The number of symmetric matrices of order \(3 \times 3\), each of whose elements is either 0 or 1, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 64
In a symmetric matrix \(a_{ij} = a_{ji}\), so once the elements on and above the diagonal are chosen, the elements below the diagonal are fixed.
Elements that can be chosen freely:
\(\text{diagonal: } a_{11}, a_{22}, a_{33}\)
\(\text{above the diagonal: } a_{12}, a_{13}, a_{23}\)
That is 6 elements, each with 2 choices (0 or 1). Hence the number of such matrices is
\(2^6 = 64\)
Why the other options are incorrect
Option (1) 512 — Incorrect. \(2^9 = 512\) counts all \(3 \times 3\) matrices with elements 0 or 1, symmetric or not.
Option (2) 8 — Incorrect. \(2^3 = 8\) counts only the choices of the diagonal elements.
Option (4) 32 — Incorrect. \(2^5 = 32\) leaves out one of the 6 free elements.