Let's solve the linear system step by step. Given: \[ \mathbf{x}' = \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix} \mathbf{x} \] Let \(\mathbf{x}(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}\). ### Step 1: Find the eigenvalues of the matrix Matrix: \[ A = \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix} \] The characteristic equation: \[ \det(A - \lambda I) = 0 \] \[ \begin{vmatrix} 0 - \lambda & 1 \\ -4 & 4 - \lambda \end{vmatrix} = 0 \] \[ (0-\lambda)(4-\lambda) - (-4)(1) = 0 \] \[ (-\lambda)(4-\lambda) + 4 = 0 \] \[ -\lambda(4-\lambda) + 4 = 0 \] \[ -\lambda(4-\lambda) + 4 = 0 \] \[ -\lambda \cdot 4 + \lambda^2 + 4 = 0 \] \[ -\lambda^2 + 4\lambda + 4 = 0 \] \[ \lambda^2 - 4\lambda - 4 = 0 \] ### Step 2: Solve the quadratic equation \[ \lambda^2 - 4\lambda - 4 = 0 \] Using the quadratic formula: \[ \lambda = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-4)}}{2} \] \[ = \frac{4 \pm \sqrt{16 + 16}}{2} \] \[ = \frac{4 \pm \sqrt{32}}{2} \] \[ = \frac{4 \pm 4\sqrt{2}}{2} \] \[ = 2 \pm 2\sqrt{2} \] ### Step 3: Find the eigenvectors Let's use \(\lambda = 2 + 2\sqrt{2}\): \[ A - \lambda I = \begin{bmatrix} -\lambda & 1 \\ -4 & 4-\lambda \end{bmatrix} \] \[ = \begin{bmatrix} -(2+2\sqrt{2}) & 1 \\ -4 & 4-(2+2\sqrt{2}) \end{bmatrix} \] \[ = \begin{bmatrix} -2-2\sqrt{2} & 1 \\ -4 & 2-2\sqrt{2} \end{bmatrix} \] Set up the eigenvector equation: \[ \begin{bmatrix} -2-2\sqrt{2} & 1 \\ -4 & 2-2\sqrt{2} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] From the first row: \[ (-2-2\sqrt{2})v_1 + v_2 = 0 \implies v_2 = (2+2\sqrt{2})v_1 \] Similarly, for \(\lambda = 2 - 2\sqrt{2}\): \[ v_2 = (2-2\sqrt{2})v_1 \] ### Step 4: Write the general solution \[ \mathbf{x}(t) = C_1 \mathbf{v}_1 e^{\lambda_1 t} + C_2 \mathbf{v}_2 e^{\lambda_2 t} \] Where: \[ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 2+2\sqrt{2} \end{bmatrix}, \lambda_1 = 2+2\sqrt{2} \] \[ \mathbf{v}_2 = \begin{bmatrix} 1 \\ 2-2\sqrt{2} \end{bmatrix}, \lambda_2 = 2-2\sqrt{2} \] But none of the options have these forms. Let's try another approach—**use the options hinting repeated eigenvalues**, or maybe the matrix is defective. But our eigenvalues are **not equal** or repeated. Let's check the options for \(e^{2t}\) and polynomial terms, suggesting a repeated root. ### Step 5: Check for repeated root solution form (if matrix is not diagonalizable) If we get a repeated eigenvalue, the solution is: \[ \mathbf{x}(t) = (C_1 \mathbf{v} + C_2 (\mathbf{v} t + \mathbf{w})) e^{\lambda t} \] But our eigenvalues are distinct. Let's **double-check the calculation**: \[ \lambda^2 - 4\lambda + 4 = 0 \] But our quadratic equation was: \[ \lambda^2 - 4\lambda - 4 = 0 \] But the original matrix: \[ A = \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix} \] So the trace is 4, determinant is 0 - (-4) = 4, but let's stick to the solution above. ### Step 6: Compare with options All options have \(e^{2t}\) or \(e^{t}\) or \(e^{2t}\), but our eigenvalues are \(2 \pm 2\sqrt{2}\). But if we factor \(\lambda^2 - 4\lambda + 4 = 0\), the roots are \(\lambda = 2\) (repeated): \[ \lambda^2 - 4\lambda + 4 = 0 \implies (\lambda - 2)^2 = 0 \] \[ \lambda = 2, 2 \] So the correct characteristic polynomial is \(\lambda^2 - 4\lambda + 4\), not \(-4\). Let's recompute the determinant: \[ \begin{vmatrix} -\lambda & 1 \\ -4 & 4-\lambda \end{vmatrix} = (-\lambda)(4-\lambda) - (-4)(1) = -4\lambda + \lambda^2 + 4 = \lambda^2 - 4\lambda + 4 \] So, eigenvalue \(\lambda = 2\) (repeated). ### Step 7: Find the eigenvector for \(\lambda = 2\): \[ A - 2I = \begin{bmatrix} -2 & 1 \\ -4 & 2 \end{bmatrix} \] Solve: \[ \begin{bmatrix} -2 & 1 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \] From the first row: \[ -2v_1 + v_2 = 0 \implies v_2 = 2v_1 \] So a basic eigenvector is \(\mathbf{v} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\). Now, for the generalized eigenvector \(\mathbf{w}\): \[ (A - 2I)\mathbf{w} = \mathbf{v} \] \[ \begin{bmatrix} -2 & 1 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \] First row: \[ -2w_1 + w_2 = 1 \implies w_2 = 1 + 2w_1 \] Second row: \[ -4w_1 + 2w_2 = 2 \] Substitute \(w_2\): \[ -4w_1 + 2(1 + 2w_1) = 2 \] \[ -4w_1 + 2 + 4w_1 = 2 \] \[ 2 = 2 \] So, \(w_1\) is arbitrary, set \(w_1 = 0\), then \(w_2 = 1\). So, generalized eigenvector: \(\mathbf{w} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\). ### Step 8: Write the general solution for repeated eigenvalue \[ \mathbf{x}(t) = \left( C_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + C_2 \left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} + t \begin{bmatrix} 1 \\ 2 \end{bmatrix} \right) \right) e^{2t} \] \[ = \left( C_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + C_2 \begin{bmatrix} t \\ 1 + 2t \end{bmatrix} \right) e^{2t} \] \[ = \left( \begin{bmatrix} C_1 \\ 2C_1 \end{bmatrix} + \begin{bmatrix} C_2 t \\ C_2 (1 + 2t) \end{bmatrix} \right) e^{2t} \] \[ = \begin{bmatrix} C_1 + C_2 t \\ 2C_1 + C_2 (1 + 2t) \end{bmatrix} e^{2t} \] ### Step 9: Match with the options The matching answer is: \[ \boxed{ \mathbf{x}(t) = \begin{bmatrix} C_1 + C_2 t \\ 2C_1 + C_2 (2t + 1) \end{bmatrix} e^{2t} } \] This matches with the **first option**. --- **Final Answer:** \[ \boxed{ \mathbf{x}(t) = \begin{bmatrix} C_1 + C_2 t \\ 2C_1 + C_2 (2t + 1) \end{bmatrix} e^{2t} } \] This is the general solution of the given linear system.