diagonalize this 2 by 2 matrix   4 2 3 -1

To diagonalize a \( 2 \times 2 \) matrix, we need to follow these steps:

Given the matrix:
\[
A = \begin{pmatrix} 4 & 2 \\ 3 & -1 \end{pmatrix}
\]
1. **Find the eigenvalues** by solving the characteristic equation:  
   \[
   \det(A - \lambda I) = 0
   \]
   where \( I \) is the identity matrix, and \( \lambda \) represents the eigenvalues.

2. **Find the eigenvectors** corresponding to each eigenvalue by solving \( (A - \lambda I) \mathbf{v} = 0 \).

3. **Form the matrix** \( P \) using the eigenvectors as columns.

4. **Construct the diagonal matrix** \( D \), where the diagonal elements are the eigenvalues.

Let's compute the eigenvalues first:

### Step 1: Find the Eigenvalues
The characteristic equation is:
\[
\det(A - \lambda I) = \det \begin{pmatrix} 4 - \lambda & 2 \\ 3 & -1 - \lambda \end{pmatrix} = 0
\]

Calculating the determinant:
\[
(4 - \lambda)(-1 - \lambda) - 2 \cdot 3 = 0
\]

Expanding this:
\[
(4 - \lambda)(-1 - \lambda) = -4 - 4\lambda + \lambda + \lambda^2 = \lambda^2 - 3\lambda - 4
\]

So:
\[
\lambda^2 - 3\lambda - 4 - 6 = 0 \quad \Rightarrow \quad \lambda^2 - 3\lambda - 10 = 0
\]

Solving this quadratic equation:
\[
\lambda = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2}
\]
\[
\lambda = \frac{3 \pm 7}{2}
\]
Thus, the eigenvalues are:
\[
\lambda_1 = \frac{3 + 7}{2} = 5, \quad \lambda_2 = \frac{3 - 7}{2} = -2
\]

### Step 2: Find the Eigenvectors
For \( \lambda_1 = 5 \), solve \( (A - 5I) \mathbf{v} = 0 \):
\[
A - 5I = \begin{pmatrix} 4 - 5 & 2 \\ 3 & -1 - 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 3 & -6 \end{pmatrix}
\]
Solving \( \begin{pmatrix} -1 & 2 \\ 3 & -6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \), gives:
\[
-1x + 2y = 0 \quad \Rightarrow \quad x = 2y
\]
Let \( y = 1 \), then \( x = 2 \). So an eigenvector corresponding to \( \lambda_1 = 5 \) is:
\[
\mathbf{v_1} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}
\]

For \( \lambda_2 = -2 \), solve \( (A + 2I) \mathbf{v} = 0 \):
\[
A + 2I = \begin{pmatrix} 4 + 2 & 2 \\ 3 & -1 + 2 \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 3 & 1 \end{pmatrix}
\]
Solving \( \begin{pmatrix} 6 & 2 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \), gives:
\[
6x + 2y = 0 \quad \Rightarrow \quad 3x + y = 0 \quad \Rightarrow \quad y = -3x
\]
Let \( x = 1 \), then \( y = -3 \). So an eigenvector corresponding to \( \lambda_2 = -2 \) is:
\[
\mathbf{v_2} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}
\]

### Step 3: Form the Matrix \( P \)
The matrix \( P \) is formed by using the eigenvectors as its columns:
\[
P = \begin{pmatrix} 2 & 1 \\ 1 & -3 \end{pmatrix}
\]

### Step 4: Construct the Diagonal Matrix \( D \)
The diagonal matrix \( D \) is:
\[
D = \begin{pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix}
\]

Thus, the matrix \( A \) can be diagonalized as:
\[
A = P D P^{-1}
\]

Would you like further details or have any questions?

### Related Questions:
1. How do you compute the inverse of a matrix like \( P \)?
2. Why do we diagonalize matrices in linear algebra?
3. What are the applications of diagonalizing a matrix?
4. How would you diagonalize a \( 3 \times 3 \) matrix?
5. What happens if a matrix does not have distinct eigenvalues?

### Tip:
Diagonalizing a matrix simplifies many matrix computations, such as raising it to large powers.

Learn how to diagonalize a 2x2 matrix by finding the eigenvalues and eigenvectors. This step-by-step guide covers key concepts in linear algebra, including the characteristic equation, eigenvalue theorem, and diagonalization. Suitable for undergraduate students studying linear algebra.

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