Question

The number of matrices of order 3 × 3, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is _________.

Answer 1

Correct Answer: 282

The number of 3×3 matrices whose entries are either 0 or 1 is calculated as follows: each entry has 2 choices (either 0 or 1). Therefore, the total number of such matrices is 29 = 512.

Next, we need to compute how many of these matrices have a sum of all entries that is a prime number. The sum of the entries is a number between 0 and 9 (inclusive) since the matrix can have 0 ones (all zero matrix) to 9 ones (all one matrix). We evaluate which of these sums are prime numbers. The prime numbers between 0 and 9 are: 2, 3, 5, 7.

We compute the number of matrices for each prime sum:

Sum = 2: Choose 2 positions from 9 to be 1, i.e., combinations of 9 taken 2 = C(9,2)=36. Sum = 3: Choose 3 positions from 9 to be 1, i.e., combinations of 9 taken 3 = C(9,3)=84.  Sum = 5: Choose 5 positions from 9 to be 1, i.e., combinations of 9 taken 5 = C(9,5)=126. Sum = 7: Choose 7 positions from 9 to be 1, i.e., combinations of 9 taken 7 = C(9,7)=36.

Adding these gives: 36 + 84 + 126 + 36 = 282

This confirms that the computed number of such matrices is 282, which is within the expected range (282,282).

Therefore, the number of 3×3 matrices with entries 0 or 1, and a prime sum is 282.

Answer 2

In a 3 × 3 order matrix there are 9 entries.
These nine entries are zero or one.
The sum of positive prime entries are 2, 3, 5 or 7.
Total possible matrices
=\(\frac{9!}{2!⋅7!}\)+\(\frac{9!}{3!⋅6!}\)+\(\frac{9!}{5!⋅4!}\)+\(\frac{9!}{2!⋅7!}\)
= 36 + 84 + 126 + 36
= 282