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Solving System of Linear Equations with matrix operations

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Robert
Robert 2014 年 1 月 20 日
コメント済み: Robert 2014 年 1 月 20 日
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 件のコメント
Amit
Amit 2014 年 1 月 20 日
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
Mischa Kim 2014 年 1 月 20 日
編集済み: Mischa Kim 2014 年 1 月 20 日
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.

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採用された回答

Mischa Kim
Mischa Kim 2014 年 1 月 20 日
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 件のコメント
Robert
Robert 2014 年 1 月 20 日
Thank you, however I am trying to solve for the 3 x 3 matrix. The two 3 x 1 matrices are known values.

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その他の回答 (1 件)

John D'Errico
John D'Errico 2014 年 1 月 20 日
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.

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