# Question:Vector equation: generate matrix for substitution (rows contain different sets of substitutions)

## Question:Vector equation: generate matrix for substitution (rows contain different sets of substitutions)

Maple 2015

I have the following problem consisting of multiple seteps.

I have a vector equation consisting of n equations with n parameters (a[n]). Usually the n <= 15. As example data I will use the case n=4.

equations := Vector[column]([ a[1], a[2], a[3],a[4]])-Vector[column]([ b[1], b[2], b[3],b[4]])=0;

The first thing I want to create is a matrix with a format 2^n x n (here: rows=16 by columns=4). The matrix only consists of ones and zeros which contains all possible combinations of ones and zeros. E.G. for n=4

subsMatrix := Matrix([[ 0 , 0 , 0 , 0 ],[ 1 , 0 , 0 , 0 ],[ 0 , 1 , 0 , 0 ],[ 0 , 0 , 1 , 0 ],[0 , 0 , 0 , 1],[1, 1 , 0 , 0],[1, 0 , 1 , 0],[1, 0 , 0 , 1],[0, 1 , 1 , 0],[0, 1 , 0 , 1],[0,0,1,1],[1, 1 , 1 , 0],[1, 1, 0 , 1],[1, 0 , 1 , 1],[0, 1 , 1 , 1],[1, 1 , 1 , 1]]);

Question 1: How do I create such a matrix for the general case? I have absolutely no idea how to achieve this with Matple

The next thing I want to do is to use the rows as substitution equations for the a[i] values, only if the value of the subsMatrix is 0. E.G. in the first case I want to set a[1]=a[2]=a[3]=a[4]=0, then a[2]=a[3]=a[4]=0, then a[1]=a[3]=a[4]=0, and so forth and save the equation as a new equation

I tried the following:

rows:=RowDimension(subsMatrix);

columns:=ColumnDimension(subsMatrix);

for i from 1 by 1 while  i<=rows do

subsEquations[i]:=equations

for j from  1 by 1 while  j<=columns do

if subsMatrix[i,j] =0 then

subsEquations[i]:= subs(a[j]=subsMatrix[i,j],subsEquations[i])

else

#do nothing if the value in the subsMatrix[i,j]=1

end if

end do:

end do:

Question 2: What is my error? Maple says the loop is indeterminate. But I don't see why it is not working.

I would be thankful if someone could help me out. I am open to other kind of strategies to this problem :).

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