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This is how I did it:
% Set up parameters. rows = 10; columns = 15; onesPerColumn = 4; % Initialize matrix. m = zeros(rows, columns, 'int32')
for col = 1 : columns % Get random order of rows. randRows = randperm(rows); % Pick out "onesPerColumn" rows that will be set to 1. rowsWithOne = randRows(1:onesPerColumn); % Set those rows only to 1 for this column. m(rowsWithOne, col) = 1; end % Display m m
% Define your matrix size and randomly pick the number of ones matSz = [20, 40]; numOnes = randi(matSz(1)); % Make your matrix myMat = false(matSz); for i = 1:matSz(2) myMat(randperm(matSz(1),numOnes), i) = true; end % Check that all went as planned sum(myMat,1)==numOnes
Since it's a RANDOM matrix, you are not guaranteed to have the same amount of one's and zero's (at least it seems logical to me). In order to get about 1/2 probability you need large matrix. You can try this:
Check out help file for rand vs. randn
I also need to make sure that this matrix is invertible.
What I currently do is check det(A)=0 & rank(A)<3 using a while loop. But sometimes the while loop runs infinitely and the script does not respond.
Is there any other way to check for non-singular matrices?
Anne, you need to define what it means to take an inverse for you binary matrix. You can treat the binary matrix as being composed of the real numbers 0 and 1 and then do an arithmetic inverse on the array, ending up with a non-binary array. Or you can treat the binary matrix as being composed of boolean values over a field with the '*' being equivalent to 'or' and '+' being equivalent to xor, and the task is then to find a second binary matrix such that matrix multiplication using those operations produces the identity matrix.
If you want the inverse to be a binary matrix instead of a real-valued matrix, please see this earlier Question: