Let A = . Find a diagonal matrix to help calculate Solution Find the eigenvalues and eigenvectors. (-1-L)(0-L) - 1*6 = 0 L^2 + L - 6 = 0 (L+3)(L-2) = 0 L = -3, 2 Let x = 1 Consider eigenvalue 3 -1(1) + 6 y = -3(1) -1 + 6y = -3 6y = -2 y = -1/3 The eigenvector is(1 -1/3) Consider eigenvalue 2 -1(1) + 6y = 2 -1 + 6y = 2 6y = 3 y = 1/2 Thus, we now have matrix P = 1 1 -1/3 1/2 Eigenvalues are -3 and 2 Thus, the diagonal matrix is -3 0 0 2 P^-1 = .6 -1.2 .4 1.2 (We can calculate this by switching the diagonal elements, taking the - of the off-diagonal, and multiplying by 1/determinant, which in this case is 5/6, so multiply by 1.2) Consider 1 1 * -3 0 * .6 .4 = -1 6 -1/3 1/2 0 2 .4 1.2 1 0 Thus, A^10 = P D^10 P^-1 = 1 1 * (-3)^10 0 * .6 .4 = -1/3 1/2 0 2 ^10 .4 1.2 1 1 * 59049 0 * .6 .4 = -1/3 1/2 0 1024 .4 1.2 35839 -69630 -11605 24234 .