1. **Problem statement:** Find the value of $m$ such that the system of equations $$\begin{cases} 2x - 3y = 2 - m \\ x + 2y = 3m + 1 \end{cases}$$ has a unique solution $(x,y)$ where $A = -\frac{y}{x}$ is an integer. 2. **Step 1: Check for unique solution condition.** The system has a unique solution if the determinant of the coefficient matrix is non-zero: $$D = \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = 2 \times 2 - (-3) \times 1 = 4 + 3 = 7 \neq 0$$ So the system has a unique solution for all $m$. 3. **Step 2: Solve the system for $x$ and $y$.** Using Cramer's rule: $$D_x = \begin{vmatrix} 2 - m & -3 \\ 3m + 1 & 2 \end{vmatrix} = (2 - m) \times 2 - (-3)(3m + 1) = 4 - 2m + 9m + 3 = 7m + 7$$ $$D_y = \begin{vmatrix} 2 & 2 - m \\ 1 & 3m + 1 \end{vmatrix} = 2(3m + 1) - 1(2 - m) = 6m + 2 - 2 + m = 7m$$ Therefore, $$x = \frac{D_x}{D} = \frac{7m + 7}{7} = m + 1$$ $$y = \frac{D_y}{D} = \frac{7m}{7} = m$$ 4. **Step 3: Find $A = -\frac{y}{x} = -\frac{m}{m+1}$.** We want $A$ to be an integer, so $$-\frac{m}{m+1} = k, \quad k \in \mathbb{Z}$$ which implies $$\frac{m}{m+1} = -k$$ 5. **Step 4: Solve for $m$.** $$m = -k(m+1) = -km - k$$ $$m + km = -k$$ $$m(1 + k) = -k$$ $$m = \frac{-k}{1 + k}$$ 6. **Step 5: $m$ must be defined and the denominator $1+k \neq 0$, so $k \neq -1$.** 7. **Step 6: Check for integer $m$ values.** Since $m$ is a parameter, it can be any real number. But the problem does not restrict $m$ to integers, only $A$ must be integer. 8. **Step 7: Find all $m$ such that $A$ is integer.** For each integer $k \neq -1$, $$m = \frac{-k}{1+k}$$ 9. **Step 8: Verify that $x = m+1 \neq 0$ to avoid division by zero in $A$.** $$x = m + 1 = \frac{-k}{1+k} + 1 = \frac{-k + 1 + k}{1+k} = \frac{1}{1+k} \neq 0$$ which is true since $k \neq -1$. **Final answer:** $$\boxed{m = \frac{-k}{1+k} \text{ for any integer } k \neq -1}$$ This gives all values of $m$ such that $A = -\frac{y}{x}$ is an integer.