We are given a system of first-order differential equations:

$\frac{dx}{dt} = 2x - 2y + z$ $\frac{dy}{dt} = 2x - 3y - 2z$ $\frac{dz}{dt} = x + 2y + 2z$

The system can be written in matrix form as:

$\frac{d}{dt} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ where $A$ is the coefficient matrix:

$A = \begin{pmatrix} 2 & -2 & 1 \\ 2 & -3 & -2 \\ 1 & 2 & 2 \end{pmatrix}$

We will solve this system using eigenvalues and eigenvectors.

Step 1: Find the eigenvalues of matrix $A$

To find the eigenvalues, solve the characteristic equation:

$\det(A - \lambda I) = 0$

Where $\lambda$ is an eigenvalue and $I$ is the identity matrix. Let's calculate the eigenvalues first.

We need to solve for the determinant of $A - \lambda I$:

$A - \lambda I = \begin{pmatrix} 2 - \lambda & -2 & 1 \\ 2 & -3 - \lambda & -2 \\ 1 & 2 & 2 - \lambda \end{pmatrix}$

The determinant is:

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

Let's compute this determinant.The eigenvalues of matrix $A$ are:

$\lambda_1 = 3, \quad \lambda_2 = -1 + 2i, \quad \lambda_3 = -1 - 2i$

These eigenvalues include a real eigenvalue $\lambda_1 = 3$ and two complex conjugate eigenvalues $\lambda_2 = -1 + 2i$ and $\lambda_3 = -1 - 2i$.

Step 2: Find the corresponding eigenvectors

Now, we will find the eigenvectors corresponding to each of these eigenvalues. Let's calculate them.The eigenvectors corresponding to the eigenvalues are:

For $\lambda_1 = 3$: $v_1 = \begin{pmatrix} -\frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}} \end{pmatrix}$

For $\lambda_2 = -1 + 2i$: $v_2 = \begin{pmatrix} -\frac{1}{2} - \frac{i}{2} \\ -\frac{1}{\sqrt{2}} \\ \frac{1}{2} + \frac{i}{2} \end{pmatrix}$

For $\lambda_3 = -1 - 2i$ (complex conjugate of $v_2$): $v_3 = \begin{pmatrix} -\frac{1}{2} + \frac{i}{2} \\ -\frac{1}{\sqrt{2}} \\ \frac{1}{2} - \frac{i}{2} \end{pmatrix}$

Step 3: General solution

The general solution to the system of differential equations is a linear combination of solutions corresponding to the eigenvalues and eigenvectors. For real eigenvalue $\lambda_1 = 3$:

$x_1(t) = c_1 v_1 e^{3t}$

For complex eigenvalues $\lambda_2 = -1 + 2i$ and $\lambda_3 = -1 - 2i$, the solution takes the form of oscillating functions (sine and cosine terms) based on Euler's formula:

$x_2(t) = e^{-t} \left( c_2 \cos(2t) + c_3 \sin(2t) \right)$

Thus, the general solution is a combination of these terms. We can now apply the initial conditions to find the constants $c_1, c_2, c_3$. Would you like to proceed with calculating these constants using the initial conditions $x(0) = 1$, $y(0) = -1$, $z(0) = 2$?