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M235A1Soln

# M235A1Soln - MATH 235 Assignment 1 Solutions Fall 2006 2...

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Unformatted text preview: MATH 235 Assignment 1 Solutions Fall 2006 2. Solve Ax = b and yA = c Where 1 —1 -1 \$1 2 A: 2 *1 —3 ,X: \$2 713: 6 7y:(ylay21y3)702(0)_272)‘ 1 0 —2 4 30 C: : To 30'th =1; ma mnjtéci mac‘Y‘Cx Mal FOUL) T‘QAAACJL) - — — 2 — — (A,WZ)~(12_\I é>w+<gﬁ> } E>RREF\/ 10-2 % oo o o m SOLJCHM Cam EL MWCQ V€C+OP 423mm (Lo, ‘ / / 32:: : +-’E(IZ) 3 { Lea/m arbthﬁcmg Paramdﬁ‘ x O l (fem) 3 To Sou. qua 9.3%; an mqufmjto 013mm. (MW: (LT or A z”; H4“ m3: WP mamnM moérrix mi row (‘QdMCSL ) (ATM: 121 o .\ to—l'+ / ’C) (—1 4 o ~2>A(ox i—Z>RREF’ “I '3 *7. Z 00 O O . HM ‘6 = = <—2> 1" Sb); Sam ‘j3 O \ or 73: (ﬁwﬂglkjg): (%I:}/o)+s(r‘/ﬂy\) (6 marks) 30V: SQJC up WAQMRRCQ Mod‘rix (A/I\ mQ‘ (“o-w reCIMCJL) (A 1» :_ \ C 2 \ C) Q ’ :2: O Q, a: a o \/ ~L \ ~11 Q <13 a \ L 2. x o o "3“ o ~2; 2 —-.7. \ O F;-*Vz*2v\ O o L L O O Y‘s-3w 3+Lq ‘ l L 2. \ O 0 <0 \ C «L V2. > rzwavﬁjrl . mail“ 0 o l \ O -L Q ﬁg \/\// 1 L 0 ~\ 9 2L Pr’ﬁ‘i‘é O l O ~ZL V2 "“\ r‘z’afl*lf‘3 0 O I \ O "t l O O ~3 {/2 3L r‘ﬁﬁ‘w O \ C) “it V; *l O C) \ I o -i , W W” I A" _I ~') L ’2“ m, A :2 D. 2; “3L \/ ~Zu ""| ‘x O *1 (5 mark5) 7. Let V = R4. Consider the subspaces: U ={(CL1,CL2,CI,3,CL4) E R4 2 a1 +202 +3613 : 0,6l1 +a2 +a3+a4 = 0}, W = {(a1,a2,a3,a4) E R4 : a1 +a4 = 0,a2 +a3 = 0} (a) Find a basis for U and a basis for W. (b) Find a basis for U H W. Exit: (C0 “ED J1?»th a \QW ~Porii) Sabre +94 ALJA‘LKM \ 01+ la2+3q =0 ) ql+a2+oi3w~cur=0 3.9.:- up W3 Nair} magi—er mci F010 FQCLM.Q_Q (I 1 3 030) do 0 " 1‘O>RREF/ \iiiO Oil-\O M @H-Mé» / °”* 0 ' / / 1~Lmna)a\om £00 UL Lo :{(I,—2,I,o), (-2, 1,0, 1))Z ’17; a “For” \A// Sabra i‘LL O\\+ / 01+Q3: O V Sash pg? MOC‘TCK amok roe.) PQCLUCJL (1 O 0‘10) ﬂFeac-LQCMRREF o\\OO To a ‘\$DF‘ 50M All Aufwl-‘Lm Q\+9\C«1+3Q3 :0 C\\+O\l—‘r Qq+abrzo \/ Ox\+O\Lt,:O a +0“, 2 l ’3 O O I ‘3’ l \ \ 1 0 M Cl 1 O O I O can 2 4&0 0‘ \O O QCDQJ C) 0 al *1 m3; a1): It *l)3{;e12 / 03 I Rana ox box/aids wer Co: Sikkim)? \/ (3momk5) / No}er AA CA/Q'krhc’ftjl aﬂﬂ‘ouc’lx mold ilk {*0 SOL/4% Cl (“v—2) ‘/ O) + Q2,(’211/O/ I): C\$<O/"/1/O)+C’Lr<—‘/ 0/ 0/ i) W ark Chart) vac/40F u QPBLJI‘WQFB veg-Er” I)“ \A/ ; ‘ I“ a \ _ , ._I Omsk 50km 70"“ CHQQICBICLf‘TOCerK-T Cit—Cl: Q3'C,L+—-‘C, 12> 5% 45¢ unw ca: § (4,4,1, I) } 8. Let p1(:c) = 1+x + 05.272, p2(x) = 1+ one + m2 and 193(95) = 04 + x + 11:2 where oz 6 R. For what value(s) of a are 191,192,193 linearly independent in P2? 50‘23 Ha LCBmoA‘Lm Q\P\+Q2\>y.+ Q3133“) / is»? 041+ 7<+ \$26) + 01(l+o<x+><2)+q3(o<+><+><l) =0 ComFoxPLné Mmdemn+s aim CH + Cxl + o< <13 : O / a\ +o<ol+ Q3:- 0 (10” + Ckl + Q3: C) \$135 up mer‘Cx mot (“ow (Roman I \ 0L 0 l O o4 O 1 0L \ o “*- O Cob-l) “~90 o Y‘ﬁﬁ“ﬂ 04 \ \ O O (beam-‘08) 0 r3”" r3"°<r| l O <>< 0 “é 0 (044) 0—04) 0) / o o (ﬂoured 0 THE“; No Mm *0 when \$¢¥Lm Kn arch/x40 OLFOU’VLCQ CuJW“CU{oe\ Sobﬂrim Q.=O\2_= (13:0 (and, no; QHSLAFE. PUPl, P3 OuA—‘L WJC kw: ﬁnal—007m => 94¢ \ 01-2: ml Pl/Pl/P3 WWO‘PH I“; <><=f~1 axo<¥l=\. \/ \/ C 5 mark5> 9. Suppose that v1, V2, . . . ,vn are linearly independent vectors in a vector space V. If X g? span {V1,V2, . . ., vn} then prove that X + v1,x + v2, . . . ,X + Vn are linearly independent vectors. 3&2: Coan Mbmod‘im ’—_— “\(';<+_qu)+%(§+"§1)+ ' ' ‘ + QR(;Z+‘OV\):’6 (I) \/ _.._‘~a. ~3- A (a\+ (11+ » ~ o who“)? + CA‘VI + Cog/1%“ +‘c«n\/m=’5 (2) \/ SLnUL i¢5meV,VM‘-3VA% {gm}, Wm. w+ 1‘an Hear Q\+Q1+~~~+c\m:o~)owrw(l)cw\\:& V S‘on {gr \/:\« Rpm .u/VN. ﬂows, (2) .4- ...z— _—>- alV,+c\lv7+~--+c\nv*o / SLWCQT/T/vl’ my?“ W. WarH E \/ H’ ONT—011‘ °‘ :C‘Am‘T-O. Show. Q\=C\l=*-°=o\n:0) {+1Q‘m (W J Hid} ’R’R/T/‘i—Fvll ’ "Iii—ﬂ?“ W Markos} cm racbchrLds. (é mowkj) MAl Question 12) Soiution: System of equations For T_1 —> T_9 is: 4T_l — T_2 — T_6 = 30 —T_l + 4T_2 — T_3 — T_5 = 20 —T_2 + 4T_3 — T_4 = 60 :T—3 f 4T—4 ‘ T_5 : T_9 f 40 _ Otwowci N i-wo 01ng T_2 T_4 + 4T_5 T_6 T_8 — O / . —T_1 — T_5 + 4T_6 — T_7 = 10 rﬁcxckks Ckﬁleﬂ. “éé ovx —T_6 + 4T_7 — T_8 = 40 ~ —T_5 — T_7 + 4T_8 — T_9 = 30 W “W‘er Y 410’” —T_4 — T_8 + 4T_9 = 70 CAFE; C1)!“ Q,L:C Set up coefficient matrix: A = [4 —1 O 0 0 —l 0 O O; —1 4 —l O —1 0 0 0 0, O —l 4 —1 0 0 0 0 O; 0 O —l 4 —1 0 O 0 —l; 0 —l O —l 4 —l O —l 0, —l O 0 O —l 4 —1 0 O, 0 O 0 O O —l 4 —l 0; 0 0 0 0 -l O —l 4 —l; O O O —l O 0 O —l 4] A = 4 -1 o 0 0 -1 0 o 0 \/ —1 4 —1 0 —l O 0 0 0 O —1 4 —1 0 0 O O 0 O 0 —1 4 —1 0 O O -1 0 —l O —l 4 —l O —1 0 —l 0 0 O —l 4 —1 0 O 0 O 0 0 0 —l 4 -l 0 O 0 0 O —1 0 —1 4 —1 0 0 0 —l 0 0 0 —l 4 Set up right hand side vector: b = [30; 20; 60; 40; 0; 10; 40; 30; 70] b = 30 20 v1/", 60 40 O 10 40 3O 70 Create augmented matrix: Ab = [A,b] Ab = 4 —l 0 O 0 —l O 0 0 30 —1 4 —l O —l 0 O 0 O 20 0 —l 4 —l 0 O O 0 0 6O \/// 0 0 —l 4 —l O O 0 —l 40 O —1 0 —l 4 —l 0 —1 0 0 —l 0 O O —1 4 —l 0 O 10 O O 0 O 0 —l 4 —1 0 40 O 0 0 0 —1 O -l 4 —l 30 MAl O O O —l 0 0 0 —1 4 70 Row reduce: rrefCAb) ans = Co1umns 1 through 6 1.0000 0 0 0 0 0 / 0 1.0000 0 0 O 0 O 0 1.0000 0 0 O 0 O 0 1.0000 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 1.0000 0 0 O 0 O 0 0 0 0 O 0 O 0 0 O 0 0 O Co1umns 7 through 10 0 0 0 17.8571 0 O 0 22.8571 0 0 0 28.5714 0 0 0 31.4286 0 0 0 25.0000 0 0 0 18.5714 1.0000 0 0 21.4286 0 1.0000 0 27.1429 0 0 1.0000 32.1429 Thus, T_1 = 17.8571, T_2 = 22.8571, ..., T_9 = 32.1429 y/// (g Marks) 0*? LFO Marks) ...
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M235A1Soln - MATH 235 Assignment 1 Solutions Fall 2006 2...

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