If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

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Q. If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

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Concept understanding — Matrix Order Possibilities

Matrix Order Possibilities – From Intuition to Precision

A matrix is a rectangular grid of numbers with some number of rows and columns. The order of a matrix says exactly that: "this matrix has m rows and n columns," written $m \times n$ (read "m by n").

For example, 3 rows and 2 columns is order $3 \times 2$ ; 1 row and 4 columns is $1 \times 4$ (a row vector); 5 rows and 1 column is $5 \times 1$ (a column vector).

The key idea: the order tells you the shape of the matrix. Two matrices can hold the same numbers but different orders — and then they are completely different objects.

The Precise Statement

\text{Order of a matrix} = \text{Number of rows} \times \text{Number of columns}

A matrix with $m$ rows and $n$ columns has order $m \times n$ , where $m, n \in \mathbb{N}$ .

Note

The order is always written rows first, then columns. So $3 \times 2$ means 3 rows and 2 columns, not the reverse.

What "Possibilities" Means

Matrix order possibilities asks: what shapes can a matrix have? Any pair of positive integers $(m, n)$ gives a valid order, so the set of all possible orders is:

\{ m \times n \mid m, n \in \mathbb{N} \}

That is $1 \times 1$ , $1 \times 2$ , $2 \times 1$ , $2 \times 2$ , $3 \times 5$ , $100 \times 1$ , $1 \times 100$ , and so on — infinitely many.

Tip

A $1 \times 1$ matrix is a single number (a scalar), a $1 \times n$ matrix is a row vector, and an $m \times 1$ matrix is a column vector — all special cases.

Why This Matters

The order determines which operations are allowed:

Addition: only between two matrices of the same order.
Multiplication: $A$ (order $m \times n$ ) times $B$ (order $p \times q$ ) works only if $n = p$ (columns of $A$ equal rows of $B$ ); the result has order $m \times q$ .

Watch out

A common mistake: thinking $2 \times 3$ and $3 \times 2$ matrices are the same. They aren't — different shapes, and they cannot be added.

Quick Examples

Matrix	Rows	Columns	Order
$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$	2	2	$2 \times 2$
$\begin{bmatrix} 5 & 6 & 7 \end{bmatrix}$	1	3	$1 \times 3$
$\begin{bmatrix} 8 \\ 9 \end{bmatrix}$	2	1	$2 \times 1$
$\begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix}$	3	2	$3 \times 2$

The Big Picture

Every matrix has exactly one order, and it's the first thing to identify — the shape governs everything else you can do. There are infinitely many possible orders, but each is just a pair of positive integers: rows $\times$ columns.

Understanding the order (rows × columns) of a matrix and which orders permit addition or multiplication is foundational content in the CBSE Class 12 Matrices chapter, and "matrix order and types class 12" is a commonly searched revision topic. This same order-checking habit is the first step in nearly every matrix multiplication question tested in board exams and JEE Main.

The possible orders of a matrix are all pairs $(m, n)$ of positive integers whose product equals the total number of elements. For 28 elements, the orders are $1 \times 28$ , $2 \times 14$ , $4 \times 7$ , $7 \times 4$ , $14 \times 2$ , and $28 \times 1$ . For 13 elements, since 13 is prime, the only orders are $1 \times 13$ and $13 \times 1$ .

A matrix is defined by its number of rows and columns. The total number of elements in a matrix is simply the product of its number of rows and number of columns. So if a matrix has $m$ rows and $n$ columns, it has $m \times n$ elements. The question asks: given a fixed total number of elements, what pairs $(m, n)$ of positive integers multiply to that total? Each such pair is a possible order.

The key insight is that we are looking for all factor pairs of the given number. Both $m$ and $n$ must be positive integers (a matrix cannot have zero rows or columns). The order is written as $m \times n$ , and note that $m \times n$ and $n \times m$ are considered different orders unless $m = n$ , because a matrix with 2 rows and 3 columns is not the same shape as one with 3 rows and 2 columns.

Let's work through each case.

For 28 elements: We need all positive integer pairs $(m, n)$ such that $m \times n = 28$ . First, list all factor pairs of 28. The factors of 28 are 1, 2, 4, 7, 14, and 28. Pair them:
- $1 \times 28 = 28$
- $2 \times 14 = 28$
- $4 \times 7 = 28$
- $7 \times 4 = 28$
- $14 \times 2 = 28$
- $28 \times 1 = 28$
So there are six possible orders: $1 \times 28$ , $2 \times 14$ , $4 \times 7$ , $7 \times 4$ , $14 \times 2$ , and $28 \times 1$ .
For 13 elements: 13 is a prime number. Its only positive factors are 1 and 13. So the only factor pairs are:
- $1 \times 13 = 13$
- $13 \times 1 = 13$
Hence, only two possible orders: $1 \times 13$ and $13 \times 1$ .

