怎么用MATLAB求解这个二次矩阵方程 如何用matlab解以下矩阵方程
\u7528matlab\u89e3\u77e9\u9635\u65b9\u7a0b1\u3001\u52a0\u51cf\u6cd5\u7684\u547d\u4ee4\u5f88\u7b80\u5355\uff0c\u76f4\u63a5\u7528\u52a0\u6216\u8005\u51cf\u53f7\u5c31\u53ef\u4ee5\u4e86\u3002\u5982\uff1ac=a+bd=a-b\u3002
2\u3001\u4e00\u822c\u4e58\u6cd5\uff1ac=a*b,\u8981\u6c42a\u7684\u5217\u6570\u7b49\u4e8eb\u7684\u884c\u6570\u3002\u5982\u679ca,b\u662f\u4e00\u822c\u7684\u5411\u91cf\uff0c\u5982a=[1,2,3] b=[3,4,5]\u70b9\u79ef\uff1adot(a,b), \u53c9\u79ef\uff1across\uff08a,b)\u5377\u79ef\uff1aconv(a,b)\u3002
3\u3001x=a\b\u5982\u679cax=b\uff0c\u5219 x=a\b\u662f\u77e9\u9635\u65b9\u7a0b\u7684\u89e3\u3002x=b/a\u5982\u679cxa=b, \u5219x=b/a\u662f\u77e9\u9635\u65b9\u7a0b\u7684\u89e3\u3002
4\u3001\u8f6c\u7f6e\u65f6\uff0c\u77e9\u9635\u7684\u7b2c\u4e00\u884c\u53d8\u6210\u7b2c\u4e00\u5217\uff0c\u7b2c\u4e8c\u884c\u53d8\u6210\u7b2c\u4e8c\u5217\uff0c\u3002\u3002\u3002x=a\u3002
5\u3001\u6c42\u9006\uff1a\u8981\u6c42\u77e9\u9635\u4e3a\u65b9\u9635\u3002\u8fd9\u5728\u77e9\u9635\u8fd0\u7b97\u4e2d\u5f88\u5e38\u7528\u3002x=inv(a)\u3002\u8fd9\u51e0\u79cd\u65b9\u5f0f\u90fd\u53ef\u4ee5\u89e3\u77e9\u9635\u65b9\u7a0b\u3002
matlab\u4e5f\u4e0d\u80fd\u89e3\u8fd9\u4e48\u590d\u6742\u7684\u65b9\u7a0b
\u624b\u52a8\u52a0\u4e0amatlab\u8f85\u52a9\u5e94\u8be5\u5e94\u8be5\u6bd4\u8f83\u597d\u89e3\u51b3\u95ee\u9898
\u9996\u5148\uff0c\u7b80\u5355\u8d77\u89c1\uff0c\u5148\u4e0d\u7528\u89d2\u5ea6\u5236\uff0c\u76f4\u63a5\u4e2d\u5f27\u5ea6\u5236
\u8bbec1=(pi*sita1)/180\uff0cc2=(pi*sita2)/180;
A1=[ cos(c1), -sin(c1), 0, cos(c1);
sin(c1), cos(c1), 0, sin(c1);
0, 0, 1, 0;
0, 0, 0, 1];
A2=[ cos(c2), -sin(c2), 0, cos(c2);
sin(c2), cos(c2), 0, sin(c2);
0, 0, 1, 0;
0, 0, 0, 1];
simplify(A1*A2)
[ cos(c1 + c2), -sin(c1 + c2), 0, cos(c1 + c2) + cos(c1)]
[ sin(c1 + c2), cos(c1 + c2), 0, sin(c1 + c2) + sin(c1)]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]
\u548cOHT\u5bf9\u6bd4
OTH =[-0.2924 -0.9563 0 0.6978;
0.9563 -0.2924 0 0.8172 ;
0 0 1 0
0 0 0 1]
\u5f97\u5230
cos(c1+c2)=-0.2924
sin(c1+c2)=0.9563
cos(c1 + c2) + cos(c1)= 0.6978
sin(c1 + c2) + sin(c1)= 0.8172
cos(c1)=0.9902
sin(c1)=0.1391
\u89e3\u5f97
c1~=8\u5ea6
c1+c2~=107\u5ea6
\u6240\u4ee5 c2=99\u5ea6
a0=[2 3;6 4];
a1=[-10 1;-20 3];
a2=[1 3;5 2];
syms p1 p2 p3 p4 p;
p=[p1 p2;p3 p4]
%二次矩阵方程是：p^2*a2+p*a1+a0=0;
eq=p^2*a2+p*a1+a0;
[p1,p2,p3,p4]=solve(eq(1,1),eq(1,2),eq(2,1),eq(2,2),p1,p2,p3,p4);
p1=double(p1);p2=double(p2);p3=double(p3);p4=double(p4);
t=1;
[m,n]=size(p1);
pp=zeros(2,2,m);
for i=1:m
if (imag(p1(i))==0)&(imag(p2(i))==0)&(imag(p3(i))==0)&(imag(p4(i))==0)
P1=p1(i,1);P2=p2(i,1);P3=p3(i,1);P4=p4(i,1);
pp(:,:,t)=[P1,P2;P3,P4];
t=t+1;
end
end
pp=subs(pp);
P=pp(:,:,1:t-1)
p =
[ p1, p2]
[ p3, p4]
P(:,:,1) =
11.1677 -5.6672
21.7147 -10.9798
P(:,:,2) =
-3.5248 4.0911
-5.8176 6.2600
带回去验算：误差10^-13
P(:,:,1)^2*a2+P(:,:,1)*a1+a0
ans =
1.0e-013 *
0.1421 -0.0711
0.5684 -0.1776
>> P(:,:,2)^2*a2+P(:,:,2)*a1+a0
ans =
1.0e-013 *
0.0711 0.1243
0.2842 0.0178
求逆
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