Question

Let S be the set of all \(\lambda \in R\) for which the system of linear equations

\(2x - y + 2z = 2\)
\(x - 2y + \lambda z = -4\)
\(x + \lambda y + z = 4\)

has no solution. Then the set S :

• A. contains more than two elements.
• B. contains exactly two elements.
• C. is a singleton.
• D. is an empty set.

Hint:

Find the values of \(\lambda\) making the coefficient determinant zero, then test consistency for each.

Answer 1

The Correct Option is B

Coefficient determinant \(\Delta = \begin{vmatrix}2&-1&2\\1&-2&\lambda\\1&\lambda&1\end{vmatrix} = 2(-2-\lambda^2) + 1(1-\lambda) + 2(\lambda+2) = -2\lambda^2 + \lambda + 1 = -(2\lambda+1)(\lambda-1)\).
\(\Delta = 0\) for \(\lambda = 1\) or \(\lambda = -\frac12\); otherwise a unique solution exists.
\(\lambda = 1\): equations \(x-2y+z=-4\) and \(x+y+z=4\) give \(y = \frac83\), \(x+z = \frac43\); but eq. 1 gives \(2(x+z) = 2+y = \frac{14}{3}\Rightarrow x+z=\frac73\). Contradiction, so no solution.
\(\lambda = -\frac12\): the third equation is \(x - \frac y2 + z = 4\); doubling gives \(2x - y + 2z = 8\), contradicting eq. 1 (\(=2\)). No solution.
So \(S = \{1, -\frac12\}\), exactly two elements.
Answer: (2) contains exactly two elements.