What happens to a Matrix after moving to the other side of the equal sign? I have the following equation where each variable is a matrix that can be multiplied
A * B = C
I have matrix A and C but matrix B is unknown and I need to get its value
the way I though about it is to make a the following
B = C / A
which means 
B = C * (1/A)
which means
B = C * A-1
where A-1 means the inverse of A
is that correct or should something happens to matrix A after moving to the other side of the equation?
 A: No, this is not correct.
There are two things going on:


*

*You have implicitly assumed that $A$ is invertible. It may be the case that this $A$ has no inverse; this is analagous to trying to divide by $0$. If $A$ does not have an inverse, then everything you've written past the first equation is nonsense.

*You have mixed up "division" with "multiplying by an inverse". In familiar situations, these are, more or less by definition, the same operation, but multiplication of matrices is not commutative. That is, there are matrices $A$ and $B$ such that $AB \neq BA$. In particular, what you should have is
$$AB = C \\
A^{-1}AB = A^{-1}C \\
IB = A^{-1}C \\
B = A^{-1}C$$
Notice how here we have $B = A^{-1}C$, not $B = CA^{-1}$. If $A^{-1}$ and $C$ do not commute, these answers will be different.
A: It is not correct. Matrices are not commutative and you can't just multiply on whatever side you feel like:
$$AB = C\implies A^{-1}AB=A^{-1}C \implies B=A^{-1}C.$$
On the other hand, $A$ might not be invertible at all, i.e. $A^{-1}$ might not exist. You need to check that before you do the above.
In general, things don't move in an equation. $A$ didn't move to the other side and magically change to $A^{-1}$. Similarly, when you write $zt+3=y\implies zt+3-y=0$, $y$ didn't move and change to $-y$. 
What happens in both cases is that we apply some function on both sides of equation. By the very definition of function, both sides must be equal. In the first case, let $f(X) = A^{-1}X$ and $$AB = C\implies f(AB) = f(C)\implies A^{-1}(AB) = A^{-1}C,$$ while in the second case $f(x) = x - y$ and $$zt+3=y\implies f(zt+3) = f(y)\implies (zt+3)-y = y- y.$$
