A matrix of `3xx3` order, should be filled either by 0 or 1 and sum of all elements should be prime number. Then the number of such matrix is equal to

Answer

To solve the problem of finding the number of 3x3 matrices filled with either 0 or 1 such that the sum of all elements is a prime number, we can follow these steps:

### Step 1: Determine the total number of elements in the matrix
A 3x3 matrix has a total of 9 elements.

### Step 2: Identify possible sums that are prime numbers
The possible sums of the elements (which can either be 0 or 1) can range from 0 (all elements are 0) to 9 (all elements are 1). The prime numbers within this range are:
- 2
- 3
- 5
- 7

### Step 3: Calculate the number of matrices for each prime sum
We will calculate the number of matrices for each prime number sum using the binomial coefficient, which counts the number of ways to choose k elements from n elements.

1. **Sum = 2**: This means 2 elements are 1 and 7 elements are 0.
\[ \text{Number of matrices} = \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36 \]

2. **Sum = 3**: This means 3 elements are 1 and 6 elements are 0.
\[ \text{Number of matrices} = \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

3. **Sum = 5**: This means 5 elements are 1 and 4 elements are 0.
\[ \text{Number of matrices} = \binom{9}{5} = \binom{9}{4} = \frac{9!}{5!(9-5)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

4. **Sum = 7**: This means 7 elements are 1 and 2 elements are 0.
\[ \text{Number of matrices} = \binom{9}{7} = \binom{9}{2} = \frac{9!}{7!(9-7)!} = \frac{9 \times 8}{2 \times 1} = 36 \]

### Step 4: Sum the number of matrices for all prime sums
Now, we will add the number of matrices for each prime sum:
\[ \text{Total matrices} = 36 + 84 + 126 + 36 = 282 \]

### Final Answer
Thus, the total number of 3x3 matrices filled with 0s and 1s such that the sum of all elements is a prime number is **282**.

---