Let S = {√n : 1 ≤ n ≤ 50 and n is odd}. Let a ∈ S and A = <table><tr><td>1</td><td>0</td><td>a</td></tr><tr><td>-1</td><td>1</td><td>0</td></tr><tr><td>-a</td><td>0</td><td>1</td></tr></table>. If Σ<sub>a∈S</sub> det(adj A) = 100λ, then λ is equal to
Step-by-Step Solution
Key Concept: The determinant of adj(A) is (det(A))^(n-1). Here n=3, so det(adj(A)) = (det(A))^2. Calculate det(A) = 1(1-0) - 0 + a(0 - (-a)) = 1 + a^2. Then det(adj(A)) = (1+a^2)^2. Sum this over all a = \sqrt{n} where n is odd and 1 \leq n \leq 50.
Given A = [[1, 0, a], [-1, 1, 0], [-a, 0, 1]]. det(A) = 1(1-0) - 0 + a(0 - (-a)) = 1 + a^2. Since A is a 3x3 matrix, det(adj A) = (det A)^(3-1) = (det A)^2 = (1+a^2)^2. Here a = \sqrt{n}, so a^2 = n. Thus, det(adj A) = (1+n)^2. S = {\sqrt{n} : 1 \leq n \leq 50 and n is odd}. The odd values of n are 1, 3, 5, ..., 49. There are 25 such values. The sum is \Sigma_{n\in {1,3,...,49}} (1+n)^2. Let k = (n+1)/2, then n = 2k-1. As n goes from 1 to 49, k goes from 1 to 25. The sum is \Sigma_{k=1}^{25} (1 + 2k - 1)^2 = \Sigma_{k=1}^{25} (2k)^2 = 4 \Sigma_{k=1}^{25} k^2 = 4 * [25 * 26 * 51 / 6] = 4 * 25 * 13 * 17 = 22100. Given 100\lambda = 22100, \lambda = 221.
Correct Answer: 2