For the given matrix A=[■(4&1@2&3)]. Find its eigenvalues and eigenvectors.

The matrix given is \( A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} \). To find the eigenvalues and eigenvectors of this matrix, we follow the steps below.

### 1. Eigenvalues
The eigenvalues \(\lambda\) are found by solving the characteristic equation:

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

Where \(I\) is the identity matrix and \(\lambda\) is a scalar (the eigenvalue). Let's compute \(A - \lambda I\):

\[
A - \lambda I = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix}
\]

Now, find the determinant of \(A - \lambda I\):

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

Expanding the determinant:

\[
(4 - \lambda)(3 - \lambda) - 2 = 12 - 7\lambda + \lambda^2 - 2 = \lambda^2 - 7\lambda + 10
\]

Set this equal to zero to find the eigenvalues:

\[
\lambda^2 - 7\lambda + 10 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
\lambda = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(10)}}{2(1)} = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm \sqrt{9}}{2} = \frac{7 \pm 3}{2}
\]

Thus, the eigenvalues are:

\[
\lambda_1 = \frac{7 + 3}{2} = 5, \quad \lambda_2 = \frac{7 - 3}{2} = 2
\]

### 2. Eigenvectors
To find the eigenvectors corresponding to each eigenvalue, we solve the equation:

\[
(A - \lambda I) \mathbf{v} = 0
\]

#### Eigenvector for \(\lambda_1 = 5\):
Substitute \(\lambda_1 = 5\) into \(A - \lambda I\):

\[
A - 5I = \begin{pmatrix} 4 - 5 & 1 \\ 2 & 3 - 5 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}
\]

Now, solve \((A - 5I) \mathbf{v} = 0\), i.e.,

\[
\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

This gives the system of equations:

\[
- v_1 + v_2 = 0
\]
\[
2v_1 - 2v_2 = 0
\]

From the first equation, \(v_1 = v_2\). Thus, the eigenvector corresponding to \(\lambda_1 = 5\) is:

\[
\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}
\]

#### Eigenvector for \(\lambda_2 = 2\):
Substitute \(\lambda_2 = 2\) into \(A - \lambda I\):

\[
A - 2I = \begin{pmatrix} 4 - 2 & 1 \\ 2 & 3 - 2 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}
\]

Now, solve \((A - 2I) \mathbf{v} = 0\), i.e.,

\[
\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

This gives the system of equations:

\[
2v_1 + v_2 = 0
\]
\[
2v_1 + v_2 = 0
\]

Both equations are the same, so we get \(v_2 = -2v_1\). Thus, the eigenvector corresponding to \(\lambda_2 = 2\) is:

\[
\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}
\]

### Summary:
- Eigenvalues: \(\lambda_1 = 5\), \(\lambda_2 = 2\)
- Eigenvector for \(\lambda_1 = 5\): \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\)
- Eigenvector for \(\lambda_2 = 2\): \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}\)

Would you like further details on any of these steps, or have any specific questions?

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Here are five related questions:
1. How do you interpret eigenvalues and eigenvectors geometrically?
2. Can a matrix have complex eigenvalues?
3. What is the significance of the determinant when finding eigenvalues?
4. How do eigenvectors relate to the diagonalization of a matrix?
5. How are eigenvalues used in solving systems of differential equations?

**Tip:** The eigenvectors corresponding to distinct eigenvalues are always linearly independent!

Explore the step-by-step process to find the eigenvalues and eigenvectors of the matrix A = [[4, 1], [2, 3]]. This problem includes concepts like the characteristic equation and the quadratic formula. Suitable for undergraduate students studying linear algebra.

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