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{\bf Question}

In this question we consider the following matrices $$ A =
\left(\begin{array}{ccc} 1 &amp; 2 &amp; 3 \\ 4 &amp; 5 &amp; 6 \end{array}
\right) \hspace{.2in}  B = \left(\begin{array}{ccc} 1 &amp; 1 &amp; 0 \\ 0
&amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1 \end{array} \right)  \hspace{.2in} C =
\left(\begin{array}{cc} 1 &amp; 2 \\ 3 &amp; 4 \\ 5 &amp; 6 \end{array}
\right) \hspace{.2in}  D = \left(\begin{array}{cc} 1 &amp; 2 \\ 3 &amp; 4
\end{array} \right)$$

Consider the following products.  In each case:
\begin{description}
\item[(i)] establish whether the quantity exists and is well
defined
\item[(ii)] if it does exist, evaluate it.
\end{description}

\begin{tabular}{llll}
\hspace{.5in}(a)\ \ $BA$\hspace{.2in}&amp;(b)\ \
$AC$\hspace{.2in}&amp;(c)\ \ $BC$\hspace{.2in}&amp;(d)\ \   $BC^T$\\
\hspace{.5in}(e)\ \ $CA$\hspace{.2in}&amp;(f)\ \
$(CB)A$\hspace{.2in}&amp;(g)\ \ $BD$\hspace{.2in}&amp;(h)\ \ $CD$
\end{tabular}


\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
$BA:$ number of columns of $B$ = 3 $\not=$ number of rows of $A$ =
2

Therefore the product does not exist
\item[(b)]
$AC:$ number of columns of $A$ = 3 = number of rows of $C$

Therefore the product does exist and gives a two by two matrix

$\left(\begin{array}{ccc} 1 &amp; 2 &amp; 3 \\ 4 &amp; 5 &amp; 6 \end{array}
\right)\left(\begin{array}{cc} 1 &amp; 2 \\ 3 &amp; 4 \\ 5 &amp; 6
\end{array} \right) = \left(\begin{array}{cc} 1+6+15 &amp; 2+8+18 \\ 4+15+30 &amp; 8+20+36
\end{array} \right) = \left(\begin{array}{cc} 22 &amp; 28 \\ 49 &amp; 64 \end{array}
\right)$
\item[(c)]
$BC:$ number of columns of $B$ = 3 = number of rows of $C$

Therefore the product does exist and gives a three by two matrix

$\left(\begin{array}{ccc} 1 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1
\end{array} \right)\left(\begin{array}{cc} 1 &amp; 2 \\ 3 &amp; 4 \\ 5 &amp; 6
\end{array} \right) = \left(\begin{array}{cc} 1+3 &amp; 2+4 \\ 5 &amp; 6
\\ 3+5 &amp; 4+6 \end{array} \right) = \left(\begin{array}{cc} 4 &amp; 6
\\ 5 &amp; 6 \\ 8 &amp; 10 \end{array} \right)$
\item[(d)]
$B$ is $3 \times 3$; $C^T$ is $2 \times 3$;  $BC^T$ does not exist
\item[(e)]
$CA:$ $C$ is $3 \times 2$ and $A$ is $2 \times 3$

Therefore the product does exist and gives a three by three matrix

$\left(\begin{array}{cc} 1 &amp; 2 \\ 3 &amp; 4 \\ 5 &amp; 6
\end{array} \right)\left(\begin{array}{ccc} 1 &amp; 2  \\ 3 &amp; 4 \\ 5 &amp; 6
\end{array} \right) = \left(\begin{array}{ccc} 1+9 &amp; 2+10 &amp; 3 + 12 \\ 3+16 &amp;
6+20 &amp; 9 + 24 \\ 5+24 &amp; 10+20 &amp; 15+36 \end{array} \right) =
\left(\begin{array}{ccc} 9 &amp; 12 &amp; 15 \\ 19 &amp; 26 &amp; 33 \\ 29 &amp; 40 &amp;
51  \end{array} \right)$

\item[(f)]
$C$ is $3 \times 2$; $B$ is $3 \times 3$; therefore $CB$ does not
exist
\item[(g)]
$B$ is $3 \times 3$; $D$ is $2 \times 2$; therefore $BD$ does not
exist
\item[(h)]
$C$ is $3 \times 2$; $D$ is $2 \times 2$; therefore $CD$ is a $3
\times 2$ matrix

$\left(\begin{array}{cc} 1 &amp; 2 \\ 3 &amp; 4 \\ 5 &amp; 6
\end{array} \right)\left(\begin{array}{cc} 1 &amp; 2 \\ 3 &amp; 4
\end{array} \right) = \left(\begin{array}{cc} 1+6 &amp; 2+8 \\ 3+12 &amp;
6+16 \\ 5+18 &amp; 10+24 \end{array} \right) = \left(\begin{array}{cc}
7 &amp; 10 \\ 15 &amp; 22 \\ 23 &amp; 34 \end{array} \right)$

\end{description}


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