The usual answer to this question is to try some experimentation with 3 x 3 matrices. A few examples will show that two 3x3 matrices, M and N, do not commute for most choices of M and N. The only way to give a better answer is to provide a technical discussion in terms of eigenvalues and eigenvectors. Such a discussion is more suitable to a text book than it is to this forum.
However, it is possible to provide a limited insight in this limited space. A key idea is that for matrix M there will exist three vectors v1 , v2 , v3 { here a vector is a 3 x 1 matrix } which are the eigenvectors of M. The idea is that
M v1 = λ1 v1 ; M v2 = λ2 v2 ; M v3 = λ3 v3 The numbers λ1 , λ2 and λ3 are the eigenvalues. In general they are three different numbers. Notice that M v1 is a 3x3 matrix multiplying a 3 x 1 matrix resulting in a 3 x 1 vector. Suppose that λ1 is a large number and that λ2 and λ3 are not . Matrix N might have the property that N v1 is a vector that is nearly perpendicular to v1 but not much longer (i.e. similar in modulus) than v1 .
Thus N M v1 will produce a large magnitude vector ( because λ1 is large) , but M N v1 will produce a small magnitude vector ( because matrix M multiplying a vector perpendicular to v1 will produce a small magnitude vector because λ2 and λ3 are small. )
This argument shows that M N acting on v1 will not produce the same vector as N M acting on the same vector. This is equivalent to the statement that M and N do not commute.
