如何找到具有固定分量的矩阵的所有可能组合

Question

Given a 3by2 matrix with constant values

B = [b1 b1;b2 b2;b3 b3] 

I need to make a code that would use the initial B matrix and place zeros progressively to any place of its current components. This means that we need factorial(6) combinations. The first combination which would be of no use, is when we replace all components with 6 zeros:

B_0 = [0 0;0 0;0 0]

The first useful output would be, placing 5 zeros in all places BUT the (1,1)

B_1 = [b1 0;0 0;0 0]

The second, would be placing zeros in all places but the (2,1):

B_2 = [0 0;b1 0;0 0]

so on for the third:

B_3 = [0 0;0 0;b1 0]

After finishing with placing all the combinations for 5 zeros, we start to add 4 zeros:

B_k = [b1 0;b2 0;0 0]
B_k+1 = [b1 0;0 0;b3 0]

etc. and then 3 zeros

B_n = [b1 0;b2 0;b3 0]
B_n+1 = [b1 0;b2 0;0 b3]

etc.

Until, we arrive to the last case of no zeros to be replaced which brings us to the initial matrix

B_6! = [b1 b1;b2 b2;b3 b3]

Answer 1

I made the assumption that there is a mistake in your question (according to my comment). So there is only 2^6 = 64 possible combinations.

In this case you can use:

m        = 6                           %vector length
[y{1:m}] = ndgrid(0:1);                %all combination of 0 and 1 with m possible permutations
y        = reshape(cat(m+1,y{:}),[],m);
res      = y.*[1 1 2 2 3 3]            %get the result

%reshape and permute
res = permute(reshape(res.',2,3,64),[2,1,3])

I've added a new dimension instead of creating 64 new variables (which is not the way to go).

result (before adding a new dimension and reshaping):

res =

   0   0   0   0   0   0
   1   0   0   0   0   0
   0   1   0   0   0   0
   1   1   0   0   0   0
   0   0   2   0   0   0
   1   0   2   0   0   0
   0   1   2   0   0   0
   1   1   2   0   0   0
   0   0   0   2   0   0
   ...