Question:

For the system to have infinitely many solutions, the coefficient matrix and the augmented matrix must have the same rank, which must be less than 3. This means the determinant of the coefficient matrix must be zero. $$\textbf{Step 1: Calculate the determinant of the coefficient matrix}$$ Let us denote the coefficient matrix as $A$: $$A = \begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix}$$ We calculate $\det(A)$ using the first row expansion: $$\det(A) = 1 \cdot \begin{vmatrix} 1 & \alpha \\ \alpha & 1 \end{vmatrix} - \alpha \cdot \begin{vmatrix} \alpha & \alpha \\ \alpha^2 & 1 \end{vmatrix} + \alpha^2 \cdot \begin{vmatrix} \alpha & 1 \\ \alpha^2 & \alpha \end{vmatrix}$$ $$\textbf{Step 2: Evaluate the 2×2 determinants}$$ $$\begin{vmatrix} 1 & \alpha \\ \alpha & 1 \end{vmatrix} = 1 - \alpha^2$$ $$\begin{vmatrix} \alpha & \alpha \\ \alpha^2 & 1 \end{vmatrix} = \alpha - \alpha^3$$ $$\begin{vmatrix} \alpha & 1 \\ \alpha^2 & \alpha \end{vmatrix} = \alpha^2 - \alpha^2 = 0$$ $$\textbf{Step 3: Substitute back into the determinant}$$ $$\det(A) = 1(1 - \alpha^2) - \alpha(\alpha - \alpha^3) + \alpha^2(0)$$ $$= 1 - \alpha^2 - \alpha^2 + \alpha^4$$ $$= 1 - 2\alpha^2 + \alpha^4$$ $$= (\alpha^2 - 1)^2$$ $$\textbf{Step 4: Apply the condition for infinitely many solutions}$$ For infinitely many solutions to exist, we require: $$\det(A) = 0$$ $$(\alpha^2 - 1)^2 = 0$$ $$\alpha^2 - 1 = 0$$ $$\alpha^2 = 1$$ $$\alpha = 1 \text{ or } \alpha = -1$$ $$\textbf{Step 5: Verify consistency with the augmented matrix}$$ We must verify that the system is consistent (not inconsistent) for both values. <b>Case 1:</b> $\alpha = 1$ $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$$ All three rows of the coefficient matrix are identical, but the right-hand side has different values. This system is <b>inconsistent</b>, so $\alpha = 1$ does not satisfy our requirement. <b>Case 2:</b> $\alpha = -1$ $$\begin{bmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$$ The first and third rows of the coefficient matrix are identical, and the first and third elements of the right-hand side are also identical