Solve the system \[ \frac{d}{d t} \vec{x}=A \vec{x} \text { where } A=\left[\begin{array}{ll} 8 & -4 \\ 9 & -4 \end{array}\right] \]

Step 1 ：The given system is a first order linear homogeneous system of differential equations. The general solution of such a system is given by \(\vec{x}(t) = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2\), where \(\lambda_1, \lambda_2\) are the eigenvalues of the matrix A and \(\vec{v}_1, \vec{v}_2\) are the corresponding eigenvectors.

Step 2 ：First, we find the eigenvalues of the matrix A. The eigenvalues are the roots of the characteristic equation, which is given by \(\text{det}(A - \lambda I) = 0\), where I is the identity matrix.

Step 3 ：The matrix \(A - \lambda I\) is given by \[A - \lambda I = \begin{bmatrix} 8 - \lambda & -4 \\ 9 & -4 - \lambda \end{bmatrix}\]

Step 4 ：The determinant of \(A - \lambda I\) is given by \((8 - \lambda)(-4 - \lambda) - (-4)(9) = \lambda^2 - 4\lambda - 20\).

Step 5 ：Setting this equal to zero gives the characteristic equation \(\lambda^2 - 4\lambda - 20 = 0\).

Step 6 ：Solving this quadratic equation for \(\lambda\) gives the roots \(\lambda_1 = 2 - 2\sqrt{6}\) and \(\lambda_2 = 2 + 2\sqrt{6}\). These are the eigenvalues of the matrix A.

Step 7 ：Next, we find the eigenvectors corresponding to these eigenvalues. For each eigenvalue \(\lambda\), the corresponding eigenvector \(\vec{v}\) is a solution to the equation \(A\vec{v} = \lambda\vec{v}\).

Step 8 ：For \(\lambda_1 = 2 - 2\sqrt{6}\), we solve the system of equations \[(8 - (2 - 2\sqrt{6}))v_1 - 4v_2 = 0\] and \[9v_1 - (4 + (2 - 2\sqrt{6}))v_2 = 0\]. This gives the eigenvector \(\vec{v}_1 = \begin{bmatrix} 2 \sqrt{6} - 2 \\ 3 \end{bmatrix}\).

Step 9 ：Similarly, for \(\lambda_2 = 2 + 2\sqrt{6}\), we solve the system of equations \[(8 - (2 + 2\sqrt{6}))v_1 - 4v_2 = 0\] and \[9v_1 - (4 + (2 + 2\sqrt{6}))v_2 = 0\]. This gives the eigenvector \(\vec{v}_2 = \begin{bmatrix} 2 \sqrt{6} + 2 \\ 3 \end{bmatrix}\).

Step 10 ：Therefore, the general solution of the system is given by \[\vec{x}(t) = c_1 e^{(2 - 2\sqrt{6}) t} \begin{bmatrix} 2 \sqrt{6} - 2 \\ 3 \end{bmatrix} + c_2 e^{(2 + 2\sqrt{6}) t} \begin{bmatrix} 2 \sqrt{6} + 2 \\ 3 \end{bmatrix}\]

Step 11 ：\boxed{\vec{x}(t) = c_1 e^{(2 - 2\sqrt{6}) t} \begin{bmatrix} 2 \sqrt{6} - 2 \\ 3 \end{bmatrix} + c_2 e^{(2 + 2\sqrt{6}) t} \begin{bmatrix} 2 \sqrt{6} + 2 \\ 3 \end{bmatrix}} is the final answer.