what is the proof for matrix multiplication being commutative

Question

I understand that if we have matrix $A$ and $B$ then $A \cdot B \neq B \cdot A$ as when you multiply the matrices in a different order, then their cells will shift in another form, thus making their multiplication equate differently in their product matrix.

but what is valid proof i can give to illustrate this

All help is much appreciated

Answer 1

A counter-example would suffice here. Take for example the following: $$\begin{align*} \pmatrix{1&0\\0&0} \pmatrix{0&1\\0&0} & = \pmatrix{0&1\\0&0}\\ & \qquad\not\parallel\\ \pmatrix{0&1\\0&0} \pmatrix{1&0\\0&0} & = \pmatrix{0&0\\0&0} \end{align*}$$

Answer 2

The only proof you need here is an example that shows you that, sometimes, $AB \neq BA$. For example, try $$ A = \pmatrix{0 & 1\\0&0}, \quad B = \pmatrix{0&0\\1&0} $$ We find $$ AB = \pmatrix{1&0\\0&0}, \quad BA = \pmatrix{0&0\\0&1} $$