how to solve system of linear equations with unorganized unknowns?
显示 更早的评论
hi everybody. I have a system of n equations n unknowns and there are two forms of this system. in the first form the system of equations is in the form of A*X=Y. in the second form the system of equations is in the form of A*X=B*Y. A and B are the matrix of constant factors. some of the unknowns are in the matrix X and the others are in the matrix Y(X and Y are columns). so the problem is that unknowns are not in one matrix so they are unorganized. i can define unknowns as zero. if there is another way to define unknowns, please let me know. so how can i solve these systems? please guide me how to solve these systems and how to define unknowns? there are several ways to organize unknowns and put them in just one matrix but i want to know if there is any way to solve the system without organizing unknowns. if all the unknowns were in one matrix then it would be easy to solve the system with linsolve in matlab.
采纳的回答
John D'Errico
2015-3-23
编辑：John D'Errico
2015-3-23
Just move all of the unknowns to the left hand side.
A*X - B*Y = 0
Think of it like
[A,-B]*[X;Y] = 0
Any knowns in that set among the a and y variables, you move to the right hand side and sum them up. They are all constants. I fail to see the problem. How would you solve it using pencil and paper? (Answer: Exactly the same way.)
My guess is you are getting confused because you think you have TWO sets of unknowns, mixed in with knowns, in both X and Y. In fact, you can do the entire procedure in only several lines of code, fully vectorized. TRY IT! You might surprise yourself. (Take a shot and try it. REALLY. See how it goes. If you still get confused, then just comment here and I'll show you how.)
5 个评论
John D'Errico
2015-3-23
Take a shot and try it. REALLY. See how it goes, on just a simple problem. If you still get confused, then just comment here, showing what you tried, and I'll show you how.
Omidreza Ghafarinejad
2015-3-23
John D'Errico
2015-3-23
编辑：John D'Errico
2015-3-23
So what? I never said the size of the array matters in the least. In fact, I have code that does similar things with many THOUSANDS of variables, some known, some unknown.
As for needing to use penalty functions, this is not remotely needed. Sorry, but not. As I said, this should all take only a very few lines, in EITHER form. So come on, MAKE AN EFFORT. TRY IT!
For example, suppose we have the problem
A = magic(3)
A =
8 1 6
3 5 7
4 9 2
B = (magic(3)-4)^2
B =
19 -5 -5
-5 19 -5
-5 -5 19
Yeah, I know, I'm not terribly creative with my matrices. Lets presume that A*X = B*Y.
Now, presume that we know the value of
X(1) == 1
X(3) == -2.5
Y(2) == 7
So, when we move everything to the left hand side combining the arrays such that
[A,-B]*[X;Y] = 0
knowns = [1 3 5];
unknowns = setdiff(1:6,knowns);
we can think of the vector of variables as
XY = NaN(6,1);
XY(knowns) = [1;-2.5;7];
Here, I've inserted NaN where we need to solve for an unknown.
OK, so I've done all the hard work here. What is the next step? If you don't think about it, you won't learn. There are three unknowns. Come on. What remains is just two lines of code.
Omidreza Ghafarinejad
2015-3-23
John D'Errico
2015-3-23
编辑：John D'Errico
2015-3-23
In the example I just gave, here is my solution...
AB = [A,-B];
XY(unknowns) = AB(:,unknowns)\(-AB(:,knowns)*XY(knowns))
XY =
1
14.625
-2.5
4.875
7
10
(As I said, two lines of code.)
As a test to confirm this is indeed a solution...
A*XY(1:3) - B*XY(4:6)
ans =
0
0
0
In the other case, where you have
A*X = Y
with some elements of both X and Y unknown, this is IDENTICAl to the above more general problem, except that here we have B=eye(numel(Y)). No penalty functions are needed. That is simply a bad idea when the solution is so trivial. Honestly, this really is not hard to do. I call it trivial, but I've been writing the linear algebra code for things like this for years now. No matter what, it is still easy.
更多回答（1 个）
Omidreza Ghafarinejad
2015-3-24
0 个投票
2 个评论
John D'Errico
2015-3-24
Errors on that order of magnitude are often the result of how you input the data. That is, if you entered the numbers to that precision, expect errors in the last digit or so, since the true numbers that were stored in MATLAB were not exactly those values.
As I said above, for the A*X=Y problem, just use B as an identity matrix, then you can solve it the same way.
Omidreza Ghafarinejad
2015-3-24
编辑：Omidreza Ghafarinejad
2015-3-24
类别
在 帮助中心 和 File Exchange 中查找有关 Operating on Diagonal Matrices 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
