Let M denote the set of all real matrices of order 3 × 3 and let S={-3,-2,-1,1,2}. Let
S₁={A=[aᵢⱼ] ∈ M: A=Aᵀ and aᵢⱼ ∈ S, ∀ i, j}, ; S₂={A=[aᵢⱼ] ∈ M: A=-Aᵀ and aᵢⱼ ∈ S, ∀ i, j}, ; S₃={A=[aᵢⱼ] ∈ M: a₁₁+a₂₂+a₃₃=0 and aᵢⱼ ∈ S, ∀ i, j} .
If n(S₁ ∪ S₂ ∪ S₃)=125 α, then α equls __________.

Solution

Maths sir · 5 steps

S₂ is empty since a skew matrix has a zero diagonal. Count S₁ and S₃, then remove the overlap.

1. S2 is empty. If A = -Aᵀ, then aᵢᵢ = -aᵢᵢ, so every diagonal entry must be 0. But 0 ∉ S, so no such matrix exists and n(S₂) = 0.
2. Count S1. A symmetric matrix is fixed by its 3 diagonal entries and the 3 entries above the diagonal. Each has 5 choices, so n(S₁) = 5⁶ = 15625.
3. Count S3. First find the diagonal triples from S with sum 0. {-3, 1, 2} gives 6 orders, {-2, 1, 1} gives 3 and {-1, -1, 2} gives 3, so there are 12 triples. The 6 off-diagonal entries are free, so n(S₃) = 12 × 5⁶ = 187500.
4. Overlap S1 and S3. A matrix in both sets is symmetric with trace 0. It has 12 choices of diagonal and 5³ choices for the entries above the diagonal, so n(S₁ ∩ S₃) = 12 × 125 = 1500.

   n(S₁ ∪ S₂ ∪ S₃) = 15625 + 187500 - 1500 = 201625

5. So. 201625 = 125 × 1613, so α = 1613.

The idea to keep. Before counting, check whether a set can exist at all. A skew-symmetric matrix always has a zero diagonal. With overlapping sets, always subtract the intersection.

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