1. **Problem statement:** Given matrices $$A=\begin{pmatrix}0.5 & 0.4 & 0.1 \\ 0.4 & 0.6 & 0 \\ 0.3 & 0.3 & 0.4\end{pmatrix},\quad B=\begin{pmatrix}0.55 & 0.45 \\ 0.3 & 0.7\end{pmatrix},\quad v_0=\begin{pmatrix}2000 \\ 2000 \\ 2000\end{pmatrix}$$ and the relations $$v_{t+1} = A^\top v_t,$$ $$(B - E)^\top x = 0,$$ with $x=\begin{pmatrix}x_1 \\ x_2\end{pmatrix}$, $x_1 + x_2 = 1$, $x_1,x_2 \geq 0$, and $$C=\begin{pmatrix}a_{11} & a_{12} - a_{13} \\ a_{22} & a_{21} + a_{23}\end{pmatrix}$$ Find the next state vector $v_1$ and solve for $x$. 2. **Step 1: Calculate $v_1 = A^\top v_0$** Transpose $A$: $$A^\top = \begin{pmatrix}0.5 & 0.4 & 0.3 \\ 0.4 & 0.6 & 0.3 \\ 0.1 & 0 & 0.4\end{pmatrix}$$ Multiply $A^\top$ by $v_0$: $$v_1 = A^\top v_0 = \begin{pmatrix}0.5 & 0.4 & 0.3 \\ 0.4 & 0.6 & 0.3 \\ 0.1 & 0 & 0.4\end{pmatrix} \begin{pmatrix}2000 \\ 2000 \\ 2000\end{pmatrix}$$ Calculate each component: $$v_1^{(1)} = 0.5 \times 2000 + 0.4 \times 2000 + 0.3 \times 2000 = 1000 + 800 + 600 = 2400$$ $$v_1^{(2)} = 0.4 \times 2000 + 0.6 \times 2000 + 0.3 \times 2000 = 800 + 1200 + 600 = 2600$$ $$v_1^{(3)} = 0.1 \times 2000 + 0 \times 2000 + 0.4 \times 2000 = 200 + 0 + 800 = 1000$$ So, $$v_1 = \begin{pmatrix}2400 \\ 2600 \\ 1000\end{pmatrix}$$ 3. **Step 2: Solve $(B - E)^\top x = 0$ with $x_1 + x_2 = 1$ and $x_1,x_2 \geq 0$** Identity matrix $E = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ Calculate $B - E$: $$B - E = \begin{pmatrix}0.55 - 1 & 0.45 - 0 \\ 0.3 - 0 & 0.7 - 1\end{pmatrix} = \begin{pmatrix}-0.45 & 0.45 \\ 0.3 & -0.3\end{pmatrix}$$ Transpose: $$(B - E)^\top = \begin{pmatrix}-0.45 & 0.3 \\ 0.45 & -0.3\end{pmatrix}$$ Set up the system: $$\begin{cases}-0.45 x_1 + 0.3 x_2 = 0 \\ 0.45 x_1 - 0.3 x_2 = 0 \\ x_1 + x_2 = 1\end{cases}$$ From the first equation: $$-0.45 x_1 + 0.3 x_2 = 0 \implies 0.3 x_2 = 0.45 x_1 \implies x_2 = \frac{0.45}{0.3} x_1 = 1.5 x_1$$ Using $x_1 + x_2 = 1$: $$x_1 + 1.5 x_1 = 1 \implies 2.5 x_1 = 1 \implies x_1 = \frac{1}{2.5} = 0.4$$ Then, $$x_2 = 1.5 \times 0.4 = 0.6$$ Check non-negativity: $x_1=0.4 \geq 0$, $x_2=0.6 \geq 0$ valid. 4. **Step 3: Construct matrix $C$** Given $$C = \begin{pmatrix}a_{11} & a_{12} - a_{13} \\ a_{22} & a_{21} + a_{23}\end{pmatrix}$$ From $A$: $$a_{11} = 0.5, \quad a_{12} = 0.4, \quad a_{13} = 0.1,$$ $$a_{21} = 0.4, \quad a_{22} = 0.6, \quad a_{23} = 0,$$ Calculate entries: $$c_{11} = 0.5$$ $$c_{12} = 0.4 - 0.1 = 0.3$$ $$c_{21} = 0.6$$ $$c_{22} = 0.4 + 0 = 0.4$$ So, $$C = \begin{pmatrix}0.5 & 0.3 \\ 0.6 & 0.4\end{pmatrix}$$ **Final answers:** $$v_1 = \begin{pmatrix}2400 \\ 2600 \\ 1000\end{pmatrix}, \quad x = \begin{pmatrix}0.4 \\ 0.6\end{pmatrix}, \quad C = \begin{pmatrix}0.5 & 0.3 \\ 0.6 & 0.4\end{pmatrix}$$