<p>For a real number \(\alpha\), if the system \[\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}\] of linear equations, has infinitely many solutions, then \(1 + \alpha + \alpha^2 =\) ___. <em>(JEE Advanced 2017)</em></p>
Step-by-Step Solution
Key Concept: For infinitely many solutions, the system must be dependent, meaning det(A) = 0 and the augmented matrix rank equals coefficient matrix rank. The circulant structure of the matrix allows factorization that reveals when det(A) = 0.
<p><strong>Step 1: Find when det(A) = 0</strong></p><p>For the circulant matrix, expand the determinant:</p><p>det(A) = 1(1 - α²) - α(α - α³) + α²(α² - α²)</p><p>= 1 - α² - α² + α⁴ = (1 - α)²(1 + α + α²)</p><p>So det(A) = 0 when α = 1 or 1 + α + α² = 0</p><p><strong>Step 2: Check consistency for each case</strong></p><p><strong>Case α = 1:</strong> Matrix becomes all 1's, RHS = [1, -1, 1]ᵀ. Row 1 gives x + y + z = 1, but row 2 gives x + y + z = -1. Contradiction! No solution.</p><p><strong>Case 1 + α + α² = 0:</strong> Then α³ = 1 (since 1 + α + α² = 0 ⟹ (α - 1)(1 + α + α²) = 0, and α ≠ 1). Verify: α = (-1 ± i√3)/2 (complex cube roots). For these values, the system reduces to dependent equations and RHS lies in column space.</p><p><strong>Step 3: Conclusion</strong></p><p>Infinitely many solutions occur when 1 + α + α² = 0.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 1