<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\begin{document}
\parindent=0pt

{\bf Question}

Using the matrices $$ A = \left(\begin{array}{ccc} 1 &amp; 2 &amp; 3 \\ 4
&amp; 5 &amp; 6 \end{array} \right) \hspace{.2in}  B =
\left(\begin{array}{ccc} 1 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1
\end{array} \right)  \hspace{.2in} C = \left(\begin{array}{cc} 1
&amp; 2 \\ 3 &amp; 4 \\ 5 &amp; 6 \end{array} \right) $$ show that the
matrices $(C^TB)A^T$ and $C^T(BA^T)$ exist and are equal

\vspace{.25in}

{\bf Answer}

\bigskip

 $\begin{array}{lll} C^T &amp; \ \mathrm{is}\ &amp; 2\times 3 \\
B &amp;\ \mathrm{is}\ &amp; 3\times 3\end{array} \Rightarrow\ \ C^TB\
\mathrm{is}\ 3\times 3$

$A\ \mathrm{is}\ 2\times 3 \Rightarrow A^T\ \mathrm{is}\ 3\times
2$

Similarly $BA^T$ is $3\times 2$ and $C^T$ is $2\times 3$
$\Rightarrow$ $C^T(BA^T)$ is $2\times 2$

Hence both exist

$C^TB = \left(\begin{array}{ccc} 1 &amp; 2 &amp; 3 \\ 4 &amp; 5 &amp; 6
\end{array} \right)\left(\begin{array}{ccc} 1 &amp; 1 &amp; 0  \\ 0 &amp; 0 &amp;
1 \\ 0 &amp; 1 &amp; 1 \end{array} \right) = \left(\begin{array}{ccc} 1 &amp;
1+5 &amp; 3+5
\\ 2 &amp; 2+6 &amp; 4+6 \end{array} \right) = \left(\begin{array}{ccc}
1 &amp; 6 &amp; 8 \\ 2 &amp; 8 &amp; 10 \end{array} \right)$

\begin{eqnarray*} (C^TB)A^T = \left(\begin{array}{ccc} 1 &amp; 6 &amp; 8 \\ 2 &amp; 8 &amp; 10
\end{array} \right)\left(\begin{array}{cc} 1 &amp; 4 \\ 2 &amp; 5 \\ 3 &amp; 6
\end{array} \right)&amp;=&amp; \left( \begin{array}{cc} 1+12+24 &amp; 4+ 30+48
\\ 2+16+30 &amp; 8+40+60 \end{array} \right)\\ &amp;=&amp; \left(
\begin{array}{cc} 37 &amp; 82 \\48 &amp; 108 \end{array} \right)
\end{eqnarray*}

$BA^T = \left(\begin{array}{ccc} 1 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp;
1\end{array} \right)\left(\begin{array}{cc} 1 &amp; 4 \\ 2 &amp; 5 \\ 3 &amp;
6 \end{array} \right)=\left(\begin{array}{cc} 1+2 &amp; 4+5 \\ 3 &amp; 6
\\ 2+3 &amp; 5+6 \end{array} \right) = \left(\begin{array}{cc} 3 &amp; 9 \\
3 &amp; 6 \\ 5 &amp; 11 \end{array} \right)$


\begin{eqnarray*} C^T(BA^T) = \left(\begin{array}{ccc} 1 &amp; 2 &amp; 3 \\ 4 &amp; 5 &amp; 6
\end{array} \right) \left(\begin{array}{cc} 3 &amp; 9 \\
3 &amp; 6 \\ 5 &amp; 11 \end{array} \right) &amp;=&amp; \left( \begin{array}{cc}
3+9+25 &amp; 9+18+55 \\ 6+12+30 &amp; 18+24+66 \end{array} \right)\\ &amp;=&amp;
\left( \begin{array}{cc} 37 &amp; 82 \\48 &amp; 108 \end{array} \right)
\end{eqnarray*}

Therefore $(C^TB)A^T = C^T(BA^T)$ as required



\end{document}
</pre></body></html>