Matrices & Determinants
Orthogonal Matrix
Grade None
Question:
<p>Let \(A = \begin{bmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{bmatrix}\). If \(A\) is orthogonal (i.e., \(AA^T = I\)), then which of the following values are correct?</p>
<p>(a) \(a = \pm \dfrac{1}{\sqrt{2}}\)</p>
<p>(b) \(b = \pm \dfrac{1}{\sqrt{6}}\)</p>
<p>(c) \(c = \pm \dfrac{1}{\sqrt{3}}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For an orthogonal matrix, both rows and columns must be unit vectors and mutually orthogonal. Use row orthonormality conditions: each row has magnitude 1 and distinct rows are perpendicular (dot product = 0).
<p><strong>Step 1: Apply row magnitude condition (each row is a unit vector)</strong></p><p>Row 1: 0² + (2b)² + c² = 1 → 4b² + c² = 1 ... (i)</p><p>Row 2: a² + b² + (-c)² = 1 → a² + b² + c² = 1 ... (ii)</p><p>Row 3: a² + (-b)² + c² = 1 → a² + b² + c² = 1 ... (iii)</p><p><strong>Step 2: Apply orthogonality condition between distinct rows</strong></p><p>Row 1 ⊥ Row 2: 0·a + 2b·b + c·(-c) = 0 → 2b² - c² = 0 → c² = 2b² ... (iv)</p><p>Row 1 ⊥ Row 3: 0·a + 2b·(-b) + c·c = 0 → -2b² + c² = 0 → c² = 2b² ... (confirms iv)</p><p>Row 2 ⊥ Row 3: a² + b·(-b) + (-c)·c = 0 → a² - b² - c² = 0 → a² = b² + c² ... (v)</p><p><strong>Step 3: Solve the system</strong></p><p>From (iv): c² = 2b²</p><p>Substitute in (i): 4b² + 2b² = 1 → 6b² = 1 → b² = 1/6</p><p>Therefore: c² = 2/6 = 1/3</p><p>From (ii): a² + 1/6 + 1/3 = 1 → a² = 1 - 1/2 = 1/2</p><p>Verify (v): a² = 1/2 = 1/6 + 1/3 = b² + c² ✓</p><p><strong>Step 4: Find abc</strong></p><p>Since a² = 1/2, b² = 1/6, c² = 1/3:</p><p>|a| = 1/√2, |b| = 1/√6, |c| = 1/√3</p><p>|abc| = (1/√2)·(1/√6)·(1/√3) = 1/√36 = 1/6</p><p>∴ Possible values: abc = ±1/6 (depending on signs of a, b, c)</p>
Correct Answer: abc