For $A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $B=\begin{bmatrix}3&1\\2&0\end{bmatrix}$, $AB=\begin{bmatrix}7&1\\2&0\end{bmatrix}$. For example, the top-left entry is $1\cdot3+2\cdot2=7$. Reversing the order gives $BA=\begin{bmatrix}3&7\\2&4\end{bmatrix}$, a different matrix.
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