a)
First we find the eigenvalues
$$det\begin{bmatrix}4-r & -3\\8 & -6-r\end{bmatrix} = (r-4)(r+6)-24 = (r^2+2r) = r(r+2)$$
$$r_1 = 0, r_2 = -2$$
The associated eigenvector for $r_1$ is:
$$r_1=0: Null\begin{bmatrix}4 & -3\\8 & -6\end{bmatrix} = Null\begin{bmatrix}4 & -3\\0 & 0\end{bmatrix} \implies 4\xi_1 = 3\xi_2 \implies \xi^{(1)} = \begin{bmatrix}3\\4\end{bmatrix} $$
The associated eigenvector for $r_2$ is:
$$r_2=-2: Null\begin{bmatrix}6 & -3\\8 & -4\end{bmatrix} = Null\begin{bmatrix}2 & -1\\0 & 0\end{bmatrix} \implies 2\xi_1 = \xi_2 \implies \xi^{(2)} = \begin{bmatrix}1\\2\end{bmatrix} $$
This gives us our general solution:
$$X(t) = c_1\begin{bmatrix}3\\4\end{bmatrix} + c_2e^{-2t}\begin{bmatrix}1\\2\end{bmatrix}$$
b)
The plot follows the same idea as 1. e) in this handout: (
www.math.toronto.edu/courses/mat244h1/20181/LN/Ch7-LN9.pdf)
The plot approaches the vector $\begin{bmatrix}3\\4\end{bmatrix}$ as t approaches infinity.
