To find the eigenvectors of the matrix \( A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \), we need to follow these steps:

### 1. **Find the Eigenvalues:**

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

\[
\det(A - \lambda I) = 0
\]

Where \( I \) is the identity matrix:

\[
A - \lambda I = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1-\lambda & 2 \\ 2 & 3-\lambda \end{bmatrix}
\]

The determinant of this matrix is:

\[
\det(A - \lambda I) = (1-\lambda)(3-\lambda) - (2)(2)
\]

\[
= \lambda^2 - 4\lambda - 1 - 4 = \lambda^2 - 4\lambda - 4
\]

Now, solve the characteristic equation:

\[
\lambda^2 - 4\lambda - 3 = 0
\]

Use the quadratic formula:

\[
\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 1 \), \( b = -4 \), and \( c = -3 \):

\[
\lambda = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}
\]

So the eigenvalues are:

\[
\lambda_1 = 2 + \sqrt{7}, \quad \lambda_2 = 2 - \sqrt{7}
\]

### 2. **Find the Eigenvectors:**

For each eigenvalue \( \lambda \), solve the equation \( (A - \lambda I)\mathbf{v} = 0 \) to find the corresponding eigenvector \( \mathbf{v} \).

#### For \( \lambda_1 = 2 + \sqrt{7} \):

\[
(A - \lambda_1 I) = \begin{bmatrix} 1-(2+\sqrt{7}) & 2 \\ 2 & 3-(2+\sqrt{7}) \end{bmatrix} = \begin{bmatrix} -1-\sqrt{7} & 2 \\ 2 & 1-\sqrt{7} \end{bmatrix}
\]

Set up the system \( (A - \lambda_1 I)\mathbf{v} = 0 \):

\[
\begin{bmatrix} -1-\sqrt{7} & 2 \\ 2 & 1-\sqrt{7} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
\]

This simplifies to:

\[
\begin{cases} 
(-1-\sqrt{7})x_1 + 2x_2 = 0 \\
2x_1 + (1-\sqrt{7})x_2 = 0 
\end{cases}
\]

We can solve for \( x_2 \) in terms of \( x_1 \) from the first equation:

\[
x_2 = \frac{1+\sqrt{7}}{2} x_1
\]

So, the eigenvector corresponding to \( \lambda_1 = 2 + \sqrt{7} \) is:

\[
\mathbf{v}_1 = \begin{bmatrix} 1 \\ \frac{1+\sqrt{7}}{2} \end{bmatrix}
\]

#### For \( \lambda_2 = 2 - \sqrt{7} \):

\[
(A - \lambda_2 I) = \begin{bmatrix} 1-(2-\sqrt{7}) & 2 \\ 2 & 3-(2-\sqrt{7}) \end{bmatrix} = \begin{bmatrix} -1+\sqrt{7} & 2 \\ 2 & 1+\sqrt{7} \end{bmatrix}
\]

Set up the system \( (A - \lambda_2 I)\mathbf{v} = 0 \):

\[
\begin{cases} 
(-1+\sqrt{7})x_1 + 2x_2 = 0 \\
2x_1 + (1+\sqrt{7})x_2 = 0 
\end{cases}
\]

We can solve for \( x_2 \) in terms of \( x_1 \) from the first equation:

\[
x_2 = \frac{1-\sqrt{7}}{2} x_1
\]

So, the eigenvector corresponding to \( \lambda_2 = 2 - \sqrt{7} \) is:

\[
\mathbf{v}_2 = \begin{bmatrix} 1 \\ \frac{1-\sqrt{7}}{2} \end{bmatrix}
\]

### Final Answer:

The eigenvectors of the matrix \( A = \

Learn how to find the eigenvalues and eigenvectors of the matrix A=[1 2; 2 3]. This problem involves solving the characteristic equation and calculating eigenvectors using matrix algebra. Suitable for advanced high school students.

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