Question:

Given the matrix $A = \begin{pmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{pmatrix}$ and the condition $AA^T = I_3$. First, determine the transpose of $A$: $A^T = \begin{pmatrix} 0 & p & p \\ 2q & q & -q \\ r & -r & r \end{pmatrix}$ Next, compute the product $AA^T$: $$AA^T = \begin{pmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{pmatrix} \begin{pmatrix} 0 & p & p \\ 2q & q & -q \\ r & -r & r \end{pmatrix}$$ $$AA^T = \begin{pmatrix} (0)(0) + (2q)(2q) + (r)(r) & (0)(p) + (2q)(q) + (r)(-r) & (0)(p) + (2q)(-q) + (r)(r) \\ (p)(0) + (q)(2q) + (-r)(r) & (p)(p) + (q)(q) + (-r)(-r) & (p)(p) + (q)(-q) + (-r)(r) \\ (p)(0) + (-q)(2q) + (r)(r) & (p)(p) + (-q)(q) + (r)(-r) & (p)(p) + (-q)(-q) + (r)(r) \end{pmatrix}$$ $$AA^T = \begin{pmatrix} 4q^2 + r^2 & 2q^2 - r^2 & -2q^2 + r^2 \\ 2q^2 - r^2 & p^2 + q^2 + r^2 & p^2 - q^2 - r^2 \\ -2q^2 + r^2 & p^2 - q^2 - r^2 & p^2 + q^2 + r^2 \end{pmatrix}$$ Since $AA^T = I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$, we equate the corresponding elements to form a system of equations: 1. $4q^2 + r^2 = 1$ 2. $2q^2 - r^2 = 0$ 3. $p^2 + q^2 + r^2 = 1$ 4. $p^2 - q^2 - r^2 = 0$ From equation (2), we have: $r^2 = 2q^2$ Substitute this expression for $r^2$ into equation (1): $4q^2 + (2q^2) = 1$ $6q^2 = 1$ $q^2 = \frac{1}{6}$ Now, substitute the value of $q^2$ back into the expression for $r^2$: $r^2 = 2 \left(\frac{1}{6}\right) = \frac{1}{3}$ Finally, substitute the values of $q^2$ and $r^2$ into equation (4) to find $p^2$: $p^2 - q^2 - r^2 = 0$ $p^2 = q^2 + r^2$ $p^2 = \frac{1}{6} + \frac{1}{3}$ $p^2 = \frac{1}{6} + \frac{2}{6}$ $p^2 = \frac{3}{6}$ $p^2 = \frac{1}{2}$ To confirm consistency, substitute $p^2, q^2, r^2$ into equation (3): $p^2 + q^2 + r^2 = \frac{1}{2} + \frac{1}{6} + \frac{1}{3} = \frac{3}{6} + \frac{1}{6} + \frac{2}{6} = \frac{6}{6} = 1$. The system of equations is consistent. The value of $|p|$ is: $|p| = \sqrt{p^2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$