Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade None
Question:
The system of homogeneous equations $\lambda x + (\lambda + 1)y + (\lambda - 1)z = 0$, $(\lambda + 1)x + \lambda y + (\lambda + 2)z = 0$, $(\lambda - 1)x + (\lambda + 2)y + \lambda z = 0$ has non trivial solution for:
Exactly 3 real values of $\lambda$
Exactly 2 real values of $\lambda$
Exactly 1 real value of $\lambda$
Infinitely many real values of $\lambda$
Step-by-Step Solution
Key Concept: Row and column operations reduce the determinant to a simple form without changing its value.
Expanding the determinant $\begin{vmatrix} \lambda & \lambda+1 & \lambda-1 \\ \lambda+1 & \lambda & \lambda+2 \\ \lambda-1 & \lambda+2 & \lambda \end{vmatrix} = 0$ by applying row and column operations to simplify, we obtain $-4(\lambda - \lambda - 1) = 0$, which yields $\lambda = -\frac{1}{2}$.
Correct Answer: 3