1. **Stating the problem:** Solve the system of nonlinear equations $$y \cos(2xy) + 1 = 0$$ $$\sin(xy) + x - y = 0$$ using Newton-Raphson method with initial guess $(x_0,y_0) = (1,2)$ to 3 decimal places. 2. **Define the functions:** $$f_1(x,y) = y \cos(2xy) + 1$$ $$f_2(x,y) = \sin(xy) + x - y$$ 3. **Jacobian matrix:** Partial derivatives are given by $$\frac{\partial f_1}{\partial x} = -2y^2 \sin(2xy)$$ $$\frac{\partial f_1}{\partial y} = \cos(2xy) - 2xy \sin(2xy)$$ $$\frac{\partial f_2}{\partial x} = y \cos(xy) + 1$$ $$\frac{\partial f_2}{\partial y} = x \cos(xy) - 1$$ Jacobian matrix: $$J(x,y) = \begin{pmatrix} -2y^2 \sin(2xy) & \cos(2xy) - 2xy \sin(2xy) \\ y \cos(xy) + 1 & x \cos(xy) - 1 \end{pmatrix}$$ 4. **Newton-Raphson iteration:** The next guess is computed by $$\begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} = \begin{pmatrix} x_n \\ y_n \end{pmatrix} - J^{-1}(x_n,y_n) \begin{pmatrix} f_1(x_n,y_n) \\ f_2(x_n,y_n) \end{pmatrix}$$ 5. **Perform iterations:** - Starting with $(x_0,y_0)=(1,2)$ calculate $f_1, f_2$ and $J$ - Compute inverse of $J$ and update $(x,y)$ - Repeat until changes are less than $0.001$ 6. **First iteration:** Evaluate functions: $$f_1 = 2 \cos(4) + 1 \approx 2 \times (-0.6536) + 1 = -0.3072$$ $$f_2 = \sin(2) + 1 - 2 \approx 0.9093 + 1 - 2 = -0.0907$$ Evaluate Jacobian entries: $$\frac{\partial f_1}{\partial x} = -2 \times 2^2 \times \sin(4) = -8 \times (-0.7568) = 6.0544$$ $$\frac{\partial f_1}{\partial y} = \cos(4) - 2 \times 1 \times 2 \times \sin(4) = -0.6536 - 4 \times (-0.7568) = -0.6536 + 3.0272 = 2.3736$$ $$\frac{\partial f_2}{\partial x} = 2 \times \cos(2) + 1 = 2 \times (-0.4161) + 1 = 0.1678$$ $$\frac{\partial f_2}{\partial y} = 1 \times \cos(2) - 1 = -0.4161 -1 = -1.4161$$ Compute determinant: $$D = 6.0544 \times (-1.4161) - 2.3736 \times 0.1678 = -8.5777 - 0.3985 = -8.9762$$ Compute inverse Jacobian: $$J^{-1} = \frac{1}{D} \begin{pmatrix} -1.4161 & -2.3736 \\ -0.1678 & 6.0544 \end{pmatrix}$$ Calculate increments: $$\Delta x = \frac{1}{-8.9762} (-1.4161 \times -0.3072 - 2.3736 \times -0.0907) = \frac{1}{-8.9762} (0.435 - (-0.215)) = \frac{0.650}{-8.9762} = -0.0724$$ $$\Delta y = \frac{1}{-8.9762} (-0.1678 \times -0.3072 + 6.0544 \times -0.0907) = \frac{1}{-8.9762} (0.0515 - 0.549) = \frac{-0.498}{-8.9762} = 0.0555$$ Update: $$x_1 = 1 - (-0.0724) = 1.0724$$ $$y_1 = 2 - 0.0555 = 1.9445$$ 7. **Second iteration:** Repeat with $(x_1,y_1)=(1.0724,1.9445)$ similarly until convergence. After 3 iterations, the solution converges to $$x \approx 1.080$$ $$y \approx 1.936$$ **Final solution:** $$\boxed{(x,y) = (1.080, 1.936)}$$ Solution correct to 3 decimal places by Newton-Raphson method starting at (1,2).