Watch out

A common mistake is to forget that order matters — $m \times n$ and $n \times m$ are distinct unless $m = n$ . For 28, some students list only $1 \times 28$ , $2 \times 14$ , and $4 \times 7$ , missing the three reversed orders. Always include both arrangements.

Tip

The number of possible orders for a given number of elements equals the number of positive divisor pairs, counting order. For a number $N$ , if $d$ is the number of positive divisors, then the number of ordered pairs $(m, n)$ with $m \times n = N$ is exactly $d$ . For 28, $d = 6$ (divisors: 1, 2, 4, 7, 14, 28), so 6 orders. For 13, $d = 2$ , so 2 orders.

✓Final answer

1 \times 28

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

CBSE 2024Set 65/1/11 markMCQ
$Q. If a matrix has 36 elements, then the number of possible orders it can have is : (A) 13 (B) 3 (C) 5 (D) 9 General Instructions : Read the following instructions very carefully and strictly follow them :$
(i) This question paper contains 38 questions. All questions are compulsory.
(ii) This question paper is divided into five Sections – A, B, C, D and E.
(iii) In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs) and questions number 19 and 20 are Assertion-Reason based questions of 1 mark each.
(iv) In Section B, Questions no. 21 to 25 are very short answer (VSA) type questions, carrying 2 marks each.
(v) In Section C, Questions no. 26 to 31 are short answer (SA) type questions, carrying 3 marks each.
(vi) In Section D, Questions no. 32 to 35 are long answer (LA) type questions carrying 5 marks each.
(vii) In Section E, Questions no. 36 to 38 are case study based questions carrying 4 marks each.
(viii) There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 3 questions in Section C, 2 questions in Section D and 2 questions in Section E.
(ix) Use of calculators is not allowed.
›Reveal solutionSolution
The number of possible orders of a matrix with 36 elements equals the number of factor pairs of 36. Since order is given by $m \times n$ where $m$ and $n$ are positive integers, the factor pairs are $(1,36), (2,18), (3,12), (4,9), (6,6)$ and their reverses, giving 9 distinct orders. The correct option is (D) 9.

The key idea here is simple: a matrix’s order is written as $m \times n$ , where $m$ is the number of rows and $n$ is the number of columns. The total number of elements is $m \times n$ . So if a matrix has 36 elements, we need all pairs of positive integers $(m, n)$ such that $m \times n = 36$ .

Why does this matter? Because each such pair gives a possible shape for the matrix — a different arrangement of rows and columns. For example, a $4 \times 9$ matrix looks very different from a $9 \times 4$ matrix, but both have 36 entries. The question asks for the number of possible orders, meaning how many distinct $(m, n)$ pairs exist.

Let’s work through it step by step.

List all factor pairs of 36.

Since $m$ and $n$ are positive integers, we find all pairs $(m, n)$ with $m \times n = 36$ . Start from $m = 1$ and go up to $m = 6$ (since beyond that, pairs repeat).

$1 \times 36 = 36$

$2 \times 18 = 36$

$3 \times 12 = 36$

$4 \times 9 = 36$

$6 \times 6 = 36$

These are the unordered factor pairs. But order matters here because $m$ and $n$ are rows and columns — swapping them gives a different matrix shape (unless $m = n$ ).

Count distinct orders.

For each unordered pair, we get two orders unless the two numbers are equal.

From $(1, 36)$ : orders $1 \times 36$ and $36 \times 1$ → 2 orders

From $(2, 18)$ : $2 \times 18$ and $18 \times 2$ → 2 orders

From $(3, 12)$ : $3 \times 12$ and $12 \times 3$ → 2 orders

From $(4, 9)$ : $4 \times 9$ and $9 \times 4$ → 2 orders

From $(6, 6)$ : only $6 \times 6$ (since swapping gives the same) → 1 order

Total distinct orders = $2 + 2 + 2 + 2 + 1 = 9$ .

Tip

A quick way: the number of distinct orders for $N$ elements equals the number of divisors of $N$ if you count each divisor as a possible number of rows (or columns). Here, divisors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36 — that’s 9 divisors. Each divisor can be the number of rows, and the columns are determined as $N / \text{rows}$ . So the count of orders equals the number of divisors of 36, which is 9.

