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CUET · MATHS · PYQ PAPER 2025

The system of equations
\(x+y+z=7\)
\(x+2 y+3 z=5\)
\(x+3 y+\lambda z=\mu\)
has a unique solution, if

  1. A \(\lambda=5\) and \(\mu=7\)
  2. B \(\lambda \neq 1\) and \(\mu=5\)
  3. C \(\lambda \neq 5\) and \(\mu\) is any real number
  4. D \(\lambda \neq 1\) and \(\mu \neq 5\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\lambda \neq 5\) and \(\mu\) is any real number

Step-by-step Solution

Detailed explanation

\(\det \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{pmatrix} = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)\) \(= 2\lambda - 9 - \lambda + 3 + 1\) \(= \lambda - 5\) \(\text{For a unique solution, } \det(A) \neq 0\) \(\lambda - 5 \neq 0 \implies \lambda \neq 5\)…