Solution:

(i) AB

= $\left[ \begin{matrix} 3 & 0 \\ -1 & 2 \\ \end{matrix} \right]\times \left[ \begin{matrix} 6 \\ 5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times 6+0\times 5 \\ \left( -1 \right)\times 6+2\times 5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 18+0 \\ -6+10 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 18 \\ 4 \\ \end{matrix} \right]$

So, AB = $\left[ \begin{matrix} 18 \\ 4 \\ \end{matrix} \right]$

(ii) BA

BA is not possible as it is not conformable for multiplication (number of columns of B is not equal to number of rows of A).