The total number of 3 × 3 matrices A having entries from the set 0, 1, 2, 3 such that the sum of all the diagonal entries of AA T is 9, is equal to _____________.

Solution

Maths sir · 3 steps

The trace of AAᵀ is the sum of squares of all nine entries, so this becomes a counting problem.

1. What the trace is. The i-th diagonal entry of AAᵀ is row i of A dotted with itself, which is aᵢ₁² + aᵢ₂² + aᵢ₃². Adding the three diagonal entries gives the sum of the squares of all nine entries.

   tr(AAᵀ) = Σᵢ₌₁³Σⱼ₌₁³ aᵢⱼ² = 9

2. Available squares. Each entry is 0, 1, 2 or 3, so its square is 0, 1, 4 or 9. Write 9 as a sum of nine such squares, then count the ways to place them in the 9 positions.

   Split of 9	Entries used	Number of matrices
   9	one 3, eight 0s	9
   4 + 1 × 5	one 2, five 1s, three 0s	9 × C(8, 5) = 9 × 56 = 504
   4 + 4 + 1	two 2s, one 1, six 0s	C(9, 2) × 7 = 36 × 7 = 252
   1 × 9	nine 1s	1

3. So. The total is 9 + 504 + 252 + 1 = 766, so the answer is 766.

The idea to keep. Whenever you see tr(AAᵀ) or tr(AᵀA), replace it with the sum of squares of every entry. Then list every way to split the total and count placements.

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