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## Solve Linear Equations using Jacobi Method (medium)

#### Example

## Solving Linear Equations Using the Jacobi Method

The Jacobi method is an iterative algorithm used for solving a system of linear equations \(Ax = b\). This method is particularly useful for large systems where direct methods like Gaussian elimination are computationally expensive.
### Algorithm Overview

For a system of equations represented by \(Ax = b\), where \(A\) is a matrix and \(x\) and \(b\) are vectors, the Jacobi method involves the following steps:
### Practical Considerations

- The method may not converge for all matrices.

- Choosing a good initial guess can improve convergence.

- Diagonal dominance of \(A\) ensures convergence of the Jacobi method.

Write a Python function that uses the Jacobi method to solve a system of linear equations given by Ax = b. The function should iterate n times, rounding each intermediate solution to four decimal places, and return the approximate solution x.

Example: input: A = [[5, -2, 3], [-3, 9, 1], [2, -1, -7]], b = [-1, 2, 3], n=2 output: [0.146, 0.2032, -0.5175] reasoning: The Jacobi method iteratively solves each equation for x[i] using the formula x[i] = (1/a_ii) * (b[i] - sum(a_ij * x[j] for j != i)), where a_ii is the diagonal element of A and a_ij are the off-diagonal elements.

**Initialization**: Start with an initial guess for \(x\).**Iteration**: For each equation \(i\), update \(x[i]\) using: \[ x[i] = \frac{1}{a_{ii}} (b[i] - \sum_{j \neq i} a_{ij} x[j]) \] where \(a_{ii}\) are the diagonal elements of \(A\), and \(a_{ij}\) are the off-diagonal elements.**Convergence**: Repeat the iteration until the changes in \(x\) are below a certain tolerance or until a maximum number of iterations is reached.

- Choosing a good initial guess can improve convergence.

- Diagonal dominance of \(A\) ensures convergence of the Jacobi method.

import numpy as np def solve_jacobi(A: np.ndarray, b: np.ndarray, n: int) -> list: d_a = np.diag(A) nda = A - np.diag(d_a) x = np.zeros(len(b)) x_hold = np.zeros(len(b)) for _ in range(n): for i in range(len(A)): x_hold[i] = (1/d_a[i]) * (b[i] - sum(nda[i]*x)) x = x_hold.copy() return np.round(x,4).tolist()

Output will be shown here.

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