Let's solve the initial value problem step by step: --- ## **Step 1: Write the System and Find Eigenvalues/Eigenvectors** The system is: \[ \mathbf{x}'(t) = \begin{bmatrix} & -1 & -1 \\ -1 & & -1 \\ -1 & -1 & \\ \end{bmatrix} \mathbf{x}(t), \quad \mathbf{x}() = \begin{bmatrix} 3 \\ 9 \\ 3 \end{bmatrix} \] Let's find the eigenvalues of the matrix \(A\): Set \(\det(A - \lambda I) = \): \[ \begin{vmatrix} -\lambda & -1 & -1 \\ -1 & -\lambda & -1 \\ -1 & -1 & -\lambda \\ \end{vmatrix} = \] Expanding the determinant, we get: \[ -\lambda \left| \begin{matrix} -\lambda & -1 \\ -1 & -\lambda \end{matrix} \right| - (-1)\left| \begin{matrix} -1 & -1 \\ -1 & -\lambda \end{matrix} \right| + (-1)\left| \begin{matrix} -1 & -\lambda \\ -1 & -1 \end{matrix} \right| \] Calculate: - First minor: \((-\lambda)(-\lambda) - (-1)(-1) = \lambda^2 - 1\) - Second minor: \((-1)(-\lambda) - (-1)(-1) = \lambda - 1\) - Third minor: \((-1)(-1) - (-1)(-\lambda) = 1 - \lambda\) So: \[ -\lambda (\lambda^2 - 1) + (1)(\lambda - 1) + (-1)(1 - \lambda) = \] \[ -\lambda^3 + \lambda + \lambda - 1 - 1 + \lambda = \] \[ -\lambda^3 + 3\lambda - 2 = \] Multiply both sides by -1: \[ \lambda^3 - 3\lambda + 2 = \] This factors as: \[ (\lambda - 1)^2 (\lambda + 2) = \] So, the eigenvalues are \(\lambda = 1\) (multiplicity 2), and \(\lambda = -2\). **Explanation:** This step finds the eigenvalues, which are necessary for writing the general solution for the system. --- ## **Step 2: Find the General Solution and Apply Initial Condition** The general solution is a linear combination of eigenvector solutions: \[ \mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + c_3 \mathbf{v}_3 e^{\lambda_3 t} \] #### For \(\lambda = 1\): Solve \((A - I)\mathbf{v} = \): \[ A - I = \begin{bmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \\ \end{bmatrix} \] Any vector with components summing to zero works. Try \(\mathbf{v}_1 = [1, -1, ]^T\) and \(\mathbf{v}_2 = [1, , -1]^T\). #### For \(\lambda = -2\): Solve \((A + 2I)\mathbf{v} = \): \[ A + 2I = \begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \\ \end{bmatrix} \] Set up the equations: \[ 2x - y - z = \\ -x + 2y - z = \\ -x - y + 2z = \\ \] This system is symmetric; setting \(x = y = z\) works. Try \([1, 1, 1]^T\). So the general solution: \[ \mathbf{x}(t) = c_1 \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} e^{t} + c_2 \begin{bmatrix} 1 \\ \\ -1 \end{bmatrix} e^{t} + c_3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} e^{-2t} \] Plug in \(t = \) and \(\mathbf{x}() = [3, 9, 3]^T\): \[ c_1 \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ \\ -1 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 9 \\ 3 \end{bmatrix} \] Add the vectors: \[ \begin{bmatrix} c_1 + c_2 + c_3 \\ - c_1 + c_3 \\ - c_2 + c_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 9 \\ 3 \end{bmatrix} \] Set up the equations: 1. \(c_1 + c_2 + c_3 = 3\) 2. \(-c_1 + c_3 = 9\) 3. \(-c_2 + c_3 = 3\) From (2): \(c_3 = 9 + c_1\) From (3): \(c_3 = 3 + c_2\) So: \(9 + c_1 = 3 + c_2 \implies c_2 = c_1 + 6\) Plug into (1): \(c_1 + (c_1 + 6) + (9 + c_1) = 3\) \(c_1 + c_1 + 6 + 9 + c_1 = 3\) \(3c_1 + 15 = 3\) \(3c_1 = -12 \implies c_1 = -4\) Then \(c_2 = c_1 + 6 = -4 + 6 = 2\) And \(c_3 = 9 + c_1 = 9 + (-4) = 5\) So: \[ \mathbf{x}(t) = -4 \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} e^{t} + 2 \begin{bmatrix} 1 \\ \\ -1 \end{bmatrix} e^{t} + 5 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} e^{-2t} \] Or, combine the two \(e^t\) terms: \[ -4 \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} + 2 \begin{bmatrix} 1 \\ \\ -1 \end{bmatrix} = \begin{bmatrix} -4 + 2 \\ 4 + \\ - 2 \end{bmatrix} = \begin{bmatrix} -2 \\ 4 \\ -2 \end{bmatrix} \] So the final solution: \[ \boxed{ \mathbf{x}(t) = \begin{bmatrix} -2 \\ 4 \\ -2 \end{bmatrix} e^{t} + 5 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} e^{-2t} } \] --- **Explanation:** This step found the eigenvectors, built the general solution, and solved for the constants using the initial value. The final answer expresses \(\mathbf{x}(t)\) in terms of exponentials and vectors. --- ## **Final Answer** \[ \boxed{ \mathbf{x}(t) = \begin{bmatrix} -2 \\ 4 \\ -2 \end{bmatrix} e^{t} + 5 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} e^{-2t} } \]