Solution
= Solution
The three columns of $N=\sqrt8\,A$ are $(\sqrt3,1,2)$, $(-\sqrt2,\sqrt6,0)$, and $(-\sqrt3,-1,2)$. Each has squared <Euclidean norm> eight. Their pairwise <dot products> are
$$
-\sqrt6+\sqrt6=0,\qquad -3-1+4=0,\qquad \sqrt6-\sqrt6=0.
$$
Hence $N^{\mathsf T}N=8I$, so $\boxed{A^{\mathsf T}A=I}$. Thus $A$ is an <orthogonal matrix>, with $A^{-1}=A^{\mathsf T}$. The <determinant> from (i) places it in the determinant-one component.