04spr-m2sols

# 04spr-m2sols - EXAM 2 Math 51, Spring 2004. You have 2...

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Unformatted text preview: EXAM 2 Math 51, Spring 2004. You have 2 hours. No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck! Name go! 1.) it‘ms ID number 1. (/30 points) 2. (/30 points) 3. (/30 points) 4. (/30 points) 5. (/30 points) Bonus (/15 points) Total (/ 150 points) “On my honor, I have neither given nor received any aid on this examination. I have furthermore abided by all other aspects of the honor code with respect to this examination.” Signature: Circle your TA’s name: Brett Parker (2 and 6) Chad Groft (3 and 7) Joe Blitzstein (4 and 8) Ryan Vinroot (ACE) Circle your section meeting time: 11:00am 1:15pm 7pm 1. Find the determinant and the inverse of the matrix 10000 01020 00103 7—2010 00001 I O o 0 O l O O O O O | o 2 O o l O O O 0 (j | O 3 O O I o o 7 —Z O | O o o O I o O O 0 0 1 o C) o (3 I l O O o 0 I O o o 0 V: O l o 2 o o \ O o C) [‘2 O O \ O 3 o O l O 0 r3 0 O O 5— O ‘7 2 O ( 0 111 +28 0 O O O l o o O O f r S— f O o o O I C) O Q 0 r1 0 ( a 0 0 Isl/g. 1/; O _%_O r1—%n+ O O I O C) O O ‘ O ’3 {aviary O O O ' 0‘7; 7/5 o 1/; o C4/5" O O O o( C) O o O | F5— 3. Find the equation of the ellipse obtained by rotating (clockwise around the origin by angle 7r/ 6) the ellipse with equation I Hint: A point [y on this new ellipse is deﬁned by the property that, if rotated counter- u clockwise around the origin, the resulting point [v :| will be on the original ellipse and thus will satisfy the equation Lzlr T CDU‘n‘ief‘ OLQQLJISQ arbqu ~6- n76: 33/: 4/2 3/1 411/3?) 62 F‘- t *‘3 may (if) “’7 A v2, 5/2 4. Let A be an m x 71 matrix. Describe the precise procedure by which you would determine bases for the column Space and the null space of A. Based on this description, preve the Rank—Nullity Theorem, which states that dimmmj) + dim(N(A)) = 71 Make sure to explain clearly all of your reasoning. To 3% «Loss e co), m it it some or A it} st it Jive Loiuwwxs oi: We? (9) mid.“ Pivoi'si To "\ [Dads gr NOD») we So we “Li” pivo‘i Vari‘RUES "‘A 4am; oi: ﬁne mﬁmLizs in Sjs‘iw Wei: (/ﬂk" :7)”; "MW goiuiim Seii’ 09m 13‘ WALL/A exfiiciiitf in ierrms 0‘: “HR 7C.“ meimiaies / 0wan i9“; SefmiivD won hinting! we cowx wriie i’Le Saiuiim St'i as Jim 569m oi: at ge‘i‘ a? UtC’infg‘r'“ ‘H‘lsi \itcirorg owe 0\ [oasis £3? k ei‘ a hows PM iii «have We. See ilm‘ CCA') w 3 \JQCJrur ‘CDI‘ eacix p‘woi chime—x in (Vii: (A) L A Q” Nag) we 0L LaéiS Vac-i0“ ior eatix ‘Cr‘be Column it“ FF€£(A). 11w: iii nmhar oi: eiww‘i; in HM ‘iwo EARS ( 0» er) 5 berries? Abﬁﬁﬂer is “we NJme off PM); A *Crfe cJuwwa/ uLigL {g Hm VWhnL-Rr o'p m/umnS. ‘ L ‘ ‘ J! -. §;nce J’L-E WUMLLF‘ 0‘; QL/nxm‘l‘: HA 0k (RS—[S IS (Mmglm/ 5. Prove that for every linear transformation T : R” —> Rm, there exists a. matrix A such that for 3.11 E’ e R”, Tm = A? QiVﬁ‘A ‘Hva hmar iW§LrMLM T, DQQ'Qi‘m‘i 0“ Mlﬂk A L7 A: TL) chm “3 3‘“ O‘H‘W W5 5, 'l'Lﬁ CJumnS (A: A 191 £9.9th as ‘LLG T0?) = A? 6 Bonus Question: Find a matrix A such that ( 1) None of \$1,142,143, . . . ,A9 is the zero matrix (2) A1” is the zero matrix. F Lei T a a m mama Md L7 A7 :TOWT .. lo a; 4 H) a A : T0 - .- 0T ‘_ ‘62 u _ I ‘ I ' 6‘0 0 ma ‘JTMSVLFM‘ng WMSfa-anlrj “lb g} All “7 Ag SIAer ET 473 a DAMPER) Vac/£01"! So Mm Veo'g’crs cavélf L4: ﬁne 'E-EPO (Wavyﬁx‘ lo 6‘ ﬂ S'o bl {5 Mn 242:2: W4M‘KT' dxr A“) S 5 ever»; 8 (ABQLWHV, M m m A 413 Lager W} Me the me .,5 3m LOW MQ LOW; A1 L” a t “L, I.“ 44% MW 14:; Pusalc‘m . A“) [5 Jug} a“ 34mg; . ) ...
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04spr-m2sols - EXAM 2 Math 51, Spring 2004. You have 2...

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