\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Find $AB$ and $BA$ where

\begin{description}
\item[(i)]
$A=\left(\begin{array}{cc} -5 & 1\\ 9 & -1\end{array}\right);\ \
B=\left(\begin{array} {cc} 3 & 3\\ 1 & -3 \end{array} \right)$

\item[(ii)]
$A=\left(\begin{array}{cc} 7 & 1\\ -3 & 6\\ 6 & -3
\end{array}\right);\ \ B=\left(\begin{array} {c} 1\\3 \end{array}\right)$
\end{description}

\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$AB=\left(\begin{array}{cc} -5 & 1\\9 & -1
\end{array}\right)\left(\begin{array}{cc} 3 & 3\\1 & -3
\end{array}\right)=\left(\begin{array}{cc} -14 & -18\\26 & 30
\end{array}\right)$

\item[(ii)]
$BA=\left(\begin{array}{cc} 3 & 3\\1 & -3
\end{array}\right)\left(\begin{array}{cc} -5 & 1\\9 & -1
\end{array}\right)=\left(\begin{array}{cc} 12 & 0\\-32 & 4
\end{array}\right)$

so $AB \ne BA$

\item[(ii)]
$AB=\left(\begin{array}{cc} 7 & 1\\-3 & 6\\6 & -3
\end{array}\right)\left(\begin{array}{c} 1\\3
\end{array}\right)=\left(\begin{array}{c} 10\\15\\-3
\end{array}\right)$

$\overbrace{3 \times \underbrace{2\ \ \ 2} \times 1}$

\item[(iii)]
$BA=\left(\begin{array}{c} 1\\3
\end{array}\right)\left(\begin{array}{cc} 7 & 1\\-3 & 6\\6 & -3
\end{array}\right)$

$\overbrace{2 \times \underbrace{1\ \ \ 3} \times 2}$

Incompatible so $BA$ does not exist!

\end{description}

\end{document}