Question

The number of real values of \(\lambda\) for which the system of linear equations
\(2x + 4y - \lambda z = 0\)
\(4x + \lambda y + 2z = 0\)
\(\lambda x + 2y + 2z = 0\)
has infinitely many solutions, is :

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

A homogeneous system has non-trivial (infinitely many) solutions iff the coefficient determinant is zero; count real roots of the resulting cubic.

Answer 1

The Correct Option is B

The system is homogeneous, so infinitely many solutions require \(\Delta = 0\).
\(\Delta = \begin{vmatrix} 2 & 4 & -\lambda \\ 4 & \lambda & 2 \\ \lambda & 2 & 2 \end{vmatrix}\)
\(= 2(2\lambda - 4) - 4(8 - 2\lambda) - \lambda(8 - \lambda^2)\)
\(= 4\lambda - 8 - 32 + 8\lambda - 8\lambda + \lambda^3 = \lambda^3 + 4\lambda - 40\).
Let \(h(\lambda) = \lambda^3 + 4\lambda - 40\). \(h'(\lambda) = 3\lambda^2 + 4 \gt 0\) for all real \(\lambda\), so \(h\) is strictly increasing and has exactly one real root (between 3 and 4 since \(h(3) = -1\), \(h(4) = 40\)).
At this root no two rows are proportional, so the rank is 2 and the system has infinitely many solutions.
Number of real values of \(\lambda\) = 1.
Answer: (2) 1