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martes, 20 de julio de 2010

Gaussian elimination

Gaussian elimination

We can solve linear equation systems, or matrices of the form:
 Ax=B

Or,
                                         [a_(11) a_(12) ... a_(1k); a_(21) a_(22) ... a_(2k); | | ... |; a_(k1) a_(k2) ... a_(kk)][x_1; x_2; |; x_k]=[b_1; b_2; |; b_k],  

If you want you can use the matrix in extended form:
                                            [a_(11) a_(12) ... a_(1k); a_(21) a_(22) ... a_(2k); | | ... |; a_(k1) a_(k2) ... a_(kk)|b_1; b_2; |; b_k][x_1; x_2; |; x_k].

You have to use elementary row operations till the matrix will be an upper triangular, like this:
                                                [a_(11)^' a_(12)^' ... a_(1k)^'; 0 a_(22)^' ... a_(2k)^'; | | ... |; 0 0 ... a_(kk)^'|b_1^'; b_2^'; |; b_k^'].

At this moment we can solve the first unknown value, and using following expression we can obtain all of them:
                                                         x_i=1/(a_(ii)^')(b_i^'-sum_(j=i+1)^ka_(ij)^'x_j).
For instance, we have following linear equations system: 
Lets write it like a matrix:

Applying Gaussian Elimination we have: 

Its a upper triangular matrix, here we know that:
Z = 2

Going back we can obtain X and Y values, because:
Y - 5Z = -9

Solving for Y:
Y=1

Next step is find X:
X + Y + 2Z = 8

where, 
X=3



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