1. **State the problem:** Simplify the expression $$\left(\frac{2v^{-3} \cdot u^{2} v^{-1}}{2u^{3} v^{4}}\right)^{-2}$$. 2. **Combine terms in the numerator:** Multiply the powers of $v$ by adding exponents: $$2v^{-3} \cdot u^{2} v^{-1} = 2u^{2} v^{-3 + (-1)} = 2u^{2} v^{-4}$$ 3. **Rewrite the fraction:** $$\frac{2u^{2} v^{-4}}{2u^{3} v^{4}}$$ 4. **Cancel common factors:** The 2's cancel out: $$\frac{\cancel{2}u^{2} v^{-4}}{\cancel{2}u^{3} v^{4}} = \frac{u^{2} v^{-4}}{u^{3} v^{4}}$$ 5. **Divide powers with the same base by subtracting exponents:** $$\frac{u^{2}}{u^{3}} = u^{2-3} = u^{-1}$$ $$\frac{v^{-4}}{v^{4}} = v^{-4-4} = v^{-8}$$ 6. **Rewrite the fraction after simplification:** $$u^{-1} v^{-8}$$ 7. **Apply the outer exponent $-2$ to each factor:** $$\left(u^{-1} v^{-8}\right)^{-2} = u^{-1 \times (-2)} v^{-8 \times (-2)} = u^{2} v^{16}$$ **Final answer:** $$u^{2} v^{16}$$