Matrices & Determinants
Orthogonal Matrix
Grade 12
Question:
<p>If matrix \( \begin{bmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{bmatrix} \) is orthogonal, then</p>
<p>\(a^2 + b^2 + c^2 = 1\)</p>
<p>\(a^2 + b^2 + c^2 = 2\)</p>
<p>\(a^2 - b^2 - c^2 = 0\)</p>
<p>\(a = \pm \dfrac{1}{\sqrt{2}}\);\( b = \pm \dfrac{1}{\sqrt{6}}\); \(c = \pm \dfrac{1}{\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: For an orthogonal matrix, both row vectors and column vectors must be orthonormal (unit length and mutually perpendicular). This means each row/column has magnitude 1, and dot product between different rows/columns equals 0.
<p><strong>Step 1:</strong> For orthogonal matrix, apply row normalization condition.</p><p>Row 1: (0)² + (2b)² + c² = 1 → <strong>4b² + c² = 1</strong></p><p>Row 2: a² + b² + c² = 1</p><p>Row 3: a² + b² + c² = 1</p><p><strong>Step 2:</strong> Apply orthogonality (perpendicularity) between rows.</p><p>Row 1 ⊥ Row 2: 0·a + 2b·b + c·(-c) = 0 → <strong>2b² - c² = 0</strong> → c² = 2b²</p><p>Row 1 ⊥ Row 3: 0·a + 2b·(-b) + c·c = 0 → <strong>-2b² + c² = 0</strong> → c² = 2b²</p><p><strong>Step 3:</strong> Solve for a, b, c.</p><p>From 4b² + c² = 1 and c² = 2b²: 4b² + 2b² = 1 → 6b² = 1 → b² = 1/6</p><p>Then c² = 2/6 = 1/3, and a² = 1 - b² - c² = 1 - 1/6 - 1/3 = 1/2</p><p>∴ <strong>a = ±1/√2, b = ±1/√6, c = ±1/√3</strong></p><p>Checking typical options: (A) a² = 1/2 ✓ (C) c² = 1/3 ✓ (D) b² = 1/6 ✓</p>
Correct Answer: A,C,D