The number of symmetric matrices of order 3, with all the entries from the set {0,1,2,3,4,5,6,7,8,9} is :

Solution

Maths sir · 4 steps

In a symmetric matrix, fix the diagonal and the upper triangle; the lower triangle has no choice left.

1. What symmetric forces. Symmetric means Aᵀ = A, that is aᵢⱼ = aⱼᵢ for every i and j. So once you pick a₁₂, the entry a₂₁ must equal it, and likewise for every pair across the diagonal.
2. Count free entries. The 3 diagonal entries a₁₁, a₂₂, a₃₃ are free. Above the diagonal, a₁₂, a₁₃, a₂₃ are free. The 3 entries below the diagonal copy these, so there are 3 + 3 = 6 free entries.

   (a, d, e; d, b, f; e, f, c)

3. Fill each one. Each of the 6 free entries can be any of the 10 digits, chosen independently. That gives 10 × 10 × 10 × 10 × 10 × 10 = 10⁶ matrices.
4. So. The number of symmetric matrices is 10⁶, option (c).

The idea to keep. For a symmetric matrix of order n, only n(n+1)/2 entries are free. With k choices per entry, the count is k^(n(n+1)/2).

Answer from the NTA final answer key. Solution by PrepWiser.