For the eigenvector associated with eigenvalue 3 12 12 22儿C C 12 4C12=-2C C12=0.5C22(again the second row offers no new information) C122+C222=1(from normalization) 0.25C22+C 1.25C C22=0.8 C22=V0.8=2v0.2, and therefore C12=v0.2 Therefore the eigenvector matrix becomes 202V02 √022V02 b first determine th -2-入0 0 2-λ From la. the solutions then become -2 -2 and 3. Next. determine the eigenvectors First the eigenvector associated with eigenvalue 3(the third root)2 For the eigenvector associated with eigenvalue 3: ë ê é û ú ù -1 2 2 2 ë ê é û ú C ù 12 C22 = 3 ë ê é û ú C ù 12 C22 -C12 + 2C22 = 3C12 -4C12 = -2C22 C12 = 0.5C22 (again the second row offers no new information) C122 + C222 = 1 (from normalization) (0.5C22) 2 + C222 = 1 0.25C222 + C222 = 1 1.25C222 = 1 C222 = 0.8 C22 = 0.8 = 2 0.2 , and therefore C12 = 0.2 . Therefore the eigenvector matrix becomes: ë ê é û ú ù -2 0.2 0.2 0.2 2 0.2 b. First determine the eigenvalues: det ë ê ê é û ú ú ù -2 - l 0 0 0 -1 - l 2 0 2 2 - l = 0 det [ ] -2 - l det ë ê é û ú -1 - l 2 ù 2 2 - l = 0 From 1a, the solutions then become -2, -2, and 3. Next, determine the eigenvectors. First the eigenvector associated with eigenvalue 3 (the third root):