since $A$ is a $2 \times 3$ matrix and $B$ is $3 \times 2$ matrix. Hence $A B$ and $B A$ are both defined and are matrices of order $2 \times 2$ and $3 \times 3,$ respectively. Note that

$AB=$ $\left[ \begin{array}{*{35}{r}}
   1 & -2 & 3  \\
   -4 & 2 & 5  \\
\end{array} \right]\left[ \begin{array}{*{35}{r}}
   2 & 3  \\
   4 & 5  \\
   2 & 1  \\
\end{array} \right]$ $=\left[ \begin{matrix}
   2-8+6 & 3-10+3  \\
   -8+8+10 & -12+10+5  \\
\end{matrix} \right]$ $=\left[ \begin{array}{*{35}{r}}
   0 & -4  \\
   10 & 3  \\
\end{array} \right]$

and  $\text{BA}=\left[ \begin{array}{*{35}{l}}
   2 & 3  \\
   4 & 5  \\
   2 & 1  \\
\end{array} \right]$  $\left[ {\begin{array}{*{20}{r}}
  1&{ – 2}&3 \\ 
  { – 4}&2&5 
\end{array}} \right] = $ $\left[ {\begin{array}{*{20}{c}}
  {2 – 12}&{ – 4 + 6}&{6 + 15} \\ 
  {4 – 20}&{ – 8 + 10}&{12 + 25} \\ 
  {2 – 4}&{ – 4 + 2}&{6 + 5} 
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
  { – 10}&2&{21} \\ 
  { – 16}&2&{37} \\ 
  { – 2}&{ – 2}&{11} 
\end{array}} \right]$

Clearly $\mathrm{AB} \neq \mathrm{BA}$

In the above example both $AB$ and $BA$ are of different order and so $AB \neq BA$. But one may think that perhaps $A B$ and $B A$ could be the same if they were of the same order. But it is not so, here we give an example to show that even if $\mathrm{AB}$ and $\mathrm{BA}$ are of same order they may not be same.