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 :

Showing the 12 most recent of 16 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.Which of the following cannot be the order of a row-matrix? (A) $2 \times 1$ (B) $1 \times 2$ (C) $1 \times 1$ (D) $1 \times n$

   ›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.

   1. 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).

   2. Option (B): $1 \times 2$

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

   3. 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.

   4. 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

   The order that cannot be that of a row-matrix is $2 \times 1$, which is option (A).

2. CBSE 2026Set 65/3/11 markMCQ
   Q.If $\begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} A = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix}$, then order of $A$ must be: (A) $3 \times 1$ (B) $1 \times 3$ (C) $1 \times 1$ (D) $3 \times 3$

   ›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

   1. 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: $3 \times 1$

   2. 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: $1 \times 3$

   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: $3 \times 3$

   4. 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

   The correct option is (D) $3 \times 3$.

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

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Number of elements in a matrix of order $3 \times 5$ $(A_{3\times5})$ will be:

   (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

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The total number of all possible matrices of order $3 \times 3$ with entry 0 or 2, is:

   (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

   $2^9 = 512$ — option (d).

6. 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.

7. 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.

   1. 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 $m = n$).

   2. 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

   The number of possible orders is $\boxed{9}$, which corresponds to option (D).

8. CBSE 2024Set A1 markMCQ
   Q.The number of all possible matrices of order $2\times 2$ with entry $0$ or $1$ is

   (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.

9. CBSE 2024Set ANNUAL1 markMCQ
   Q.Total numbers of possible matrices of order $3 \times 3$ with each entry $2$ or $0$, are:

   (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

   $512$ — option (d).

10. 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

11. 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

    Possible orders: $1\times14,\ 14\times1,\ 2\times7,\ 7\times2$

12. CBSE 2023Set ANNUAL1 markMCQ
    Q.If the order of a matrix is $m \times n$, then the number of elements in it are:

    (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

    (c) $mn$.