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NCERT Exemplar · Q52

Q.Total number of possible matrices of order 3×33 \times 3 with each entry 22 or 00 is
(A) 99
(B) 2727
(C) 8181
(D) 512512

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Each of the 9 entries in a 3×33 \times 3 matrix can be chosen independently from 2 options (2 or 0). So the total number of such matrices is 29=5122^9 = 512, which matches option (D).

The key insight here is that a matrix is defined entirely by its entries. For a 3×33 \times 3 matrix, there are exactly 9 positions to fill. The problem says each entry must be either 2 or 0 — that’s two choices per position, with no other restrictions.

  1. Count the positions.

    A 3×33 \times 3 matrix has 3 rows and 3 columns, so 3×3=93 \times 3 = 9 entries. Each entry is independent of the others.

  2. Choices per entry.

    For each of the 9 positions, you can put either a 2 or a 0. That’s 2 possibilities per slot.

  3. Apply the multiplication principle.

    When choices are independent, the total number of combinations is the product of the number of choices for each slot.

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

  1. Compute 292^9. 29=5122^9 = 512 …

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