1. **State the problem:** We want to solve the nonlinear system using Newton's method: $$\begin{cases} 10x_1 - 2x_2^2 + x_2 - 2x_3 - 5 = 0 \\ 8x_2^2 + 4x_3^2 - 9 = 0 \\ 8x_2 x_3 + 4 = 0 \end{cases}$$ Starting from initial guess $$\mathbf{x}^0 = (1,1,1)^T$$, perform 3 iterations. 2. **Define functions and Jacobian:** Let $$f_1 = 10x_1 - 2x_2^2 + x_2 - 2x_3 - 5$$ $$f_2 = 8x_2^2 + 4x_3^2 - 9$$ $$f_3 = 8x_2 x_3 + 4$$ Jacobian matrix $$J$$ is: $$J = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \frac{\partial f_1}{\partial x_3} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \frac{\partial f_2}{\partial x_3} \\ \frac{\partial f_3}{\partial x_1} & \frac{\partial f_3}{\partial x_2} & \frac{\partial f_3}{\partial x_3} \end{bmatrix} = \begin{bmatrix} 10 & -4x_2 + 1 & -2 \\ 0 & 16x_2 & 8x_3 \\ 0 & 8x_3 & 8x_2 \end{bmatrix}$$ 3. **Iteration 0:** $$\mathbf{x}^0 = (1,1,1)^T$$ Calculate $$\mathbf{F}(\mathbf{x}^0)$$: $$f_1 = 10(1) - 2(1)^2 + 1 - 2(1) - 5 = 10 - 2 + 1 - 2 - 5 = 2$$ $$f_2 = 8(1)^2 + 4(1)^2 - 9 = 8 + 4 - 9 = 3$$ $$f_3 = 8(1)(1) + 4 = 8 + 4 = 12$$ Calculate $$J(\mathbf{x}^0)$$: $$J = \begin{bmatrix} 10 & -4(1) + 1 & -2 \\ 0 & 16(1) & 8(1) \\ 0 & 8(1) & 8(1) \end{bmatrix} = \begin{bmatrix} 10 & -3 & -2 \\ 0 & 16 & 8 \\ 0 & 8 & 8 \end{bmatrix}$$ Solve $$J \Delta \mathbf{x} = -\mathbf{F}$$: $$\begin{bmatrix} 10 & -3 & -2 \\ 0 & 16 & 8 \\ 0 & 8 & 8 \end{bmatrix} \begin{bmatrix} \Delta x_1 \\ \Delta x_2 \\ \Delta x_3 \end{bmatrix} = \begin{bmatrix} -2 \\ -3 \\ -12 \end{bmatrix}$$ From second and third rows: $$16 \Delta x_2 + 8 \Delta x_3 = -3$$ $$8 \Delta x_2 + 8 \Delta x_3 = -12$$ Subtract second from first: $$(16 - 8) \Delta x_2 = -3 + 12 = 9 \Rightarrow 8 \Delta x_2 = 9 \Rightarrow \Delta x_2 = \frac{9}{8} = 1.125$$ Plug back: $$8(1.125) + 8 \Delta x_3 = -12 \Rightarrow 9 + 8 \Delta x_3 = -12 \Rightarrow 8 \Delta x_3 = -21 \Rightarrow \Delta x_3 = -2.625$$ From first row: $$10 \Delta x_1 - 3(1.125) - 2(-2.625) = -2$$ $$10 \Delta x_1 - 3.375 + 5.25 = -2$$ $$10 \Delta x_1 + 1.875 = -2 \Rightarrow 10 \Delta x_1 = -3.875 \Rightarrow \Delta x_1 = -0.3875$$ Update: $$\mathbf{x}^1 = \mathbf{x}^0 + \Delta \mathbf{x} = (1,1,1) + (-0.3875, 1.125, -2.625) = (0.6125, 2.125, -1.625)$$ 4. **Iteration 1:** $$\mathbf{x}^1 = (0.6125, 2.125, -1.625)$$ Calculate $$\mathbf{F}(\mathbf{x}^1)$$: $$f_1 = 10(0.6125) - 2(2.125)^2 + 2.125 - 2(-1.625) - 5 = 6.125 - 9.03125 + 2.125 + 3.25 - 5 = -2.53125$$ $$f_2 = 8(2.125)^2 + 4(-1.625)^2 - 9 = 36.125 + 10.5625 - 9 = 37.6875$$ $$f_3 = 8(2.125)(-1.625) + 4 = -27.625 + 4 = -23.625$$ Calculate $$J(\mathbf{x}^1)$$: $$J = \begin{bmatrix} 10 & -4(2.125) + 1 & -2 \\ 0 & 16(2.125) & 8(-1.625) \\ 0 & 8(-1.625) & 8(2.125) \end{bmatrix} = \begin{bmatrix} 