If AAT = I3, then |p| is:

Engineering

Mathematics

Multiplication of Matrices

Question

If $Let A = (\begin{matrix} 0 & 2 q & r \\ p & q & - r \\ p & - q & r \end{matrix})$ AA^T = I₃, then |p| is:

$\frac{1}{\sqrt{6}}$

$\frac{1}{\sqrt{5}}$

$\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{3}}$

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${AA}^{T} = [\begin{matrix} 0 & 2 q & r \\ p & q & - r \\ p & - q & r \end{matrix}] [\begin{matrix} 0 & p & p \\ 2 q & q & - q \\ r & - r & r \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

⇒ 4q² + r² = 1; r² = 2q² ; p² + q² + r² = 1

$\Rightarrow q^{2} = \frac{1}{6}, r^{2} = \frac{1}{3} \Rightarrow p^{2} = \frac{1}{2} \Rightarrow | p | = \frac{1}{\sqrt{2}}$