Question

2. Find the eigenvalues of A= [ 10 -2 1 -5 -2 2 3 -5 1 3 5 and verify that the eigenvectors that mutually orthogonal.

Answer 1

Given matix is A= -5 1 1 10 -2 22 -5. 3 5 we have to find eigenvalue of A for this consider det (AndI) so a I lood | -2 5 -2 -5 1 1 27 3 3 5-0 L |(10-){(2-0) (5-0) -9) +24-265-4+15) -56-6'+5(270)) =0

d3 + 17d² _42dzo & d3 - 178+42dzo A d ( d²-174+42) = 0 5 dl Cd-3) Cd-14)) == 0 d (-3)Cd-14)=0 B Z tod= 0,3, 14 inchezig arre eigenvalue of A - For di=0 we find eigenvector consider (A-dI) V = 0 :r lo-di -2 -5 Ilv -2 2- 3 | -5 3 5-di] L v3 CAST -5 31 V2 - so we have homogeneous system of lineare equation, we solve it by Guassiq Jimination. Augumated mamix of above system is 110 -2 -510 1-2 2 3 _15_3_51.

R → Gori | -V न | 0 | -2 2 3 | 0 | |-5 350 २. > R+224 →ps+5 । । - न । 075 20 ___० 2 5 /- | 0 K2- (F) | ना 08 | 0 0 ना । 514 | ० 5 | 1 2 Ra - K - 2 Ra म ना -/4 । ७ Ooo +पित) bo GO __

from above matrix we have vo N, - + z = 0 a Vz+ 5 5 V3=0 a t Here so is free vorable -- put Vart S = + = h tt & Vo = { vz = - 57 Tat-574 -5t tau L V I put tat Via L l is eigenvestor Associated -5161 with eigenvalue_d=0 ALL I Now, I with we find elgen value eigen vector Associated dos do = 3 consider CA ada 1) V2= dat To-do -2 -5 Tu -2 2-d2 3 821 L-5 3 15-dz val

10-3 -2 as -2 -5 2 3 3 3 3 5- TV 1 vel lao 0 000 17_-2 -5 TviTTO -2 - 1 3 Vol و اک۔ الافعال we solve this system to find eigenvector Val | Augumated mapix of above system is 1 7 -2 - 5 -2 7 3 -5lo 3 To 2 Tos - Å (12) 24 II. .-247 5577 10 T-s 3 - 2 To Re > R2 F22 Rg & Rgt 5R TI -217 -51710 o't117 10 17+ to

Re 11 - (-2) 22 -247 -5/+10] R₂ (7) 3 L -27 -57% R₂ -To R3-R2 -217 -5/#lo 1 * R4 – (-43) B Lowo OL - TO Looool From above vi -Val = ---Wal-va! o & va is 0 ) vi= O__ = Vz!= Vz' free variable pat uz=t

: Ve= vist veit v T t T Th | Val P ati I vå' J Lt J Lut put t=1 Vaa l is eigenvector corrosponding to eigenvalue d2=3 T Inow we find to eigenvalue corrosponding eigen vector dz=14 consider ICA-d3I) V2 = 5 lorda -2 2-d3 3 -5 3 5-dal -5

I 10-14 -2 -5 -2 -5 Tvill 2-143 F 31 5-14 / Lv 1 -2 -12 3 || Val we solve this system of linear egn to find eigenector v = |v." Lus! Augumated of above system of egn is 1-4 -2 -12 3 -5 LOT 3 lo -9 lor L-5 o - Å » (4) R 514187 -2 Los 72 30 3 glol Ref R₂ IL 16 O + To 1/2 R2 Retar R₂ + 5 R 5/4 lo VITO 1/4 o (1) ke

IL 1.6 V į 514 the 000 R → R3 (4) T / 514 lol lo 712 o Loooo ke R - R2 10 312 10 o L te lo from above V" + V" = 0 & vind is free venoble put valt V," = -3t ra" = ht -3/2t Vot EE Lkoll +

put tot Lis eigenvector nosociated F with eigenvalye. Aos = 16 ale 7 V= 1-5/4 LI are A eigenvector Of Herre Vivoa 15 Szetta-4+1= 7th LEI a co A Vie V2=0 V. Vo = - +LO - + Luba -Vee It T-3/ o V3 = -5/4|| --the-17450 V V2 L

L V. v3=0 vo Vezo Voo V3=0 . Observe that Leve U₂=0 Li v e and va to each others are mautually cothogonal