## Answers

A- AAT ☺ 2 3 -2 35-3 -2 -3 2 - -2 Charateristic equation of AAT 2-1 3 3 -2 5-) -3 -3 2-7 - 13+972-27=0 , (^-0.228 069-8.772)= 0 1:0, 1:0.228, i = 8.772 Vio values the Eigen are The (-1.5902] Eigen vectift = 8.712 is V2 = The Eigen Vect for x = 0.228 is | .2573 is V3 = for Ergen 4.5907 ) = (-0.469 2.1284 2 1284 1-253 18923 1-8923 1.8923 The Eigen vects for x = 0 1:] vectosals (-1,-1.5909, 1) lengit LaV7C+59073757= 2:1284 So, normalizing gives 2 Cairnsis 4898, - 2,730,468 fa eigen Vectn-2 (, 1-2523,1). Lengua LH**8.25187+= 1.5923 So. normalizing gives uz = ((0.5 255,0.6646, 0.5285 fa eigen vecta-3 (1,0,1 length 1 = tot r= 1.4142 so normalizing guves uz ATA [:] [:)-(3-1) charadui of Ata are by Calculating stic Equation | APA-17)20 1 3-2-4 129742 = 0 (1-0-228)(N-8.722) -0 9,-0.228 4,8.132 au Values Eigen -)=6*2,002 / 1.4142 1.4142 1.4142 obtained Ergen values -4 6-7 .) the 2 of ATA

V,: 2- -0.693 1.2162 C 1.2167 * Ergen Nectors for. A= 8.772 (-0.693 Eigen Vectes for p = 0.228 [ 1:43 for Eigen vectorial (-0.693, 1) lengla L = V(-0.693)+ 12 = 1.2167 So, normalizing gives V, =( =(-0.5096,0-8219) fn Ergen vectra (1.443, i) length L= V (1.243) 12 = 1.2556 So, normalizing vecta gurs = (0.8219, 0.5696 1 st Solution *-0.6698 0.52850.2001 √8.722 2 Ves found using formula Vi 0.8219 -0.5692 d) zad solution fo.

.
5696 E 8.772 :: V{x,y) 5694 $219 V: 1.443 1.7556 1-2557 2.9618 o 0.4775 50.9473 06644 0.4698 0.528 0.70H AT. Ui Ve -0.5696 -0.8222 2.9618 o 0.4275 ca Se 0.228 8.8219 0.5696 l Avi e is found using formula via lia Ji -0.4698 0.5284 0.2014 -0.7473 -0.6645 0.4698 -0.5284 0-707) ist solution A=oEvT Verify ist solution is possible verfy and solution A=u&ut and solution is possible

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