M235A5Soln

M235A5Soln - MATH 235 Assignment 5 Solutions Fall 2006 1...

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Unformatted text preview: MATH 235 Assignment 5 Solutions Fall 2006 1 —1~3 0 2 3. LetVbe the null space ofAWhereAz 0 1 2 ‘1 4 and let 2 1 0 3 4 3 2 1 5 6 5 1 V: 3 —2 1 (21) Find a basis for V. (b) Find the distance from V to V. Hm “A "I" saw : (a) Le": x:(x\,x1>< x xskéNm/ai ISJH/ / A 3? t . TO ,9 ohm ! 1,1113 MCCL" \‘X f: POM) VECLM.C5L l"! "3 0131131 [Owiouréiib 012*l%§0 Mumyouzozgo mar 2103+‘O 0001—210 32.156113 ooooo}o \/ .. 1 ~ X1 l ""l‘ 3R5: Sounmvmi I? -2_ "2 r) x; =2: 5 1 + 1 o ) g,~'ce1;\ xg O 1 X5 0 \ \ _L+m/ M} 0. ba/J/Lfl “pOY‘szg ‘1 *2 *1 2 g 2 o 2. O s “10er Kg” 6W Or+nganA (3 marks» \o) M axm‘ma, QM 17 40v (,3 awe/“Jag H‘C/‘m Frog‘s/mu / To COMPWM FOCAVLV) Om or‘Jrkogo-mug £0? V LE}: “0‘ = (WA 1722(3) m \ O O 2. O I m w>= “xv: v. + was: / V <Vl’ | 4V11V1> 5 u 6 ....>- “a. “A. ” O \/ '- ~— + M V : V -\/ —-_ j e ‘ 15 1 ‘ l ‘1 “f 5 5 i) A, A : l ’ O W, \/ PCOJVW3 3 \ z 2 :2 “2 a) \ ‘\ 2 amcl \\?~pvajv(V)}} = 3 \/ (H‘MCWkS) 9. (a) A vector x = (x1,x2,x3) E R3 has the following properties: 2 V6, x is orthogonal to (1, —1, 1)7 and the angle between x and (1,1,0) is 0 z 7r/6. Find the vector (or vectors) x which satisfy these properties. (b) Find a non—zero vector in C2 which is orthogonal to both (i, 1) and (1 + i, 2), or show that none exists. 591%: (a) We, km JAmos}; MHz-)2 => xf+xfg+x§=éj m 4i/cll—l/1)>=o =‘> ><\~><l+><3=0J (2) \/ ma 62‘, (\,\,o)> =11?<‘nllu,\,o)nm<9 =1> ><\+><2‘:3 (3) From X1: 3"X| amcl 'Jq‘m (7‘)! X5: 3‘1)“- M C0 Hug Cbmckcli‘mh‘L / almaofl’ x‘z— 3x\ +l=o => (xx—Zlbc—lfio 11w», xsl or ><\=l Mal 3?: (2) l}?l)\é‘ \,1)\)r/ (lo) Leflc “26W =(z,, 23/ mm H marksl (‘2‘, (gm-=0 Mal (E,(t+i,1)>=0- l/ ~EZI +21: 0 ancl (\~L)E| + 221:0 Frm ‘lllrS-l: asbuaiio-m 2;: Ci. Sulm‘l‘i‘llxl‘t‘ng \/ 4%»; v;~1‘0 4h Axconol chucq‘imx gin/s (I—Uz, +2Lz‘: o :> 2‘: 21:0 HJLAQL/ fig: (0,0) no non—g/U‘o \/ U‘CQJVOF [:2 MA K9. C3 mamkS) 11. Let W be the subspace of R4 spanned by (1,0, —1, 2) and (2,1,0,—1) and take the inner product to be the standard dot product. (a) Find the vector in W which is closest to the vector (4, 3, 2, 1). (b) Find a basis for Wi and the dimension of Wi. (c) Find the unique W1 E W and W2 6 Wi such that (4, 3, 2, 1) 2 W1 + W2. 