]], numbers: [[0,0]]
Output: [[1]]
Input: Initial Matrix: [[1, 2, 3]], numbers: [[0,0], [0, 1]]
Output: [[1, 1, 4]]
Notes
Input/Output rules apply
You can assume input matrix will never be empty
You can assume coordinates are always going to be valid
Input coord in test cases is given as (row, column). If you need it to be (x, y) you can swap the values. If so, please state that in your answer