Let A = \begin{pmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}, a, b \in \mathbb{R}. If for some n \in \mathbb{N}, A^n = \begin{pmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{pmatrix} then n + a + b is equal to.
Step-by-Step Solution
Key Concept: The matrix A is an upper triangular matrix. Compute A^2, A^3, and generalize to A^n. Compare the resulting matrix with the given A^n to find n, a, and b.
Let A = I + N, where N = [[0, a, a], [0, 0, b], [0, 0, 0]]. Note that N^2 = [[0, 0, ab], [0, 0, 0], [0, 0, 0]] and N^3 = 0. Using binomial expansion, A^n = (I + N)^n = I + nN + n(n-1)/2 * N^2. Calculating this gives A^n = [[1, na, na + n(n-1)ab/2], [0, 1, nb], [0, 0, 1]]. Comparing with the given matrix, nb = 96, na = 48, and na + n(n-1)ab/2 = 2160. From nb = 96 and na = 48, we get b = 2a. Substituting into the third equation: 48 + n(n-1)a(2a)/2 = 2160 => 48 + n(n-1)a^2 = 2160 => n(n-1)a^2 = 2112. Since na = 48, a = 48/n. Substituting: n(n-1)(48/n)^2 = 2112 => (n-1)/n * 2304 = 2112 => (n-1)/n = 2112/2304 = 11/12. Thus n = 12. Then a = 48/12 = 4 and b = 2(4) = 8. So n + a + b = 12 + 4 + 8 = 24.
Correct Answer: 1