Let A = [a<sub>ij</sub>] be a 3x3 matrix such that a<sub>ij</sub> = 2<sup>i-j</sup> for all i, j. Then the matrix A<sup>n</sup> for any positive integer n is equal to:
Step-by-Step Solution
Key Concept: Calculate A^2 and A^3 to identify a pattern for A^n. The matrix A is defined as a_ij = 2^(i-j), which results in a matrix where each row is a geometric progression.
The matrix A is given by A = [[2^0, 2^-1, 2^-2], [2^1, 2^0, 2^-1], [2^2, 2^1, 2^0]] = [[1, 1/2, 1/4], [2, 1, 1/2], [4, 2, 1]]. Calculating A^2 = A * A = [[1+1+1, 1/2+1/2+1/2, 1/4+1/4+1/4], [2+2+2, 1+1+1, 1/2+1/2+1/2], [4+4+4, 2+2+2, 1+1+1]] = [[3, 3/2, 3/4], [6, 3, 3/2], [12, 6, 3]] = 3A. By induction, A^n = 3^(n-1) A.
Correct Answer: 1