线性方程组的迭代解法——雅可比迭代法

Posted 2020-11-26 guliangt

　　1.代码

%%雅可比迭代法（此迭代法对于病态矩阵的解不理想）
%%线性方程组M*X = b，M是方阵，X0是初始解向量,epsilon是控制精度
function JIM = Jacobian_iteration_method(M,b,X0,epsilon)
[m,n] = size(M);
d = diag(M);L = zeros(m,n);U = zeros(m,n);D = zeros(m,n);
delta = 0;ub = 100;X = zeros(m,ub);X(:,1) = X0;X_delta = X;X_end = zeros(m,1);k_end = 0;k = 1;e = floor(abs(log(epsilon)));
for i = 1:1:m
    for j = 1:1:n
        if i > j
            L(i,j) = -M(i,j);
        elseif i < j
            U(i,j) = -M(i,j);
        elseif i == j
            D(i,j) = d(i);
        end     
    end
end
B = inv(D)*(L+U);
f = inv(D)*b;
X_real = inv(M)*b;
for k = 1:1:ub
    X_delta(:,k) = X(:,k)-X_real;
    delta = norm(X_delta(:,k),2);
    if delta < epsilon
         break
    end
    X(:,k+1) = B*X(:,k)+f;
end
disp(‘迭代次数为：‘);
k
disp(‘迭代解为：‘);
JIM=vpa([X(:,k)],e);
end

　　2.例子

clear all
clc
for i = 1:8
    for j = 1:8
        if i == j
            M(i,j) = 2.1;
        elseif i - j == 1
            M(i,j) = 1;
        elseif j - i == 1
            M(i,j) = -1;
        else
            M(i,j) = 0;
        end
    end
end
b = [1 2 3 4 4 3 2 1]‘;
X0 = [1 1 1 1 1 1 1 1]‘;
epsilon = 1e-4;

S = Jacobian_iteration_method(M,b,X0,epsilon)
M

　　结果为

迭代次数为：
k =
    73
迭代解为：
S =
  1.07162282
  1.25046562
  1.69755831
  1.81519199
  1.50952631
 0.985363364
  0.57873237
 0.200592913
ans =
    1.0716
    1.2504
    1.6975
    1.8152
    1.5096
    0.9853
    0.5787
    0.2006