Let A=[1, a, a; 0, 1, b; 0, 0, 1], a, b ∈ ℝ. If for some
n ∈ ℕ, Aⁿ=[1, 48, 2160; 0, 1, 96; 0, 0, 1] then n+a+b is equal to ____________.

Solution

Maths sir · 4 steps

Split A into I plus a nilpotent part N; since N³ = 0, the binomial expansion stops after three terms.

1. Split the matrix. Write A = I + N with N = (0, a, a; 0, 0, b; 0, 0, 0). Then N² has a single non-zero entry, ab, in the top-right corner, and N³ = O.

   Aⁿ = I + nN + n(n-1)/2N² = (1, na, na + n(n-1)/2ab; 0, 1, nb; 0, 0, 1)

2. Match entries. Comparing with the given matrix: na = 48 and nb = 96, so ab = (48 × 96)/n² = 4608/n². The top-right entry gives 48 + n(n-1)/2 × 4608/n² = 2160.
3. Solve for n. This simplifies to 2304(n-1)/n = 2112, so 2304n - 2304 = 2112n, giving 192n = 2304 and n = 12. Then a = 48/12 = 4 and b = 96/12 = 8.
4. So. n + a + b = 12 + 4 + 8 = 24.

The common slip: 25. Dropping the na term from the top-right entry gives n = 16, a = 3, b = 6 and a sum of 25.

The idea to keep. For an upper triangular matrix with 1s on the diagonal, write it as I + N. Since I commutes with N and N dies after a few powers, Aⁿ comes from a short binomial expansion.

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