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Exercise 3.1 · Q10

Q.Find the value of the following: The number of all possible matrices of order 3×33 \times 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512

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A 3×33 \times 3 matrix has 9 entries, each can be 0 or 1 independently. So the total number of such matrices is 29=5122^9 = 512. The correct option is (D).

The key idea here is simple but powerful: a matrix is just a rectangular arrangement of numbers. When we talk about "all possible matrices" with entries restricted to 0 or 1, we are really counting the number of ways to fill each of the 9 positions with one of two choices.

This is a direct application of the fundamental principle of counting (also called the multiplication principle). If you have nn independent choices, each with kk possibilities, the total number of outcomes is knk^n. There is no need for Inclusion-Exclusion here because the choices are independent — no restrictions overlap or exclude each other.

Let’s walk through it step by step.

  1. Identify the number of entries.

    A 3×33 \times 3 matrix has 3 rows and 3 columns. That gives 3×3=93 \times 3 = 9 entries in total. Each entry is a separate "slot" to be filled.

  2. Choices per entry.

    The problem says each entry can be either 0 or 1. That’s exactly 2 choices per entry.

  3. Apply the multiplication principle.

    Since the choice for one entry does not affect the choice for another, we multiply the number of possibilities for each of the 9 entries:

    2×2×2×…(9 times)=292 \times 2 \times 2 \times \dots \text{(9 times)} = 2^9 …

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