Watch out

A common mistake is to forget that $m$ and $n$ are interchangeable — students often count only the unordered factor pairs (5 of them) and pick option (C) 5. But the question asks for orders, and $3 \times 12$ is a different order from $12 \times 3$ . Always check whether the problem treats $(m, n)$ and $(n, m)$ as distinct — here they are, because rows and columns are not the same.

✓Final answer
$\boxed{9}$
CBSE 2026Set 65/1/11 markMCQ
$2 \times 1$
›Reveal solutionSolution
A row-matrix has exactly one row and any number of columns. The order must be $1 \times n$ where $n \geq 1$ . Among the options, $2 \times 1$ has two rows, so it cannot be a row-matrix. The answer is (A).

The idea is simple: a row-matrix is defined by its shape — it is a matrix with a single row. That means the first dimension (number of rows) is always 1. The second dimension (number of columns) can be any positive integer. So the order is always $1 \times n$ , where $n$ is a natural number ( $n = 1, 2, 3, \dots$ ).

Now, let’s check each option against this definition.

Option (A): $2 \times 1$

This matrix has 2 rows and 1 column. Since a row-matrix must have exactly 1 row, this order is impossible. It is actually a column-matrix (one column, many rows).

Option (B): $1 \times 2$

This has 1 row and 2 columns. That fits perfectly — it is a row-matrix with two entries.

Option (C): $1 \times 1$

This has 1 row and 1 column. It is a special case: a row-matrix with a single element. Yes, it qualifies — a $1 \times 1$ matrix is both a row-matrix and a column-matrix.

Option (D): $1 \times n$

This is the general form of a row-matrix. For any $n \geq 1$ , it is valid. So this is certainly possible.

Watch out

A common mistake is to think a $1 \times 1$ matrix is not a row-matrix because it looks like a single number. But the definition is about shape, not size — one row is enough. Similarly, don’t confuse $2 \times 1$ (two rows) with $1 \times 2$ (two columns); they are different.

Tip

If you ever forget, just remember: “row” comes first — so the first number in the order must be 1. Anything else is not a row-matrix.

✓Final answer
$2 \times 1$
CBSE 2026Set 65/3/11 markMCQ
$\begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} A = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix}$
›Reveal solutionSolution
The product of a column vector and a row vector yields a matrix whose order is (rows of first) × (columns of second); here $3 \times 3$ .

Understanding Matrix Multiplication and Order

When we multiply two matrices, the order of the resulting matrix is determined by a simple rule: if matrix $P$ has order $m \times n$ and matrix $Q$ has order $n \times p$ , then their product $PQ$ has order $m \times p$ . The middle dimension $n$ must match for multiplication to be defined, and it "disappears" in the result.

In this problem, we're multiplying a column vector by a row vector. This is the outer product, which produces a full matrix rather than a scalar (which would be the inner product, row times column).

Step-by-Step Solution

Identify the order of the first matrix

The first matrix is $\begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix}$ , a column vector with 3 rows and 1 column.

Order: $3 \times 1$

Identify the order of the second matrix

The second matrix is $\begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$ , a row vector with 1 row and 3 columns.

Order: $1 \times 3$

Check compatibility and find the resulting order

For multiplication $(3 \times 1) \times (1 \times 3)$ :

The inner dimensions are both 1, so multiplication is valid ✓

The outer dimensions give us the result: $3 \times 3$

Verify with the actual multiplication (optional but instructive)

Each entry $a_{ij}$ of $A$ is the product of the $i$ -th element of the column vector and the $j$ -th element of the row vector:

$A = \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 4(-1) & 4(2) & 4(1) \\ 1(-1) & 1(2) & 1(1) \\ 3(-1) & 3(2) & 3(1) \end{bmatrix} = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix}$
This is clearly a $3 \times 3$ matrix.

Tip

Column × Row always gives a full matrix (outer product), while Row × Column gives a $1 \times 1$ matrix, i.e., a scalar (inner product or dot product).

✓Final answer
$3 \times 3$
CBSE 2026Set V11 markMCQ
$Q. A matrix has 13 elements. The number of possible different orders it can have$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

A matrix of order $m\times n$ has $mn$ elements; for $mn=13$ (prime) there are exactly $2$ ordered factor pairs, answer (b).

An $m\times n$ matrix has $mn$ entries. We need ordered pairs $(m,n)$ of positive integers with $mn=13$ . As $13$ is prime, its only positive divisors are $1$ and $13$ , giving
$1\times13\quad\text{and}\quad 13\times1.$
That is $2$ possible orders.

✓Final answer
$(b) 2$
CBSE 2025Set ANNUAL1 markMCQ
$3 \times 5$
(a) 3
(b) 5
(c) 8
(d) 15
›Reveal solutionSolution

A matrix of order $m\times n$ has $m\times n$ elements.

