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Unformatted text preview: Name: 1 2n Snug/d kg 1 Math 21b Final Exam Tuesday, January 14‘“, 2002 Please circle your section: Tom Judson Andy Engelward Andy Engelward
Katherine Visnjic (CA) Jakub Topp (CA) Erin Aylward (CA)
MWF 910 MWFlOl] MWF 1112 —
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12  12 : You have three'hours to take this ﬁnal exam. Pace yourself by keeping track of how many problems
you have left to go and how much time remains. You don't have to answer the problems in any particular order. So move on to another problem if you ﬁnd you're stuck and that you are spending too
much time on one problem. To receive ﬁJll credit on a problem, you will need to justify your answers carefully  unsubstantiated
answers, even if correct, will receive little or no credit (except if the directions for that question K
speciﬁcally say no justiﬁcation is necessary, such as the TmefFalse). Please be sure to write neatly  illegible answers will also receive little or no credit. If more Space is needed, use the back of the previous page to continue your work. Be sure to make a
note of that so that the grader knows where to ﬁnd your answers. You are allowed one page of notes on it during the test, but you are not allowed to use any other
references or calculators during this test. Good luck! Focus and do we!!! Question 1. (20 points total) True or False (2 points each) No justiﬁcation is necessary, simply circle T or F for each statement. 6 F (a) If A is a 2 x 2 matrix with det(A) = 0, then one column of A is a multiple of the other.
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4t .L c. t. c‘ :0) M“) 15% +mo Minivans ILA'1; “wif‘lrkﬁ '5 E‘GL yTKIr O)
a F (b) The matrix ATA is symmetric for all matrices A. q IM'l‘It‘k '5 SUMMCEF‘I'L "+ 11w‘f‘5 H: own #‘wlhﬁ'
T_ T as and: art) __pc4/ was; s.
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a F (c) If A andB are invertible n x n matrices, then AB and BA are similar matrices.
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use: «a hawﬁbca “tax S van SMS‘EM
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at,“ {5 (FE) B" = M as" = M w..._. mh'iinr M'i'mkaé T F (d) IfA is an invertible matrix whose eigenvalues are all positive, then the eigenvalues of A2
must be the same as the eigenvalues of A. Mo‘l' +nurv 13; {ﬂasnuqluu if A} =rL Sn‘nPb TEAL K
STHII‘C a; The. tljeﬁuqiu‘b at A) 4," gr {ﬁfhmuL if A: ‘ikcn. A's eigenm(u:6 1ft, i 1"} EA] EVT
4*: LL 53] 4“ wmlm I «J (sf4V T @ (e) If A is an 2 x 2 matrix with determinant 6, then the determinant of 2A must equal 12.
8Q. i/ ant. (“out 5‘: A L & mere—Ans “but. Jgkrm.‘a\=n‘i— b a: cellar" A) 5;», mqu'flj't‘ﬁj £01k Fiﬁv93 Le. / WES
a... lemma (:3 « 413V 03 ‘1‘, s. in (EMT—1‘4 T a) (D If 171 and i"2 are both eigenvectors for an n x n matrix A, then the sum, 17, + 17, , must also
be an eigenvector of A. Md" “Well? 1 TC :7: n‘J \7: H ﬁjtnuLLTIWP‘S HIE1N “ha:
4—) ‘Hn‘v‘k 3‘5, ﬁ+ﬁl art“ «is: be. (A 55trw'fCllbr“ Of’xwvﬂk no. C*~c»‘lel— A: I 3,7cﬁﬂfbﬁi 5;: E]
5 SM 9‘33"“! .'3 an noggnuexj'or 5 MT— “ ejandar“t ﬁ+fi=ijj r “J ZR] x 5‘ Eh]
/‘\ a F (3) HA is anyf'ﬁ‘l " mam the“ detMTA) cannot be negative.
Jar (HT/4) = Ciaf—‘(HTJ alarm), ear M647): 91.1204)
5:. = Qatari5): "Lela C3 '20 T @ (h) There are invertible 3 x 3 matrices A and S such that SIAS = A.
mi {3055th “' oletbr Mil~13 / «(Us (wars) = Arcs“) Jet—CA") his) = ﬁgs Lem LrS
24;“), Lur MK—ﬂ)=JLt(I3)J_gtC4}:—cl¢@
so etamﬁwtntétj, s. ctr2+ :6, Ln“ hm ,4 an“?
In; twu‘rb'btg.
