Given the system of equations $\mathbf{x}' = A \mathbf{x}$ and a known solution $\mathbf{x}_1 = e^t \begin{pmatrix} \cos(4t) - 3\sin(4t) \\ 2\cos(4t) + 2\sin(4t) \end{pmatrix}$, we need to find the general solution $\mathbf{x}(t)$ that satisfies the initial condition $\mathbf{x}(0) = \begin{pmatrix} 7 \\ -2 \end{pmatrix}$.

Step 1: Recognize the form of the general solution

For a system $\mathbf{x}' = A \mathbf{x}$, the general solution typically consists of linear combinations of particular solutions. Since $\mathbf{x}_1(t)$ is one such solution, the general solution can be written as: $\mathbf{x}(t) = c_1 \mathbf{x}_1(t)$ where $c_1$ is a constant to be determined from the initial condition.

Step 2: Evaluate the solution at $t = 0$

We substitute $t = 0$ into $\mathbf{x}_1(t)$ to find $\mathbf{x}_1(0)$: $\mathbf{x}_1(0) = e^0 \begin{pmatrix} \cos(0) - 3\sin(0) \\ 2\cos(0) + 2\sin(0) \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ Thus, at $t = 0$, the solution is: $\mathbf{x}(0) = c_1 \mathbf{x}_1(0) = c_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix}$

Step 3: Apply the initial condition

We are given that $\mathbf{x}(0) = \begin{pmatrix} 7 \\ -2 \end{pmatrix}$. Using this condition, we can solve for $c_1$: $c_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ -2 \end{pmatrix}$ This gives the system of equations: $c_1 = 7$ $2c_1 = -2$ However, the system is inconsistent, meaning there might be a mistake in the setup or the expression for $\mathbf{x}_1(t)$. Could you confirm the exact structure of $\mathbf{x}_1(t)$ or provide any additional details?