Matrices & Determinants
System of linear equations with no solution
Grade 12
Question:
<p>The number of values of <em>k</em> for which the system of linear equations,<br>\((k+2)x+10y=k\)<br>\(kx+(k+3)y=k-1\)<br>has no solution, is</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>infinitely many</p>
Step-by-Step Solution
Key Concept: A system has no solution when the coefficient matrix is singular (determinant = 0) but the augmented matrix has a different rank. Use the condition: det(A) = 0 AND the system becomes inconsistent.
<p><strong>Step 1:</strong> Write the coefficient matrix and find when det(A) = 0.</p><p>det(A) = (k+2)(k+3) - 10k = k² + 5k + 6 - 10k = k² - 5k + 6 = (k-2)(k-3)</p><p>det(A) = 0 when k = 2 or k = 3</p><p><strong>Step 2:</strong> Check k = 2: Equations become 4x + 10y = 2 and 2x + 5y = 1.</p><p>The second equation is exactly half the first (2x + 5y = 1 is the first divided by 2). These are consistent—infinitely many solutions, NOT no solution.</p><p><strong>Step 3:</strong> Check k = 3: Equations become 5x + 10y = 3 and 3x + 6y = 2.</p><p>From equation 1: 5x + 10y = 3. From equation 2: 3x + 6y = 2, or multiplying by 5/3: 5x + 10y = 10/3.</p><p>This gives 3 = 10/3, which is a contradiction. The system has NO SOLUTION.</p><p><strong>Step 4:</strong> Only k = 3 gives no solution.</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: A