T Gigi) If ﬁl,ﬁz,...,§m isaneigenbasis for bothA and B, thenA andBmust be similar matrices. Na'i AER—£154th "" :4: M3 LuJ ﬂue, Sawvs. mammalu «viii. rm we, cons4w [m ~ 1:2 :1
at)“ LuS hgmbalus 1.,e.._ inﬂ— 7113 an; Jet: ;b m?" sail».th a; S" IL: ll
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L._._1 T F (j) Suppose 0 is an eigenvalue for two. :3 x n matrices, A and B. Suppose the geometric
multiplicity of 0 is the same for both A and B, then A and .3 must also have the same rank. 8 malt—maliu'ij) mwii‘ialic'ii‘ .p O Cir—L?“ CherMl),
J '= “HIM3,. same i?» AIR 5' quUL Lola Iii/R aHL fink“NJ hen 3: hank :3" ﬂ  nu Question 2 (10 points total) Find the determinants for each of the following matrices (2 points each). Be sure to show all your work, and justify your answers (i.e. just writing down “0,” even if it’s correct, will not be considered a
complete answer). 1 2 3 ' 1 You. CAIN ink3 T‘Lt .Rﬁq‘lt L pm?
0 3 3 3 , J “A 3 __ x
(a) be?!“ ail43; CJNLLLC r Incaxv its cf»: mu... {rd—ST l 2 0 .J h
2 0 2 2 *3 Citnflj +Lt. 3’ tinf ‘1 Ofmmnﬁ InnIL Sum/Jimmq
rt J'sr but) dumn‘i) Se =' O i
1 0 U ] .715: ‘IS GIN«051‘ q lww +r‘9mcjuler xviJr'ch ."F
(b) 0 —1 3 0 jun it.qu 11.1 is? an H” was) hm 5'qu The
M] ed e. I i
1 ‘—I 0 0 l. “Jr 3 NUS \jlu [ahf a : “Nihk EMIRE"
2 0 0 0 o 1 ‘5 0
e1 Lu‘ +0 ‘6. I. o a I MawJ mimic. :53“ "MJC— De Dovi SwaflS, ﬁlm ‘hHL JL'RJ‘MIkﬂﬁI (J'sinij “{ muJ 2‘4... JdUmQx—st [IS F6 (or‘coad .115; {ﬂ MT— “: P‘ A.
. _ '$ MAE?! L3“: 11 ' “I153 mix«‘1’ — (c) The 2 x 2 matrix representing a 210 degree rotation counterc ookwme. lands can}: a,“ M715“ Wan—x : [corn ~55...) sakn LSlc pLj‘trMvkﬁu—f— = Lale — (— 5.3x) = Loft“! +311'1x: ‘1. (cl) The n x 71 matrix representing a dilation by a factor of 10. go '6"; M‘HIK licks “a; [lawn 1g“ "'4‘ H.“ .1. JJ Er fan
so W n ml; (e) The 5 x 5 matrix representing a transformation TUE) that has the effect of swapping the ﬁrst two
standard basis vectors, and that has no effect on the other standard basis vectors, i.e. T(é‘1) = 51,
T(§2)=é‘1,andT(é})=§i,fori>2. a. D
l O 0
0 ma. ‘ in LCS 3' Ts {w e O :5 r 0 O
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O l
0 O o l H.”
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OD {QM I? I 5.. ﬁres11¢? = —I Question 3. (6 points total) Find all values for k such that the following homogeneous linear system has nontrivial solutions (i .e. nonzero solutions).
1: + 3y — 22 = 0
2x + y + 33 = 0 5x  5y + k2 = 0
c; "hos Sjsfem. ‘l‘o luxu— nanTmb‘xl Sah'hing Mn +7~L Mk 0? no mamLars mm». mst I“. a L Les:
ﬂu" 3 #L qwjmﬁll mf'rrg l—n Pro; «rahd JHJT LIL was 5.2m; x=j=2=0) S. calm“: ml“:
’2 a :9 ~l r 3 ‘1
1 l 3 —>Lcr) a 0 —§ ‘1 ~(r)
5' —5 k * $611) 0 ~20 JotIo ‘f(IE)
/ 3 l
¢—> [o J ?’r] We‘re mil" r; A no?
0 0 k4? Dell; JamT T)": talew huff—— 71; our 1w... mac c 3
PotJ i0 0 E—r‘a’] Is, [a o 0])
slut. fr 1c Ls Luis a, rmlc (3/ 0M:S¢~ LVle #0/ ﬂak“ I‘mC: "L32 134’. Satx L:. is 'h—‘L +P:;:wl i=3 = ' (Duel{J MALIn. fiat 147‘ .g 6 it W113 Question 4. (10 points total) (a) ('7 points) Find a basis for the orthogonal complement of the subspace of R4 spanned by vectors 1 —4 3
_3 , 6 , and 7 .
