Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>If the system of linear equations,</p><p>\(x + ky + 3z = 0\)</p><p>\(3x + ky - 2z = 0\)</p><p>\(2x + 4y - 3z = 0\)</p><p>has a non-zero solution \((x, y, z)\), then \(\dfrac{xz}{y^2}\) is equal to</p>
<p>(1) 30</p>
<p>(2) \(-10\)</p>
<p>(3) 10</p>
<p>(4) \(-30\)</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have non-zero solutions, the coefficient matrix determinant must equal zero. Find k from det = 0, then use the resulting system to find the relationship between x, y, z.
<p><strong>Step 1:</strong> For non-zero solution to exist, the coefficient determinant must be zero:</p><p>det|1 k 3 |</p><p> |3 k -2 | = 0</p><p> |2 4 -3 |</p><p><strong>Step 2:</strong> Expanding along row 1:</p><p>1(k(-3) - (-2)(4)) - k(3(-3) - (-2)(2)) + 3(3(4) - k(2)) = 0</p><p>1(-3k + 8) - k(-9 + 4) + 3(12 - 2k) = 0</p><p>-3k + 8 + 5k + 36 - 6k = 0</p><p>-4k + 44 = 0</p><p><strong>k = 11</strong></p><p><strong>Step 3:</strong> Substitute k = 11 in the original system:</p><p>x + 11y + 3z = 0 ... (1)</p><p>3x + 11y - 2z = 0 ... (2)</p><p>2x + 4y - 3z = 0 ... (3)</p><p><strong>Step 4:</strong> From (2) - (1): 2x - 5z = 0 ⟹ x = (5z)/2</p><p><strong>Step 5:</strong> Substitute in (3): 2(5z/2) + 4y - 3z = 0</p><p>5z + 4y - 3z = 0 ⟹ 2z + 4y = 0 ⟹ y = -z/2</p><p><strong>Step 6:</strong> Calculate xz/y²:</p><p>xz/y² = [(5z/2)·z]/[(-z/2)²] = (5z²/2)/(z²/4) = (5z²/2)·(4/z²) = 10</p><p>∴ Answer: <strong>C (10)</strong></p>
Correct Answer: C