MATLAB 矩阵问题
我有一个方程组（总共 5 个），有 5 个未知数。我已将它们放入矩阵中以尝试解决，但我不确定结果是否正确。基本上设置是 AX = B，其中 A、X 和 B 是矩阵。 A 是 5x5，X 是 1x5，B 是 5x1。
当我使用 MATLAB 使用公式 X = A\B 求解 X 时，它会向我发出警告：
Matrix is singular to working precision.
并为所有 5 个 X 未知数给出 0，但如果我说 X = B\A 事实并非如此，并给出了 5 个 X 未知数的值。
有人知道我做错了什么吗？如果这很重要，这就是我的 X 矩阵的样子：
X= [1/C3; 1/P1; 1/P2; 1/P3; 1/P4]
其中 C3、P1、P2、 P3、P4 是我的未知数。
I have a system of equations (5 in total) with 5 unknowns. I've set these out into matrices to try solve, but I'm not sure if this comes out right. Basically the setup is AX = B, where A,X, and B are matrices. A is a 5x5, X is a 1x5 and B is a 5x1.
When I use MATLAB to solve for X using the formula X = A\B, it gives me a warning:
Matrix is singular to working precision.
and gives me 0 for all 5 X unknowns, but if I say X = B\A it doesnt, and gives me values for the 5 X unknowns.
Anyone know what I'm doing wrong? In case this is important, this is what my X matrix looks like:
X= [1/C3; 1/P1; 1/P2; 1/P3; 1/P4]
Where C3, P1, P2, P3, P4 are my unknowns.
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您的矩阵是奇异，这意味着它的行列式为 0。这样的方程组无法为您提供足够的信息来找到唯一的解决方案。我在你的问题中看到的一件奇怪的事情是 X 是 1x5，而 B 是 5x1。这不是提出问题的正确方式。 X 和 B 都必须为 5x1。如果您想知道，这不是 Matlab 的东西 - 这是线性代数的东西。此
[5x5]*[1x5]是非法的。此[5x5]*[5x1]生成[5x1]结果。此[1x5]*[5x5]生成[1x5]结果。首先检查代数，然后检查行列式（Matlab 中的det函数）是否为 0。Your matrix is singular, which means its determinant is 0. Such system of equations does not give you enough information to find a unique solution. One odd thing I see in your question is that X is 1x5 while B is 5x1. This is not a correct way of posing the problem. Both X and B must be 5x1. In case you're wondering, this is not a Matlab thing - this is a linear algebra thing. This
[5x5]*[1x5]is illegal. This[5x5]*[5x1]produces a[5x1]result. This[1x5]*[5x5]produces a[1x5]result. Check you algebra first, and then check whether the determinant (detfunction in Matlab) is 0.所以，接下来的事情就是弄清楚为什么
A是单数。 （请注意，在具有平方和单数A的情况下，您可能想要求解A x = b，但仅在
b在A的范围空间内。）也许你可以写出你的矩阵
A和向量b（因为它只是5x5）？或者解释一下你是如何创建它的。这可能会提供一些线索，说明为什么A未满秩，或者为什么b不在A的范围空间中。So, the next thing is to figure out why
Ais singular. (Note that it's possible that you'd want to solveA x = bin cases with square and singular
A, but they'd only be in cases wherebis in the range space ofA.)Maybe you can write your matrix
Aand vectorbout (since it's only 5x5)? Or explain how you create it. That might give a clue as to whyAisn't full rank or as to whybisn't in the range space ofA.