<p>If \(S\) is the set of distinct values of \(b\) for which the following system of linear equations:<br>\(x + y + z = 1\)<br>\(x + ay + z = 1\)<br>\(ax + by + z = 0\)<br>has no solution, then \(S\) is</p>
Step-by-Step Solution
Key Concept: A system has no solution when the coefficient matrix and augmented matrix have different ranks. Use row reduction to find conditions where rank(A) ≠ rank(A|B), then identify which value of b makes this happen.
<p><strong>Step 1:</strong> Write the augmented matrix and perform row operations.</p><p>$$\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 1 & a & 1 & | & 1 \\ a & b & 1 & | & 0 \end{bmatrix}$$</p><p><strong>Step 2:</strong> Apply R₂ → R₂ - R₁ and R₃ → R₃ - aR₁:</p><p>$$\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & a-1 & 0 & | & 0 \\ 0 & b-a & 1-a & | & -a \end{bmatrix}$$</p><p><strong>Step 3:</strong> For no solution, we need rank(A) ≠ rank(A|B).</p><p>When a = 1: Second row becomes [0 0 0 | 0]. Then third row: [0 b-1 0 | -1]</p><p>For no solution: we need 0·x + 0·y + 0·z = -1 (impossible), which requires b - 1 = 0, so <strong>b = 1</strong>.</p><p><strong>Step 4:</strong> Verify: When a = 1 and b = 1, the system reduces to equations that are contradictory (0 = -1), confirming no solution.</p><p>∴ Answer: S = {1} or the single value b = 1</p>
Correct Answer: C