an answer for my yesterday question about determinant of matrix

Question

Yesterday and about five months ago, I asked two similar questions but nobody could answer me; maybe because my question was ambiguous. But, I myself tried and I could finally find my desired answer. But, now my question is that is there a command to shorten the following and does not have the problem of memory?

My yesterday question: assume that you have an $n \times n$ matrix $A$; here for simplicity I suppose that $n=2$. Some of specified entries of this matrix is zero and the others "which we know which ones", are nonzero but the nonzero entries can have 3 different cases; for example for the $2 \times 2$ matrix, $a_{11}=a_{22}=0$ and $a_{12},a_{2,1} \neq 0 $ and can be 2,3 or 5. Now, I want a code to show me all possible values for the determinant of $A$ (that here are $-4,-6,-9,-10,-15,-25)$; My code is as follows:

a = Tuples[{2, 3, 5}, {1, 2}];

b = Table[0, {2}, {2}];

For[i = 1, i <= Length[a], i++,
 {b[[1, 2]] = a[[i]][[1]][[1]]; b[[2, 1]] = a[[i]][[1]][[2]]; 
  Print[Det[b]]}]

which gives

-4

-6

-10

-6

-9

-15

-10

-15

-25

My quesion:Now, my question is that if $A$ has for example 20 nonzero entries, this code doesn't help; because of the first line which now will be

 a = Tuples[{2, 3, 5}, {1, 20}];

and the memory will be fulled.

Bests,

Answer 1

Try this

a = {2, 3, 5};
n=7; (*example 7 by 7 matrix*)
b = Table[0, {n}, {n}];
m=20; (*with 20 nonzero entries*)
For[i=0, i<Length[a]^m, i++,(*3^20 iterations*)
 j=IntegerDigits[i,Length[a],m]+1;(*build 20 subscripts*)
 {b[[1,2]], b[[2,1]],b[[2,7]],(*...all 20 positions*)} = Map[a[[#]]&,j];(*assign 20 values*) 
 Print[Det[b]]
]