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We are given the matrix $$ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 3 & 2 & 3 \end{bmatrix}. $$ To find the eigenvalues, we solve the characteristic equation $\det(A - \lambda I) = 0$. $$ A - \lambda I = \begin{bmatrix} 1-\lambda & 1 & 1 \\ 1 & 2-\lambda & 1 \\ 3 & 2 & 3-\lambda \end{bmatrix}. $$ The determinant is $$ \det(A - \lambda I) = \begin{vmatrix} 1-\lambda & 1 & 1 \\ 1 & 2-\lambda & 1 \\ 3 & 2 & 3-\lambda \end{vmatrix}. $$ Expanding along the first row: $$ (1-\lambda)\begin{vmatrix} 2-\lambda & 1 \\ 2 & 3-\lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 3 & 3-\lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2-\lambda \\ 3 & 2 \end{vmatrix}. $$ ___ ______ ______ _______ _____.
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