Eigen values of matrix $\begin{bmatrix} 2 & 1 & 1 \\ 2 & 3 & 4 \\ -1 & -1 & -2 \end{bmatrix}$ are:
Step-by-Step Solution
Key Concept: The characteristic equation is found by setting $\det(A - \lambda I) = 0$ and factoring to find eigenvalues.
The characteristic polynomial is $\begin{vmatrix} 2-\lambda & 1 & 1 \\ 2 & 3-\lambda & 4 \\ -1 & -1 & -2-\lambda \end{vmatrix} = 0$, which simplifies to $\lambda^3 - 3\lambda^2 + \lambda - 3 = 0$, factoring as $\lambda^2(\lambda - 3) - (\lambda - 3) = 0$, giving eigenvalues $\lambda = \pm 1, 3$.
Correct Answer: 1