matlab求解线性方程组

直接解法

Ax=l，求x。利用逆矩阵、矩阵除法、x=inv（A'A）*A'l

lind.m文件如下，

%求解线性方程组
% x1 + 2x2 + 3x3 + 9x4 = 5
%2x1 + 2x2 + 5x3 + 4x4 = 2
%3x1 + 5x2 + x3  + 5x4 = 3
%7x1 + 4x2 + 2x3 -10x4 = 8
A=[1 2 3 9;2 2 5 4;3 5 1 5;7 4 2 -10];
b=[5 2 3 8]';
x=A\b
c=inv(A)*b
d=inv(A'*A)*A'*b

输出

>> lind.m

x =

    4.9227
   -3.2373
   -1.1680
    1.1173


c =

    4.9227
   -3.2373
   -1.1680
    1.1173


d =

    4.9227
   -3.2373
   -1.1680
    1.1173

Warning: Direct access of structure fields returned by a function call (e.g.,
 call to lind) is not allowed. See MATLAB 7.10 Release Notes, "Subscripting Into Function Return Values" for details. 
??? Attempt to reference field of non-structure array.

求线性方程组通解

lind.m内容如下

%求解线性方程组通解
%  X1  +5 *X2 -1 *X3  -1 *X4 = -1
%  X1  -2 *X2 +1 *X3  +3 *X4 =  3
%3*X1  +8 *X2 -1 *X3  +1 *X4 = 1
%  X1  -9 *X2 +3 *X3  +7 *X4 = 7
%系数矩阵和增广矩阵的秩都为2，说明有多组解，且基础解系4-2=2
%求基础解系用Z=null(A,'r')，可获得标准化的基础解系。特解采用广义逆函数pinv(A)，
%可求得A的广义逆矩阵，采用x0=pinv(A)*b求得特解，通解为x=k1*x1+k2*x2+...+kn*xn+x0
A=[1 5 -1 -1 ;
    1 -2 1 3;
    3 8 -1 1;
    1 -9 3 7];
b=[-1 3 1 7]';
r1=rank(A)%系数矩阵的秩
r2=rank(A,b)%增广矩阵的秩
Z=null(A,'r')%求基础解系
x0=pinv(A)*b%求得特解
%验证该解
k1=rand(1);
k2=rand(1);
x=k1*Z(:,1)+k2*Z(:,2)+x0
err=norm(A*x-b)

>> lind.m

r1 =

     2


r2 =

     2


Z =

   -0.4286   -1.8571
    0.2857    0.5714
    1.0000         0
         0    1.0000


x0 =

    0.3785
   -0.0876
    0.1873
    0.7530


x =

   -0.1805
    0.0994
    0.2848
    1.0315


err =

  1.8971e-015

矛盾方程组最小二乘解

%矛盾方程组的最小二乘解
%原始数据表
% x 1     1     2      2     2     3       3     4     4    4   5    6
% y 14.8  15.9  20.2   20.0  18.55   22.2  20.9  21  18.3 20.7 16.1 14.75
% 分别一次

A=[  1  1   2  2  2     3       3     4     4    4   5    6]';
t=ones(size(A));
A1=[t,A];
y= [14.8  15.9  20.2  20.0 18.55   22.2  20.9  21  18.3 20.7 16.1 14.75]';
%x=inv(A'*A)*A'*y
x1=pinv(A1)*y%线性回归方程系数
n=length(A);
rmse1=sqrt(sum((y-A1*x1).^2)/n)%均方误差
xsq=A.^2;
A2=[t,A,xsq]
x2=pinv(A2)*y%线性回归方程系数
rmse2=sqrt(sum((y-A2*x2).^2)/n)%均方误差
xmin=min(A);
xmax=max(A);
nx1=xmin:0.2:xmax;
ny1=x1(1)+x1(2)*nx1;
nx2=nx1;
ny2=x2(1)+x2(2)+x2(3)*nx2.^2;
plot(A',y,'*',nx1,ny1,'+b',nx2,ny2,'r')

输出

>> lind.m

x1 =

   18.8820
   -0.0861


rmse1 =

    2.5105


A2 =

     1     1     1
     1     1     1
     1     2     4
     1     2     4
     1     2     4
     1     3     9
     1     3     9
     1     4    16
     1     4    16
     1     4    16
     1     5    25
     1     6    36


x2 =

   10.3924
    6.3994
   -0.9793


rmse2 =

    1.0960