To find the general solution of the system $\mathbf{x}'(t) = \mathbf{A} \mathbf{x}(t)$ for the given matrix $\mathbf{A}$, we need to find the eigenvalues and eigenvectors of the matrix $\mathbf{A}$.

Given: $\mathbf{A} = \begin{pmatrix} -5 & -1 \\ 26 & 5 \end{pmatrix}$

Step 1: Find the Eigenvalues

The eigenvalues $\lambda$ are found by solving the characteristic equation $\det(\mathbf{A} - \lambda \mathbf{I}) = 0$.

$\mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} -5 - \lambda & -1 \\ 26 & 5 - \lambda \end{pmatrix}$

The determinant is: $\det(\mathbf{A} - \lambda \mathbf{I}) = (-5 - \lambda)(5 - \lambda) - (-1)(26) = 0$

$(\lambda + 5)(\lambda - 5) + 26 = 0$

$\lambda^2 - 25 + 26 = 0$

$\lambda^2 + 1 = 0$

$\lambda^2 = -1$

$\lambda = \pm i$

The eigenvalues are $\lambda_1 = i$ and $\lambda_2 = -i$.

Step 2: Find the Eigenvectors

For $\lambda_1 = i$:

$\mathbf{A} - \lambda_1 \mathbf{I} = \begin{pmatrix} -5 - i & -1 \\ 26 & 5 - i \end{pmatrix}$

To find the eigenvector $\mathbf{v}_1$, we solve:

$\begin{pmatrix} -5 - i & -1 \\ 26 & 5 - i \end{pmatrix} \begin{pmatrix} v_{1,1} \\ v_{1,2} \end{pmatrix} = \mathbf{0}$

We can take:

$v_{1,1} = 1$ $v_{1,2} = 5 + i$

So, the eigenvector $\mathbf{v}_1$ corresponding to $\lambda_1$ is:

$\mathbf{v}_1 = \begin{pmatrix} 1 \\ 5 + i \end{pmatrix}$

For $\lambda_2 = -i$:

$\mathbf{A} - \lambda_2 \mathbf{I} = \begin{pmatrix} -5 + i & -1 \\ 26 & 5 + i \end{pmatrix}$

To find the eigenvector $\mathbf{v}_2$, we solve:

$\begin{pmatrix} -5 + i & -1 \\ 26 & 5 + i \end{pmatrix} \begin{pmatrix} v_{2,1} \\ v_{2,2} \end{pmatrix} = \mathbf{0}$

We can take:

$v_{2,1} = 1$ $v_{2,2} = 5 - i$

So, the eigenvector $\mathbf{v}_2$ corresponding to $\lambda_2$ is:

$\mathbf{v}_2 = \begin{pmatrix} 1 \\ 5 - i \end{pmatrix}$

Step 3: Write the General Solution

The general solution to the system is a linear combination of the solutions corresponding to each eigenvalue and eigenvector.

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

$\mathbf{x}(t) = c_1 e^{it} \begin{pmatrix} 1 \\ 5 + i \end{pmatrix} + c_2 e^{-it} \begin{pmatrix} 1 \\ 5 - i \end{pmatrix}$

Using Euler's formula $e^{i \theta} = \cos(\theta) + i \sin(\theta)$, we can rewrite the general solution in terms of real functions:

$\mathbf{x}(t) = c_1 (\cos(t) + i \sin(t)) \begin{pmatrix} 1 \\ 5 + i \end{pmatrix} + c_2 (\cos(t) - i \sin(t)) \begin{pmatrix} 1 \\ 5 - i \end{pmatrix}$

Combining the terms and simplifying gives the general real solution.

Final General Solution

$\mathbf{x}(t) = c_1 \begin{pmatrix} \cos(t) \\ 5 \cos(t) - \sin(t) \end{pmatrix} + c_2 \begin{pmatrix} \sin(t) \\ 5 \sin(t) + \cos(t) \end{pmatrix}$

Would you like more details or have any questions about this solution? Here are some related questions you might be interested in:

1. How do you find eigenvalues and eigenvectors of a 3x3 matrix?
2. What is the significance of eigenvalues and eigenvectors in linear algebra?
3. How do you solve systems of differential equations using eigenvalues and eigenvectors?
4. What is the Cayley-Hamilton theorem and how is it used?
5. How do you diagonalize a matrix?

Tip: When solving differential equations with complex eigenvalues, using Euler's formula helps convert the complex solutions into real-valued functions, which are often easier to interpret and apply.