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The problem in the image involves finding the eigenvalues and eigenvectors of a 2x2 matrix \( A \). I'll walk you through the steps based on the image provided.

### Step 1: Matrix \( A \)
The matrix \( A \) given is:
\[
A = \begin{pmatrix} 2 & 2 \\ 1 & 3 \end{pmatrix}
\]

### Step 2: Finding the Eigenvalues
To find the eigenvalues \( \lambda \), we solve the characteristic equation:
\[
\det(A - \lambda I) = 0
\]
Here, \( I \) is the identity matrix:
\[
I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
So, \( A - \lambda I \) becomes:
\[
A - \lambda I = \begin{pmatrix} 2 - \lambda & 2 \\ 1 & 3 - \lambda \end{pmatrix}
\]
The determinant of this matrix is:
\[
\det(A - \lambda I) = (2 - \lambda)(3 - \lambda) - (2)(1)
\]
Expanding the determinant:
\[
\det(A - \lambda I) = (2 - \lambda)(3 - \lambda) - 2 = (6 - 5\lambda + \lambda^2) - 2 = \lambda^2 - 5\lambda + 4
\]
Setting the determinant to zero to find the eigenvalues:
\[
\lambda^2 - 5\lambda + 4 = 0
\]
This quadratic equation can be factored as:
\[
(\lambda - 1)(\lambda - 4) = 0
\]
So, the eigenvalues are:
\[
\lambda_1 = 1, \quad \lambda_2 = 4
\]

### Step 3: Finding the Eigenvectors

#### Eigenvector for \( \lambda_1 = 1 \):
Substitute \( \lambda = 1 \) into \( (A - \lambda I)\mathbf{v} = 0 \):
\[
(A - I)\mathbf{v} = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0
\]
This simplifies to the system of equations:
\[
x_1 + 2x_2 = 0
\]
The solution is:
\[
\mathbf{v}_1 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}
\]

#### Eigenvector for \( \lambda_2 = 4 \):
Substitute \( \lambda = 4 \) into \( (A - \lambda I)\mathbf{v} = 0 \):
\[
(A - 4I)\mathbf{v} = \begin{pmatrix} -2 & 2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0
\]
This simplifies to the system of equations:
\[
-2x_1 + 2x_2 = 0
\]
The solution is:
\[
\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}
\]

### Final Answer
- The eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = 4 \).
- The corresponding eigenvectors are:
  - For \( \lambda_1 = 1 \), the eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} -2 \\ 1 \end{pmatrix} \).
  - For \( \lambda_2 = 4 \), the eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \).

Would you like more details, or do you have any questions?

### Related Questions:
1. How do you verify if a vector is an eigenvector of a matrix?
2. What is the geometric interpretation of eigenvalues and eigenvectors?
3. Can a matrix have complex eigenvalues and eigenvectors? If yes, when?
4. What does it mean if a matrix has repeated eigenvalues?
5. How are eigenvalues used in diagonalization of matrices?
6. What are some applications of eigenvalues and eigenvectors in real-world problems?
7. How do you find eigenvalues and eigenvectors for larger matrices (e.g., 3x3)?
8. Can a non-square matrix have eigenvalues and eigenvectors?

### Tip:
Eigenvectors corresponding to different eigenvalues are linearly independent, which is useful in matrix diagonalization.

Learn how to find eigenvalues and eigenvectors of a 2x2 matrix step-by-step with a detailed solution. This problem involves matrix algebra, determinant calculation, and solving characteristic equations. Suitable for advanced high school students.

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