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 = $
Step-by-Step Solution
Key Concept: For a system Ax = b to have infinitely many solutions, the coefficient matrix A must be singular (det(A) = 0) AND the augmented matrix must have the same rank as A. Here, expanding det(A) = det([1, α, α²; α, 1, α; α², α, 1]) and setting it equal to zero gives the condition on α, then verify consistency with the RHS vector [1, -1, 1].
Expand the given determinant equation $D=0$ and factor to obtain $(1-a^2)(1+a+a^2-2a^2-a) = 0$, which simplifies to $(1-a^2)^2 = 0$, giving $a = \pm 1$. For $a=1$, the system of linear equations has no solutions. For $a=-1$, we have $1+a+a^2 = 1$.
Correct Answer: 1