Match the following for the system of linear equations <br>λx + y + z = 1, <br>x + λy + z = λ, <br>x + y + λz = λ<sup>2</sup>. <br><table><thead><tr><th>Column-I</th><th>Column-II</th></tr></thead><tbody><tr><td>(A) λ = 1</td><td>(P) unique solution</td></tr><tr><td>(B) λ ≠ 1</td><td>(Q) infinite solutions</td></tr><tr><td>(C) λ ≠ 1, λ ≠ -2</td><td>(R) no solution</td></tr><tr><td>(D) λ = -2</td><td>(S) finite many solutions</td></tr></tbody></table>
A → Q; B → P, R; C → P; D → R
A → Q; B → P; C → P, Q, S; D → R, T
A → P, Q; II → R; III → Q, S; IV → T
I → P; II → Q; III → R, S; IV → T
Step-by-Step Solution
Key Concept: The system of linear equations can be analyzed using the determinant of the coefficient matrix. If \Delta \neq 0, there is a unique solution. If \Delta = 0, the system may have infinite solutions or no solution depending on the consistency of the equations.
The coefficient matrix is A = [[\lambda, 1, 1], [1, \lambda, 1], [1, 1, \lambda]]. The determinant \Delta = \lambda(\lambda^2 - 1) - 1(\lambda - 1) + 1(1 - \lambda) = \lambda(\lambda-1)(\lambda+1) - (\lambda-1) - (\lambda-1) = (\lambda-1)(\lambda^2 + \lambda - 2) = (\lambda-1)(\lambda+2)(\lambda-1) = (\lambda-1)^2(\lambda+2). If \lambda \neq 1 and \lambda \neq -2, \Delta \neq 0, so there is a unique solution. If \lambda = 1, the equations become x+y+z=1, x+y+z=1, x+y+z=1, which represents the same plane, so there are infinite solutions. If \lambda = -2, the equations are -2x+y+z=1, x-2y+z=-2, x+y-2z=4. Adding these gives 0 = 3, which is impossible, so there is no solution.
Correct Answer: A → Q; B → P, R; C → P; D → R