Let $A=\left(\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right)$. If…
Let $A=\left(\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right)$. If $\mathrm{AA}^{\mathrm{T}}=\mathrm{I}_{3},$ then $|\mathrm{p}|$ is:
$\frac{1}{\sqrt{5}}$
$\frac{1}{\sqrt{3}}$
$\frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{6}}$
Solution
$A=\left[\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right]$
$\begin{aligned} \therefore & A \cdot A^{T}=\left[\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right] \times\left[\begin{array}{ccc}0 & p & p \\ 2 q & q & -q \\ r & -r & r\end{array}\right] \\ &=\left[\begin{array}{ccc}4 q^{2}+r^{2} & 2 q^{2}-r^{2} & -2 q^{2}+r^{2} \\ 2 q^{2}-r^{2} & p^{2}+q^{2}+r^{2} & p^{2}-q^{2}-r^{2} \\ -2 q^{2}+r^{2} & p^{2}-q^{2}-r^{2} & p^{2}+q^{2}+r^{2}\end{array}\right] \end{aligned}$
Given, $A A^{T}=I$
$\therefore \quad 4 q^{2}+r^{2}=p^{2}+q^{2}+r^{2}=1$
$\Rightarrow \quad p^{2}-3 q^{2}=0$ and $r^{2}=1-4 q^{2}$
and $2 q^{2}-r^{2}=0 \Rightarrow r^{2}=2 q^{2}$
$\begin{array}{ll}\therefore \quad & p^{2}=\frac{1}{2}, q^{2}=\frac{1}{6} \text { and } r^{2}=\frac{1}{3} \\ \therefore \quad & |p|=\frac{1}{\sqrt{2}}\end{array}$