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  • Let <img alt="A=\begin{pmatrix} 0 & 2q & r\\ p & q & -r\\ p& -q & r \end{pmatrix}.\: If\: \: AA^{T}=I_{3},\: Then \: \left | p \right |\: is\: :" src="https://learn.car

Let A=\begin{pmatrix} 0 & 2q & r\\ p & q & -r\\ p& -q & r \end{pmatrix}.\: If\: \: AA^{T}=I_{3},\: Then \: \left | p \right |\: is\: :

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According to the question, $A$ is an orthogonal matrix, and thus its rows (or columns) form an orthonormal set. So, the dot product of each row with itself is 1, and the dot product of different rows is 0.

From the dot product of the first row with itself - $0^2 + (2q)^2 + r^2 = 4q^2 + r^2 = 1$. 
From the dot product of the second row with itself - $p^2 + q^2 + r^2 = 1$. 
From the dot product of the first and second rows-  $0 \cdot p + 2q \cdot q + r \cdot (-r) = 2q^2 - r^2 = 0$, 
So $r^2 = 2q^2$. 
Substituting into the first equation, $4q^2 + 2q^2 = 6q^2 = 1 \Rightarrow q^2 = \frac{1}{6} \Rightarrow r^2 = \frac{1}{3}$. 
Now, using the second equation: $p^2 + \frac{1}{6} + \frac{1}{3} = 1 \Rightarrow p^2 + \frac{1}{2} = 1 \Rightarrow p^2 = \frac{1}{2} \Rightarrow |p| = \frac{1}{\sqrt{2}}$.

Posted by

Saniya Khatri

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