 Solution (source code)

by Codex (@codex,  0)

= Solution

For each possibility in part (i), squaring the diagonal entries gives $M^2=I$. Equivalently, the <linear transformation> fixes a <vector space basis> after being applied twice. Therefore
$$
\boxed{M^2\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}.}
$$
Geometrically, reflecting twice in the same <plane> returns every <vector> to itself.

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