Key Concepts and Formulas

• Combinations (Binomial Coefficient): The number of ways to choose $k$ items from a set of $n$ distinct items, where the order does not matter, is given by $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
• Prime Number: A natural number greater than 1 that has no positive divisors other than 1 and itself.
• Combination Identity: $\binom{n}{k} = \binom{n}{n-k}$. This property is useful for simplifying calculations.

Step-by-Step Solution

Step 1: Understanding the Problem and Matrix Structure

We are given a $3 \times 3$ matrix where each entry is either 0 or 1. The goal is to find the number of such matrices where the sum of all entries is a prime number. Since each entry is either 0 or 1, the sum of all entries is simply the number of 1s in the matrix. A $3 \times 3$ matrix has 9 entries.

Step 2: Determining the Possible Range for the Sum of Entries

The sum of the entries can range from 0 (all entries are 0) to 9 (all entries are 1). Therefore, the possible sums are the integers from 0 to 9 inclusive.

Step 3: Identifying Prime Numbers within the Possible Range

We need to identify the prime numbers between 0 and 9.

• 0 is not a prime number.
• 1 is not a prime number.
• 2 is a prime number.
• 3 is a prime number.
• 4 is not a prime number.
• 5 is a prime number.
• 6 is not a prime number.
• 7 is a prime number.
• 8 is not a prime number.
• 9 is not a prime number.

Thus, the possible prime sums are 2, 3, 5, and 7.

Step 4: Calculating the Number of Matrices for Each Prime Sum

For each possible prime sum, we need to find the number of ways to place that many 1s into the 9 available positions. We use the combination formula $\binom{n}{k}$, where $n=9$ and $k$ is the number of 1s (prime sum).

• Case 1: Sum of Entries = 2 We need to choose 2 positions out of 9 to place the 1s. $\binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36$

• Case 2: Sum of Entries = 3 We need to choose 3 positions out of 9 to place the 1s. $\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$

• Case 3: Sum of Entries = 5 We need to choose 5 positions out of 9 to place the 1s. Using the identity $\binom{n}{k} = \binom{n}{n-k}$, we have $\binom{9}{5} = \binom{9}{4}$. $\binom{9}{5} = \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

• Case 4: Sum of Entries = 7 We need to choose 7 positions out of 9 to place the 1s. Using the identity $\binom{n}{k} = \binom{n}{n-k}$, we have $\binom{9}{7} = \binom{9}{2}$. $\binom{9}{7} = \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36$

Step 5: Calculating the Total Number of Matrices

The total number of matrices is the sum of the number of matrices for each prime sum: $\text{Total} = \binom{9}{2} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7} = 36 + 84 + 126 + 36 = 282$

Common Mistakes & Tips

• Remember that 1 is not a prime number.
• Using the property $\binom{n}{k} = \binom{n}{n-k}$ simplifies calculations when $k > n/2$.
• Double-check your arithmetic calculations to avoid errors.

Summary

We found the possible prime sums (2, 3, 5, 7) for a $3 \times 3$ matrix with entries 0 or 1. We then calculated the number of matrices for each prime sum using combinations and summed these values to find the total number of matrices.

The final answer is $\boxed{282}$.