Solving a tridiagonal linear system
3 = 7 10x 2 +9x 3 = 47 Now subtract 5 times equation #2 from equation #3, to eliminate x 2 from equation #3: 3x 1 +4x 2 = 11 +2x 2 +1x 3 = 7 +4x 3 = 12 The original tridiagonal equations now have a format known as upper triangular. Upper triangular systems are easy to solve. Now we do the back solve step. x 3 = 12=4 x 2 = (7 1x 3)=2 x 1 = (11 ... ................
................
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- a production problem
- 2 3 introduction to ees
- 3 5 finite differences fast poisson solvers
- disguised quadratic equations
- systems of linear equations in three variables
- a guide to writing your rst cfd solver
- solving literal equations
- the three moment equation for continuous beam analysis
- mathematica tutorial differential equation solving with
- solving a tridiagonal linear system
Related searches
- solving systems of linear equations calculator
- solving a system of equations using matrices
- solving a system of equations with matrices
- solving 3 variable linear equations
- solving systems of linear equations by substitution
- solving systems of linear equations
- solving a linear equation calculator
- solving systems of linear inequalities calculator
- solving one variable linear equations
- solving systems of linear equations worksheet
- solving a system of equations in matlab
- solving a system of equations by substitution