Matrices & Determinants
Orthogonal Matrix
Grade 12
Question:
<p>Let \(A = \begin{pmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{pmatrix}\). If \(AA^T = I_3\), then \(|p|\) is:</p>
<p>\(\dfrac{1}{\sqrt{6}}\)</p>
<p>\(\dfrac{1}{\sqrt{5}}\)</p>
<p>\(\dfrac{1}{\sqrt{3}}\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: Since AA^T = I₃, matrix A is orthogonal, meaning its rows (or columns) are orthonormal. Use row orthonormality conditions: each row has unit length and distinct rows are perpendicular.
<p><strong>Step 1:</strong> For AA^T = I₃, each row of A must have unit length.</p><p>Row 1: (0)² + (2q)² + r² = 1 → 4q² + r² = 1 ... (i)</p><p>Row 2: p² + q² + r² = 1 ... (ii)</p><p>Row 3: p² + q² + r² = 1 ... (iii)</p><p><strong>Step 2:</strong> Rows must be orthogonal (dot product = 0).</p><p>Row 1 · Row 2: 0·p + 2q·q + r·(-r) = 0 → 2q² - r² = 0 → r² = 2q² ... (iv)</p><p>Row 1 · Row 3: 0·p + 2q·(-q) + r·r = 0 → -2q² + r² = 0 → r² = 2q² ✓ (consistent)</p><p>Row 2 · Row 3: p² + q·(-q) + (-r)·r = 0 → p² - q² - r² = 0 → p² = q² + r² ... (v)</p><p><strong>Step 3:</strong> Substitute (iv) into (i): 4q² + 2q² = 1 → 6q² = 1 → q² = 1/6</p><p>From (iv): r² = 2q² = 2/6 = 1/3</p><p><strong>Step 4:</strong> From (ii): p² + 1/6 + 1/3 = 1 → p² + 1/2 = 1 → p² = 1/2</p><p><strong>Step 5:</strong> Verify with (v): p² = q² + r² = 1/6 + 1/3 = 1/2 ✓</p><p>∴ |p| = √(1/2) = 1/√2 = <strong>√2/2</strong></p>
Correct Answer: D