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{\bf Question}

Using the matrices $$ A = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 4
& 5 & 6 \end{array} \right) \hspace{.2in}  B =
\left(\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 1
\end{array} \right)  \hspace{.2in} C = \left(\begin{array}{cc} 1
& 2 \\ 3 & 4 \\ 5 & 6 \end{array} \right) $$ show that the
matrices $(C^TB)A^T$ and $C^T(BA^T)$ exist and are equal

\vspace{.25in}

{\bf Answer}

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 $\begin{array}{lll} C^T & \ \mathrm{is}\ & 2\times 3 \\
B &\ \mathrm{is}\ & 3\times 3\end{array} \Rightarrow\ \ C^TB\
\mathrm{is}\ 3\times 3$

$A\ \mathrm{is}\ 2\times 3 \Rightarrow A^T\ \mathrm{is}\ 3\times
2$

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

$C^TB = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6
\end{array} \right)\left(\begin{array}{ccc} 1 & 1 & 0  \\ 0 & 0 &
1 \\ 0 & 1 & 1 \end{array} \right) = \left(\begin{array}{ccc} 1 &
1+5 & 3+5
\\ 2 & 2+6 & 4+6 \end{array} \right) = \left(\begin{array}{ccc}
1 & 6 & 8 \\ 2 & 8 & 10 \end{array} \right)$

\begin{eqnarray*} (C^TB)A^T = \left(\begin{array}{ccc} 1 & 6 & 8 \\ 2 & 8 & 10
\end{array} \right)\left(\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 & 6
\end{array} \right)&=& \left( \begin{array}{cc} 1+12+24 & 4+ 30+48
\\ 2+16+30 & 8+40+60 \end{array} \right)\\ &=& \left(
\begin{array}{cc} 37 & 82 \\48 & 108 \end{array} \right)
\end{eqnarray*}

$BA^T = \left(\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 1 &
1\end{array} \right)\left(\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 &
6 \end{array} \right)=\left(\begin{array}{cc} 1+2 & 4+5 \\ 3 & 6
\\ 2+3 & 5+6 \end{array} \right) = \left(\begin{array}{cc} 3 & 9 \\
3 & 6 \\ 5 & 11 \end{array} \right)$


\begin{eqnarray*} C^T(BA^T) = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6
\end{array} \right) \left(\begin{array}{cc} 3 & 9 \\
3 & 6 \\ 5 & 11 \end{array} \right) &=& \left( \begin{array}{cc}
3+9+25 & 9+18+55 \\ 6+12+30 & 18+24+66 \end{array} \right)\\ &=&
\left( \begin{array}{cc} 37 & 82 \\48 & 108 \end{array} \right)
\end{eqnarray*}

Therefore $(C^TB)A^T = C^T(BA^T)$ as required



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