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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

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    $\begingroup$ Do you mean the proof for matrix multiplication being non-commutative? $\endgroup$ – Omnomnomnom Mar 4 '15 at 22:11
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A counter-example would suffice here. Take for example $$\begin{align*} \pmatrix{1&0\\0&0} \pmatrix{0&1\\0&0} & = \pmatrix{0&1\\0&0}\\ & \qquad\quad\small\diagdown\hspace{-6 pt}||\\ \pmatrix{0&1\\0&0} \pmatrix{1&0\\0&0} & = \pmatrix{1&0\\0&0} \end{align*}$$

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  • $\begingroup$ Does anyone know how to a) remove the manual spacing hack of the $\ne$ and b) rotate said symbol by $90^\circ$? $\endgroup$ – AlexR Mar 4 '15 at 22:13
  • $\begingroup$ for the spacing you might be able to do something with the align environment. Don't know about rotating the symbol though. $\endgroup$ – Omnomnomnom Mar 4 '15 at 22:15
  • $\begingroup$ @Omnomnomnom It's already aligned. array will do this, but not quite worth it if rotatebox doesn't work :( $\endgroup$ – AlexR Mar 4 '15 at 22:16
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    $\begingroup$ @String Thanks, I tweaked it to \small\diagdown\hspace{-6pt}||: $\small\diagdown\hspace{-6pt}||$ cf. $\displaystyle\ne$ $\endgroup$ – AlexR Mar 4 '15 at 22:45
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    $\begingroup$ @AlexR: That actually looks quite nice. But to have graphicx among supported packages would be even nicer, though! $\endgroup$ – String Mar 4 '15 at 22:50
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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 $$ AB = \pmatrix{1&0\\0&0}, \quad BA = \pmatrix{0&0\\0&1} $$

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