Let's solve part (i) and (ii) for the example given: The system is: \[ \begin{cases} x'(t) = -2x - 2y \\ y'(t) = -x - 3y \end{cases} \] --- ### **Step 1: Find the associated coefficient matrix \(A\).** Write the system in matrix form: \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -2 & -2 \\ -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \] **Explanation:** The coefficients of \(x\) and \(y\) in each equation become the entries of the matrix \(A\). - First row: coefficients from the \(x'\) equation: \(-2\) (for \(x\)), \(-2\) (for \(y\)) - Second row: coefficients from the \(y'\) equation: \(-1\) (for \(x\)), \(-3\) (for \(y\)) **Answer for Step 1:** \[ A = \begin{pmatrix} -2 & -2 \\ -1 & -3 \end{pmatrix} \] --- ### **Step 2: Compute the set of eigenvalues of \(A\). Show supporting work.** The eigenvalues \(\lambda\) are found by solving: \[ \det(A - \lambda I) = 0 \] \[ A - \lambda I = \begin{pmatrix} -2 - \lambda & -2 \\ -1 & -3 - \lambda \end{pmatrix} \] \[ \det(A - \lambda I) = (-2 - \lambda)(-3 - \lambda) - (-2)(-1) = 0 \] \[ = (\lambda + 2)(\lambda + 3) - 2 = 0 \] \[ = \lambda^2 + 5\lambda + 6 - 2 = 0 \] \[ = \lambda^2 + 5\lambda + 4 = 0 \] Solve the quadratic: \[ \lambda = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2} \] So, \[ \lambda_1 = \frac{-5 + 3}{2} = -1 \] \[ \lambda_2 = \frac{-5 - 3}{2} = -4 \] **Explanation:** We use the determinant of \(A - \lambda I\) to set up a characteristic polynomial, which we solve using the quadratic formula. --- ## **Final Answers:** 1. **The associated coefficient matrix is:** \[ A = \begin{pmatrix} -2 & -2 \\ -1 & -3 \end{pmatrix} \] 2. **The eigenvalues of \(A\) are:** \[ \boxed{-1 \text{ and } -4} \]

Let's proceed step-by-step to fully solve the problem, covering all requested parts for the given system: \[ \begin{cases} x'(t) = -2x - 2y \\ y'(t) = -x - 3y \end{cases} \] --- ### **Part (i): Find the coefficient matrix \(A\).** **Work:** The system can be written in matrix form as: \[ \mathbf{X}'(t) = A \mathbf{X}(t) \] where \[ \mathbf{X}(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} \] Matching coefficients from the system: \[ A = \begin{bmatrix} -2 & -2 \\ -1 & -3 \end{bmatrix} \] **Answer:** \[ \boxed{ A = \begin{bmatrix} -2 & -2 \\ -1 & -3 \end{bmatrix} } \] --- ### **Part (ii): Compute the eigenvalues of \(A\).** **Work:** Eigenvalues \(\lambda\) satisfy: \[ \det(A - \lambda I) = 0 \] \[ A - \lambda I = \begin{bmatrix} -2 - \lambda & -2 \\ -1 & -3 - \lambda \end{bmatrix} \] Determinant: \[ (-2 - \lambda)(-3 - \lambda) - (-2)(-1) = 0 \] Calculate: \[ (\lambda + 2)(\lambda + 3) - 2 = 0 \] \[ (\lambda^2 + 5\lambda + 6) - 2 = 0 \] \[ \lambda^2 + 5\lambda + 4 = 0 \] Quadratic formula: \[ \lambda = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm \sqrt{9}}{2} = \frac{-5 \pm 3}{2} \] Eigenvalues: \[ \lambda_1 = \frac{-5 + 3}{2} = -1 \] \[ \lambda_2 = \frac{-5 - 3}{2} = -4 \] **Answer:** \[ \boxed{ \text{Eigenvalues: } \lambda_1 = -1, \quad \lambda_2 = -4 } \] --- ### **Part (iii): Find the eigenvectors corresponding to each eigenvalue.** **Work for \(\lambda = -1\):** Solve: \[ (A + I) \mathbf{v} = 0 \] \[ A + I = \begin{bmatrix} -2 + 1 & -2 \\ -1 & -3 + 1 \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ -1 & -2 \end{bmatrix} \] Set up: \[ \begin{bmatrix} -1 & -2 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} \] From the first row: \[ -1 v_1 - 2 v_2 = 0 \Rightarrow v_1 = -2 v_2 \] Let \(v_2 = 1\), then \(v_1 = -2\). Eigenvector for \(\lambda = -1\): \[ \boxed{ \mathbf{v}^{(1)} = \begin{bmatrix} -2 \\ 1 \end{bmatrix} } \] --- **Work for \(\lambda = -4\):** Solve: \[ (A + 4I) \mathbf{v} = 0 \] \[ A + 4I = \begin{bmatrix} -2 + 4 & -2 \\ -1 & -3 + 4 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ -1 & 1 \end{bmatrix} \] From the first row: \[ 2 v_1 - 2 v_2 = 0 \Rightarrow v_1 = v_2 \] Choose \(v_2 = 1\), then \(v_1 = 1\). Eigenvector for \(\lambda = -4\): \[ \boxed{ \mathbf{v}^{(2)} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} } \] --- ### **Part (iv): Find the associated real-valued solutions to this example with initial conditions \(\#(0) = 1\) and \(\#(0) = 2\).