Difference between revisions of "ApCoCoA1:LinBox.REF"
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Revision as of 10:15, 13 September 2019
LinBox.REF
Computes a row echelon form of a matrix.
Syntax
LinBox.REF(M:MAT, CompRREF:BOOL):MAT LinBox.REF(M:MAT, P:INT, CompRREF:BOOL):MAT
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 allows you to compute a (reduced) row echelon form of M over a finite field. If you want to use the first version without the parameter P, the components of the input matrix M must castable to type ZMOD and your current working ring must be a finite field. The second version of this function lets you compute a (reduced) row echelon form of M mod P and the components of M must be of type INT.
The parameter CompRREF lets you specify if you want to compute a row echelon form or the reduced row echelon form of M. If CompRREF is set to TRUE, the reduced row echelon form will be computed, and if it is set to FALSE, a row echelon form where all pivot elements are equal to one will be computed.
The return value of both functions is the computed (reduced) row echelon form of M.
@param M A matrix whose (reduced) row echelon form to compute. If parameter P is given, the components of M must be of type INT. Otherwise, they must be of type ZMOD.
@param CompRREF Set to TRUE if you want to compute the reduced row echelon form of M or to FALSE otherwise.
@return A (reduced) row echelon form of M.
The following parameter is optional.
@param P An integer value. If P is specified, the (reduced) row echelon form computation will be carried out over the ring Z/pZ.
Example
Use ZZ/(239)[x]; M := Mat([[1, 2, 3], [4, 5, 6], [7, 8, 9], [11, 12, 13]]); M; LinBox.REF(M, FALSE); Mat([ [1, 2, 3], [4, 5, 6], [7, 8, 9], [11, 12, 13] ])  Mat([ [1 % 239, 2 % 239, 3 % 239], [0 % 239, 1 % 239, 2 % 239], [0 % 239, 0 % 239, 0 % 239], [0 % 239, 0 % 239, 0 % 239] ]) 
Example
Use QQ[x,y]; M := Mat([[ 1, 1, 2], [200, 3000, 1], [2, 5, 17], [1, 1, 1]]); M; LinBox.REF(M, 17, TRUE); Mat([ [1, 1, 2], [200, 3000, 1], [2, 5, 17], [1, 1, 1] ])  Mat([ [1, 0, 0], [0, 1, 0], [0, 0, 1], [0, 0, 0] ]) 