The set of all real values of \(\lambda\) for which the matrix \(\begin{bmatrix}\lambda & 1 & 1\\ 1 & \lambda & 1\\ 1 & 1 & \lambda\end{bmatrix}\) is not invertible is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\{1, -2\}\)
A square matrix is not invertible exactly when its determinant is zero. Expanding along the first row:
\(\Delta = \lambda(\lambda^{2} - 1) - 1(\lambda - 1) + 1(1 - \lambda)\)
\(\Delta = \lambda^{3} - \lambda - \lambda + 1 + 1 - \lambda\)
\(\Delta = \lambda^{3} - 3\lambda + 2\)
Since \(\lambda = 1\) gives \(1 - 3 + 2 = 0\), \((\lambda - 1)\) is a factor:
\(\lambda^{3} - 3\lambda + 2 = (\lambda - 1)(\lambda^{2} + \lambda - 2)\)
\(\lambda^{3} - 3\lambda + 2 = (\lambda - 1)^{2}(\lambda + 2)\)
So \(\Delta = 0\) only for \(\lambda = 1\) and \(\lambda = -2\).
Why the other options are incorrect
Option (1) \(\{-1, 2\}\) — Incorrect. The signs are reversed: \(\lambda = -1\) gives \(\Delta = 4\) and \(\lambda = 2\) gives \(\Delta = 4\).
Option (2) \(\{1\}\) — Incorrect. \(\lambda = -2\) is also a root: \(-8 + 6 + 2 = 0\).
Option (4) \(\{-2\}\) — Incorrect. \(\lambda = 1\) is also a root: all three rows become equal.