You can use point jacobi or gauss siedel to solve this problem. Write a function

Question

You can use point jacobi or gauss siedel to solve this problem.

Write a function that implements a Point-Jacobi iterative method to solve for the solution to the system of equations Ax = b. Your function should accept the input matrix A, right-hand-side vector b, the initial solution guess x0 and maximum approximate relative error || epsilon a || infinity as arguments, and return the value of the solution vector xi as its output. You can use MATLAB, C, C++ or Fortran to code the solution to this problem. You will receive zero credit if you use a built-in linear algebra function to compute the solution to the system. In MATLAB, your function declaration should look something like this:

Explanation / Answer

function x_i =pointjacobi(A,b,x_0,eps_a_max);

format compact

n = length(b);

Error_eval = ones(n,1);

%% Check if the matrix A is diagonally dominant

for i = 1:n

j = 1:n;

j(i) = [];

B = abs(A(i,j));

Check(i) = abs(A(i,i)) - sum(B); % Is the diagonal value greater than the remaining row values combined?

if Check(i) < 0

fprintf('The matrix is not strictly diagonally dominant at row %2i ',i)

end

end

%% Start the Iterative method

iteration = 0;

while max(Error_eval) > eps_a_max

iteration = iteration + 1;

Z = x_0; % save current values to calculate error later

for i = 1:n

j = 1:n; % define an array of the coefficients' elements

j(i) = []; % eliminate the unknow's coefficient from the remaining coefficients

Xtemp = x_0; % copy the unknows to a new variable

Xtemp(i) = []; % eliminate the unknown under question from the set of values

x_0(i) = (b(i) - sum(A(i,j) * Xtemp)) / A(i,i);

end

Xsolution(:,iteration) = x_0;

Error_eval = sqrt((x_0 - Z).^2);

end

x_i=Xsolution(:,iteration);

% sample input

% A = [-6 2 1 2 1;

% 3 8 -4 1 0;

% -1 1 4 10 1;

% 3 -4 1 9 2;

% 2 0 1 3 10];% coefficients matrix

% b = [3;4;-2 ;12;1];% constants vector

% n = length(b);

% x_0 = zeros(n,1);

% eps_a_max=0.001;

% pointjacobi(A,b,x_0,eps_a_max)

%output:

%

% x_i =

% -1.1222

% -1.3887

% -4.2461

% 1.4952

% 0.3005

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