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(D) The system has infinitely many solutions if the determinant of the coefficient matrix $D = 0$ and the augmented matrix satisfies the consistency c

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$1$

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(D) The system has infinitely many solutions if the determinant of the coefficient matrix $D = 0$ and the augmented matrix satisfies the consistency condition.First,calculate the determinant $D$:$D = \begin{vmatrix} 1 & \alpha & \alpha^2 \ \alpha & 1 & \alpha \ \alpha^2 & \alpha & 1 \end{vmatrix} = 1(1 - \alpha^2) - \alpha(\alpha - \alpha^3) + \alpha^2(\alpha^2 - \alpha^2) = 1 - \alpha^2 - \alpha^2 + \alpha^4 = \alpha^4 - 2\alpha^2 + 1 = (\alpha^2 - 1)^2$.Setting $D = 0$ gives $(\alpha^2 - 1)^2 = 0$,so $\alpha^2 = 1$,which means $\alpha = 1$ or $\alpha = -1$.Case $1$: If $\alpha = 1$,the system becomes:$x + y + z = 1$$x + y + z = -1$$x + y + z = 1$This is inconsistent because $1 
eq -1$,so there are no solutions.Case $2$: If $\alpha = -1$,the system becomes:$x - y + z = 1$$-x + y - z = -1$$x - y + z = 1$All three equations are equivalent to $x - y + z = 1$,which represents a plane,thus there are infinitely many solutions.Therefore,$\alpha = -1$.Then,$1 + \alpha + \alpha^2 = 1 + (-1) + (-1)^2 = 1 - 1 + 1 = 1$.

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