801'}: (a) Lei “Li: (|,O,—l,l) ‘31: (2, |, O,~l) ammo) v=CLfl3/Z)i).mx+bv~eclnr~ \/ WL’CLk CLow')‘, +0 '{7 (A: {gnaw/(<7) T0 Com vie: ro‘ (7) am of‘JrLo on») ‘DW ‘QPW (/1 1: MA 31:1 ML OF‘HLOSOROJ. \/ M; Ppojwwyz 4&ufi j“ +4’\7,u1§ “J; <uu U‘l> 431161> A O“). - 5 ’1 v“ (3 mokaS) :7. 561—11, +l(: u, ~ (4,, 3—, 3-, g) / Let 7<~=C>(‘,Xll)<3)XL+\€\A/'L WJF SOCh‘S—Fé 4;,a,\7: <n>Z/‘L:;‘>=O *0 “bk Sykm 22(\ + X; _ z X. \ — , . (7(2): S(‘2>+%(éz>;3,+€flz,m/a [:ijgrwl‘w: >5; I 0 l /-2 ‘/ / 4 O l ,. ‘ J. .. (3 Morlj‘s) amd, dam(\A/)*Z Codd 0330 WA:&Lm(RH)= Jto Okfiflm. (c) Frm (a), 17:7,= progwrvk wag—r1”): m“, :13: “32$”. = (cu—3, gag.) \/ (2 marks) 13. Using the Gram—Schmidt procedure and the standard inner product in C4, construct an orthonormal set of Vectors from the set: 1 1 0 0 i 0 1 0 X1: 0 3X2: 2 7X3-. 0 5X4 0 0 0 1 7; SO"): “\JLSL GFMA“.SC)'\Wx/Cdf{: PPOULOLUJ‘E, aha/n \/—_\7;:.>—2-‘ .A. A 2' ‘ y; .15.. \/“v‘1“ “We; (3 wa- 5 = :2 4mm o, o o T/Zzi: ... (Kym A __ Gal-593M, “‘ZI‘ZST' l I ~ \/ WNW @357; 1 ¥ ‘ O t l L Ll-u/ Lfb : __ (“(4) L __ .1. I L} 6‘ A" L‘— ‘ ———— /q \/ ~ 0 l O \‘3 1+ gl/q J 3 \7‘ ‘ O O l C\ ’v‘; ’7 - (fllvfi - 4L 571%: Jim/>3, (vim) <—\?3//:73/> 3 1 1 i , f 1+/ : 8 O L __ O \ Ch L“ ~__Lrl{3 A; *1; “ ""‘ W q. - w“ (‘1’ —. _ '3 j\/ ._ . o 1 g 18 0 H1 ~2 2y”, L* a L 6‘ LH/ia CZL' ‘ ‘ .3 A/Al—Jhl Thug/am orPLofionJ kWQ’FQ: WMEO/gvllvl/vj‘lvh} Md am, Or— norrvxal Lo gwm I ,_:§. \ ~51 (v \ \/ 7:..Vi__.\..t "Jaw ,Lgf)’ “ w “ ) L “:37 - ' “77H Vi g llvlll “7 0 wk "(7/ H": \/ .4:- jl ) -2; \/ M : 3 1; J”- L‘\" m(& uL+z :l :‘I’: L 3’ my” W i; M ” ‘g :u (2/ marks) MAS Question 14: So1ution The coefficient matrix, A, is given by: A = [0.5 0.01; —12.5 1.25] 2 A: The corresponding diagona1izing matrices P & D are found using: [P. D] = eigCA) P = \/// —0.0400 -0.0200 —0.9992 —0.9998 D = 0.7500 0 E \I/ 0 1.0000 0.5000 0.0100 f; —12.5000 1.2500 DAk as k tends to infinity approaches: DD=[00;Ol] DD 0 O 0 1 The initia1 popu1ations are: 51 = [40; 1500] 51 = 40 1500 The 1imiting popu1ations are Found from: SL = P"DD*inv(P)*Sl SL = 20 1000 Thus, the 1imiting hawk popu1ation is 20 whi1e the Timiting mouse popu1ation is 1,000.\// (1? NWCRrLCS‘) (19m) 05;: 3§mark35 Page 6 ...
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This note was uploaded on 01/05/2012 for the course MATH 235 taught by Professor Celmin during the Fall '08 term at Waterloo.

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M235A5Soln - MATH 235 Assignment 5 Solutions Fall 2006 1...

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