= Solution
To invert the <matrix>, row-reduce the augmented <matrix> $(M\mid I)$. Subtracting row one from rows two and three gives left rows $(0,-4,1)$ and $(0,0,2)$, so the pivots are nonzero and back-substitution is possible. The resulting <matrix inverse> is
$$
\boxed{M^{-1}=\begin{pmatrix}
3/8&1/4&3/8\\1/8&-1/4&1/8\\-1/2&0&1/2
\end{pmatrix}}.
$$
Multiplying by $M$ verifies the identity <matrix>. This inverse is not $M^T$; indeed
$$
M^TM=\begin{pmatrix}3&-1&0\\-1&11&0\\0&0&2\end{pmatrix}\ne I.
$$
Thus \b[$M$ is not <orthogonal>].
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