# Linear Algebraic System Solution for Known and constrained variables

1. May 20, 2013

### Ronankeating

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,

2. May 20, 2013

### Stephen Tashi

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 $x_5 = x_{12}$ is expressed by the equation $x_5 - x_{12} = 0$. Of course the "target" $x_{13} = 25$ 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 $x_{13}$ 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. May 21, 2013

### SteamKing

Staff Emeritus
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:

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. May 21, 2013

### Ronankeating

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. May 21, 2013

### SteamKing

Staff Emeritus
For other types of constraint, one possible source of a solution would be to study a branch of mathematics called 'linear programming'.