Let's address **Exercise 4.6 (2)** step by step, as you requested: --- ### **Step 1: Restate Theorem 4.5 for the Given Matrix** **Theorem 4.5** states that a matrix function \(\Phi(t)\) is a matrix of solutions to \(\dot{X} = AX\) if and only if: - (a) \(\Phi(t)\) is a solution of \(\dot{X} = AX\), i.e., \(\dot{\Phi}(t) = A\Phi(t)\) - (b) \(\det(\Phi(t_0)) \neq 0\) for some \( t_0 \) (i.e., \(\Phi(t_0)\) is invertible) For the given system, the matrix \(A\) is: \[ A = \begin{bmatrix} 0 & 1 \\ -1 & -2 \end{bmatrix} \] **Explanation:** We need to find a fundamental matrix solution \(\Phi(t)\) for this \(A\) and check both conditions (a) and (b) of the theorem. --- ### **Step 2: Find the Fundamental Matrix Solution and Verify the Conditions** #### **2.1: Find the General Solution** Let’s find the general solution to \(\dot{X} = AX\): **Find eigenvalues:** \[ \det(A - \lambda I) = \begin{vmatrix} -\lambda & 1 \\ -1 & -2-\lambda \end{vmatrix} = (-\lambda)(-2-\lambda) - (-1)(1) = \lambda(\lambda + 2) + 1 = \lambda^2 + 2\lambda + 1 = (\lambda + 1)^2 \] So, \(\lambda = -1\) (double root). **Find eigenvector for \(\lambda = -1\):** \[ (A + I)v = 0 \implies \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \Rightarrow v_1 + v_2 = 0 \implies v = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] **Find generalized eigenvector \(w\):** \[ (A + I)w = v \implies \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] \[ w_1 + w_2 = 1 \implies w_2 = 1 - w_1 \] \[ -w_1 - w_2 = -1 \implies w_1 + w_2 = 1 \] So let’s choose \(w_1 = 0, w_2 = 1\), so \(w = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\). **General solution:** \[ X(t) = c_1 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{-t} \left[ t \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right] \] So a fundamental matrix is: \[ \Phi(t) = e^{-t} \begin{bmatrix} 1 & t \\ -1 & -t + 1 \end{bmatrix} \] **Explanation:** We constructed a fundamental matrix solution \(\Phi(t)\) for the given \(A\). #### **2.2: Verify Theorem 4.5 Conditions** - **Condition (a):** Compute \(\dot{\Phi}(t)\) and check if \(\dot{\Phi}(t) = A\Phi(t)\). - **Condition (b):** \(\det(\Phi(0)) = \det \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} = (1)(1) - (0)(-1) = 1 \neq 0\). Thus, both conditions of Theorem 4.5 are satisfied for this fundamental matrix solution. --- ## **Final Answer** **We have verified Theorem 4.5 for the system with** \[ A = \begin{bmatrix} 0 & 1 \\ -1 & -2 \end{bmatrix} \] **by constructing a fundamental matrix solution and checking both required conditions.**