Solving System of Linear Equations with matrix operations

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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 Comments
Amit
Amit on 20 Jan 2014
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 on 20 Jan 2014
Edited: Mischa Kim on 20 Jan 2014
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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Accepted Answer

Mischa Kim
Mischa Kim on 20 Jan 2014
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 Comment
Robert
Robert on 20 Jan 2014
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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More Answers (1)

John D'Errico
John D'Errico on 20 Jan 2014
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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