使用Matlab实现：Jacobi、Gauss-Seidel迭代
例题
方程组 { 5 x 1 + 2 x 2 + x 3 = − 12 − x 1 + 4 x 2 + 2 x 3 = 20 2 x 1 − 3 x 2 + 10 x 3 = 3 \begin{cases} 5x_1 + 2x_2 + x_3 = -12\\ -x_1 + 4x_2 + 2x_3 = 20\\ 2x_1 - 3x_2 + 10x_3 = 3\\ \end{cases} ⎩⎪⎨⎪⎧5x1+2x2+x3=−12−x1+4x2+2x3=202x1−3x2+10x3=3 求解，当 m a x ∣ x i ( k + 1 ) − x i ( k ) ∣ ≤ 1 0 − 5 max|x_i^{(k + 1)} - x_i^{(k)}| \leq 10^{-5} max∣xi(k+1)−xi(k)∣≤10−5 时候迭代终止。
以下解答过程，上标表示迭代次数，下标表示序号。
Jacobi迭代
定义变量：
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D = diag(a_{11}, a_{22}, ..., a_{nn}),\\ L = \left[\begin{array}{cccccc} 0\\ -a_{21} & 0\\ ...\\ -a_{i1} & ... & -a_{i,i-1} & 0\\ ...\\ -a_{n1} & ... & -a_{n,i-1} & ... & -a_{n,n-1} & 0\\ \end{array}\right],\\ U = \left[\begin{array}{cccccc} 0 &-a_{12} & ... & -a_{1,i} & ... & -a_{1,n}\\ ...\\ & & 0 & -a_{i-1,i} & ... & -a_{i-1,n}\\ ...\\ & & & & 0 & -a_{n-1,n}\\ ...\\ & & & & & 0\\ \end{array}\right]
D=diag(a11,a22,...,ann),L=⎣⎢⎢⎢⎢⎢⎢⎡0−a21...−ai1...−an10......−ai,i−1−an,i−10...−an,n−10⎦⎥⎥⎥⎥⎥⎥⎤,U=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎡0.........−a12...0−a1,i−ai−1,i......0−a1,n−ai−1,n−an−1,n0⎦⎥⎥⎥⎥⎥⎥⎥⎥⎤
其矩阵迭代形式为：
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x^{(k+1)} = B_J \cdot x^{(k)} + f_J\\ B_J = D^{-1} \cdot (L + U), \quad f_J = D^{-1} \cdot b
x(k+1)=BJ⋅x(k)+fJBJ=D−1⋅(L+U),fJ=D−1⋅b。
写出分量形式：
{ x 1 ( k + 1 ) = 1 5 ( − 12 − 2 x 2 ( k ) − x 3 ( k ) ) x 2 ( k + 1 ) = 1 4 ( 20 + x 1 ( k ) − 2 x 3 ( k ) ) x 3 ( k + 1 ) = 1 10 ( 3 − 2 x 1 ( k ) + 3 x 2 ( k ) ) \begin{cases} x_1^{(k + 1)} = \frac15 (-12 - 2 x_2^{(k)} - x_3^{(k)} )\\ x_2^{(k + 1)} = \frac14 (20 + x_1^{(k)} - 2x_3^{(k)} )\\ x_3^{(k + 1)} = \frac1{10} (3 - 2 x_1^{(k)} + 3x_2^{(k)} )\\ \end{cases} ⎩⎪⎨⎪⎧x1(k+1)=51(−12−2x2(k)−x3(k))x2(k+1)=41(20+x1(k)−2x3(k))x3(k+1)=101(3−2x1(k)+3x2(k))
写成矩阵形式：
[ x 1 ( k + 1 ) x 2 ( k + 1 ) x 3 ( k + 1 ) ] = [ 0 − 0.4 − 0.2 0.25 0 − 0.5 − 0.2 0.3 0 ] ⋅ [ x 1 ( k ) x 2 ( k ) x 3 ( k ) ] + [ − 2.4 5 0.3 ] \left[\begin{array}{c} x_1^{(k + 1)}\\ x_2^{(k + 1)}\\ x_3^{(k + 1)}\\ \end{array}\right] = \left[\begin{array}{cccc} 0 & -0.4 & -0.2\\ 0.25 & 0 & -0.5\\ -0.2 & 0.3 & 0\\ \end{array}\right] \cdot \left[\begin{array}{c} x_1^{(k)}\\ x_2^{(k)}\\ x_3^{(k)}\\ \end{array}\right] + \left[\begin{array}{c} -2.4\\ 5\\ 0.3\\ \end{array}\right] ⎣⎢⎡x1(k+1)x2(k+1)x3(k+1)⎦⎥⎤=⎣⎡00.25−0.2−0.400.3−0.2−0.50⎦⎤⋅⎣⎢⎡x1(k)x2(k)x3(k)⎦⎥⎤+⎣⎡−2.450.3⎦⎤
取初始向量： x 0 = ( 0 , 0 , 0 ) T x^0 = (0, 0, 0)^T x0=(0,0,0)T，依次按照上式进行迭代。使用Matlab进行编程求解。
a=[0,-0.4,-0.2;0.25,0,-0.5;-0.2,0.3,0];
b = [-2.4;5;0.3];
x = [0;0;0];
xx = a * x + b;
i = 0;
while norm(x - xx, inf) >= 1e-5
x = xx;
xx = a * x + b;
i = i +1;
end
以上代码，最终 x = x i , x x = x ( i + 1 ) x = x^{i}, xx = x^{(i + 1)} x=xi,xx=x(i+1) ，最终迭代次数位 i + 1 i + 1 i+1 次，如果你需要看到更长的小数位置，可以使用以下Matlab代码，表示使用15位浮点或定点数。
format long g
运行结果为：
即精确解为 x = ( − 4 , 3 , 2 ) T x = (-4,3,2)^T x=(−4,3,2)T 。
Gauss-Seidel迭代
其矩阵迭代形式为：
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x^{(k+1)} = B_G \cdot x^{(k)} + f_G\\ B_G = (D - L) ^{-1} \cdot U, \quad f_G = (D - L) ^{-1} \cdot b
x(k+1)=BG⋅x(k)+fGBG=(D−L)−1⋅U,fG=(D−L)−1⋅b
使用Matlab编程求解：
d = [5,0,0;0,4,0;0,0,10];
l = [0,0,0;1,0,0;-2,3,0];
u = [0,-2,-1;0,0,-2;0,0,0];
b = [-12;20;3];
t = inv(d - l);
bg = t * u;
fg = t * b;
x = [0;0;0];
xx = [-2.4;4.4;2.1];
i = 0;
while norm(x - xx, inf) >= 1e-5
x = xx;
xx = bg * x + fg;
i = i +1;
end
运行结果为：
同样求得精确解为 x = ( − 4 , 3 , 2 ) T x = (-4,3,2)^T x=(−4,3,2)T 。