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{\bf Question}

Decide which of the following matrices can be added, and which can
be multiplied.  Carry out the calculations whenever possible.


$$ A = \left( \begin{array}{rrr} 1 &amp; {-1} &amp; {-2} \\ 0 &amp; 2 &amp; 1 \\
\end{array} \right);\
B = \left(\begin{array}{rr} 1 &amp; {-1} \\ {0} &amp; 2
\\ 1 &amp; -3 \\ \end{array} \right);\
C = \left( \begin{array}{rrr} 4 &amp; {-1} &amp; 0 \\ 3 &amp; {-2} &amp; 1 \\ 5 &amp;
{-6} &amp; {-7} \end{array} \right);\
 D = \left( \begin{array}{rrr} 3 &amp; {-4} &amp; {7} \\ {-2} &amp; 1 &amp; 6 \\
\end{array} \right). $$

For each of the matrices, write down its transpose and say which
of the transposed matrices can be multiplied.


{\bf Answer}

Any matrix can be added to itself: this has the effect of doubling
each entry.  For example:

$$A+A=\left( \begin{array} {rrr} {2} &amp; {-2} &amp; {-4}\\ {0} &amp; {4} &amp;
{2}
\end{array} \right)$$
Otherwise, only $A$ and $D$ can be added, with $$A+D=\left(
\begin{array} {rrr} {4} &amp; {-5} &amp; {5}\\ {-2} &amp; {3} &amp; {7}
\end{array} \right).$$
Possibilities for multiplication are: $$AB=\left(
\begin{array} {rr} {-1} &amp; {3}\\
{1} &amp; {1}
\end{array} \right);\ \
AC=\left(\begin{array} {rrr} {-9} &amp; {13} &amp; {13}\\ {11} &amp; {-10} &amp;
{-5} \end{array} \right)$$

$$BA=\left(\begin{array} {rrr} {1} &amp; {-3} &amp; {-3}\\ {0} &amp; {4} &amp;
{2}\\ {1} &amp; {-7} &amp; {-5}
\end{array} \right);\ \
BD=\left(\begin{array} {rrr} {5} &amp; {-5} &amp; {1}\\ {-4} &amp; {2} &amp;
{12}\\ {9} &amp; {-7} &amp; {-11}
\end{array} \right)$$

$$CB=\left(\begin{array} {rr} {4} &amp; {-6}\\ {4} &amp; {-10}\\ {-2} &amp;
{4}
\end{array} \right);\ \
CC=C^2=\left(\begin{array} {rrr} {13} &amp; {-2} &amp; {-1}\\ {11} &amp; {-5}
&amp; {-9}\\ {-33} &amp; {49} &amp; {43} \end{array} \right)$$

$$DB=\left(\begin{array} {rr} {10} &amp; {-32}\\ {4} &amp; {-14}
\end{array} \right);\ \
DC=\left(\begin{array} {rrr} {35} &amp; {-37} &amp; {-53}\\ {25} &amp; {-36} &amp;
{-41} \end{array} \right)$$

Transposed matrices:

$$A^T=\left(\begin{array} {rr} {1} &amp; {0}\\ {-1} &amp; {2}\\ {-2} &amp; {1}
\end{array} \right);\ B^T=\left(\begin{array} {rrr} {1} &amp; {0} &amp; {1}\\ {-1} &amp; {2} &amp; {-3}
\end{array} \right);$$ $$C^T=\left(\begin{array} {rrr} {4} &amp; {3} &amp; {5}\\ {-1} &amp; {-2}
&amp; {-6}\\ {0} &amp; {1} &amp; {-7} \end{array} \right);\
D^T=\left(\begin{array} {rr} {3} &amp; {-2}\\ {-4} &amp; {1}\\ {7} &amp; {6}
\end{array} \right)$$

Possibilities for multiplying: $$A^TB^T,\ B^TA^T,\ B^TC^T,\
B^TD^T,\ C^TA^T,\ C^TC^T,\ D^TB^T,\ C^TD^T.$$

\underline {Note}: $A^TB^T=(BA)^T=\left(\begin{array} {rrr} {1} &amp;
{-3} &amp; {-3}\\ {0} &amp; {4} &amp; {2}\\ {1} &amp; {-7} &amp; {-5} \end{array}
\right)^T=\left(\begin{array} {rrr} {1} &amp; {0} &amp; {1}\\ {-3} &amp; {4} &amp;
{-7}\\ {-3} &amp; {2} &amp; {-5} \end{array} \right)$



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