We are asked to find the eigenvalues and eigenvectors of the matrix [ A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 1 \ 3 & 2 & 3 \end{bmatrix}. ] The eigenvalues (\lambda) satisfy the characteristic equation: [ \det(A - \lambda I) = 0. ] So, [ A - \lambda I = \begin{bmatrix} 1-\lambda & 1 & 1 \ 1 & 2-\lambda & 1 \ 3 & 2 & 3-\lambda \end{bmatrix}. ] The determinant is: [ \det(A - \lambda I) = (1-\lambda)\begin{vmatrix}2-\lambda & 1 \ 2 & 3-\lambda\end{vmatrix} * 1\begin{vmatrix}1 & 1 \ 3 & 3-\lambda\end{vmatrix} - 1\begin{vmatrix}1 & 2-\lambda \ 3 & 2\end{vmatrix}. ] Compute each minor: 1. (\begin{vmatrix}2-\lambda & 1 \ 2 & 3-\lambda\end{vmatrix} = (2-\lambda)(3-\lambda) - 2\cdot1 = 6 -5\lambda + \lambda^2 -2 = \lambda^2 -5\lambda +4) 2. (\begin{vmatrix}1 & 1 \ 3 & 3-\lambda\end{vmatrix} = 1(3-\lambda) - 1\cdot 3 = 3 - \lambda -3 = -\lambda) 3. (\begin{vmatrix}1 & 2-\lambda \ 3 & 2\end{vmatrix} = 1\cdot 2 - 3(2-\lambda) = 2 -6 +3\lambda = 3\lambda -4) So determinant: [ \det(A - \lambda I) = (1-\lambda)(\lambda^2 -5\lambda +4) -1(-\lambda) +1(3\lambda-4) ] [ = (1-\lambda)(\lambda^2 -5\lambda +4) +