Question

The sum of distinct values of \(\lambda\) for which the system of equations
\((\lambda - 1)x + (3\lambda + 1)y + 2\lambda z = 0\)
\((\lambda - 1)x + (4\lambda - 2)y + (\lambda + 3)z = 0\)
\(2x + (3\lambda + 1)y + 3(\lambda - 1)z = 0\)
has non-zero solutions, is ________ .

Hint:

A homogeneous system has non-zero solutions iff the coefficient determinant is zero; simplify using \(R_2 - R_1\) and \(C_3 + C_2\).

Answer 1

Correct Answer: 3

Non-zero solutions exist iff \(\Delta = \begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \\ \lambda-1 & 4\lambda-2 & \lambda+3 \\ 2 & 3\lambda+1 & 3(\lambda-1) \end{vmatrix} = 0\).
\(R_2 \to R_2 - R_1\): second row becomes \((0,\ \lambda - 3,\ 3 - \lambda) = (\lambda - 3)(0, 1, -1)\).
So \(\Delta = (\lambda - 3)\begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \\ 0 & 1 & -1 \\ 2 & 3\lambda+1 & 3\lambda-3 \end{vmatrix}\).
\(C_3 \to C_3 + C_2\): third column becomes \((5\lambda+1,\ 0,\ 6\lambda-2)\). Expanding along row 2 (only the middle entry 1 is non-zero): \((\lambda-1)(6\lambda-2) - 2(5\lambda+1) = 6\lambda^2 - 18\lambda\).
Thus \(\Delta = (\lambda - 3)\cdot 6\lambda(\lambda - 3) = 6\lambda(\lambda - 3)^2\).
\(\Delta = 0\) gives \(\lambda = 0\) or \(\lambda = 3\). Distinct values: 0 and 3; their sum is 3.
Answer: 3