—4 —2 6
3 3 —4
.L
._ l "7' '3 ‘
SD ; A '— 3 6 mm Ht'm IsakIL] ‘Rr .37 :51 .1: ‘11: J‘JE’T— ﬂawLl "t. T.l"3"‘1‘3 [—3—‘13 [anf 6 —:L 3]+‘1CI) —> [o G 4'? 15'130‘) o a e. 1' _§(:[) ( > (—3 v 3 +3017] a o {14, x,=~5‘5+?{13
o O o o o X3=Sl {1:15 F i/,
£6 : 5 _3 +t 57.1 J S. A Last}, “9 an. I O .. o l orbxojami Cue’LIOWT l; Dwtn *5” c,
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O .1 (b) (3 points) Suppose that A is a symmetric 6 x 6 matrix such that the image of A is equal to a 2
dimensional plane in R6. Is it possible to determine whether or not 0 is an eigenvalue for A? If so, is it
also possible to determine both the algebraic and geometric multiplicities of 0'? If it is possible, then
ﬁnd these, if it is not possible, then explain why not. 5. 1.; mum): A =rc~~k0¢L I, “(A Howe,
< j 1‘) ﬂail“: /
{Amr (4‘ :5 JJMMJh—L, Ii? in; am. tajeabusil; (mag «a «(7:1)me
NJ Wadn; MHlﬁpl.;.7l;¢ «04.... Wu, 5. 911'. MsKﬁPlfoiv .19
75a; 0 eigenvalue: din} Gear(ﬁgu=‘/== «lﬁebru—e Mugﬁpfiufj A: Q;
6:"; jes a I": an eaﬁmwiw— 1GP Question 5. (12 points total) (a) (4 points) Let 1’“ be the linear space of polynomials of degree n or less. Let T : P3 —> R1 be deﬁned by T(p(x)) = xap"(x) , where p(x) is a polynomial in P3. Show that Tis a linear
transformation. CLﬁcjc: I; 7—(PCX)+7JX)J Z 7—(PﬁxyLﬂl—(TCAU) ﬁr {57,6 ' TKP‘WTW: “3 000%“) if : XVPW WW)
= x3 PHCKJ + {2 crﬂcid : {Var is 711:9sz = I: Tdfmj ‘Rr all he It?) We {)2 ?
me.” 771‘ (900) : K3(kP(X))//=~ {3 k ‘9'le = 12641315?!”
= RTGWU w” 5., 31.: 773,100): >89"sz is a 1.1m edema.” (b) (2 points) Find a basis for the kernel of transformation Tand determine its dimension. Pix.) '3 jen‘i' is 0 L3 TQM) = o f ml “5 “air: p”dx)=oj. the. M5 :4“ Pal=a><+b (4:,— 11;th Mamet;
P’Yxiio/ : npﬂl);0J 71...; a L—srs (.1, a... we a? ‘7‘ r; 55% L4 {‘«XL
an JAEg2“ .3 ; Question 5 continued. . . . . . 1
(c) (6 pomts) Conmder the hnear space V consisting of all 2 x 2 matnces for which the vector: I] is an eigenvector. Find a basis for this space and determine its dimension. L” ’4: Z: iL :1 1/} am
A 1—1!) : 2:; gr :we. v.1“; kt Question 6. (12 points total) 2 l I D l 1 (a) (6 points) The matrix A = 1 2 1 has eigenvectors 1 , 0 and l . Find an orthogonal
1 l 2 — 1 — l 1 matrix S, and a diagonal matrix D, so that A = SDS" CJxLILk 'Hu’. ﬁt‘jmdafuaé B? 'hv. 3 bkjenu—LCHWS ,' II 
A, : Q‘vﬁlo‘g—= 1) A Z {0] ) CUHLAI. 3" I 43‘”.
_l ___] “I  “J A’[::) =3 eUAL‘L '5 L/ . Si 1L0 an orkcjgﬂq(
l' MILh“ six 3) w; cI'q‘ 7— )“ﬂ— c. M +14% V‘CLT‘OF'S/ we. MLJ +0 orMMMIIa (m cm.» Sdmﬂt) 3hr, Muck}
JwwLuLf' MT The. 3"] Emir—+134 I: oft—r.ij L “h: M ‘C‘hff‘
{—W' (“mm mm Jﬂ‘emt emu.“ Qr « gamma. mm:
SirIL. 4“...)qu WJLwixr +0 Buck Omar) §~ we. nuJ +1. G—S
1‘1”; HT“ #1.: 6 wafers: I : a O
[I : i— h. QM U: the o .. [o] .[Vﬁ ) VJ;
I d: “I / 2' hi ‘I l/Ji yJ—i /ra
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50 OM LL Nib L I S
S j J yr: "/52 V3}
J/J'i "I/JZ :45 Question 6 continued. 3 12
— 4 — 6
equal to 2. By diagonalizing B calculate 3'0 (note 2‘“ = 1,024) [mg bbcnmruLS' O "J A (‘pwm Jlf=Qj Ylx _5m
6] wt 9 m (b) (6 points) Consider the matrix B = [ Note that B has determinant equal to 0 and trace So 13 el‘jtnﬂdori : t EILku (AIL—B) :5; so Ma. S: [:31] a.“ veg}?! (.J 9%: ;, gzsassg an; gummy
= r: a [Z :q [2: :1 Question 7. (6 points total) Suppose that if, , 1'52 , . . . , it are eigenvectors of an n x :1 matrix A. Let Vbe the subspace spanned by
171,115”... , i2} . Show that if if is a vector in V, themii is in Vas well.
