Matrices & Determinants
Homogeneous system of equations
Grade 12
Question:
<p><strong>For Problems 19–21</strong><br>Given that the system of equations \(x = cy + bz\), \(y = az + cx\), \(z = bx + ay\) has nonzero solutions and at least one of the \(a, b, c\) is a proper fraction.<br>System has solution such that</p>
<p>\(x:y:z \equiv (1-2a^2):(1-2b^2):(1-2c^2)\)</p>
<p>\(x:y:z \equiv \dfrac{1}{1-2a^2}:\dfrac{1}{1-2b^2}:\dfrac{1}{1-2c^2}\)</p>
<p>\(x:y:z \equiv \dfrac{a}{1-a^2}:\dfrac{b}{1-b^2}:\dfrac{c}{1-c^2}\)</p>
<p>\(x:y:z \equiv \sqrt{1-a^2}:\sqrt{1-b^2}:\sqrt{1-c^2}\)</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have nonzero solutions, the coefficient matrix determinant must equal zero. This creates a characteristic equation in terms of a, b, c that determines when nontrivial solutions exist.
<p><strong>Step 1:</strong> Rewrite the system in standard form:</p><p>x - cy - bz = 0</p><p>-cx + y - az = 0</p><p>-bx - ay + z = 0</p><p><strong>Step 2:</strong> Write the coefficient matrix and set determinant to zero:</p><p>|1 -c -b|</p><p>|-c 1 -a| = 0</p><p>|-b -a 1|</p><p><strong>Step 3:</strong> Expand the determinant (using first row):</p><p>1(1 - a²) + c(-c + ab) - b(ac + b) = 0</p><p>1 - a² - c² + abc - abc - b² = 0</p><p>1 - a² - b² - c² = 0</p><p><strong>Step 4:</strong> Therefore: <strong>a² + b² + c² = 1</strong></p><p>This is the constraint equation. Since at least one of a, b, c is a proper fraction (< 1), and their squares sum to 1, the system has nonzero solutions satisfying this relationship.</p><p>∴ Answer: D</p>
Correct Answer: D