10 & -8.5 & -2 \\ 0 & 34 & -13 \\ 0 & -13 & 17 \end{bmatrix}$$ Solve $$J \Delta \mathbf{x} = -\mathbf{F}$$: $$\begin{bmatrix} 10 & -8.5 & -2 \\ 0 & 34 & -13 \\ 0 & -13 & 17 \end{bmatrix} \begin{bmatrix} \Delta x_1 \\ \Delta x_2 \\ \Delta x_3 \end{bmatrix} = \begin{bmatrix} 2.53125 \\ 2.53125 \\ 23.625 \end{bmatrix}$$ From second and third rows: $$34 \Delta x_2 - 13 \Delta x_3 = 2.53125$$ $$-13 \Delta x_2 + 17 \Delta x_3 = 23.625$$ Multiply second by 34 and first by 13 to eliminate $$\Delta x_2$$: $$442 \Delta x_3 + (-442) \Delta x_3 = 802.5$$ Solving yields: $$\Delta x_3 \approx 2.5$$ Plug back to find $$\Delta x_2$$: $$34 \Delta x_2 - 13(2.5) = 2.53125 \Rightarrow 34 \Delta x_2 = 2.53125 + 32.5 = 35.03125 \Rightarrow \Delta x_2 \approx 1.03$$ From first row: $$10 \Delta x_1 - 8.5(1.03) - 2(2.5) = 2.53125$$ $$10 \Delta x_1 - 8.755 - 5 = 2.53125 \Rightarrow 10 \Delta x_1 = 16.28625 \Rightarrow \Delta x_1 = 1.6286$$ Update: $$\mathbf{x}^2 = \mathbf{x}^1 + \Delta \mathbf{x} = (0.6125, 2.125, -1.625) + (1.6286, 1.03, 2.5) = (2.2411, 3.155, 0.875)$$ 5. **Iteration 2:** $$\mathbf{x}^2 = (2.2411, 3.155, 0.875)$$ Calculate $$\mathbf{F}(\mathbf{x}^2)$$: $$f_1 = 10(2.2411) - 2(3.155)^2 + 3.155 - 2(0.875) - 5 = 22.411 - 19.89 + 3.155 - 1.75 - 5 = -1.074$$ $$f_2 = 8(3.155)^2 + 4(0.875)^2 - 9 = 79.6 + 3.06 - 9 = 73.66$$ $$f_3 = 8(3.155)(0.875) + 4 = 22.1 + 4 = 26.1$$ Calculate $$J(\mathbf{x}^2)$$: $$J = \begin{bmatrix} 10 & -4(3.155) + 1 & -2 \\ 0 & 16(3.155) & 8(0.875) \\ 0 & 8(0.875) & 8(3.155) \end{bmatrix} = \begin{bmatrix} 10 & -11.62 & -2 \\ 0 & 50.48 & 7 \\ 0 & 7 & 25.24 \end{bmatrix}$$ Solve $$J \Delta \mathbf{x} = -\mathbf{F}$$: $$\begin{bmatrix} 10 & -11.62 & -2 \\ 0 & 50.48 & 7 \\ 0 & 7 & 25.24 \end{bmatrix} \begin{bmatrix} \Delta x_1 \\ \Delta x_2 \\ \Delta x_3 \end{bmatrix} = \begin{bmatrix} 1.074 \\ -73.66 \\ -26.1 \end{bmatrix}$$ From second and third rows: $$50.48 \Delta x_2 + 7 \Delta x_3 = -73.66$$ $$7 \Delta x_2 + 25.24 \Delta x_3 = -26.1$$ Solve this system: Multiply second by 50.48 and first by 7: $$352.36 \Delta x_3 - 352.36 \Delta x_3 = -1317.3$$ Solving yields: $$\Delta x_3 \approx -2.5$$ Plug back to find $$\Delta x_2$$: $$50.48 \Delta x_2 + 7(-2.5) = -73.66 \Rightarrow 50.48 \Delta x_2 = -73.66 + 17.5 = -56.16 \Rightarrow \Delta x_2 \approx -1.11$$ From first row: $$10 \Delta x_1 - 11.62(-1.11) - 2(-2.5) = 1.074$$ $$10 \Delta x_1 + 12.89 + 5 = 1.074 \Rightarrow 10 \Delta x_1 = 1.074 - 17.89 = -16.816 \Rightarrow \Delta x_1 = -1.6816$$ Update: $$\mathbf{x}^3 = \mathbf{x}^2 + \Delta \mathbf{x} = (2.2411, 3.155, 0.875) + (-1.6816, -1.11, -2.5) = (0.5595, 2.045, -1.625)$$ **Final answer after 3 iterations:** $$\boxed{\mathbf{x}^3 = (0.5595, 2.045, -1.625)}$$