Let A=pmatrix0&2q&r&q&-r&-q&rpmatrix.

Question

Let $A=\begin{pmatrix}0&2q&r\p&q&-r\p&-q&r\end{pmatrix}$. If $AA^T=I_3$, then $|p|$ is:

Accepted Answer

$AA^T=\begin{pmatrix}4q^2+r^2 & 2q^2-r^2 & -2q^2+r^2\ 2q^2-r^2 & p^2+q^2+r^2 & p^2-q^2-r^2\ -2q^2+r^2 & p^2-q^2-r^2 & p^2+q^2+r^2\end{pmatrix}=I_3$.

From the diagonal entries: $4q^2+r^2=1$ and $p^2+q^2+r^2=1$; from an off-diagonal entry: $2q^2-r^2=0\Rightarrow r^2=2q^2$.

Substituting into $4q^2+r^2=1$: $4q^2+2q^2=1\Rightarrow q^2=\dfrac16$, so $r^2=\dfrac13$, and $p^2=1-q^2-r^2=1-\dfrac16-\dfrac13=\dfrac12$.

$|p|=\dfrac1{\sqrt2}$.

Solutions · 1