Understanding the Problem
- We are given a 2x2 matrix 'A' with eigenvalues e^(iπ/5) and e^(iπ/6).
- We need to find the smallest value of n for which A^n = I, where I is the identity matrix.

Finding the Eigenvalues
- The eigenvalues of a 2x2 matrix can be found by solving the characteristic equation det(A - λI) = 0, where λ is the eigenvalue.
- For matrix A, the characteristic equation is det(A - λI) = 0.
- Substituting the given eigenvalues e^(iπ/5) and e^(iπ/6) into the characteristic equation, we can find the matrix A.

Matrix A
- Let the matrix A be [[a, b], [c, d]].
- Using the given eigenvalues, we can write the characteristic equation as det(A - λI) = 0.
- Solving the characteristic equation, we get the matrix A.

Finding the Power of Matrix A
- Since A has distinct eigenvalues, it is diagonalizable.
- A can be diagonalized as A = PDP^(-1), where D is the diagonal matrix with eigenvalues on the diagonal.
- A^n = PD^nP^(-1).
- Since D is a diagonal matrix, D^n is also a diagonal matrix with eigenvalues raised to the power n.
- We need to find the smallest value of n such that A^n = I, which means D^n = I.

Smallest Value of n
- The smallest value of n for which D^n = I is the least common multiple of the orders of the eigenvalues.
- The order of e^(iπ/5) is 10 and the order of e^(iπ/6) is 12.
- The smallest value of n = LCM(10, 12).
Therefore, the smallest value of n for which A^n = I is the least common multiple of the orders of the eigenvalues, which is 60.