<p>The system of equations \[\alpha x - y - z = \alpha - 1\] \[x - \alpha y - z = \alpha - 1\] \[x - y - \alpha z = \alpha - 1\] has no solution if \(\alpha\) is</p>
Step-by-Step Solution
Key Concept: A system has no solution when the coefficient matrix is singular (determinant = 0) but the augmented matrix has different rank. Notice all three equations have identical RHS, so subtract equations to find when the system becomes inconsistent.
<p><strong>Step 1:</strong> Write coefficient matrix A and find det(A):</p><p>$$A = \begin{pmatrix} \alpha & -1 & -1 \\ 1 & -\alpha & -1 \\ 1 & -1 & -\alpha \end{pmatrix}$$</p><p><strong>Step 2:</strong> Add all three rows: Row sum = $(\alpha+1-1, -1-\alpha-1, -1-1-\alpha) = (\alpha, -\alpha-2, -\alpha-2)$. Factor: $(R_1 + R_2 + R_3)$ shows structure. Computing det(A) by subtracting R₁ from R₂ and R₃:</p><p>$$\det(A) = \begin{vmatrix} \alpha & -1 & -1 \\ 1-\alpha & -\alpha+1 & 0 \\ 1-\alpha & 0 & -\alpha+1 \end{vmatrix}$$</p><p>$$= (\alpha-1)\begin{vmatrix} \alpha & -1 & -1 \\ -(\alpha-1) & \alpha-1 & 0 \\ -(\alpha-1) & 0 & \alpha-1 \end{vmatrix} = (\alpha-1)^2(\alpha+2)$$</p><p><strong>Step 3:</strong> det(A) = 0 when $\alpha = 1$ or $\alpha = -2$.</p><p><strong>Step 4:</strong> Check consistency:</p><p><strong>If α = 1:</strong> All equations become $x - y - z = 0$. System has infinitely many solutions.</p><p><strong>If α = -2:</strong> Equations are: $-2x - y - z = -3$, $x + 2y - z = -3$, $x - y + 2z = -3$. Adding all three: $0 = -9$ (contradiction). System has no solution.</p><p><strong>∴ Answer: B (α = -2)</strong></p>
Correct Answer: B