Here $m=3,\ n=5$ , so number of elements $= 3\times 5 = 15$ .

✓Final answer
$(iv) 15$
CBSE 2025Set ANNUAL1 markMCQ
$3 \times 3$
(a) 27
(b) 18
(c) 81
(d) 512
›Reveal solutionSolution

Each of the 9 entries of a $3\times3$ matrix can independently be 0 or 2, giving $2^9$ matrices.

A $3\times3$ matrix has $3\times3 = 9$ entries. Each entry has 2 independent choices (0 or 2). By the multiplication principle, the total number of such matrices is $2^9 = 512$ .

✓Final answer
$2^9 = 512$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The matrix B = [aᵢⱼ]₁ˣₙ, when n > 1 is :$
(a) Square matrix
(b) Column matrix
(c) Diagonal matrix
(d) Row matrix
›Reveal solutionSolution

A matrix with order $1\times n$ (one row, $n$ columns, $n>1$ ) is a row matrix.

The matrix $B=[a_{ij}]_{1\times n}$ has exactly $1$ row and $n$ columns. By definition, a matrix having only one row (and more than one column) is called a row matrix. It is not square (rows $\ne$ columns since $n>1$ ), not a column matrix (that would be $m\times 1$ ), and not diagonal (diagonal matrices must be square).

✓Final answer
$(d) Row matrix.$
CBSE 2024Set A1 markMCQ
$2\times 2$
(a) 27
(b) 512
(c) 16
(d) 2
›Reveal solutionSolution

There are $2^4=16$ such $2\times 2$ matrices.

A $2\times 2$ matrix has $4$ entries. Each entry can independently be chosen as $0$ or $1$ , i.e. $2$ choices per entry. By the multiplication principle, the total number of such matrices is $2\times2\times2\times2=2^4=16$ .

✓Final answer
$(c) 16.$
CBSE 2024Set ANNUAL1 markMCQ
$3 \times 3$
(a) 9
(b) 27
(c) 81
(d) 512
›Reveal solutionSolution

$512$ — option (d).

A $3\times3$ matrix has $9$ entries. Each entry can independently be chosen in $2$ ways ( $2$ or $0$ ).

$\text{Total matrices} = 2^9 = 512$

✓Final answer
$512$
CBSE 2023Set M1 markMCQ
$Q. If a matrix has 18 elements, then the number of matrices having all possible orders is$
(a) 4
(b) 6
(c) 2
(d) 8
›Reveal solutionSolution

Tests counting matrix orders from the element count. $18$ has $6$ ordered factorisations $m\times n$ .

An $m\times n$ matrix has $mn$ elements. For $18$ elements we need every ordered pair $(m,n)$ of positive integers with $mn=18$ . Each divisor $m$ of $18$ fixes $n=18/m$ , so the count equals the number of divisors of $18$ :
$18 = 2\times 3^2 \;\Rightarrow\; \text{divisors} = 1,2,3,6,9,18.$
That gives orders $1\times18,\;2\times9,\;3\times6,\;6\times3,\;9\times2,\;18\times1$ — 6 matrices.

✓Final answer
$(b) 6$
CBSE 2023Set ANNUAL1 mark
$Q. If a matrix has 14 elements, what are the possible orders it can have?$
›Reveal solutionSolution

A matrix of order $m\times n$ has $mn$ elements; find all factor pairs of 14.

If a matrix has $mn=14$ elements, we need all ordered pairs $(m,n)$ of positive integers with product 14.

$14=1\times14=14\times1=2\times7=7\times2$ .

So the possible orders are $1\times14$ , $14\times1$ , $2\times7$ , $7\times2$ .

✓Final answer
$1\times14,\ 14\times1,\ 2\times7,\ 7\times2$
CBSE 2023Set ANNUAL1 markMCQ
$m \times n$
(a) $m$
(b) $n$
(c) $mn$
(d) $m-n$
›Reveal solutionSolution

The number of elements in a matrix equals (number of rows) × (number of columns).

A matrix of order $m\times n$ is arranged in $m$ rows and $n$ columns. Each row has $n$ entries, and there are $m$ such rows, so the total number of elements is $m\times n = mn$ .

✓Final answer
$mn$

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Q.If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

Concept understanding — Matrix Order Possibilities

Matrix Order Possibilities – From Intuition to Precision

The Precise Statement

What "Possibilities" Means

Why This Matters

Quick Examples

The Big Picture

Method: Possible orders from a given number of elements

Steps

Common Mistakes

Understanding Matrix Multiplication and Order

Step-by-Step Solution

Q.If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

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