Non-commutativity of matrix multiplication

Counterexample

Let $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, as real $2 \times 2$ matrices. Then $$AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad BA = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.$$ Thus $AB \ne BA$. This shows that matrix multiplication of $n \times n$ matrices is not commutative for $n = 2$.

Remarks

The same example, extended with zero rows and zero columns, shows that $M_n(R)$ is not commutative for each $n \ge 2$ and each nonzero ring $R$. Two matrices $A$ and $B$ with $AB = BA$ are said to commute. Some pairs commute, for example a matrix and the identity matrix, or two diagonal matrices.

Explanations

L1 human

For numbers, $2 \cdot 3 = 3 \cdot 2$, but for matrices the order of the factors matters. Here $AB$ and $BA$ are different matrices. A reason is that $AB$ means "first apply $B$, then $A$", and two actions done in a different order can give different results.