Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A² is 1, then the possible number of such matrices is :

Solution

Maths sir · 4 steps

Write A with three unknowns, square it, and the diagonal sum becomes a sum of squares equal to 1.

1. Set up A. A symmetric 2 × 2 matrix has equal off-diagonal entries, so take A = (a, b; b, c) with a, b, c integers.
2. Diagonal of A squared. The top-left entry of A² is a² + b² and the bottom-right entry is b² + c². Their sum must be 1.

   a² + 2b² + c² = 1

3. Integer solutions. Every term is a non-negative integer. If b ≠ 0, then 2b² ≥ 2, which is too big, so b = 0. Then a² + c² = 1, so one of a, c is ± 1 and the other is 0.
4. So. (a, c) can be (1, 0), (-1, 0), (0, 1) or (0, -1), with b = 0. That gives 4 matrices, option (b).

The idea to keep. For integer entries, a sum of squares equal to a small number leaves very few cases. Count the signs and positions separately, and watch the coefficient on each square.

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