<p>Given \(A = \begin{pmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{pmatrix}\) and \(AA^T = I_3\), find \(|p|\).</p>
<p>(1) \(\dfrac{1}{\sqrt{3}}\)</p>
<p>(2) \(\dfrac{1}{2}\)</p>
<p>(3) \(\dfrac{1}{\sqrt{6}}\)</p>
<p>(4) \(\dfrac{1}{\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: Since AA^T = I₃, matrix A is orthogonal, meaning its rows (or columns) are orthonormal vectors. Use the orthonormality conditions: row dot product with itself equals 1, and different rows are orthogonal.
<p><strong>Step 1: Apply orthonormality condition to Row 1</strong></p><p>Row 1: (0, 2q, r). For orthonormality: 0² + (2q)² + r² = 1</p><p>∴ 4q² + r² = 1 ... (i)</p><p><strong>Step 2: Apply orthonormality condition to Row 2</strong></p><p>Row 2: (p, q, -r). For orthonormality: p² + q² + r² = 1 ... (ii)</p><p><strong>Step 3: Apply orthonormality condition to Row 3</strong></p><p>Row 3: (p, -q, r). For orthonormality: p² + q² + r² = 1 ... (iii)</p><p>Note: (ii) and (iii) are identical, confirming consistency.</p><p><strong>Step 4: Apply orthogonality between Row 1 and Row 2</strong></p><p>(0)(p) + (2q)(q) + (r)(-r) = 0</p><p>∴ 2q² - r² = 0 → r² = 2q² ... (iv)</p><p><strong>Step 5: Substitute (iv) into (i)</strong></p><p>4q² + 2q² = 1 → 6q² = 1 → q² = 1/6</p><p>∴ r² = 2(1/6) = 1/3</p><p><strong>Step 6: Substitute into (ii) to find p²</strong></p><p>p² + 1/6 + 1/3 = 1</p><p>p² + 1/6 + 2/6 = 1 → p² + 3/6 = 1 → p² = 1/2</p><p>∴ |p| = 1/√2 = √2/2</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D