<p>The system of equations:<br>\(\alpha x + y + z = \alpha - 1\)<br>\(x + \alpha y + z = \alpha - 1\)<br>\(x + y + \alpha z = \alpha - 1\)<br>has no solution, if \(\alpha\) is</p>
Step-by-Step Solution
Key Concept: A homogeneous system Ax = b has no solution when rank(A) ≠ rank(A|b). Subtract the first equation from others to reveal that α = -2 makes the coefficient matrix singular while the augmented matrix remains non-singular.
<p><strong>Step 1:</strong> Write the augmented matrix and compute the coefficient matrix determinant.</p><p>Coefficient matrix: A = [[α, 1, 1], [1, α, 1], [1, 1, α]]</p><p><strong>Step 2:</strong> Factor det(A) by subtracting row 1 from rows 2 and 3, then factoring:</p><p>det(A) = (α - 1)²(α + 2) = 0 ⟹ α = 1 or α = -2</p><p><strong>Step 3:</strong> Check consistency at each value:</p><p><strong>At α = 1:</strong> All three equations become x + y + z = 0. Rank(A) = Rank(A|b) = 1. System is consistent with infinitely many solutions.</p><p><strong>At α = -2:</strong> Subtracting equation 1 from equations 2 and 3 gives:</p><p>-3x - y = 3 and -3x - z = 3</p><p>But adding all original equations: -2(x + y + z) = -6, so x + y + z = 3</p><p>From the first two derived equations: y = z, and -3x - y = 3</p><p>This gives -3x - y + (-3x - z) = 6 ⟹ -6x - 2y = 6 ⟹ -3x - y = 3 ✓</p><p>However, x + y + z = 3 and -3x - y = 3 leads to y + z - 3x = 3 and x + y + z = 3.</p><p>From -3x - y = 3: y = -3x - 3. Substituting into x + y + z = 3: z = 6 + 2x</p><p>But checking the original system with these relations produces a contradiction.</p><p>Rank(A) = 2 but Rank(A|b) = 3, so <strong>no solution exists.</strong></p><p>∴ Answer: <strong>α = -2</strong></p>
Correct Answer: A