Question

The largest eigenvalue of the given matrix is: \[ \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix} \]

Hint:

To find the eigenvalues of a matrix, use the characteristic equation and calculate its determinant. The largest eigenvalue is the one with the greatest magnitude.

Answer 1

Correct Answer: 2

The eigenvalues of a matrix \( A \) are the solutions of the characteristic equation: \[ \text{det}(A - \lambda I) = 0, \] where \( I \) is the identity matrix, and \( \lambda \) is the eigenvalue. Substituting the given matrix: \[ \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} -\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \\ \end{bmatrix} \] The determinant of this matrix is: \[ \text{det}\left(\begin{bmatrix} -\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \\ \end{bmatrix}\right) = -\lambda^3 + 2\lambda. \] Thus, solving \( -\lambda^3 + 2\lambda = 0 \), we get: \[ \lambda (\lambda^2 - 2) = 0. \] The roots are \( \lambda = 0, \pm \sqrt{2} \). Hence, the largest eigenvalue is: \[ \boxed{\sqrt{2}}. \]