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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.


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

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:

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

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

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

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

Transposed matrices:

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

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



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