when u have small sys like this u can :
this one let's say
and when u have u can get ...
Can someone tell me where is my mistake, my solution doesn't agree with the book's answers.
1) Use Gaussian elimination, solve the equation for x and y:
2R2 - R1 -> R1
R1 - R2 -> R2
correct answer is x=2.2 and y=-1.6
Gaussian Elimination method is based on basic transformations of linear equations... (hope u know that.... there's 3 basic transformation)
now... if u look at sys : (sys No 1)
now we assume that and first equation of sys No 1 multiply with and we add it to the second equation and multiply with and we add to the 3rd equation and so on .... we get new sys (No 2) :
and so on...
using same procedure after repetition u get sys
witch is equivalent to sys No 1 ....
if sys have unique solution.... if sys have infinite solutions . But if in any repetition u get coefficients at least one equations are zeros and free member different from zero it means that sys have no solutions....
as for this small sys u have... first one multiply with witch in this case is ... and add to second and u have solution
Reply to Paymemoney at post # 3: yeKciM essentially DID use Gaussian elimination - but also back substitution. When doing Gaussian elimination, you have one high-level choice to make: do you go all the way to the identity matrix (in the square matrix case), or do you stop when your matrix is upper triangular, and then do back substitution? Numerical analysis suggests that Gaussian elimination with back substitution is the most efficient exact method known for solving a system of equations.
[EDIT]: The back substitution method also has the advantage in being able to work in more cases, such as when the original matrix is not invertible. In that case, you won't be able to get all the way to the identity matrix, and you'll need to use back substitution to find out the parametric dependence for your infinite solution set (unless, of course, your system is inconsistent, in which case all bets are off.)
hence it's ur decision how to eliminate things... and that can be quick way (if u do it right) or...
but using Gaussian algorithm u have correct way every time
You did all the same mathematical operations as are required for Gaussian elimination. About the only difference is that you broke out of the matrix notation. There's overlap between elimination and Gaussian elimination.
Incidentally, if you have thousands of equations, you won't use Gaussian elimination, due to round-off error. You'll use some iterative non-exact technique such as Gauss-Seidel or Jacobi iteration, which, believe it or not, will converge to the correct answer faster than if you used Gaussian elimination. It has to do with being a contraction mapping in a Banach space.
reply to Ackbeet
I'm sorry for wrong interpretation, my English's poor... especially when it comes to some "math words", and while translating it to English everything goes to .... I hope i'll do better in time
P.S. ur absolutely right Thank you, because ur last post i'll be more cautious, when it comes to translation and posting