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LinAlg.EF

Computes a row echelon form of a matrix over F_2 with record keeping.
Syntax
          
LinAlg.EF(M:LIST, L1:LIST, L2:LIST):LIST

          

Description
Please note: The function(s) explained on this page is/are using the ApCoCoAServer. You will have to start the ApCoCoAServer in order to use it/them.

This function computes a row echelon form of a matrix M defined over the field F_2. It allows to keep record of the order of rows inside the matrix using the parameter L1. If the matrix contains some rows which are already in echelon form then the parameter L2 represent them with 0 and all others with 1.



Example
Use ZZ/(2)[x,y];
M := [
      [1, 1, 0, 1, 0],
      [0, 1, 1, 0, 1],
      [1, 0, 1, 0, 0], 
      [1, 1, 1, 0, 1]
     ];
-- order of lists in M
L1:=[1, 2, 3, 4]; 

-- 0 for lists which are already in echelon form and 1 for those to be reduced.
L2:=[0, 0, 1, 1];

LinAlg.EF(M, L1, L2);

-- CoCoAServer: computing Cpu Time = 0
-------------------------------
[ [[1, 1, 0, 1, 0], [0, 1, 1, 0, 1], [0, 0, 1, 1, 1], [0, 0, 0, 1, 1]], [1, 2, 4, 3], [0, 0, 1, 1]]

-- The last two lists represent the new order of lists in M.


Example
Use ZZ/(2)[x,y];
M := [
      [0, 1, 0, 1, 0],
      [0, 1, 0, 0, 1],
      [1, 0, 1, 1, 0], 
      [1, 1, 0, 0, 1]
     ];
-- order of lists in M
L1:=[1, 2, 3, 4]; 

-- 0 for lists which are already in echelon form and 1 for those to be reduced.
L2:=[0, 0, 1, 1];

LinAlg.EF(M, L1, L2);

-- CoCoAServer: computing Cpu Time = 0.015
-------------------------------
[[[1, 0, 1, 1, 0], [0, 1, 0, 1, 0], [0, 0, 1, 0, 1], [0, 0, 0, 1, 1]], [3, 1, 4, 2], [1, 0, 1, 0]]
-------------------------------

-- The last two lists represent the new order of lists in M.



See Also