Set \begin{align*} A= \begin{bmatrix} 1 & 3\\ -1 & 2 \end{bmatrix} && B= \begin{bmatrix} 4&0\\ 5&1 \end{bmatrix}. \end{align*}
Then we have two square, $2\times 2$ matrices, so Definition MM allows us to multiply them in either order. We find \begin{align*} AB= \begin{bmatrix} 19 & 3\\ 6 & 2 \end{bmatrix} && BA= \begin{bmatrix} 4 & 12\\ 4 & 17 \end{bmatrix} \end{align*} and $AB\neq BA$. Not even close. It should not be hard for you to construct other pairs of matrices that do not commute (try a couple of $3\times 3$'s). Can you find a pair of non-identical matrices that do commute?