(Note, Fl, 17}, . . . , it isn’t necessarily an eigenbasis, as I: might be less than n) :43 f {S Tn \f “\Lﬁ IikCL \/ IS SP‘rnnLc/ _.. ‘ "4 .1, v * A _L
3 l/ V}; “7 Vt ) ’hm X = lel +CLVL+ ....+CK\/K)
gr SH“: CUCA’N’ Ck/ JM‘ 2 :ﬁrétﬁ+~+Lk7lm) ...J
:cxA‘VI+"'+CKA—VKJ
(\va 37mm. 'Vxe. an; 4*” etherwud‘ors 'Cnr A (Sdeaer Hm mPe—d‘z’t EBMMIWS TL!» A324: : C (IQ713+“ + Cde‘VKJ
:(ICW +  5:, {3 aka «2 )MW mmhiha‘ﬁam 6? ﬂat 72/ Sa :5 ﬂISo I; V Question 8. (8 points total) (a) (4 points) Given that (x — 1)(x — 2)(x — 3) = x3 — 6x2 + 11x — 6 , ﬁnd all solutions to the
differential equation
2f"—12f'+22f’ = 12f S», sake. al—‘"’11F“+A>A"—m£=aJ
or F"—él‘"+IlJF'—6\C he,
dmrmluaHA Poljmmt‘ql A: 7;“; )ltml. Jﬁcﬂmlgl ﬁfwﬁn
:3. ><3— éscnux —<—3/ “In mtg 1,1, all 3 (3.1“ L3
m annual); 5.. 5L 3qu( hum" LS jIﬂ 5 it: 3t
(93 Rd=C.E’__ +C3_E‘_ FCBE. 4;, coal, ca mm: (b) (4 points) Find the solution to the differential equation in part (a) such that f (O) = 2, f '(0) = 2 and f"(0):=0.
( t. It: 3?
[/lthj £2.45): g9; +<;_,_Q, + (3%;
I 3L1: 3?:—
D‘Qn Clﬁti—lCLQ’ +3C—3Q’
M: 31‘
{wit} = C'Qt +Hc3~E +7 C31 ants 90) = (—l + CA'l' I l "T C: + M3f¥ 3C3 = l
Fall/O) .": C... “rt1C; 'l" :0 Question 9. (10 points total) 0 1
Let A = [ b ] where b and c are real numbers. Consider the continuous dynamical system
— — c
E = A;
d! (a) (4 points) What inequality involving 2: and c ensures that the solution to the system will have a
phase portrait composed of trajectories spiraling inwards towards the origin? anﬂsil'bri'la'c. Pblewmiml : 9L1" IL  1: +59 _¢ 1 CL__.{L
L PLﬁﬂi. looks ll'lce @ ﬂr‘ww‘l SPT‘Hlllﬁ‘j foamJ3 curjun)
“ughELL real PalCf) Eijmmlwld m1. HLQA 1.3.; USQNUAIM‘S tau—c. “Miami 50 Q; — < 0/ a“) C. i) o 9.“;6 hf: 2
(a) (6 points) Solve this continuous dynamical system ifb = 4, C = 5’ and 55(0) : [ 1] (your answer should be a closed formula for 56(0) 30 wt.— NJ 'lo L‘amuJuej ‘. “Ci ‘1‘75 ___$”t lr'ie'. Question 10. (6 points total) Find a symmetric 2 x 2 matrix, A, with the following properties: 3
(i) f = [1] is an eigenvector forA (ii) the sum of the two eigenvalues of A equals 0 (iii) the determinant of A equals 1. 11m Sun“ 0? ttjtﬁuqfuq S c) ‘Mphes
k') * k Cnf —“=  I‘MPL'ES l‘C =r “ k1
{a k .s 1 .F  z, a“ GM, sailing 1’ ENAIQL: I _.__
—_ QAJA_' M j mus (m «or1 l / 3H3] endJove. ‘'  1' ...
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 Fall '03
 JUDSON
 Differential Equations, Linear Algebra, Algebra, Equations, Eigenvalue, eigenvector and eigenspace, Emoticon, ASCII art, standard basis vectors

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