Whenever the 1st row is 0 then its determent is 0, and similarly if any 2 or more rows are linearly dependent then its |det|=0
1st row should not be all zero- hence #Cases=(2^4-1)
2nd row should be non-zero and not linearly depending on 1st row- So #Cases=(2^4-2)
3rd row, it cannot be either of those rows(i.e 1st or 2nd row) OR the sum (sum of 1st and 2nd row) OR all 0 entry in 3rd row, so #Cases=(2^4-4)
4th row, it cannot be either of those rows( i.e 1st , 2nd or 3rd row)- #Cases=3
OR cannot be the linear combination of any two row of 1st ,2nd and 3rd – #Cases=3C2=3
OR cannot be the linear combination of all row 1st ,2nd and 3rd – #Cases=1
OR cannot be the all entry are zero— #Cases=1
So total number of cases for 4th row = (2^4-8)
Hence total possible matrix = (2^4-1).(2^4-2).(2^4-4).(2^4-8)= 15*14*12*8= 20160
A should be the answer.