1. **State the problem:** We want to find the general solution of the system of differential equations $$\vec{y}' = A \vec{y}$$ where $$A = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 4 \\ 1 & 0 & -4 \end{bmatrix}$$ using the method of eigenvectors. 2. **Recall the method:** The general solution to $$\vec{y}' = A \vec{y}$$ is given by $$\vec{y}(t) = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2 + c_3 e^{\lambda_3 t} \vec{v}_3$$ where $$\lambda_i$$ are eigenvalues and $$\vec{v}_i$$ are corresponding eigenvectors of matrix $$A$$. 3. **Find eigenvalues:** Solve the characteristic equation $$\det(A - \lambda I) = 0$$. Calculate: $$A - \lambda I = \begin{bmatrix} -1-\lambda & 1 & 0 \\ 0 & -1-\lambda & 4 \\ 1 & 0 & -4-\lambda \end{bmatrix}$$ The determinant is: $$\det(A - \lambda I) = (-1-\lambda) \begin{vmatrix} -1-\lambda & 4 \\ 0 & -4-\lambda \end{vmatrix} - 1 \begin{vmatrix} 0 & 4 \\ 1 & -4-\lambda \end{vmatrix} + 0$$ Calculate minors: $$= (-1-\lambda)((-1-\lambda)(-4-\lambda) - 0) - 1(0 \cdot (-4-\lambda) - 4 \cdot 1)$$ Simplify: $$= (-1-\lambda)((-1-\lambda)(-4-\lambda)) + 4$$ Expand: $$(-1-\lambda)(-4-\lambda) = ( -1)(-4) + (-1)(-\lambda) + (-\lambda)(-4) + (-\lambda)(-\lambda) = 4 + \lambda + 4\lambda + \lambda^2 = \lambda^2 + 5\lambda + 4$$ So determinant: $$= (-1-\lambda)(\lambda^2 + 5\lambda + 4) + 4 = -(1+\lambda)(\lambda^2 + 5\lambda + 4) + 4$$ Expand: $$= -(\lambda^2 + 5\lambda + 4) - \lambda(\lambda^2 + 5\lambda + 4) + 4 = -\lambda^2 - 5\lambda - 4 - \lambda^3 - 5\lambda^2 - 4\lambda + 4$$ Combine like terms: $$= -\lambda^3 - 6\lambda^2 - 9\lambda - 4 + 4 = -\lambda^3 - 6\lambda^2 - 9\lambda$$ Set equal to zero: $$-\lambda^3 - 6\lambda^2 - 9\lambda = 0$$ Divide both sides by $$-\lambda$$ (assuming $$\lambda \neq 0$$): $$\lambda^2 + 6\lambda + 9 = 0$$ 4. **Solve quadratic:** $$\lambda^2 + 6\lambda + 9 = (\lambda + 3)^2 = 0$$ So eigenvalues are: $$\lambda_1 = 0, \quad \lambda_2 = -3, \quad \lambda_3 = -3$$ 5. **Find eigenvectors:** - For $$\lambda_1 = 0$$, solve $$A \vec{v} = 0$$: $$\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 4 \\ 1 & 0 & -4 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ From first row: $$-v_1 + v_2 = 0 \Rightarrow v_2 = v_1$$ From second row: $$-v_2 + 4 v_3 = 0 \Rightarrow -v_1 + 4 v_3 = 0 \Rightarrow v_3 = \frac{v_1}{4}$$ From third row: $$v_1 - 4 v_3 = 0 \Rightarrow v_1 - 4 \cdot \frac{v_1}{4} = 0 \Rightarrow 0 = 0$$ (consistent) Eigenvector for $$\lambda_1=0$$ is: $$\vec{v}_1 = v_1 \begin{bmatrix} 1 \\ 1 \\ \frac{1}{4} \end{bmatrix}$$ - For $$\lambda_2 = \lambda_3 = -3$$, solve $$ (A + 3I) \vec{v} = 0$$: $$A + 3I = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 4 \\ 1 & 0 & -1 \end{bmatrix}$$ Solve: $$\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 4 \\ 1 & 0 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ From first row: $$2 v_1 + v_2 = 0 \Rightarrow v_2 = -2 v_1$$ From second row: $$2 v_2 + 4 v_3 = 0 \Rightarrow 2(-2 v_1) + 4 v_3 = 0 \Rightarrow -4 v_1 + 4 v_3 = 0 \Rightarrow v_3 = v_1$$ From third row: $$v_1 - v_3 = 0 \Rightarrow v_1 - v_1 = 0$$ (consistent) Eigenvector for $$\lambda = -3$$ is: $$\vec{v}_2 = v_1 \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$$ Since $$\lambda = -3$$ has multiplicity 2, check for generalized eigenvector if needed, but here we have one eigenvector. 6. **Write general solution:** $$\vec{y}(t) = c_1 e^{0 \cdot t} \begin{bmatrix} 1 \\ 1 \\ \frac{1}{4} \end{bmatrix} + c_2 e^{-3 t} \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} + c_3 t e^{-3 t} \vec{w}$$ where $$\vec{w}$$ is the generalized eigenvector satisfying $$ (A + 3I) \vec{w} = \vec{v}_2$$. 7. **Find generalized eigenvector $$\vec{w}$$:** Solve: $$ (A + 3I) \vec{w} = \vec{v}_2 = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$$ Let $$\vec{w} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}$$, then $$\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 4 \\ 1 & 0 & -1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$$ From first row: $$2 w_1 + w_2 = 1$$ From second row: $$2 w_2 + 4 w_3 = -2$$ From third row: $$w_1 - w_3 = 1$$ From third row: $$w_1 = 1 + w_3$$ Substitute into first row: $$2(1 + w_3) + w_2 = 1 \Rightarrow 2 + 2 w_3 + w_2 = 1 \Rightarrow w_2 = 1 - 2 w_3 - 2$$ $$w_2 = -1 - 2 w_3$$ Substitute $$w_2$$ into second row: $$2(-1 - 2 w_3) + 4 w_3 = -2 \Rightarrow -2 - 4 w_3 + 4 w_3 = -2$$ Simplifies to: $$-2 = -2$$ (always true) So $$w_3$$ is free parameter, choose $$w_3 = 0$$ for simplicity: Then $$w_1 = 1$$, $$w_2 = -1$$, $$w_3 = 0$$. Generalized eigenvector: $$\vec{w} = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$$ 8. **Final general solution:** $$\boxed{\vec{y}(t) = c_1 \begin{bmatrix} 1 \\ 1 \\ \frac{1}{4} \end{bmatrix} + c_2 e^{-3 t} \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} + c_3 t e^{-3 t} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}}$$ where $$c_1, c_2, c_3$$ are arbitrary constants determined by initial conditions.