Let A = begin{bmatrix} 0 & 2q & r p & q & -r | vedclass

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(C) Given that $AA^T = I_3$,$A$ is an orthogonal matrix. Thus,the product $AA^T$ is: $\begin{bmatrix} 0 & 2q & r \ p & q & -r \ p & -q & r \end{bmat

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

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(C) Given that $AA^T = I_3$,$A$ is an orthogonal matrix. Thus,the product $AA^T$ is: $\begin{bmatrix} 0 & 2q & r \ p & q & -r \ p & -q & r \end{bmatrix} \begin{bmatrix} 0 & p & p \ 2q & q & -q \ r & -r & r \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ Performing matrix multiplication: Row $1$ $\cdot$ Col $1$: $0^2 + (2q)^2 + r^2 = 4q^2 + r^2 = 1$ (Eq. $1$) Row $2$ $\cdot$ Col $2$: $p^2 + q^2 + (-r)^2 = p^2 + q^2 + r^2 = 1$ (Eq. $2$) Row $2$ $\cdot$ Col $3$: $p^2 - q^2 - r^2 = 0$ (Eq. $3$) Adding (Eq. $2$) and (Eq. $3$): $2p^2 = 1 \implies p^2 = \frac{1}{2} \implies |p| = \frac{1}{\sqrt{2}}$.

Let $A = \begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix}$. If $AA^T = I_3$,then $|p|$ is

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