Question:
Write out the multiplication table for the following set of matrices over $\mathbb Q$:
$$A=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)$$
$$B=\left(\begin{array}{ll}
-1 & 0 \\
0 & 1
\end{array}\right)$$
$$C=\left(\begin{array}{ll}
1 & 0 \\
0 & -1
\end{array}\right)$$
$$D=\left(\begin{array}{ll}
-1 & 0 \\
0 & -1
\end{array}\right)$$
My Attempt: I am not sure what the question is asking. Is it just asking to do a 4×4 multiplication table and multiply each of the matrix by each other? For example:
\begin{array}{|c|c|c|c|}
\hline
& A& B & C & D \\ \hline
A& & &\\ \hline
B& & &\\ \hline
C& & &\\ \hline
D& & &\\ \hline
\end{array}
If that is what it's asking, should I multiply row x column or column x row?
Best Answer
Could you provide some more context? Without further context, your guess seems reasonable. In this special case, the matrices are commutative, i.e. AB = BA (this is easy to see:
$$\left(\begin{array}{ll} a & 0 \\ 0 & b \end{array}\right) \times \left(\begin{array}{ll} c & 0 \\ 0 & d \end{array}\right) = \left(\begin{array}{ll} ac & 0 \\ 0 & bd \end{array}\right) = \left(\begin{array}{ll} ca & 0 \\ 0 & db \end{array}\right) = \left(\begin{array}{ll} c & 0 \\ 0 & d \end{array}\right) \times \left(\begin{array}{ll} a & 0 \\ 0 & b \end{array}\right) $$)
Therefore, you only have to compute half of that table.
simply using that $Q\backslash\{0\}$ is a commutative group with 1.