How to make a matrix invertable when determinant is zero? I am analyzing the forces in a truss, and have found 15 unknowns. Consequently I have 15 equations. 
I have put these equations in Excel & Matlab, but found that when I try to take the inverse of my $15\times 15$ matrix, the determinant is $0$.
What should I do to make my matrix invertable? 
I have exhausted all my ideas at this point.
 A: If you have grounds for believing that there should be a unique answer, then you must have made a mistake in setting up your equations.  At least one of the equations follows from or is contradicted by 
the others, and some constraint has been left out.
A: Such a system might still be solvable. The usual technique is called "row reduction" or "Gauss-Jordan elimination". Matlab probably does it, Wolfram Alpha does and other sites too. 
A: It is not invertible.
At least one of the equations is a scalar multiple of another linear combination of the others.
There is not only one solution. If it is solvable then there are infinitely many.  You can row reduce it and then determine the set of solutions.
There is an example that illustrates the technique here:
https://www.math.hmc.edu/calculus/tutorials/linearsystems/
This is an online row reducer (But there is probably another wolfram alpha tool that will just solve the equations). 
https://www.wolframalpha.com/input/?i=row+reduce+%7B%7B2,1,0,-3%7D,%7B3,-1,0,1%7D,%7B1,4,-2,-5%7D%7D&lk=3
Edit:
But, I've done some more research and there is something called "regularization" that you might want to check out. This is regularization applied to matrices. It probably doesn't apply yo what you're working on.:
https://en.wikipedia.org/wiki/Tikhonov_regularization
As the article states, this appears to be "artificial."
I want to add that a physical problem might have an infinite number of solutions and still be meaningful. In such cases, there are probably realistic and unrealistic solutions.   
