To classify the critical point $(-1, 1)$ of the system of differential equations, we need to linearize the system at the critical point and analyze the eigenvalues of the resulting Jacobian matrix. Here are the steps to follow:

1. Write down the system of differential equations: \frac{dx}{dt} = x - 2y + 3 \\ \frac{dy}{dt} = x - y + 2 \end{cases}$$
2. Find the critical points by setting $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$: x - 2y + 3 = 0 \\ x - y + 2 = 0 \end{cases}$$ Solving this system: $$x - y + 2 = 0 \implies x = y - 2$$ Substitute $$x = y - 2$$ into $$x - 2y + 3 = 0$$: $$(y - 2) - 2y + 3 = 0 \implies -y + 1 = 0 \implies y = 1$$ Thus, $$x = y - 2 = 1 - 2 = -1$$ The critical point is $$(-1, 1)$$.
3. Linearize the system at the critical point $(-1, 1)$. Compute the Jacobian matrix: \frac{\partial (dx/dt)}{\partial x} & \frac{\partial (dx/dt)}{\partial y} \\ \frac{\partial (dy/dt)}{\partial x} & \frac{\partial (dy/dt)}{\partial y} \end{pmatrix}$$ Evaluate each partial derivative: $$\frac{\partial (dx/dt)}{\partial x} = 1, \quad \frac{\partial (dx/dt)}{\partial y} = -2$$ $$\frac{\partial (dy/dt)}{\partial x} = 1, \quad \frac{\partial (dy/dt)}{\partial y} = -1$$ The Jacobian matrix $$J$$ at $$(-1, 1)$$ is: $$J = \begin{pmatrix} 1 & -2 \\ 1 & -1 \end{pmatrix}$$
4. Find the eigenvalues of the Jacobian matrix by solving the characteristic equation $\det(J - \lambda I) = 0$: 1 - \lambda & -2 \\ 1 & -1 - \lambda \end{pmatrix} = 0$$ $$(1 - \lambda)(-1 - \lambda) - (-2)(1) = 0$$ $$\lambda^2 - \lambda - 1 + 2 = 0$$ $$\lambda^2 - \lambda + 1 = 0$$ Solve this quadratic equation using the quadratic formula: $$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a = 1$$, $$b = -1$$, and $$c = 1$$: $$\lambda = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$$ $$\lambda = \frac{1 \pm \sqrt{1 - 4}}{2}$$ $$\lambda = \frac{1 \pm \sqrt{-3}}{2}$$ $$\lambda = \frac{1 \pm i\sqrt{3}}{2}$$
5. Classify the critical point based on the eigenvalues. Since the eigenvalues are complex conjugates with a positive real part $\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\right)$, the critical point $(-1, 1)$ is classified as a spiral. Specifically, because the real part is positive, it is an unstable spiral.

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Here are 8 related questions that might help you expand your understanding:

1. How do the real parts of eigenvalues affect the stability of a critical point?
2. What is the significance of complex eigenvalues in the phase portrait of a system?
3. How can you determine if a spiral is stable or unstable?
4. What are the other types of critical points and their characteristics?
5. How do you draw the phase portrait for a system with a spiral critical point?
6. How can you linearize a non-linear system of differential equations?
7. What is the relationship between the trace and determinant of the Jacobian and the type of critical point?
8. How does the presence of a non-zero constant vector $\mathbf{f}$ affect the classification of critical points?

Tip: When dealing with systems of differential equations, always check the Jacobian matrix at the critical points. The eigenvalues of the Jacobian provide valuable insights into the nature and stability of the critical points.