1 markMCQSection A

The number of all possible matrices of order $3 \times 2$ with each entry 1 or 2 is :

Q: The number of all possible matrices of order $3 \times 2$ with each entry 1 or 2 is :

The correct option is **D) 64**. **Working:** - A $3 \times 2$ matrix has $3 \times 2 = 6$ entries (positions). - Each position can be filled in 2 ways (either 1 or 2). - By the fundamental principle of counting, total matrices = $2^6 = 64$.

Algebra (Matrices)

Matrices — counting matrices

Options

(A)6

(B)16

(D)64

Official Answer

The correct option is D) 64.

Working:

A $3 \times 2$ matrix has $3 \times 2 = 6$ entries (positions).
Each position can be filled in 2 ways (either 1 or 2).
By the fundamental principle of counting, total matrices = $2^6 = 64$ .

order of a matrix3 x 2 matrixnumber of entriesfundamental principle of counting2^664choices per entry

Marking Scheme

11 mark: recognising 6 entries and computing $2^6 = 64$ (option D).

Hint

Count the entries ( $m \times n$ ), then raise the number of choices per entry (2) to that power.

Quick Oral Answer

A $3 \times 2$ matrix has six positions and each can be 1 or 2, so by the counting principle there are $2^6 = 64$ matrices.

Analysis & Explanation

This tests the order of a matrix combined with the multiplication principle of counting.

Concept:

A matrix of order $m \times n$ has exactly $m \cdot n$ entries.
If each entry has k possible values, the number of distinct matrices is $k^{m \cdot n}$ .

Why D is correct:

Here $m \cdot n = 3 \times 2 = 6$ and $k = 2$ , so $2^6 = 64$ .

Why the distractors are wrong:

A (6): counts only the number of entries, not the arrangements.
B (16): equals $2^4$ , mistaking the order for $2 \times 2$ .
C (24): a spurious product (e.g. $4!$ or $6 \times 4$ ), unrelated to the counting rule.

Common Mistakes

1Taking the order as $2 \times 2$ or $3 \times 3$ instead of $3 \times 2$ , giving 16 or 512.
2Adding instead of using powers ( $2 \times 6 = 12$ ) rather than $2^6$ .
3Reporting 6 (the number of entries) as the final answer.

Interesting Facts

The same counting logic underlies binary data storage — 6 binary cells also have $2^6 = 64$ possible states.

For an $m \times n$ matrix with k entry-choices, the count $k^{mn}$ grows explosively; a $3 \times 3$ matrix of 1s and 2s already has 512 possibilities.

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Frequently Asked Questions

Why is it 2 raised to the power 6 and not 6 raised to 2?

Each of the 6 positions independently has 2 choices, so we multiply 2 by itself 6 times: $2^6$ . The base is the number of choices; the exponent is the number of positions.

How many entries does a $3 \times 2$ matrix have?

Order $m \times n$ means 3 rows and 2 columns, so $3 \times 2 = 6$ entries in total.