Question

Let $A=\left(\begin{array}{ccc}0& 2q& r\\ p& q& -r\\ p& -q& r\end{array}\right)$. If $A{A}^T=I_3$, then $|p|$ is:

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{5}}$
• D. $\frac{1}{\sqrt{6}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}$

$A{A}^T=I_3$
$\Rightarrow \left[\begin{array}{ccc}0& 2q& r\\ p& q& -r\\ p& -q& r\end{array}\right]\left[\begin{array}{ccc}0& p& p\\ 2q& q& -q\\ r& -r& r\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}4q^2+r^2& 2q^2-r^2& -2q^2+r^2\\ 2q^2-r^2& p^2+q^2+r^2& p^2-q^2-r^2\\ -2q^2+r^2& p^2-q^2-r^2& p^2+q^2+r^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow p^2-q^2-r^2=0\dots \left(1\right)$
$\Rightarrow p^2+q^2+r^2=1\dots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$
$\Rightarrow p^2=\frac{1}{2}$
$\Rightarrow p=\pm \frac{1}{\sqrt{2}}$