1. **Stating the problem:** We are given a system of linear differential equations: $$\begin{cases} x' = 2x + y + 2z \\ y' = x + 2y + 2z \\ z' = x + y + 3z \end{cases}$$ We want to analyze this system using matrix methods and find the general solution in terms of $x(t)$, $y(t)$, and $z(t)$. 2. **Matrix form:** Write the system as $$\mathbf{x}' = A\mathbf{x}$$ where $$\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$$ and $$A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{bmatrix}.$$ This converts the system into a vector differential equation. 3. **Finding eigenvalues:** To solve, find eigenvalues $\lambda$ by solving $$\det(A - \lambda I) = 0,$$ where $I$ is the identity matrix. Calculate: $$\det\begin{bmatrix} 2-\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix} = 0.$$ 4. **Determinant expansion:** Expanding the determinant: $$ (2-\lambda) \begin{vmatrix} 2-\lambda & 2 \\ 1 & 3-\lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 1 & 3-\lambda \end{vmatrix} + 2 \begin{vmatrix} 1 & 2-\lambda \\ 1 & 1 \end{vmatrix} = 0.$$ Calculate minors: $$ (2-\lambda)((2-\lambda)(3-\lambda) - 2) - 1(1(3-\lambda) - 2) + 2(1 \cdot 1 - 1(2-\lambda)) = 0.$$ Simplify: $$ (2-\lambda)((2-\lambda)(3-\lambda) - 2) - (3-\lambda - 2) + 2(1 - (2-\lambda)) = 0.$$ 5. **Simplify further:** Calculate $(2-\lambda)(3-\lambda)$: $$ (2-\lambda)(3-\lambda) = 6 - 2\lambda - 3\lambda + \lambda^2 = \lambda^2 - 5\lambda + 6.$$ So, $$ (2-\lambda)(\lambda^2 - 5\lambda + 6 - 2) - (3 - \lambda - 2) + 2(1 - 2 + \lambda) = 0,$$ which is $$ (2-\lambda)(\lambda^2 - 5\lambda + 4) - (1 - \lambda) + 2(\lambda - 1) = 0.$$ 6. **Expand:** $$ (2-\lambda)(\lambda^2 - 5\lambda + 4) = 2\lambda^2 - 10\lambda + 8 - \lambda^3 + 5\lambda^2 - 4\lambda = -\lambda^3 + 7\lambda^2 - 14\lambda + 8.$$ So the equation becomes: $$ -\lambda^3 + 7\lambda^2 - 14\lambda + 8 - 1 + \lambda + 2\lambda - 2 = 0,$$ which simplifies to $$ -\lambda^3 + 7\lambda^2 - 11\lambda + 5 = 0.$$ 7. **Multiply both sides by -1:** $$ \lambda^3 - 7\lambda^2 + 11\lambda - 5 = 0.$$ 8. **Solve cubic equation:** Try rational roots $\pm1, \pm5$. Test $\lambda=1$: $$1 - 7 + 11 - 5 = 0,$$ so $\lambda=1$ is a root. Divide polynomial by $(\lambda - 1)$: $$ \lambda^3 - 7\lambda^2 + 11\lambda - 5 = (\lambda - 1)(\lambda^2 - 6\lambda + 5).$$ 9. **Solve quadratic:** $$ \lambda^2 - 6\lambda + 5 = 0,$$ using quadratic formula: $$ \lambda = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}.$$ So, $$ \lambda = 5 \text{ or } 1.$$ Eigenvalues are $\lambda_1 = 1$, $\lambda_2 = 1$, $\lambda_3 = 5$. 10. **Find eigenvectors:** For each eigenvalue, solve $(A - \lambda I)\mathbf{v} = 0$ to find eigenvectors. 11. **General solution:** The solution is a linear combination: $$ \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 t e^{\lambda_1 t} \mathbf{v}_2 + c_3 e^{\lambda_3 t} \mathbf{v}_3,$$ where $c_1, c_2, c_3$ are constants determined by initial conditions. **Summary:** We transformed the system into matrix form, found eigenvalues by solving the characteristic polynomial, and outlined the method to find eigenvectors and the general solution.