Question

1. For the matrix A = 2 -2 1 1 [1 3 3] 1 -1] answer the following: (a) Find the eigenvalues 24.. iz of A in decreasing order. (6) Find the corresponding eigenvectors. (c) Is A diagonalizable? Why? If yes, find the matrices P and D such that p-AP = D.

Answer 1

- Solution As I l 1 @ consider | A-tIl=0 3 2- -2 Int 3 J CO . =) (2-t) (1-t) (cat) - 3(2-t) +2|(-1-t)-1] 13 C3-cint)) = 0 = (2-t) Cut(1-2) - 6t3t - 4-2t to t3t zo 0 - -> = =) 2 = (2-t) (int) (al-t) tut-4 = 0 (2-t) (t-1) (tt) + 4 (t-1) = cta (4 + (2-t) (ltt)) = 0 (t-1) cut ittti- t2 =0 (t-1) cott-t².) =0 ct ) & & (2+t)= 0 tol, 3, -2 ir x1=3 X2=1, Az =-2 are eigenvalues of A.

- page no a 6 we have to find eigenvalues corres - -ponding to X222 0 consider the system (A-d, I) X =0 for 2,=3 - 2 3 1 34 1 [o o i perform 3 -4 R2 - .23 Rath, R₂ R3th 17 -2 To - 3 4 L perform R.2 - R₂ & 4 R3, R, - - R, o 1 J1.23 J ol perform Rh- & R - Ro-2hz 2 Tio $3410 o 4 - 820 Too ol la lo ) we get xyrt as free variable a=t, n= = $t

page No 4 1 . Al is eigenreetor correspon ng to eigenvalue 6 consider the system for consider (A - z I) X 3 = -2 =0 I = 1 4 2 3 12 10 L' 3 Perform - R s R3-₂ loj R - RI-4 R2 - - To -14 (24 o] 3 1220 Lo o o ll 23 J Lol Perform : RK ² - TI 3 175x7 lol o – 14 -1 182 lo 0.0 0 23 perform a R2 4 4 4 22 o lo IV/14 22 Lo o o es z = t as frere variable 2 - - 1 t & a- 3 t t - - t 14

page no. 1 -1/14 -1/14 L t x2 = © Y A is diagonalizable A has 3 Because If then it is distinct eigenvalues =) A is diagenalizable matrix has a distinet eigenvalures diagenalizable] Now , TI - | | | -1824 - Lu Because i7 -7 -1/14 I are U -114 JIJI!] [ eigenveeters of A . . . - 12 YIO. 45 p = -12 516 Y . o thes 1415 DE PAP-[12 Yo 25 12 -2 3 2 S16 -1/3 II II To 14/15 14/15 1 3 - I .1 -11/14 -/14 1 3 1 0 0 o 1 0 o 0 -2