Q.If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Concept: Matrix Order Possibilities — The order of a matrix is given by $m \times n$, where $m$ is the number of rows and $n$ is the number of columns. The total number of elements is $m \times n$. So, for a given number of elements, all factor pairs $(m, n)$ give possible orders.

Reasoning:

1. For 24 elements: find all positive integer pairs $(m, n)$ such that $m \times n = 24$.
2. The factor pairs of 24 are: $(1,24), (2,12), (3,8), (4,6), (6,4), (8,3), (12,2), (24,1)$.
3. For 13 elements: since 13 is prime, the only factor pairs are $(1,13)$ and $(13,1)$.

The key idea is simple: a matrix with $m$ rows and $n$ columns has exactly $m \times n$ elements. So if you know the total number of elements, you’re looking for all pairs of positive integers $(m, n)$ whose product equals that number. That’s it — no hidden trick, just factorisation.

Let’s walk through it.

1. Understand what “order” means.

   The order of a matrix is written as $m \times n$, where $m$ is the number of rows and $n$ is the number of columns. Both $m$ and $n$ must be positive integers (you can’t have zero or negative rows). The total number of elements is $m \times n$.

2. For 24 elements: find all factor pairs.

   We need all pairs $(m, n)$ such that $m \times n = 24$. Start listing systematically:

   • $1 \times 24$
   • $2 \times 12$
   • $3 \times 8$
   • $4 \times 6$
   • $6 \times 4$ (this is a different order from $4 \times 6$ — rows and columns swap)
   • $8 \times 3$
   • $12 \times 2$
   • $24 \times 1$

   That’s 8 possible orders. Notice that we include both $m \times n$ and $n \times m$ because a $4 \times 6$ matrix is not the same as a $6 \times 4$ matrix — they have different shapes.

   Tip

   A quick way: list all divisors of 24 in increasing order. For each divisor $d$, the other factor is $24/d$. Then write both $(d, 24/d)$ and $(24/d, d)$ — but be careful not to double-count when $d = 24/d$ (which doesn’t happen here because 24 is not a perfect square).

3. For 13 elements: handle the prime case.

   13 is a prime number. Its only positive divisors are 1 and 13. So the only factor pairs are:

   • $1 \times 13$
   • $13 \times 1$

   That gives exactly 2 possible orders. A $1 \times 13$ matrix is a row vector (1 row, 13 columns), and a $13 \times 1$ matrix is a column vector (13 rows, 1 column). Both are valid matrices.

   Watch out

   A common mistake is to think that a matrix must have at least 2 rows and 2 columns. That’s not true — a $1 \times n$ or $m \times 1$ matrix is perfectly fine. It’s still a matrix, just a special case (a row or column vector).

4. Summarise the results.

   For 24 elements: 8 possible orders.

   For 13 elements: 2 possible orders.

Method: Listing All Possible Orders From a Given Number of Elements

Use this whenever you're told a matrix has $N$ elements and asked for its possible orders. The number of elements is the product of rows and columns, so the task is finding factor pairs of $N$.

Steps

Step 1: Write the element-count relation.

A matrix of order $m\times n$ has exactly $m\times n$ elements. So possible orders correspond to ordered pairs of positive integers $(m,n)$ with $mn=N$.

Step 2: List every ordered factor pair of $N$.

Find all divisors $d$ of $N$; each gives the order $d\times(N/d)$. Count $(m,n)$ and $(n,m)$ separately — a $2\times9$ matrix differs from a $9\times2$ one.

Step 3: Handle special counts (primes) and the $1\times N$ cases.

If $N$ is prime, the only orders are $1\times N$ and $N\times1$. Rows and columns must be positive integers, so no fractional shapes appear; the number of orders equals the number of divisors of $N$.

Common Mistakes

Mistake 1: Counting $m\times n$ and $n\times m$ as one order.

Why it's wrong: a $3\times8$ matrix and an $8\times3$ matrix have different shapes and are distinct orders. Correct approach: list both directions for each factor pair.

Mistake 2: Forgetting the $1\times N$ and $N\times1$ orders.

Why it's wrong: a single-row or single-column matrix is still a valid matrix. Correct approach: include $1$ as a divisor, giving the row/column-vector orders.

Mistake 3: Trying non-integer factorisations for a prime count.

Why it's wrong: with $13$ elements you cannot use $m=2,\,n=6.5$ — orders must be whole numbers. Correct approach: for a prime $N$, only $1\times N$ and $N\times1$ exist.