Question

6. Let A= 0.7 0.1 0.3 0.9 Find P and D such that A= PDP-1, where D is a diagonal matrix.
This is a two-parts of the same question please answer both

Answer 1

() And bin Given Aero9 0.3 09 of A, let For eigenvalues TA-Io 0 - 0 3 04-/- (7) (-) - 0.03 Eo 3) 063 0.7) - 09 > + 2. -0030 ? 2) 16 +0.60 디로 16 서음 10 lo 2) on²-16 | 4 6 co ₂) 5 - 0 + 3 20 1 5²-5- 3 +32o 51) - 3 (4) 0 (-) (5-3) 20 서0의 여 55-320 을 8 왕 9 920.6

(2) let X=[] be the eigenvecter Corresponding the eigenvalue i=1, then (A-II)X1= 10:02 07-1 0.3 0:9-][39] [] -0.3 he co] [52] [3] 3 0.3 -0.34 to.1 12 zo and 0.3 04-0171220 which given 0.324 = 0.1712 ve T be a て lo (2 34E re fu=1 Ezzle your Hence X1= [5] Also, be the eigenvector Let X2= [] corresponding the ergenvalus 1=0.6, then

(3) (A -0.61) X220 0.9-0.6 0.1 50296 1J-[:] los *1*1[] ]-[:] 0.3 0.14 tool Y Z 6.34 + 0.34 on equating 0.14 toly zo and 0.34to.32 2o which gines ol4 = -01 2 Y = - Z of Y = 1 then Y2=-1 =) Hence X2= [-] Then matrix pa [3] AS New Pl= -1-3 -> IP1=-4

We have O 8 1 | (4) a b e d 으 then -d -b adj P= 년 a = . adjp= Then P= I adj P IPI - 3 PE -1 - 1 야 -3 문 New Apa at - 1 개 고 - 3 070-3 07-0에 4 -3 6-3 4 21 6 3 - 04 To ] ] Jes 7 [ ] [ 2] 3 3 ~0.6 ~0.6 +0.6 -10 - o - 6 BAPE CD ) 이 S 0.6

(5) Hence D= AS [6 86] 0.6 from (1) PAP: р => PCPAP) P'= POP → (PPT) A (pp)) = ppp! c) A= PPP (2) Ans (7) foom (2) A = PDP A² = (PDPIJ? = (PDP') (POP) = PDCPP). OP A² = PD²P -) A²=P [ . p (0.6)2 p! EPDO 2 (6)

(6) Similarly А Po² p = ho p[ develop P-[o] AK = Pokp' =p Now lum K100 AK 11 um (PDk py) kyo = Plum ok ok j pt Koo p · Plum kyo [ S er]) K Since & cl, then um umer Good =0 then um AK - Kyoo P[bo] 1/4 = [] "][b][ 3/4 4/4

(7) lim Ak ha 1 il s] 1/4 34 내 Kyo 3 1/4 ht h/s 윤 314 Nw 0 0 Um Ak KA0o h// ht / sh/s AILe 0.2 0 0 +2 1 밤 24 + 90 30 h 1/4 314 \ Ang