Matrices & Determinants
System of linear equations
Grade Class 12
Question:
Match the following for the system of linear equations<br>λx + y + z = 1, x + λy + z = λ, 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; (D) → R
(A) → P; (B) → Q; (C) → R; (D) → S
(A) → R; (B) → P; (C) → Q; (D) → S
Step-by-Step Solution
Key Concept: Analyze the determinant of the coefficient matrix and the consistency of the system for different values of \lambda.
The system is: <br>\lambda x + y + z = 1<br>x + \lambda y + z = \lambda<br>x + y + \lambda z = \lambda<sup>2</sup><br>The determinant of the coefficient matrix is \Delta = \lambda(\lambda<sup>2</sup> - 1) - 1(\lambda - 1) + 1(1 - \lambda) = \lambda(\lambda-1)(\lambda+1) - (\lambda-1) - (\lambda-1) = (\lambda-1)(\lambda<sup>2</sup> + \lambda - 2) = (\lambda-1)(\lambda+2)(\lambda-1) = (\lambda-1)<sup>2</sup>(\lambda+2).<br>If \lambda = 1, the equations become x+y+z=1, x+y+z=1, x+y+z=1, which represents the same plane, hence infinite solutions (Q).<br>If \lambda = -2, the determinant is 0. Checking consistency: x+y+z=1, x-2y+z=-2, x+y-2z=4. Subtracting equations shows inconsistency, hence no solution (R).<br>If \lambda \neq 1 and \lambda \neq -2, \Delta \neq 0, so unique solution (P).<br>Thus, (A) \to Q, (B) \to P, R, (C) \to P, (D) \to R.
Correct Answer: 1