Solutions First determine the eigenvalue 1-2 (-1-)(2-)-22=0 2+λ-2λ+12-4=0 (-3)X+2)=0 Next, determine the eigenvectors. First the eigenvector associated with eigenvalue -2 12‖Cr 22LC21 C11+2C21=-2C11 C11=-2C21(Note: The second row offers no new information, e.g. 2C11 2C21=-2C21) C112+C212= 1(from normalization (-2C212+C212=1 C212=0.2 C21=V0. 2, and therefore C1=-2vo1 Solutions 1. a. First determine the eigenvalues: det ë ê é û ú -1 - l 2 ù 2 2 - l = 0 (-1 - l)(2 - l) - 22 = 0 -2 + l - 2l + l2 - 4 = 0 l2 - l - 6 = 0 (l - 3)(l + 2) = 0 l = 3 or l = -2. Next, determine the eigenvectors. First, the eigenvector associated with eigenvalue -2: ë ê é û ú ù -1 2 2 2 ë ê é û ú C ù 11 C21 = -2 ë ê é û ú C ù 11 C21 -C11 + 2C21 = -2C11 C11 = -2C21 (Note: The second row offers no new information, e.g. 2C11 + 2C21 = -2C21) C112 + C212 = 1 (from normalization) (-2C21) 2 + C212 = 1 4C212 + C212 = 1 5C212 = 1 C212 = 0.2 C21 = 0.2 , and therefore C11 = -2 0.2