If A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}, then A^{20} + (adj A) is equal to
A^{20} + adj(A) = A^{20} + A
A^{20} + adj(A) = A^{20} - A
A^{20} + adj(A) = 2A^{20} - A
A^{20} + adj(A) = 2A^{20} + A
Step-by-Step Solution
Key Concept: Calculate A^2, A^3 to find the pattern for A^n, then compute adj(A) and combine.
Given A = [[1,0,0],[2,1,0],[3,2,1]]. A^2 = [[1,0,0],[4,1,0],[12,4,1]]. A^3 = [[1,0,0],[6,1,0],[33,6,1]]. By induction, A^n = [[1,0,0],[2n,1,0],[n(2n+1),2n,1]]. adj(A) = [[1,0,0],[-2,1,0],[1,-2,1]]. Adding A^20 and adj(A) leads to the result.
Correct Answer: 3