1
$\begingroup$

I have a system of Linear Simultaneous Equations with n variables. I have solved the equations for particular values of co-efficients of variables & R.H.S. values. Now I am maintaining the co-efficients of variables same & just changing the R.H.S. values. So is there a method of computing the solution of this set of equations from the solution of previous set of equations?

In short, $E_1$: $[A][X] = [B]$ Now equations are solved & solution set $[X]$ is computed. $E_2$: $[A][X] = [C]$ Now how do I compute new solution set $[X]$ using previous solution set? If there is a method to do this quickly compared to again solving the equations then it will be very beneficial to me. Thank You.

$\endgroup$
  • $\begingroup$ If you had found the inverse of A, while calculating X, then you can directly use it in the next case as well. $\endgroup$ – lsp Apr 5 '13 at 10:06
  • $\begingroup$ sorry but I am calculating by matrix reduction method (Gauss-Jordan) which does not require inverse of matrix. $\endgroup$ – Cool_Coder Apr 5 '13 at 10:14
2
$\begingroup$

If you are going to solve $A \, x = b$ with Gauss-Jordan, you can directly solve for both right-hand sides instead. That is, you do Gauss-Jordan with $(a \, b)$ (that is the matrix with the two columns $a$ and $b$) on the right-hand side and perform the usual steps.

Btw: Similarly, you could solve for $A \, X = I$, where $I$ is the identity matrix. Then, you end with $X = A^{-1}$.

$\endgroup$
  • $\begingroup$ oops that was stupid of me to not realise that. Thanks for reminding me :) $\endgroup$ – Cool_Coder Apr 5 '13 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.