The system of equations begin{cases} alpha x | vedclass

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(C) For a system of linear equations to have no solution or infinitely many solutions,the determinant of the coefficient matrix $D$ must be zero.$D =

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$-2$

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(C) For a system of linear equations to have no solution or infinitely many solutions,the determinant of the coefficient matrix $D$ must be zero.$D = \begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix} = 0$Expanding the determinant:$\alpha(\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha) = 0$$\alpha(\alpha - 1)(\alpha + 1) - 1(\alpha - 1) - 1(\alpha - 1) = 0$$(\alpha - 1)[\alpha(\alpha + 1) - 2] = 0$$(\alpha - 1)(\alpha^2 + \alpha - 2) = 0$$(\alpha - 1)(\alpha + 2)(\alpha - 1) = 0$$(\alpha - 1)^2(\alpha + 2) = 0$So,$\alpha = 1$ or $\alpha = -2$.If $\alpha = 1$,the equations become $x + y + z = 0$,which has infinitely many solutions.If $\alpha = -2$,the equations become:$-2x + y + z = -3$$x - 2y + z = -3$$x + y - 2z = -3$Adding these three equations gives $0 = -9$,which is a contradiction. Thus,there is no solution when $\alpha = -2$.

The system of equations $\begin{cases} \alpha x + y + z = \alpha - 1 \\ x + \alpha y + z = \alpha - 1 \\ x + y + \alpha z = \alpha - 1 \end{cases}$ has no solution,if $\alpha$ is

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