I began this problem by evaulating $x' = Ax$. Let $$ x' = \begin{bmatrix}{x_1} \\ {x_2}\end{bmatrix}.$$ Then we have $$x'_1=-3x_1 + 2x_2, $$ $$x'_2=-x_1 - x_2, $$ $$x_1(0)=1,$$ $$x_2(0)=-2.$$ From the first equation we have $x_2= \frac{1}{2}x'_1+ \frac{3}{2}x_1$. Substituting this into the second equation yields: $$x''_1+4x'_1-2x_1=0,$$ The characteristic equation of this is $m^2+4m-2=0$, which has the solutions $m = -2 \pm \sqrt{6}$.
However, I am confused how to proceed to finding the final solution. Any guidance is greatly appreciated!
We have
$$\begin{align}x' &= -3x + 2 y \\ y' &= - x- y \end{align}$$
Taking the derivative of the first equation and substituting to isolate $x$
$$x'' = -3 x' + 2 y' = -3 x'+2(-x-y) = -3x' + 2(-x -\dfrac{1}{2}(x' + 3 x)) = - 4 x' - 5x$$
We now have
$$x'' + 4 x' + 5 x = 0, x(0) = 1, x'(0) = -7$$
This leads to
$$x(t) = e^{-2t}(\cos(t) - 5 \sin(t))$$
From the first equation, we know
$$y(t) = \dfrac{1}{2}(x' + 3 x) = -e^{-2 t} (3 \sin (t)+2 \cos (t))$$
We can also solve this using eigenvalues and eigenvectors and many other methods (undetermined coefficients, Laplace transforms, matrix exponential, diagonalization...).