Linear Algebraic System Solution for Known and constrained variables

In summary: But if you're looking for an algorithm that solves this type of problem for you automatically, I don't know of one.
  • #1
Ronankeating
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Dear All,

If I have the linear algebraic system where its composed as of matrices in that form K*X=F, what column/row operations should I perform if I want to solve it where some of the X variables are known (targeted values) or if I want to solve when variables are constrained relative to each other.

For example, say that I want to solve linear algebraic equations(K*X=F) when its constrained (e.g. X5=X12=X125=X128) or when it has targeted values such as (X13=25; X23=5.4; X33=13 etc...)

Your help will be appreciated!

Regards,
 
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  • #2
That's a good question! Off the top of my head, the brute force approach would be to add more equations and enlarge the matrix K and the vector F so they have more columns. That would work if you have software that solve systems where there are more equations than unknowns. For example [itex] x_5 = x_{12} [/itex] is expressed by the equation [itex] x_5 - x_{12} = 0 [/itex]. Of course the "target" [itex] x_{13} = 25 [/itex] is itself an equation.

However, I don't know that the average software routine for solving linear equations can handle the problem efficiently. For example, if it uses Gaussian elimination, I don't know that it would perform the operations in an order that would make the answer for [itex] x_{13} [/itex] exactly equal to 25. If the software computes the answer by finding the pseudoinverse of a non-square matrix then roundoff errors might be a problem.

Obviously the proper way to do things would be to use substitutions and rewrite your original system of equations so its has fewere variables. You seem to be asking for a matrix oriented method of implementing the substitutions. Off hand, I don't know one. I'll have to think about it.
 
  • #3
Your question about how to apply known values of X when solving the matrix equation K*X = F has been discussed recently in this thread:

https://www.physicsforums.com/showthread.php?t=691663

In short, using the known values of X, the matrices K and F can be modified without adding additional equations such that the original matrix equation can be solved using elimination or whatever solution method you choose. The attached pdf in the thread will illustrate the procedure.
 
  • #4
SteamKing said:
Your question about how to apply known values of X when solving the matrix equation K*X = F has been discussed recently in this thread:

https://www.physicsforums.com/showthread.php?t=691663

Yes SteamKing you've answered this question in the shown thread and thank you for your help, but if you remember I was looking additionally for the constrained type solution also where I couldn't find entry point.

What inspires me actually to ask those question again, I still do have belief deep down inside my brain that I'm following very old methods for this, instead of contemporary and robust ones.
 
  • #5
For other types of constraint, one possible source of a solution would be to study a branch of mathematics called 'linear programming'.
 

FAQ: Linear Algebraic System Solution for Known and constrained variables

What is a linear algebraic system?

A linear algebraic system is a collection of linear equations that are used to solve for unknown variables. The system can be represented in matrix form and solved using various methods, such as Gaussian elimination or matrix inversion.

What are known and constrained variables?

Known variables are values that are given in the linear system and do not need to be solved for. Constrained variables are variables that have certain restrictions or conditions placed on them, often in the form of equations or inequalities. These constraints help to narrow down the possible solutions for the system.

How is a linear algebraic system solved for known and constrained variables?

To solve for known and constrained variables in a linear algebraic system, the system needs to be set up in matrix form and then solved using various methods. The known variables can be placed in a separate matrix, while the constrained variables can be represented as coefficients in the system of equations. The system can then be solved using techniques such as Gaussian elimination or matrix inversion.

What are some applications of solving linear algebraic systems for known and constrained variables?

Linear algebraic systems with known and constrained variables are used in various fields, including engineering, physics, and economics. They can be used to model and solve real-world problems, such as optimizing production processes, predicting stock market trends, or analyzing the behavior of electrical circuits.

Can a linear algebraic system have multiple solutions for known and constrained variables?

Yes, a linear algebraic system can have multiple solutions for known and constrained variables. This is known as an underdetermined system, where there are more unknown variables than equations. In this case, the system will have an infinite number of solutions, which can be represented by a set of equations with free variables.

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