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(B) For $\alpha=1$, the system reduces to a homogeneous system which is always consistent. So, $\alpha 
eq 1$. For $\alpha 
eq 1$, we calculate the

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$-2$

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(B) For $\alpha=1$, the system reduces to a homogeneous system which is always consistent. So, $\alpha 
eq 1$. For $\alpha 
eq 1$, we calculate the determinant $D$: $D = \begin{vmatrix} \alpha & -1 & -1 \ 1 & -\alpha & -1 \ 1 & -1 & -\alpha \end{vmatrix} = \begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix}$. Applying $R_1 ightarrow R_1 + R_2 + R_3$, we get: $D = \begin{vmatrix} \alpha+2 & \alpha+2 & \alpha+2 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix} = (\alpha+2) \begin{vmatrix} 1 & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix}$. Applying $C_2 ightarrow C_2 - C_1$ and $C_3 ightarrow C_3 - C_1$: $D = (\alpha+2) \begin{vmatrix} 1 & 0 & 0 \ 1 & \alpha-1 & 0 \ 1 & 0 & \alpha-1 \end{vmatrix} = (\alpha+2)(\alpha-1)^2$. Now, calculate $D_1$: $D_1 = \begin{vmatrix} \alpha-1 & -1 & -1 \ \alpha-1 & -\alpha & -1 \ \alpha-1 & -1 & -\alpha \end{vmatrix} = (\alpha-1) \begin{vmatrix} 1 & -1 & -1 \ 1 & -\alpha & -1 \ 1 & -1 & -\alpha \end{vmatrix}$. Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$: $D_1 = (\alpha-1) \begin{vmatrix} 1 & -1 & -1 \ 0 & 1-\alpha & 0 \ 0 & 0 & 1-\alpha \end{vmatrix} = (\alpha-1)(1-\alpha)^2 = (\alpha-1)^3$. For the system to be inconsistent, we need $D=0$ and $D_1 
eq 0$. $D=0$ implies $\alpha = -2$ or $\alpha = 1$. Since $\alpha 
eq 1$, we check $\alpha = -2$. For $\alpha = -2$, $D=0$ and $D_1 = (-2-1)^3 = -27 
eq 0$. Thus, the system is inconsistent for $\alpha = -2$.

If the system of equations $\begin{bmatrix} \alpha & -1 & -1 \\ 1 & -\alpha & -1 \\ 1 & -1 & -\alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \alpha-1 \\ \alpha-1 \\ \alpha-1 \end{bmatrix}$ is inconsistent, then $\alpha=$

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