Question

2. Find a general solution of X' = AX if 3 -1 (1A A1 = 12 = 2 1 [ 1] 3 -1 (2) A= 1 Xi = 1; 12 = 13 = 2 1 (3) A = 1 Xi = 0; 12 = 13 = 5 1 (4) A= 5 -4 0 0 2 0 2 5 0 0 0 3 1 0-1 1 1 0 0 2 2 - 1 0 1 0 11 = 1; 12 = 13 = 2 (5) A = 11 = 12 = 13 = 1

Answer 1

6. A= [?] X's AX given H has eigen values x, = id=2 Now fine eigen veetor for lad IA-2I)V70 = 11:21:1 Her Hq to Hg = Vi=( Now find generalized generalized eigen vector ival for A=2 -) ((A-231) V- Vi C-1) -1 n 16) - V Hy - Ho al it ng if Hg =1

General Solution', t و ) = ۱+) Creavit ) و + اندام (۷) ) و At e at t e =) Xct (41) (1) ام و ۱۰}} = Ye e 1 X+1 = (( ar past + ( e علم مع ) + + ( +و tert e د ۱+) « ) - ( 2 ) احم و + ع (+) و عو Ittient G, CG Constant و 4 ) 3 اح ا۔ اے Given eigen valves = و = ا-۸ Now fine eigen vectors for dial i Aga Mga-2 ور

for А, - | (А- А, ) V – о *(-){:: :) 3) - ) и, - изто к, - Из – о V, - И 3 и, М [ :) М2 So", () for 7. - 2 - 3 (А-8T) ve o ( :)... :: -) и, 2 и 4 из и, - и, и; - из

Solution Set * * * *(1-01 Let ų = 1; H₂= o ; V, - (1 Hg =0 i H₂=1 LA V₂= (!) General Solution det dit Vt عوام X14) = cie Vgt (se V₃ Xit = 12*(;)+5883)+5+44)

cm A 5 -4 . o 0 og of F Given eigen valores 1,50 ; la = d;=5 Now find eigen vector for Azo A-OIVO 5 -4 2 Duco 512 Q f >> 25+ NV350 Ugt [113=0 if H3=1; V, -blue for das W-dg I) Y CO -) O 0 -4 -5 O 2 ul u 0 o 0 2 o o o 0

Het 243=0 Hq zo if H3=1 ; Vaz 2 *:( o How fare generelized generelized eigen vector V for A325 -) | A-₂ I) V = Va o 0 0 -4 - 5 2 2 0 no 0 чо -2 0 O 0 0 Hi - Shot 243=0 - 40g = -2 = > Hoa -lo Hi= S-243 if Uzali V₂= tog for

General Solutions X = GX, + CGXg+ (3X3 It where c e V, Y2 = Agt V, е det X3= = ( Yt + V3) @ Soj ot ( е lo X + Ģe 2 sel) (13) (1) + C lol st st -test lest ll + C3 + Ge 2 Ist e too testtest