 Find the general solution to
x′ = Ax
for
A =
1 −3
1 1

To find the general solution to the system of differential equations given by \( \mathbf{x}' = A\mathbf{x} \) where

\[
A = \begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix},
\]

we follow these steps:

### 1. **Find the Eigenvalues of \( A \)**:
The eigenvalues \( \lambda \) of matrix \( A \) are found by solving the characteristic equation:

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

where \( I \) is the identity matrix and \( \det \) denotes the determinant. The characteristic equation is:

\[
\det\left(\begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\right) = \det\begin{pmatrix} 1 - \lambda & -3 \\ 1 & 1 - \lambda \end{pmatrix} = 0
\]

Calculate the determinant:

\[
(1 - \lambda)(1 - \lambda) - (-3)(1) = (\lambda - 1)^2 + 3 = \lambda^2 - 2\lambda + 4 = 0
\]

This quadratic equation is:

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

### 2. **Solve for Eigenvalues**:
To solve the quadratic equation:

\[
\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)}
\]

\[
\lambda = \frac{2 \pm \sqrt{4 - 16}}{2}
\]

\[
\lambda = \frac{2 \pm \sqrt{-12}}{2}
\]

\[
\lambda = 1 \pm i\sqrt{3}
\]

So the eigenvalues are:

\[
\lambda_1 = 1 + i\sqrt{3}, \quad \lambda_2 = 1 - i\sqrt{3}
\]

### 3. **Find the Eigenvectors**:
For each eigenvalue \( \lambda_i \), solve \( (A - \lambda_i I)\mathbf{v} = 0 \) to find the corresponding eigenvector \( \mathbf{v}_i \).

For \( \lambda_1 = 1 + i\sqrt{3} \):

\[
\left(\begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix} - (1 + i\sqrt{3}) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\right) \mathbf{v} = 0
\]

\[
\begin{pmatrix} -i\sqrt{3} & -3 \\ 1 & -i\sqrt{3} \end{pmatrix} \mathbf{v} = 0
\]

The eigenvector \( \mathbf{v}_1 \) can be found by solving this system. Similarly, we find \( \mathbf{v}_2 \) for \( \lambda_2 = 1 - i\sqrt{3} \).

### 4. **Form the General Solution**:
The general solution is given by:

\[
\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2
\]

Given the complex eigenvalues, the general solution can also be written in terms of real functions using Euler's formula:

\[
\mathbf{x}(t) = e^{t} \left[ c_1 \begin{pmatrix} \text{Re}(\mathbf{v}_1) \\ \text{Im}(\mathbf{v}_1) \end{pmatrix} \cos(\sqrt{3}t) + c_2 \begin{pmatrix} \text{Re}(\mathbf{v}_1) \\ \text{Im}(\mathbf{v}_1) \end{pmatrix} \sin(\sqrt{3}t) \right]
\]

Here, \( \text{Re}(\mathbf{v}_1) \) and \( \text{Im}(\mathbf{v}_1) \) are the real and imaginary parts of the eigenvector \( \mathbf{v}_1 \).

### Final General Solution:
\[
\mathbf{x}(t) = e^t \left[ c_1 \mathbf{v}_r \cos(\sqrt{3}t) + c_2 \mathbf{v}_i \sin(\sqrt{3}t) \right]
\]

where \( \mathbf{v}_r \) and \( \mathbf{v}_i \) are real vectors derived from the real and imaginary parts of the eigenvectors.

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

---

**Relative Questions**:
1. How do you find eigenvectors for a given matrix?
2. What is Euler's formula, and how is it applied in solving differential equations?
3. Can eigenvalues be real, and what does that imply for the solution?
4. How do the real and imaginary parts of eigenvalues affect the general solution?
5. What are the physical interpretations of complex eigenvalues in differential systems?

**Tip**: When solving systems of differential equations with complex eigenvalues, always express your final solution using real functions (sine and cosine) to make it easier to interpret and apply.

Find the general solution to x′ = Ax for A = [1, -3; 1, 1]

Explore how to solve the system of differential equations x' = Ax, where A = [1, -3; 1, 1]. Learn the step-by-step process to find eigenvalues, eigenvectors, and form the general solution using Euler's formula. Suitable for college-level students studying linear algebra and differential equations.

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