Q.If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

The key idea is that 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 the product $m \times n$.

Step 1: For 18 elements, we need all pairs of positive integers $(m, n)$ such that $m \times n = 18$.

Step 2: The factor pairs of 18 are: $1 \times 18$, $2 \times 9$, $3 \times 6$, $6 \times 3$, $9 \times 2$, $18 \times 1$.

Step 3: For 5 elements, since 5 is prime, the only factor pairs are $1 \times 5$ and $5 \times 1$.

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. Each such pair gives a possible order.

Why does this matter? Because the order tells you the shape of the matrix — how many rows and columns it has. Two matrices with the same number of elements can look completely different: a $2 \times 9$ matrix is wide and short, while a $9 \times 2$ matrix is tall and narrow. Both have 18 elements, but they are not the same order.

Let’s work through both parts.

1. For 18 elements:

   We need all positive integer pairs $(m, n)$ such that $m \times n = 18$.

   Start by listing the factor pairs of 18:

   • $1 \times 18$
   • $2 \times 9$
   • $3 \times 6$
   • $6 \times 3$
   • $9 \times 2$
   • $18 \times 1$

   Each of these is a valid order. Notice that the order $(m, n)$ is different from $(n, m)$ — a $3 \times 6$ matrix is not the same as a $6 \times 3$ matrix. So we count both.

   Tip

   A quick way to generate all orders: find all divisors of 18. For each divisor $d$, the pair is $(d, 18/d)$. Since 18 has 6 divisors (1, 2, 3, 6, 9, 18), you get 6 orders.

2. For 5 elements:

   Now $m \times n = 5$. The number 5 is prime — its only positive divisors are 1 and 5. So the only factor pairs are:

   • $1 \times 5$
   • $5 \times 1$

   That’s it. There is no $2 \times 2.5$ or anything like that — rows and columns must be whole numbers.

   Watch out

   A common mistake is to think that a $1 \times 5$ matrix and a $5 \times 1$ matrix are the same. They are not: one is a row vector (1 row, 5 columns), the other is a column vector (5 rows, 1 column). Both have 5 elements, but their orders are different.

So the possible orders are simply all the ways to write the total number of elements as a product of two positive integers, where the first integer is the number of rows and the second is the number of columns.

Method: Finding Possible Matrix Orders From the Element Count

Use this whenever you're given the number of elements $N$ and asked for all possible orders. Since a matrix of order $m\times n$ holds $m\times n$ elements, the orders are the factor pairs of $N$.

Steps

Step 1: Relate elements to order.

The element count satisfies $m\times n=N$ with $m,n$ positive integers. Each such ordered pair $(m,n)$ is a possible order.

Step 2: Enumerate the ordered factor pairs of $N$.

List every divisor $d$ of $N$; each yields the order $d\times(N/d)$. Count $(m,n)$ and $(n,m)$ separately — they are different shapes.

Step 3: Deal with primes and the vector cases.

If $N$ is prime, only $1\times N$ and $N\times1$ exist. Always include the divisor $1$, which gives the row- and column-vector orders; fractional shapes are never allowed.

Common Mistakes

Mistake 1: Merging $m\times n$ and $n\times m$ into one order.

Why it's wrong: $2\times9$ and $9\times2$ are distinct shapes and count separately. Correct approach: list both directions of every factor pair.

Mistake 2: Omitting the $1\times N$ / $N\times1$ orders.

Why it's wrong: single-row and single-column matrices are valid. Correct approach: treat $1$ as a divisor and include those orders.

Mistake 3: Attempting non-integer factors for a prime count.

Why it's wrong: with $5$ elements you cannot split rows/columns fractionally; only $1\times5$ and $5\times1$ work. Correct approach: for a prime $N$, give exactly those two orders.