Finding Possible Matrix Orders with Prime Numbers Less Than 30

The question asks us to determine the number of different orders possible for matrices that can be formed using all prime numbers less than 30 as their elements. To solve this, we first need to identify these prime numbers and count how many there are.

Identifying Prime Numbers Less Than 30

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. We list the prime numbers that are strictly less than 30:

• 2
• 3
• 5
• 7
• 11
• 13
• 17
• 19
• 23
• 29

Counting these numbers, we find there are exactly 10 prime numbers less than 30.

Total Number of Elements in the Matrix

Since the matrices must comprise all these prime numbers as elements, the total number of elements in any such matrix will be 10.

Determining Possible Matrix Orders (m x n)

A matrix of order \( m \times n \) has \( m \) rows and \( n \) columns. The total number of elements in a matrix of order \( m \times n \) is given by the product \( m \times n \). In our case, the total number of elements is 10. Therefore, we need to find all possible pairs of positive integers \( (m, n) \) such that \( m \times n = 10 \).

We need to find the pairs of factors of 10. The factors of 10 are 1, 2, 5, and 10. The possible pairs \( (m, n) \) are:

• If \( m = 1 \), then \( n \) must be 10, since \( 1 \times 10 = 10 \). This gives the order \( 1 \times 10 \).
• If \( m = 2 \), then \( n \) must be 5, since \( 2 \times 5 = 10 \). This gives the order \( 2 \times 5 \).
• If \( m = 5 \), then \( n \) must be 2, since \( 5 \times 2 = 10 \). This gives the order \( 5 \times 2 \).
• If \( m = 10 \), then \( n \) must be 1, since \( 10 \times 1 = 10 \). This gives the order \( 10 \times 1 \).

Each distinct pair \( (m, n) \) represents a distinct order for the matrix. The possible orders are \( 1 \times 10 \), \( 2 \times 5 \), \( 5 \times 2 \), and \( 10 \times 1 \).

Counting the Number of Different Orders

Let's list the distinct orders we found:

• \( 1 \times 10 \)
• \( 2 \times 5 \)
• \( 5 \times 2 \)
• \( 10 \times 1 \)

There are 4 different possible orders for a matrix containing 10 elements. Therefore, 4 matrices of different orders are possible with elements comprising all prime numbers less than 30.

The total number of different orders is the count of these unique pairs \( (m, n) \), which is 4.

Revision Table: Matrix Orders and Elements

Additional Information: Understanding Matrix Dimensions

The order, or dimension, of a matrix is a fundamental property. It tells us the structure of the matrix, specifically how many rows and columns it contains. An \( m \times n \) matrix has \( m \) horizontal rows and \( n \) vertical columns. The order is always given as "rows by columns".

For a fixed number of elements, say \( N \), the possible orders \( (m, n) \) are determined by finding all pairs of positive integers \( (m, n) \) such that their product \( m \times n \) is equal to \( N \). Each unique pair \( (m, n) \) corresponds to a distinct matrix order. For example, a matrix with 6 elements could have orders \( 1 \times 6 \), \( 6 \times 1 \), \( 2 \times 3 \), or \( 3 \times 2 \). That's 4 possible orders for 6 elements.

In our problem, the number of elements is 10. The pairs of positive integer factors of 10 are (1, 10), (2, 5), (5, 2), and (10, 1). Each pair represents a unique order:

• \( 1 \times 10 \): A row matrix with 1 row and 10 columns.
• \( 2 \times 5 \): A matrix with 2 rows and 5 columns.
• \( 5 \times 2 \): A matrix with 5 rows and 2 columns.
• \( 10 \times 1 \): A column matrix with 10 rows and 1 column.

Since there are 4 such unique pairs, there are 4 different possible matrix orders.