在系统方程linsolve一致性问题
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嗨
运行这段代码时,我得到一个简单的解决方案。
x =兰德(1、3);
swc (1:3) = x + 1;
信谊(x1, x2) x3;
eqn11 = swc (1,1) = = x1 + 1;
eqn12 = swc (1、2) = = x2 + 1;
eqn13 = swc (1、3) = = x3 + 1;
[A, B] = equationsToMatrix ([eqn11、eqn12 eqn13], [x1, x2, x3]);
X = linsolve (A, B)
X =
1069854916266911/1125899906842624
77565677658637/2251799813685248
987964467329731/2251799813685248
x”
ans =
(0,0,0)
但是当我运行这个我获得一个错误:
x =兰德(1、3);
swc (1:3) = x + 1;
swc (1:3) = x + 2;
swc (1:3) = x + 3;
信谊(x1, x2) x3;
eqn11 = swc (1,1) = = x1 + 1;
eqn12 = swc (1、2) = = x2 + 1;
eqn13 = swc (1、3) = = x3 + 1;
eqn21 = swc (2, 1) = = x1 + 2;
eqn22 = swc (2, 2) = = x2 + 2;
eqn23 = swc (2、3) = = x3 + 2;
eqn31 = swc (3,1) = = x1 + 3;
eqn32 = swc (3 2) = = x2 + 3;
eqn33 = swc (3、3) = = x3 + 3;
[A, B] = equationsToMatrix ([eqn11、eqn21 eqn31, eqn12, eqn22, eqn32, eqn13, eqn23, eqn33], [x1, x2, x3]);
X = linsolve (A, B)
警告:解决方案不存在,因为系统是不一致的。
>在symengine
在信谊/ privBinaryOp(第1034行)
在信谊/ linsolve(第63行)
X =
正
正
正
问题是故意在这个小例子线条平行3×3 x的,因此返回相同的值,所以我们应该发现自己的依赖,但一致的方程组。(后链接
https://ionamaths.weebly.com/uploads/1/4/2/0/14204419/consistencyanddependency.pdf
)
为什么我不获得简单的x与前面的例子? ? ?
谢谢!
