Final 15-16 Spring.pdf - Name Last Name Signature Time 16...

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Unformatted text preview: Name Last Name . .: Signature \ Time : 16: 00 Student N0 Duration I 120 minutes 5 QUESTIONS ON 4 PAGES TOTAL 100 POINTS METU Northern Cyprus Campus Math 260 Linear Algebra Final Exam 31.05.2016 Show your work and justify all your answers! Q1 (15 p-) T0”: y) 25): Find the inverse T'1 of the linear transformation T : R3 —> R3 defined as (4m+2y+3z, 5x—- y, 19+ 2). PM— A’ HES m: [‘1 3~ 3].N,|¢MM(M=—W§-i0=1. 2" : AILZ“ :c:='5—1A'3=5-’ .._’- 3 __ : (I 3 : __ q a :_ .. 11 -4, An. 01) ‘f 1‘41} 0 I [I Q2 (10+10=20 p.) This question has two UNRELATED parts. (a) )Find a. basis for the subspace U= {p(m) I p(— x): ap(:c)} C P4011). Nola M B: {x x3j‘5. 71. whfi ffx) woman—0M +0311 Max’s/u 'Hc“ Clo-Q,“ Mg: '19,): mxi— 41,, —a, x_q,_x ",4: x 3'49“ 179% as =01: 011*" ",0 MB 5;. ff»): 0,2: 4-61, x3 e Span (’13), “a“, s a 4-1.»: 12%. (b) Show that the subsets A— — {v1 = (1, 1,4),112 = (— 1 0,3),113 — (1, —1, —10)} and B = {1:11—— (7, 3 0) wt): (2 1 1), 1113: (0, l, 7)} span the same subspace of R3. Then find this subspace (i. 6. find an equation defining this subspace}. 41:4 11L: Fri'z. 40,3 A:[—403}Rflgl”[011] ”[0171’MBJ 4 4 4o 2,-2, 0-11-11: 0 o 0 symflmgfl: C4, ray-3)) (0;!) 1U. Smflxé) 130 2130 3-39; 10—2: 10—} 5:[;~11]}:—Lg,~ 017-} M 01?]~[o¢;] 01 4 01 1' 00 0 00 0 :2) SPMCB) <11 (4 or?) {0,1,9}: sped/1) mom.“- 5,..(13)={_('1 j? wars/1) ’l,/~e {R} = 5% 13+2=0jx15 q EFL-c 111112311111 TE; 0‘ML W “2(52'43 1). Q3 (15 p.) Find the values of a. and b so that there are infinitely many solutions to the below non-homogeneous linear system in R3: .1: +z=4 z—y+2z=a -2:+by =—3 * 915‘! 101:9 14-150- Rvewv 0—! 12%] ~ -1 c 05-} gm, of, 1.1%”192- [« c5 1 3 five“) Fr 9; [Ll-l o o QHH) -l ~[4-9}{ M o -l 1 Ila-q (“Der-93 ~ 0 -(3H) 0 EQ-QMHJ—l‘ta—k) o a 3-H MM 0 O éfli LFGI'IIJé é dzf. fi'f’bwc. Lave gnaw “£me W -®1+(q-‘0€=0 => 0.41:! :9- 3 Justus HMWWIOL: -1-—1-c—o-_— 0, 949“") .4 —- [VU’g —_- ~1—1-(—:):-o, 444W!) 4 5'0. =b b+1=0:’1+(a~—h\b 1C“ “‘4'”; 53 Hwy; are, ‘MFFM3W13 mafia Soluh'nqj. Q4 (20 p.) Use the Nilpotent Structure Theorem to find a. Jordan basis F and the Jordan canonical form J = M f: (T) of the linear transformation T e .8012“), T(m,y,z,w) =(33—y+z-—-719,91:—3y-7z——w,4£-—8w,2z—4w). Pwlv K\=MLT). W M'— 3 -l 1 —1— 3 -l 1—1 3 -l 1 F} 3.4:. 3 .-J -‘+ -I Kg'32|~ O O "(O 2.0 N O O 1 '2. ~ 0 o q -8 O. o 1—2. 0 o o o 00 a-‘l ”V“: 00 o o o 00 0 Q5 (30 1).) Find a Jordan basis F and the Jordan canonical form J = M f: (T) of the linear transformation T E £0136), Ta: :3 :1:- 3: m m = a: +3; n+1? 113—33 —a:4+ass -:z:3—a:5 23:5—2 . ( I: '21 37 4| 5: 5) I. y i 1‘ ? l 6 110000 Htoooo:4.‘___* E 11 000 {1—1-0 0 oo “”3“” AsficCTM 0-1-100 0 =3AM= o—l-H-o oo =d'*D (ti—2H3 000401 coo-i401 mg, 00-! 040 o 0-: o 440 “Thbflkj 0000.2...1 0000:.44' ' ) =—l ,(THJGkUXfiL, 3.4-1!” 4:,” X5) 4:, I215), VJ"={xlexL=x,=-x5=x‘=oj; @4073?) 2 (Mafia, Inn, kn.) ~74. 4-K», .1115, twig), V.“ :1”. :xst33X5- =0); Cr+93(i‘)= (Hannah, lam-I'm, «5-5!» ~1n,x,+ 1.1., 2.1:, ) V.,,,=h.=¥._=x, so); CHD‘N) = (lint, Hum, fiommh, 433.4%.) m, ’ma‘flz, 18. Ha), V_M={ x,=x1=oL CLO“; vi: [0, 0,1, 0,0,0) 9 \Lw/ V4,, . Tunfi‘hjvfi; (a, o, a, 0,4, 0) my“: ((9, 0,0, 0,0) -2.) :CHDJVI': (0,0, air-7.) 0, o). J For 3:1 we. Low. CIT-90?)=(-x,+x..,X.-¥;xx,—511,,—3x.+x5,-x,—3x;,129--3x‘) ...
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