Is division of matrices possible? Is it possible to divide a matrix by another? If...

There is a way to performa sort of division , but I am not sure if it is the way you are looking for. For motivation ,consider the ordinary real numbers R . We have that for two real numbers, $\frac{x}{y}$ is really the same as multiplying x and ${\displaystyle y^{-1}=\frac{1}{y}.}$ We call $y^{-1}$ the inverse of y, and note that it has the property that $y-1=1.$
The same goes for different algebraic structures. That is, for two elements x, y in this algebraic structure we define $\frac{x}{y}$ as ${\displaystyle x{y}^{-1}}$ (under some operation). Most notably, we have a notion of division in any division ring (hence the name!) . It turns out that if you consider invertible ${\displaystyle n\times n}$ matrices with addition and ordinary matrix multiplication, there is a sensible way to define division since every invertible matrix has well, an inverse. So just to help you grip what an inverse is, say that you have a $2\times 2$ matrix
The inverse of A is then given by
$A^{-1}=\frac{1}{(ad-bc)}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
and you should check that $A{A}^{-1}=E$, the identity matrix. Now, for two matrices B and A, $\frac{B}{A}=B{A}^{-1}$

For ordinary numbers ${\displaystyle \frac{a}{b}}$ means the solution to the equation $xb=a$. This is the same as $bx=a$, but since matrix multiplication is not commutative, there are two different possible generalizations of "division" to matrices.
If B is invertible, then you can form ${\displaystyle A{B}^{-1}\text{or}{B}^{-1}A}$, but these are not in general the same matrix. They are the solutions to $XB=A$ and $BX=A$ respectively.
If B is not invertible, then $XB=A$ and $BX=A$ may have solutions, but the solutions will not be unique. So in that situation speaking of "matrix division" is even less warranted.

Normally, matrix division is defined as $\frac{A}{B}=A{B}^{-1}$ where $B^{-1}$ stands for the inverse matrix of B. In the case where the inverse doesn't exist the so called pseudoinverse may be used.