1. **State the problem:** Solve the system of differential equations given by $\mathbf{X}' = A\mathbf{X}$ where $$A = \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix}.$$ 2. **Recall the method:** For a system $\mathbf{X}' = A\mathbf{X}$, solutions are found by solving the characteristic equation $\det(A - \lambda I) = 0$ to find eigenvalues $\lambda$. Then find eigenvectors for each eigenvalue to form the general solution. 3. **Find the characteristic polynomial:** $$\det\left(\begin{bmatrix}1 - \lambda & -2 \\ 2 & -3 - \lambda\end{bmatrix}\right) = (1 - \lambda)(-3 - \lambda) - (-2)(2).$$ 4. **Simplify:** $$= (1 - \lambda)(-3 - \lambda) + 4 = -(3 + \lambda) + \lambda(3 + \lambda) + 4 = (1 - \lambda)(-3 - \lambda) + 4.$$ Expanding: $$= (1)(-3 - \lambda) - \lambda(-3 - \lambda) + 4 = -3 - \lambda + 3\lambda + \lambda^2 + 4 = \lambda^2 + 2\lambda + 1.$$ 5. **Solve characteristic equation:** $$\lambda^2 + 2\lambda + 1 = 0.$$ This factors as: $$ (\lambda + 1)^2 = 0,$$ so the eigenvalue is $\lambda = -1$ with multiplicity 2. 6. **Find eigenvectors:** Solve $(A - (-1)I)\mathbf{v} = 0$: $$\left(\begin{bmatrix}1 & -2 \\ 2 & -3\end{bmatrix} + I\right)\mathbf{v} = \begin{bmatrix}2 & -2 \\ 2 & -2\end{bmatrix} \mathbf{v} = 0.$$ This reduces to $2v_1 - 2v_2 = 0 \Rightarrow v_1 = v_2.$ Eigenvector can be chosen as $\mathbf{v} = \begin{bmatrix}1 \\ 1\end{bmatrix}.$ 7. **General solution:** Since eigenvalue is repeated, the general solution is $$\mathbf{X}(t) = c_1 e^{-t} \begin{bmatrix}1 \\ 1\end{bmatrix} + c_2 t e^{-t} \begin{bmatrix}1 \\ 1\end{bmatrix}.$$ **Final answer:** $$\boxed{\mathbf{X}(t) = e^{-t} \begin{bmatrix}1 \\ 1\end{bmatrix} (c_1 + c_2 t)}.$$