用matlab解矩阵方程 用matlab解矩阵一般方法
\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
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\u7ea6\u5b9a\uff1a
a=[1,3,5;2,4,6;7,9,8] b=[9,6,4;3,4,5;2,3,4]
\u5de5\u5177/\u539f\u6599
matlab
\u65b9\u6cd5/\u6b65\u9aa4
\u52a0\u548c\u51cf\uff1a
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c=a+b
d=a-b
\u4e58\u6cd5\uff1a
\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\uff1a dot(a,b),
\u53c9\u79ef\uff1a cross\uff08a,b)
\u5377\u79ef\uff1a conv(a,b)
\u9664\u6cd5\uff1a\u4e00\u822c\u5728\u89e3\u7ebf\u6027\u65b9\u7a0b\u7ec4\u65f6\u4f1a\u7528\u5230\u3002
x=a\b \u5982\u679cax=b\uff0c\u5219 x=a\b\u662f\u77e9\u9635\u65b9\u7a0b\u7684\u89e3\u3002
x=b/a \u5982\u679cxa=b, \u5219x=b/a\u662f\u77e9\u9635\u65b9\u7a0b\u7684\u89e3\u3002
\u8f6c\u7f6e\uff1a
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x=a.'
\u6c42\u9006\uff1a
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x=inv(a)
要注意的是矩阵顺序要正确排好
看起来解法是对的。你可以用得到的X进行验证：看看A*X*B=C是否成立。
如果验证正确，那肯定就不是矩阵计算的问题了。
绛旓細a=[1 0;0 2];b=[1;3];syms x1 x2;y=a*[x1;x2]+b;s=solve(y(1),y(2),'x1','x2');s.x1 %w2鍊 s.x2 %w2鍊
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