Let S = { ( - 1, a; 0, b );a,b ∈ { 1,2,3,....100} } and let Tₙ = { A ∈ S:Aⁿ⁽ⁿ⁺¹⁾ = I}. Then the number of elements in ⋂ₙ₌₁¹⁰⁰ Tₙ is ___________.

Solution

Maths sir · 4 steps

The power n(n+1) is always even, so just square A and see when it becomes I.

1. The power is even. n(n+1) is a product of two consecutive integers, so it is even for every n. If A² = I, then every even power of A is I too.
2. Fix b first. A is upper triangular, so the bottom-right entry of Aᵐ is bᵐ. We need bᵐ = 1 with b a positive integer, which forces b = 1.

   A² = (-1, a; 0, b)(-1, a; 0, b) = (1, a(b-1); 0, b²)

3. Set b to 1. With b = 1 the top-right entry a(b-1) becomes 0, so A² = I for every a. Then Aⁿ⁽ⁿ⁺¹⁾ = (A²)^(n(n+1)/2) = I for all n from 1 to 100.
4. So. Every matrix with b = 1 lies in all the Tₙ, and a can be any of 1, 2, …, 100. The intersection has 100 elements.

The idea to keep. For high powers of a matrix, find a small power that gives I first. For a triangular matrix, the diagonal of Aᵐ is the diagonal entries raised to m, which pins down those entries quickly.

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