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

Answer 1

The Correct Option is A

Explanation:
Given:$A=\left(\begin{array}{ccc}0& 2q& r\\ p& q& -r\\ p& -q& r\end{array}\right)$We have to find the value of $|p|$ if ${AA}^T=I_3$Consider:${AA}^T=I_3$$\left(\begin{array}{ccc}0& 2q& r\\ p& q& -r\\ p& -q& r\end{array}\right)\left(\begin{array}{ccc}0& p& p\\ 2q& q& -q\\ r& -r& r\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$$\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)$After comparing, we get$p^2+q^2+r^2=1$....(i)$p^2-q^2-r^2=0$....(ii)Adding equation (i) and (ii)$2p^2=1$$p^2=\frac{1}{2}$$p=\frac{1}{\sqrt{2}}$Hence, the correct option is (A).