ExamBro
ExamBro
CUET · MATHS · PYQ PAPER 2025

The values of \(\lambda\) for which the system of equation \(x+2 y+z=14,-x+y+z=10, x+\lambda y+z=2\) has unique solution is :

  1. A \(R -\{2\}\), where R is set of real numbers.
  2. B \(-2 \leq \lambda \leq 2\)
  3. C \(-4<\lambda<4\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(R -\{2\}\), where R is set of real numbers.

Step-by-step Solution

Detailed explanation

\(A = \begin{vmatrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & \lambda & 1 \end{vmatrix}\) \(\det(A) = 1(1 - \lambda) - 2(-1 - 1) + 1(-\lambda - 1)\) \(\det(A) = 1 - \lambda + 4 - \lambda - 1 = 4 - 2\lambda\) For a unique solution, \(\det(A) \neq 0\) \(4 - 2\lambda \neq 0\)…