- Many physical situations are modelled with a large set of linear equations.
- The equations will depend on the geometry and certain external factors that will determine the right-hand sides.
- If we want the solution for many different values of these right-hand sides, it is inefficient to solve the system from the start with each one of the right-hand-side values using the equivalent of the coefficient matrix is preferred.
- Suppose we have solved the system by Gaussian elimination. We now know the equivalent of A: .
- Consider now that we want to solve with some new -vector. We can write
- The product of and is a vector, call it . Now, we can solve for from and this is readily done because is lower-triangular and we get by forward-substitution. Call the solution .
- Going back to the original , we see that, from , we can get from , which is again readily done by back-substitution ( is upper-triangular).
- i.e., Solve , where we already have its and matrices:
Suppose that the -vector is
. We first get from by forward substitution:
and use it to compute from :
- Now, if we want the solution with a different -vector;
we just do to get
and then use this in to find the new :
2004-12-28