Question

3. Find a general solution of the system X' = AX with 3 0 0 -3 4 (1) A (2) A = 0 2 0 (3) A 6 -5 4 0 1 -3 0 3 -5 0 5 0 0 -1 1 (4) A [1 7 4 -5 3 (5) A = 3 -5 -3 5 5 -1 3 (6) A 1 0 0 0 2 2 0 0 0 3 3 0 0 0 4 4

Answer 1

(3) Az -3 -4 6 eigen vakues of A are LA-d Il=0 -3-d da= ', dp=4 6 - 5-d eisen vectors. (A-d, I) 0,=0 ***/ [:37*** 3000 02-(3) 3 simi Party general solution -az nit ) = c, eat (2) + Caet (1) 3 O O O 0 4 6 eigen values of a care IA-d Il=o O 3-d O 2-d o = d.= 1, da=2, d3=3 4 similarly, ces we did in I exambte so ucetues of 01-10) 2- U3= 2 0 2(L) = Ciet co84 (?) 3t + 3e

O - 5 0 3 5 O eigen ucpues =) | A-d Il=0 - di=-1, da = -5, d3=-3 3 - 5-d 0 O 5 -1-d and eisen vectors are 0-(9).03(%) of GM sa soljet (?)."() +63(33) 03- 7 -5 { 3 eigen vakues = 1 Ad Il=0 7-d -5 34/- => di = 5-42, 3-dl da=5+42 Now (A-d, I) Ui=0 2+40 [* Vi=0 4 -2+hi

00 solving 0/ (+) роо 15-uit t'o rese (cosyt - 2 sin ue) L'ad') -e547 nese Cana losut -aicos ut - isinut - Qsinut 2 Losut di sinut Losut -asinut 2 losut 1) + -2 losut- sinut ***) -2 Sinut ی را 2(t)= Cierit ut caetato) a geneset sote tion. A- 3 -5 -3 3 eigen values dis-2, da-ad3=3 eigen vectors 01 - -(2) (6) 6) - 6,279)co2"(3)# 68" (4) &H= C, e at/o