用MATLAB对高斯赛德尔迭代法仿真,怎样把解x1,x2,x3输出在同一坐标系中,怎样输出每一次迭代值.

你没把G-S方法迭代的流程图法出来,其实表4.2中的迭代数据是有问题的（少了一步）.
可以生成你所给图中数据的程序：
A = [4 -1 1; 4 -8 1; -2 1 5];
b = [7; -21; 15];
x0 = zeros(3,1);
eps=1.0e-7;
D=diag(diag(A));%求A的对角矩阵
L=-tril(A,-1);%求A的下三角阵
U=-triu(A,1);%求A的上三角阵
G=(D-L)U;
f=(D-L)b;
Y = [1 zeros(1,3)];
y=G*x0+f;
n=1;%迭代次数
while norm(y-x0)>=eps
x0=y;
y=G*x0+f;
n=n+1;
Y(n,:) = [n y'];
end

% 生成图4.2
plot(Y(:,1),Y(:,2:4))
axis([1 12 0 5])
grid on
set(gca,'xtick',2:2:12,'ytick',0:5)

% 输出表4.2
fprintf('%it%.14ft%.14ft%.14fn',Y.')

1 0.00000000000000 0.00000000000000 0.00000000000000
2 1.87500000000000 3.93750000000000 2.96250000000000
3 1.99375000000000 3.99218750000000 2.99906250000000
4 1.99828125000000 3.99902343750000 2.99950781250000
5 1.99987890625000 3.99987792968750 2.99997597656250
6 1.99997548828125 3.99998474121094 2.99999324707031
7 1.99999787353516 3.99999809265137 2.99999953088379
8 1.99999964044189 3.99999976158142 2.99999990386047
9 1.99999996443024 3.99999997019768 2.99999999173256
10 1.99999999461628 3.99999999627471 2.99999999859157
而正确的迭代过程应该是这样的（图形及数据输出部分相同,略）：
A = [4 -1 1; 4 -8 1; -2 1 5];
b = [7; -21; 15];
x0 = zeros(3,1);
eps=1.0e-7;
D=diag(diag(A));%求A的对角矩阵
L=-tril(A,-1);%求A的下三角阵
U=-triu(A,1);%求A的上三角阵
G=(D-L)U;
f=(D-L)b;
% Y = [1 zeros(1,3)];
n=1;%迭代次数
while true
Y(n,:) = [n x0'];
y=G*x0+f;
if norm(y-x0)n=n+1;
x0=y;
end
输出数据如下（比你给的表格中多了第二行）：
1 0.00000000000000 0.00000000000000 0.00000000000000
2 1.75000000000000 3.50000000000000 3.00000000000000
3 1.87500000000000 3.93750000000000 2.96250000000000
4 1.99375000000000 3.99218750000000 2.99906250000000
5 1.99828125000000 3.99902343750000 2.99950781250000
6 1.99987890625000 3.99987792968750 2.99997597656250
7 1.99997548828125 3.99998474121094 2.99999324707031
8 1.99999787353516 3.99999809265137 2.99999953088379
9 1.99999964044189 3.99999976158142 2.99999990386047
10 1.99999996443024 3.99999997019768 2.99999999173256