When are multiplication on matrices commutative? 
*

*According to me multiplication on matrices are commutative only when 
(i) The given matrices are equal
(ii) When the matrices are diagonal matrices and of same order.
(iii) When a suitable identity matrix is being used as prefactor or postfactor


Are there any other possibilities when multiplication of matrices commutative?
2.If there is a matrix A and if assume $A^m$ is equal to B then 
$A^n*A=B=A*A^n$
How is this possible?Is there any proof for it? What logic is being used here?
 A: Yes, there are several other ways for matrices to commute, e.g., sitting in a proper upper-right corner of $M_n(K)$. By a theorem of Jacobson, the maximal dimension of a commutative subalgebra in $M_n(K)$ is given by $[n^2/4]$. There the maximal commutative subalgebras are also described explicitly. There are several posts here at MSE on this topic, e.g. this one:
Can any complex $n\times n$ matrix be in a maximal commuting set of $\lfloor n^2/4\rfloor + 1$ matrices?
There is a whole book on the subject by D. A. Suprunenko, see here. Also, there are several interesting open problems, see the article by C. Procesi,
Invariants of Commuting Matrices.
A: *

*There many possibilites. For instance any two matrices of the form $\left[\begin{smallmatrix}r\cos\theta&-r\sin\theta\\r\sin\theta&r\cos\theta\end{smallmatrix}\right]$ commute.

*Because bith of them are equal to$$\overbrace{A\times A\times\cdots\times A}^{n+1\text{ times}}.$$
A: An easy way to create a lot of commuting pairs of matrices is to take a square matrix $M$, and two polynomials $f$ and $g$ with coefficients in your field.
Then $A=f(M)$ and $B=g(M)$ satisfy $AB=BA$.
