To solve this problem, we need to find the eigenvalues and eigenvectors of the given matrix:

$A = \begin{pmatrix} 3 & 1 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 3 \end{pmatrix}$

Step 1: Finding Eigenvalues

The eigenvalues $\lambda$ are found by solving the characteristic equation:

$\text{det}(A - \lambda I) = 0$

Where $I$ is the identity matrix and $\lambda$ is the eigenvalue. Let’s calculate $A - \lambda I$:

$A - \lambda I = \begin{pmatrix} 3 - \lambda & 1 & 1 \\ 2 & 4 - \lambda & 2 \\ 1 & 1 & 3 - \lambda \end{pmatrix}$

Now, we compute the determinant of this matrix:

$\text{det}(A - \lambda I) = \text{det} \begin{pmatrix} 3 - \lambda & 1 & 1 \\ 2 & 4 - \lambda & 2 \\ 1 & 1 & 3 - \lambda \end{pmatrix}$

Step 2: Compute the determinant

Expanding this determinant:

$\text{det}(A - \lambda I) = (3 - \lambda) \begin{vmatrix} 4 - \lambda & 2 \\ 1 & 3 - \lambda \end{vmatrix} - 1 \begin{vmatrix} 2 & 2 \\ 1 & 3 - \lambda \end{vmatrix} + 1 \begin{vmatrix} 2 & 4 - \lambda \\ 1 & 1 \end{vmatrix}$

Simplify the 2x2 determinants:

• $\begin{vmatrix} 4 - \lambda & 2 \\ 1 & 3 - \lambda \end{vmatrix} = (4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10$
• $\begin{vmatrix} 2 & 2 \\ 1 & 3 - \lambda \end{vmatrix} = 2(3 - \lambda) - 2 = 6 - 2\lambda - 2 = 4 - 2\lambda$
• $\begin{vmatrix} 2 & 4 - \lambda \\ 1 & 1 \end{vmatrix} = 2(1) - (4 - \lambda) = 2 - 4 + \lambda = \lambda - 2$

Now substitute these back into the determinant:

$\text{det}(A - \lambda I) = (3 - \lambda)(\lambda^2 - 7\lambda + 10) - (4 - 2\lambda) + (\lambda - 2)$

Simplify the expression: