Consider the dynamical system $$\dot{r}=r(3-2r-s)$$ $$\dot{s}=s(2-r-s)$$
Find and classify the equilibrium points of the system.
I have found the equilibrium points to be $(0,0), (0,2), (\frac{3}{2},0)$ and $(1,1)$ simply by equating the given equations to $0$ and solving.
What we have been taught in my lectures is how to classify equilibrium points of a dynamical system in the form
$$\dot{x}=ax+by$$ $$\dot{y}=cx+dy$$
by writing the system in matrix form and using the method of finding the eigenvalues to determine the nature of the points (this is quite straightforward but let me know if I should expand on it).
How would I use this method to classify the points in the dynamical system in question?
We find the Jacobian of the system as
$$J(r, s) = \begin{bmatrix} \dfrac{\partial r'}{\partial r} & \dfrac{\partial r'}{\partial s} \\ \dfrac{\partial s'}{\partial r} & \dfrac{\partial s'}{\partial s} \end{bmatrix} = \begin{bmatrix} -4 r-s+3 & -r \\ -s & -r-2 s+2 \\\end{bmatrix}$$
Your critical points are correct. Next, we evaluate the eigenvalues of each critical point and have
$$J(0,0) \implies \lambda_{1,2} = 2, 3 \\ J(0,2) \implies \lambda_{1,2} = -2,1 \\ J(3/2,2) \implies \lambda_{1,2} = -3,\dfrac{1}{2} \\ J(1,1) \implies \lambda_{1,2} = \dfrac{1}{2} \left(-\sqrt{5}-3\right),\dfrac{1}{2} \left(\sqrt{5}-3\right)$$
It is worth mentioning to be careful for marginal points when using linearization. Marginal cases are those that have at least one eigenvalue with zero real part. These are centers, degenerate nodes, and non-isolated fixed points. The smallest perturbation will ”knock” them into a different type of behavior. We do not have any of these in this particular problem.
Next, you can classify those knowing the approach you mention.
We can compare that analysis to the phase portrait for the nonlinear system