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

To solve this problem, we need to confirm the condition that matrix $A{A}^T$ equals the identity matrix $I_3$. This means that matrix $A$ is orthogonal, which implies that the columns of $A$ must be perpendicular unit vectors. Let us compute each component.

1. Matrix $A$ is given by: $\left(\begin{array}{rrr}0& 2q& r\\ p& q& -r\\ p& -q& r\end{array}\right)$
2. To find $A{A}^T$, compute: $\left(\begin{array}{ccc}0& 2q& r\\ p& q& -r\\ p& -q& r\end{array}\right)T$
3. Calculate $A$ times $A{T}^{}$:

1. The first column vector:

    
           Dot product with itself:
           
             $p{=}^1$
           
         

2. The second column vector also has to be a unit vector and orthogonal to the first, leading us to: $\mathrm{2q^2}+\mathrm{q^2}=1$
3. Simplifying gives: $\mathrm{3q^2}=1$ , hence $\left|q\right|=\frac{1}{\sqrt{3}}$

1. Calculating $\left|p\right|=$, we find: $\mathrm{2p^2}=1$
2. Solving gives: $\left|p\right|=\frac{1}{\sqrt{2}}$
3. Thus, the correct answer is: Option: $\frac{1}{\sqrt{2}}$