Show that $\left[ {\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}} \right] \left[ {\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}} \right] \ne \left[ {\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}} \right] \left[ {\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}} \right]$

Let $A = \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]$ and $B = \left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]$.
First,calculate the product $AB$:
$AB = \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right] \left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]$
$= \left[\begin{array}{cc}5(2) + (-1)(3) & 5(1) + (-1)(4) \\ 6(2) + 7(3) & 6(1) + 7(4)\end{array}\right]$
$= \left[\begin{array}{cc}10 - 3 & 5 - 4 \\ 12 + 21 & 6 + 28\end{array}\right] = \left[\begin{array}{cc}7 & 1 \\ 33 & 34\end{array}\right]$
Next,calculate the product $BA$:
$BA = \left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right] \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]$
$= \left[\begin{array}{cc}2(5) + 1(6) & 2(-1) + 1(7) \\ 3(5) + 4(6) & 3(-1) + 4(7)\end{array}\right]$
$= \left[\begin{array}{cc}10 + 6 & -2 + 7 \\ 15 + 24 & -3 + 28\end{array}\right] = \left[\begin{array}{cc}16 & 5 \\ 39 & 25\end{array}\right]$
Since $\left[\begin{array}{cc}7 & 1 \\ 33 & 34\end{array}\right] \ne \left[\begin{array}{cc}16 & 5 \\ 39 & 25\end{array}\right]$,it is shown that $AB \ne BA$.