** **(Note:** The initial conditions seem to refer to initial values of the solution components; assuming \(\mathbf{X}(0) = \begin{bmatrix} x(0) \\ y(0) \end{bmatrix}\).) --- **General solution:** \[ \mathbf{X}(t) = c_1 e^{\lambda_1 t} \mathbf{v}^{(1)} + c_2 e^{\lambda_2 t} \mathbf{v}^{(2)} \] \[ = c_1 e^{-1 t} \begin{bmatrix} -2 \\ 1 \end{bmatrix} + c_2 e^{-4 t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] --- **Case 1: \(\mathbf{X}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\)** Set \(t=0\): \[ \begin{bmatrix} 1 \\ 2 \end{bmatrix} = c_1 \begin{bmatrix} -2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] Leading to the system: \[ -2 c_1 + c_2 = 1 \quad (1) \] \[ c_1 + c_2 = 2 \quad (2) \] Solve: From (2): \(c_2 = 2 - c_1\) Substitute into (1): \[ -2 c_1 + (2 - c_1) = 1 \Rightarrow -2 c_1 + 2 - c_1 = 1 \] \[ -3 c_1 + 2 = 1 \Rightarrow -3 c_1 = -1 \Rightarrow c_1 = \frac{1}{3} \] Then, \[ c_2 = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \] **Solution:** \[ \boxed{ \mathbf{X}(t) = \frac{1}{3} e^{-t} \begin{bmatrix} -2 \\ 1 \end{bmatrix} + \frac{5}{3} e^{-4 t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} } \] --- **Case 2: \(\mathbf{X}(0) = \begin{bmatrix} 0 \\ 0 \}\)** (If initial condition is zero, the solution is trivial: \(\mathbf{X}(t) = \mathbf{0}\).) --- ### **Part (v): Create two solution graphs** - **Graph 1:** Plot \(y(t)\) vs. \(x(t)\) (phase plane). - **Graph 2:** Plot \(x(t)\) and \(y(t)\) separately versus \(t\). --- ### **Part (vi): Graph the phase portrait** The phase portrait involves plotting the trajectories of the system in the \(xy\)-plane. The eigenvalues are both negative, indicating stable nodes. - Eigenvalues: \(-1, -4\) - Eigenvectors: \(\begin{bmatrix} -2 \\ 1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\) Trajectories spiral into the origin along the directions of these eigenvectors. --- ### **Part (vii): Highlight the specific solution trajectory** From initial condition \(\mathbf{X}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\), the trajectory can be traced on the phase portrait, starting at \((1, 2)\) and approaching \((0,0)\). --- ### **Part (viii): Classify the equilibrium point at the origin** Since both eigenvalues are real and negative, the equilibrium at the origin is a **stable node**. --- ## **Summary:** | Part | Description | Result | |---------|----------------|---------| | (i) | Coefficient matrix \(A\) | \(\begin{bmatrix} -2 & -2 \\ -1 & -3 \end{bmatrix}\) | | (ii) | Eigenvalues | \(\lambda = -1,\ -4\) | | (iii) | Eigenvectors | \(\mathbf{v}^{(1)} = [-2, 1]\), \(\mathbf{v}^{(2)} = [1, 1]\) | | (iv) | General solution with initial \(\mathbf{X}(0) = [1, 2]\) | \(\frac{1}{3} e^{-t} [-2, 1] + \frac{5}{3} e^{-4 t} [1, 1]\) | | (v) | Graphs and (vi) | (Graphical tasks, typically plotted using software) | | (vii) | Solution trajectory highlighted | Corresponds to initial condition \([1, 2]\) | | (viii) | Equilibrium classification | **Stable node** | --- **Final note:** For actual graphs, use graphing software or hand sketches based on eigenvectors and eigenvalues.