Is division of matrices possible?

For ordinary numbers $\frac{a}{b}$ 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 $AB^{-1}$ 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.

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 $\mathbb{R}$ . We have that for two real numbers, $x/y$ is really the same as multiplying x and $y^{-1}=1/y$. We call $y^{-1}$ the inverse of y, and note that it has the property that $yy^{-1}=1.$

The same goes for different algebraic structures. That is, for two elements x,y in this algebraic structure we define $x/y$ as $xy^{-1}$ (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 $n \times n$ 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 2x2 matrix $$A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$ The inverse of A is then given by $$A^{-1} = \dfrac{1}{(ad-bc)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$ and you should check that $AA^{-1}=E$, the identity matrix. Now, for two matrices $B$ and $A$, $B/A = BA^{-1}$.

Normally, matrix division is defined as $\frac{A}{B}=AB^{-1}$ 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.