In general, matrix multiplication is not commutative. There are some exceptions, however, most notably the identity matrices (that is, the n by n matrices I_n which consist of 1s along the main diagonal and 0 for all other entries, and which act as the multiplicative identity for matrices)
In general, when taking the product of two matrices A and B, where A is a matrix with m rows and n columns and B is a matrix with n rows and p columns, the resultant matrix AB will possess m rows and p columns. The multiplication cannot occur at all if the number of columns in A is not equal to the number of rows in B, so if AB exists, the only way for BA to exist at all would be if p=m, thus making matrix A a mxn matrix and matrix B a nxm matrix. However, if p=n, it is still quite possible for AB != BA
Recall how each entry in the matrix product is determined. When matrices A and B are multiplied into matrix AB, each entry in the new matrix is formed from the entries in the old matrix. Specifically, to find matrix entry AB_(ij) (that is, the entry in the i row and j column of matrix AB), we take the dot product of row i of matrix A, and column j of matrix B.
AB_(ij) = sum_(k=1)^n A_(ik) B_(kj)
As an example, consider the 3x3 matrices A and B, with
A = ((a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(31),a_(32),a_(33))) and B = ((b_(11), b_(12), b_(13)),(b_(21), b_(22), b_(23)),(b_(31), b_(32), b_(33))).
Then AB_(2,3) is simply the dot product of the second row of A and the third column of B, or (a_(21)\ a_(22)\ a_(23))\ .\ ((b_(13)),(b_(23)),(b_(33))) = (a_(21)*b_(13)) + (a_(22)*b_(23)) + (a_(23)*b_(33))
However, BA_(2,3) would be the dot product of row 2 of matrix B and column 3 of matrix A, or...
(b_(21)\ b_(22)\ b_(23))\ .\ ((a_(13)),(a_(23)),(a_(33))) = (b_(21)*a_(13)) + (b_(22)* a_(23))+(b_(23)*a_(33))
If matrix multiplication were commutative, we would expect BA_(2,3) = AB_(2,3) (among other things). Since these expressions may very well not be equal according to our work thus far, we can safely conclude that matrix multiplication is not necessarily commutative.
