[Scilab-users] a linear equation

[fujimoto2005]

Dear all
I found a solution.
I get the column echelon form of A  by X=rref(A')'.
Then the rows' numbers with the non-zero pivot of X are the independent
rows' numbers of A.

In this problem 26th row and 27th row are redundant.
26th rows are represented by linear combinations of 24th row and 1st-23rd
rows.
27th rows are represented by a linear combination of 25th row and 1st-23rd
rows.

Best regards.



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