1. **Problem Statement:** Find the fundamental matrix $\Phi(t)$ for the adjoint system of the linear system $\dot{x} = Ax$, where $$A = \begin{bmatrix} 1 & 3 & 8 \\ -2 & 2 & 1 \\ -3 & 0 & 5 \end{bmatrix}$$ 2. **Recall:** The adjoint system corresponds to $$\dot{y} = -A^T y$$ where $A^T$ is the transpose of $A$. 3. **Step 1: Compute $A^T$** $$A^T = \begin{bmatrix} 1 & -2 & -3 \\ 3 & 2 & 0 \\ 8 & 1 & 5 \end{bmatrix}$$ 4. **Step 2: Form the matrix for the adjoint system:** $$-A^T = \begin{bmatrix} -1 & 2 & 3 \\ -3 & -2 & 0 \\ -8 & -1 & -5 \end{bmatrix}$$ 5. **Step 3: Find the fundamental matrix $\Phi(t)$ for $\dot{y} = -A^T y$** The fundamental matrix is given by the matrix exponential: $$\Phi(t) = e^{-A^T t}$$ 6. **Step 4: Explanation** - The matrix exponential $e^{Mt}$ for a matrix $M$ is defined by the power series: $$e^{Mt} = I + Mt + \frac{(Mt)^2}{2!} + \frac{(Mt)^3}{3!} + \cdots$$ - Here, $M = -A^T$. - Computing $e^{-A^T t}$ exactly involves diagonalization or Jordan form of $-A^T$. 7. **Step 5: Summary** - The fundamental matrix for the adjoint system is: $$\boxed{\Phi(t) = e^{-A^T t}}$$ - Where $A^T$ is the transpose of $A$. This matrix $\Phi(t)$ satisfies $\dot{\Phi}(t) = -A^T \Phi(t)$ and $\Phi(0) = I$. **Note:** Explicit closed form requires eigen-decomposition of $-A^T$, which is beyond the scope here but can be computed numerically or symbolically if needed.