Then number of all possible matrices of order 3 X 3 with each entry 0 and 1 is : (A) 27 (B) 18 (C) 81 (D) 512 The correct answer is (D) But I don't understand the question and the answer as well!! Can anyone please explain me this question and how to get and answer? :)

Dear Chanchal Kumar,

Single space can be filled in 2 ways!!!!! So, 9 spaces can be filled in 2⁹ ways!!!!!!!

 

         Good Luck!!!!!!!!

 

 

          Plz. approve my answer by clicking 'Yes' given below!!!!!!! Plz. Plz. Plz.!!!!!!! Don't forget!!!!!!!!

firstly, your Q. is incomplete. how can it start with then?

the rest asks you the all possible matrices you can have by entering only two digitsi.e. 1 or 0

Now for each entry we have 2 choices .i.e it can be filled in two ways.And there are 9 place as it is a 3*3 matrix

Thus the total no. of ways of filling all the entries are 2*2*2*2*2*2*2*2*2=512 and thats your answer

if you like the answer,please approve it 

Dear Student,
Please find below the solution to your problem.

The number of possible entries in a matrix is 3*3 = 9
Every choices has 2 entries 0,1.
The total number of choices is 2^9
= 512.

Thanks and Regards