Let S be the set containing all 3 × 3 matrices with entries from {-1,0,1}. The total number of matrices A ∈ S such that the sum of all the diagonal elements of Aᵀ A is 6 is ____________.

Solution

Maths sir · 4 steps

The trace of AᵀA is the sum of squares of all nine entries, so just count non-zero entries.

1. Diagonal of ATA. The (j,j) entry of AᵀA is column j of A dotted with itself, Σᵢ aᵢⱼ². Adding over j = 1, 2, 3 uses every entry of A once.

   tr(AᵀA) = Σᵢ₌₁³Σⱼ₌₁³ aᵢⱼ²

2. Each entry squared. For aᵢⱼ ∈ {-1, 0, 1}, aᵢⱼ² is 1 when the entry is ± 1 and 0 when it is 0. So the trace equals 6 exactly when 6 of the 9 entries are non-zero and the other 3 are zero.
3. Count the matrices. Choose the 6 non-zero positions in C(9, 6) = 84 ways. Each of those 6 entries can be +1 or -1, giving 2⁶ = 64 choices.
4. So. The number of matrices is 84 × 64 = 5376.

The common slip: 84. Counting only the positions of the non-zero entries gives 84. Each of the 6 non-zero entries can still be +1 or -1, so multiply by 2⁶.

The idea to keep. Whenever you see the trace of AᵀA or AAᵀ, read it as the sum of squares of every entry of A. Then it becomes a simple counting problem.

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