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

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 $\overrightarrow{x}(t)=c_1{e}^{{\lambda }_{1}t}{\overrightarrow{v}}_1+c_2{e}^{{\lambda }_{2}t}{\overrightarrow{v}}_2$, where ${\lambda }_1,{\lambda }_2$ are the eigenvalues of the matrix A and ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2$ 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 $$A-\lambda I=\left[\begin{array}{cc}8-\lambda & -4\\ 9& -4-\lambda \end{array}\right]$$

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 $\overrightarrow{v}$ is a solution to the equation $A\overrightarrow{v}=\lambda \overrightarrow{v}$.

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 ${\overrightarrow{v}}_1=\left[\begin{array}{c}2\sqrt{6}-2\\ 3\end{array}\right]$.

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 ${\overrightarrow{v}}_2=\left[\begin{array}{c}2\sqrt{6}+2\\ 3\end{array}\right]$.

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

Step 11 ：\boxed{\vec{x}(t) = c_1 e^{(2 - 2\sqrt{6}) t} $$\left[\begin{array}{c}2\sqrt{6}-2\\ 3\end{array}\right]$$ + c_2 e^{(2 + 2\sqrt{6}) t} $$\left[\begin{array}{c}2\sqrt{6}+2\\ 3\end{array}\right]$$} is the final answer.