Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>The number of values of <em>k</em>, for which the system of equations:</p><p>\((k+1)x + 8y = 4k\)</p><p>\(kx + (k+3)y = 3k - 1\)</p><p>has no solution, is</p>
<p>(1) infinite</p>
<p>(2) 1</p>
<p>(3) 2</p>
<p>(4) 3</p>
Step-by-Step Solution
Key Concept: A system of linear equations has no solution when the coefficient matrix is singular (determinant = 0) but the augmented matrix has a different rank. This occurs when the lines are parallel but not coincident.
<p><strong>Step 1:</strong> For no solution, the coefficient matrix determinant must be zero.</p><p>det(A) = (k+1)(k+3) - 8k = k² + 4k + 3 - 8k = k² - 4k + 3 = (k-1)(k-3) = 0</p><p>This gives k = 1 or k = 3.</p><p><strong>Step 2:</strong> Check consistency for each value.</p><p><strong>For k = 1:</strong></p><p>Equations become: 2x + 8y = 4 and x + 4y = 2</p><p>Second equation × 2: 2x + 8y = 4 (identical to first)</p><p>These are dependent equations (infinite solutions), so k = 1 does NOT give no solution.</p><p><strong>For k = 3:</strong></p><p>Equations become: 4x + 8y = 12 and 3x + 6y = 8</p><p>First equation ÷ 4: x + 2y = 3</p><p>Second equation ÷ 3: x + 2y = 8/3</p><p>The ratios of coefficients are equal but constants differ: parallel and distinct lines.</p><p>This system has NO SOLUTION.</p><p><strong>Step 3:</strong> Only k = 3 satisfies the condition.</p><p>∴ Answer: <strong>1</strong> (or B if B represents 1 value of k)</p>
Correct Answer: B