<p>If the system of equations \(\alpha x + y + z = \alpha - 1\), \(x + \alpha y + z = \alpha - 1\), \(x + y + \alpha z = \alpha - 1\) has non-zero solutions, then the condition on \(\alpha\) is:</p>
<p>(1) \(\alpha = 1\) or \(\alpha = -2\)</p>
<p>(2) \(\alpha = 0\)</p>
<p>(3) \(\alpha = 2\)</p>
<p>(4) \(\alpha = -1\)</p>
Step-by-Step Solution
Key Concept: For a homogeneous-like system to have non-trivial solutions, the coefficient matrix determinant must equal zero. Rewrite the system by subtracting the constant term to recognize when the solution space is non-trivial.
<p><strong>Step 1:</strong> Write the coefficient matrix and note the symmetric structure:</p><p>The system is: αx + y + z = α - 1, x + αy + z = α - 1, x + y + αz = α - 1</p><p><strong>Step 2:</strong> Subtract (α - 1) from RHS and rewrite as:</p><p>α·x + y + z - (α - 1) = 0, x + α·y + z - (α - 1) = 0, x + y + α·z - (α - 1) = 0</p><p><strong>Step 3:</strong> For non-trivial solutions, det(A) = 0 where A = [α, 1, 1; 1, α, 1; 1, 1, α]</p><p><strong>Step 4:</strong> Calculate det(A) by adding all rows to the first row:</p><p>det(A) = (α + 2)·det[1, 1, 1; 1, α, 1; 1, 1, α] = (α + 2)[(α - 1)² - (1 - 1)]</p><p>= (α + 2)(α - 1)²</p><p><strong>Step 5:</strong> Set det(A) = 0: (α + 2)(α - 1)² = 0</p><p>Therefore: <strong>α = -2 or α = 1</strong></p><p><strong>Verification:</strong> When α = 1, all three equations become x + y + z = 0, giving infinitely many solutions. When α = -2, the system has rank 2, giving a 1-dimensional solution space.</p><p>∴ Answer: A (α = -2 or α = 1)</p>
Correct Answer: A