1. **Problem statement:** We have a trade matrix $$A = \begin{pmatrix} 0.15 & 0.5 & 0.25 \\ 0.35 & 0.3 & 0.45 \\ 0.5 & 0.2 & 0.3 \end{pmatrix}$$ and budgets vector $$\vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$ with the condition $$A\vec{x} = \vec{x}$$ representing balanced deficit-free trade. Given $$x_1 = 2000$$, find $$x_2$$ and $$x_3$$. 2. **Formula and explanation:** The equation $$A\vec{x} = \vec{x}$$ means $$\vec{x}$$ is an eigenvector of $$A$$ corresponding to eigenvalue 1. Writing out the system: $$ \begin{cases} 0.15 x_1 + 0.5 x_2 + 0.25 x_3 = x_1 \\ 0.35 x_1 + 0.3 x_2 + 0.45 x_3 = x_2 \\ 0.5 x_1 + 0.2 x_2 + 0.3 x_3 = x_3 \end{cases} $$ Rearranged: $$ \begin{cases} 0.15 x_1 + 0.5 x_2 + 0.25 x_3 - x_1 = 0 \\ 0.35 x_1 + 0.3 x_2 + 0.45 x_3 - x_2 = 0 \\ 0.5 x_1 + 0.2 x_2 + 0.3 x_3 - x_3 = 0 \end{cases} $$ Simplify each: $$ \begin{cases} -0.85 x_1 + 0.5 x_2 + 0.25 x_3 = 0 \\ 0.35 x_1 - 0.7 x_2 + 0.45 x_3 = 0 \\ 0.5 x_1 + 0.2 x_2 - 0.7 x_3 = 0 \end{cases} $$ 3. **Substitute known value:** $$x_1 = 2000$$, so: $$ \begin{cases} -0.85 \times 2000 + 0.5 x_2 + 0.25 x_3 = 0 \\ 0.35 \times 2000 - 0.7 x_2 + 0.45 x_3 = 0 \\ 0.5 \times 2000 + 0.2 x_2 - 0.7 x_3 = 0 \end{cases} $$ Calculate constants: $$ \begin{cases} -1700 + 0.5 x_2 + 0.25 x_3 = 0 \\ 700 - 0.7 x_2 + 0.45 x_3 = 0 \\ 1000 + 0.2 x_2 - 0.7 x_3 = 0 \end{cases} $$ 4. **Rewrite system:** $$ \begin{cases} 0.5 x_2 + 0.25 x_3 = 1700 \\ -0.7 x_2 + 0.45 x_3 = -700 \\ 0.2 x_2 - 0.7 x_3 = -1000 \end{cases} $$ 5. **Solve first two equations for $$x_2$$ and $$x_3$$:** From first: $$0.5 x_2 = 1700 - 0.25 x_3 \Rightarrow x_2 = \frac{1700 - 0.25 x_3}{0.5} = 3400 - 0.5 x_3$$ Substitute into second: $$-0.7 (3400 - 0.5 x_3) + 0.45 x_3 = -700$$ $$-2380 + 0.35 x_3 + 0.45 x_3 = -700$$ $$-2380 + 0.8 x_3 = -700$$ $$0.8 x_3 = 1680$$ $$x_3 = \frac{1680}{0.8} = 2100$$ 6. **Find $$x_2$$:** $$x_2 = 3400 - 0.5 \times 2100 = 3400 - 1050 = 2350$$ 7. **Check with third equation:** $$0.2 \times 2350 - 0.7 \times 2100 = 470 - 1470 = -1000$$ which matches the right side. **Final answer:** $$x_2 = 2350$$ and $$x_3 = 2100$$.