Solving System of Linear Equations with matrix operations

Question

Robert 2014-1-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/112949-solving-system-of-linear-equations-with-matrix-operations

评论： Robert 2014-1-20

I have a system of linear equations for a problem that can be generically expressed as Ax = b. Where b and A are both 3x1 matrices. I need to solve for the coefficients of x, given that x is a 3x3 matrix. I am having trouble figuring out how to do so, I have tried the backslash operator but without any luck. Thank you

4 个评论
显示 2更早的评论隐藏 2更早的评论

Amit 2014-1-20

You are very clear indeed. However, like I said, you are trying to solve 9 variables using 3 equation. Unless there are some other constrains you did not mentioned, you will get infinite solutions.

In simple ways, you can pick 6 random numbers from real number space, and solve for the rest 3 unknowns in your 3x3 matrix.

Mischa Kim 2014-1-20

编辑：Mischa Kim 2014-1-20

Careful, indeed. You are most likely working on a specfic set of problems (isotropic, linear, e.g.) for which a set of constraints needs to be satisfied.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Mischa Kim 2014-1-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/112949-solving-system-of-linear-equations-with-matrix-operations#answer_121444

在 MATLAB Online 中打开

Typically, x = A \ b works well. Make sure b is a column vector. Try:

A = rand(3);
b = [1; 2; 3];
x = A \ b;

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Robert 2014-1-20

Thank you, however I am trying to solve for the 3 x 3 matrix. The two 3 x 1 matrices are known values.

请先登录，再进行评论。

Answer 2

John D'Errico 2014-1-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/112949-solving-system-of-linear-equations-with-matrix-operations#answer_121534

As has been pointed out, this is impossible to do, at least if you want a unique solution. Time for you to do some reading in linear algebra.

Effectively the problem reduces to wanting to solve for 9 unknowns, but with only 3 pieces of information, so 3 stresses and corresponding strains. Actually, you gain a bit, since we know the matrix must be symmetric. So really there are only 6 unknowns, but even so, this is still insufficient information to learn the complete 3x3 matrix, and to do so uniquely.

Sorry, but merely wanting to do something is not sufficient for it to happen, even if you want it badly enough. Else there would be leagues and leagues of people all having won the lottery.