Question

Exercises 3.7 In Exercises 1-12, determine the e-values 4 e-vectors. [ 3-2 4] 5.4-[ -[] 7.1-3, . T 3 -1-1] [i 1-1] [1 1 -2] (9. A = -12 0 5 10. A = 10 2 -1 11. A= 0 2 -1 L 4-2-1) Lo o i Lo o 1 In Exercises 13-18, use condition (5) to determine whether the given matrix Q is orthogonal. 6 67

question 9. find the eigen value and vector

Answer 1

given matrix 3 - न2 4. -2 ड। - but N is eigen value of thin ___dat / A - CT) -0 13- - - -12 - 0 4 - - - oy__(3-4) -d(-1=a.) + n? +ा राता- 10 __- 124 + 46१ = 0 -_(3-4) { A+ ++ 107 + {124 - 2 } - 240

(3-d) f + d +10} + 8 -32 = 0 3d² +30 + 30 d ²100 + 8 -32 = 0 or Papi?. M3+ $ -2=0 or 3-2d²-1+2=0 (da 2) 160-2) = 0 (0-2) (0 - 1) = 0 or __ (-2) 11-1) 6+1)=0 => d=2,1,-1 tenu the eigen value of malrix A If ziti antata How we will find eigen vector corresponding to Eelgen value त 2,1,

for d=2 A --(A-21]x=0 l -77294710 -12- 2 5 x2 = 1 4 -2 -3 F to Change in augmented matrix and thin reduca in row nudued form os, TTT TOT rict Tol T-12 -2 sono - 14 - 7ol I 4 -2 -o o -2 - To | 0 2170 L o oTo Henna - x - x₂ = 0 2Xq_+213 0 2 _X9 € - 2967 - 47522-242 -- Quote charat X2 ET

for al TA - I ] Y=0 r21 -Ily, 1-12 - S 1 4 -2 -2 11 Zz Change in augmented matrix and then reduce in row neduad form of P2 Tayo + 12 also nlo -7 to 14 -2 -20° Honu 24 - Y₂ - Yz = 0 -742 + y 20 - 3 = -722 24 = Iz tyz - % -7% 2y = -642 Jis-342 chool Y 2 = 1 5. 3 7 then y =

for d = 1 [ Ata] =0 - 1 172 1 5 || 22 = 101 4 2013! change in augmented matrix of T4-1-10714 -1 an -12 Ton To -2 zol + -2 0 101 LO 1.-110]. lo - 2 2 Ol lo o o o J thene 42 - Z hy = 6 - 2222 +22=0 - dr = 42 = 22 + 2 q2 = 222 4231 Ethan choo -22 = 1 ---- 22 let Zoll

Henu eigenwetor corresponding d = 2 T2 eigen wetor -A-37 comosponting d = 1 eigen wetor cor. deel H