Let A be a 3x3 matrix and det(A) = 2. If n = det(adj(adj(.....(adj(A))))), where adj is applied 2024 times, then the remainder when n is divided by 9 is equal to ________.
Step-by-Step Solution
Key Concept: The determinant of the adjoint of a 3x3 matrix A is (det(A))^(3-1) = (det(A))^2. Applying this property repeatedly, if f(k) = det(adj_k(A)), then f(k) = (det(A))^(2^k). For k=2024, n = 2^(2^2024). We need to find 2^(2^2024) mod 9. Since 2^6 = 64 = 1 mod 9, we need the exponent 2^2024 mod 6. 2^2024 mod 3 is (-1)^2024 = 1. 2^2024 mod 2 is 0. By CRT, 2^2024 mod 6 = 4. Thus, 2^(2^2024) = 2^4 mod 9 = 16 mod 9 = 7.
Let |A| = 2. For a 3x3 matrix, |adj(A)| = |A|^(3-1) = |A|^2. Let A_k be the matrix after applying adj k times. Then |A_k| = |A_{k-1}|^2. This gives |A_k| = |A|^(2^k). For k=2024, n = 2^(2^2024). We need n mod 9. Since 2^6 = 64 = 1 mod 9, we find 2^2024 mod 6. 2^2024 = 2 * 2^2023. 2^2023 mod 3 = (-1)^2023 = -1 = 2 mod 3. So 2^2024 = 2 * 2 = 4 mod 6. Thus n = 2^4 mod 9 = 16 mod 9 = 7.
Correct Answer: 7