Question
Solve the system of equations
by using the Gauss Jacboi and Gauss Seidel method. The exact solution of the system is
Perform the required number of iterations so that the same accuracy is obtained by both the methods. What conclusions can you draw from the results obtained?
Answer :
Word Count : 766
We are given a system of 4 linear equations: $$ \begin{aligned} 8x_1 - x_2 + 2x_3 &= 4 \quad\quad \text{(1)}\\ -3x_1 + 11x_2 - x_3 + 3x_4 &= 23 \quad \text{(2)}\\ - x_2 + 10x_3 - x_4 &= -13 \quad \text{(3)}\\ -2x_1 + x_2 - x_3 + 8x_4 &= 13 \quad \text{(4)} \end{aligned} $$ Initial guess: $$ \mathbf{x}^{(0)} = \begin{bmatrix}0 & 0 & 0 & 0\end{bmatrix}^T $$ Exact solution: $$ \mathbf{x} = \begin{bmatrix}1 & 2 & -1 & 1\end{bmatrix}^T $$ --- ## 1. Gauss-Jacobi Method We rearrange the system to isolate each variable: $$ \begin{aligned} x_1 &= \frac{1}{8}(4 + x_2 - 2x_3) \\ x_2 &= \frac{1}{11}(23 + 3x_1 + x_3 - 3x_4) \\ x_3 &= \frac{1}{10}(-13 + x_2 + x_4) \\ x_4 &= \frac{1}{8}(13 + 2x_1 - x_2 + x_3) \end{aligned} $$ --- ### Iteration 1 (Jacobi) Using $x^{(0)} = [0, 0, 0, 0]$: $$ \begin{aligned} x_1^{(1)} &= \frac{1}{8}(4 + 0 - 2(0)) = \frac{4}{8} = 0.5 \\ x_2^{(1)} &= \frac{1}{11}(23 + 3(0) + 0 - 3(0)) = \frac{23}{11} \approx 2.0909 \\ x_3^{(1)} &= \frac{1}{10}(-13 + 0 + 0) = -1.3 \\ x_4^{(1)} &= \frac{1}{8}(13 + 2(0) - 0 + 0) _________ _________ _______ ________ __________ ________ _______ ___ ___.
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We are given a system of 4 linear equations: $$ \begin{aligned} 8x_1 - x_2 + 2x_3 &= 4 \quad\quad \text{(1)}\\ -3x_1 + 11x_2 - x_3 + 3x_4 &= 23 \quad \text{(2)}\\ - x_2 + 10x_3 - x_4 &= -13 \quad \text{(3)}\\ -2x_1 + x_2 - x_3 + 8x_4 &= 13 \quad \text{(4)} \end{aligned} $$ Initial guess: $$ \mathbf{x}^{(0)} = \begin{bmatrix}0 & 0 & 0 & 0\end{bmatrix}^T $$ Exact solution: $$ \mathbf{x} = \begin{bmatrix}1 & 2 & -1 & 1\end{bmatrix}^T $$ --- ## 1. Gauss-Jacobi Method We rearrange the system to isolate each variable: $$ \begin{aligned} x_1 &= \frac{1}{8}(4 + x_2 - 2x_3) \\ x_2 &= \frac{1}{11}(23 + 3x_1 + x_3 - 3x_4) \\ x_3 &= \frac{1}{10}(-13 + x_2 + x_4) \\ x_4 &= \frac{1}{8}(13 + 2x_1 - x_2 + x_3) \end{aligned} $$ --- ### Iteration 1 (Jacobi) Using $x^{(0)} = [0, 0, 0, 0]$: $$ \begin{aligned} x_1^{(1)} &= \frac{1}{8}(4 + 0 - 2(0)) = \frac{4}{8} = 0.5 \\ x_2^{(1)} &= \frac{1}{11}(23 + 3(0) + 0 - 3(0)) = \frac{23}{11} \approx 2.0909 \\ x_3^{(1)} &= \frac{1}{10}(-13 + 0 + 0) = -1.3 \\ x_4^{(1)} &= \frac{1}{8}(13 + 2(0) - 0 + 0) _________ _________ _______ ________ __________ ________ _______ ___ ___.
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Related Questions
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