我基于这篇论文使用Matlab实现了Jacobi迭代，代码如下：

function x = jacobi(A, b)
% Executes iterations of Jacobi's method to solve Ax = b.

assert(size(A, 1) == size(A, 2))
assert(size(A, 1) == size(b, 1))

% matrix with all zeros except the diagonal elements,
% which are those of A.
D = diag(diag(A));

% matrix A but with its diagonal elements equal zero
R = D - A;

tol = 1.e-8; 
max_num_of_iter = 1000;

% initialize relative error to large value
relative_error = inf; 
num_of_iter = 1;

% initial guess
x_prev = zeros(size(b, 1), 1);

% iterate until convergence or maximum number of iterations is reached
while relative_error > tol && num_of_iter < max_num_of_iter
    x = D \ (b + R * x_prev);

    relative_error = norm(x - x_prev, inf) / norm(x, inf);

    x_prev = x;

    num_of_iter = num_of_iter + 1;
end

现在我不明白为什么x使用 找到迭代D \ (b + R * x_prev)，更具体地说，为什么我们有一个+符号而不是一个-登录D \ (b + R * x_prev)？

根据[维基百科]( https://en.wikipedia.org/wiki/Jacobi_method，雅可比方法如下：

$$\mathbf {x} ^{(k+1)}=D^{-1}(\mathbf {b} -R\mathbf {x} ^{(k)})$$

其中是一个矩阵，除了对角线元素是的对角线元素外，它的元素都是零。中对应元素相同的矩阵。$D$$n \times n$$A$$R = D - A$$A$

显然，如果我们想解决

$$Ax = b$$

和 然后

$$A = L + D + U =\\ D + L + U = \\ D + R \Rightarrow \\ A - D = R = \\ L + U$$

$$(D + L + U) x = b \\ Dx + Lx + Ux = b \Rightarrow\\ Dx = b - Lx - Ux \Rightarrow \\ Dx = b - (L + U) x$$

既然我们已经定义了，那么$L + U = R$

$$Dx = b - (L + U) x \Rightarrow \\ Dx = b - Rx \Rightarrow \\ x = D^{-1} \times (b - Rx)$$

所以迭代方法的原始公式提供维基百科似乎与我的计算一致。问题是，如果我使用 a而不是我找不到正确的解决方案。我怎么知道？我正在使用（即高斯消除）来检查解决方案。$\mathbf {x} ^{(k+1)}=D^{-1}(\mathbf {b} -R\mathbf {x} ^{(k)})$-+x = A \ b