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**Unformatted text preview: **Name: 5 g; ﬂS Ner’ SAee/t Math 21b Midterm 1 Thursday, October 24th, 2002 Please circle your section: Tom Judson Andy Engelward Andy Engelward
Katherine Visnjic (CA) Jakub Topp (CA) Erin Aylward (CA)
MWF 9-10 MWF 10-11 MWF11-12 Question Score
1
2 3 7 5
m You have two hours to take this midterm. 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 full 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
speciﬁcally say no justiﬁcation is necessary, such as the True/False). 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 3 by 5 inch ﬁle card with formulas on it during the test, but you are not allowed
to use any other notes, or calculators during this test. Good luck! Focus and do well! Question 1. (18 points total) True or False (3 points each) No justiﬁcation is necessary, simply circle T or F for each statement. a F (a) If A is invertible then the number of rows of A must equal the rank of A [9‘ Ami-”la WA{ ,4. .3- nrq anal (~ng(4 zr/
5° Pars/C(14) MAST expm/ n = 1? oil: rows cl: F (b) If A has a nonzero kernel, thenA IS not invertible A) a. a A mvd‘ﬂtc Mﬁf NRA):
an) {4- A- Auu'ﬂlx. 5»le; lay—(4)::- I20 3
S o I? A lxsg nan-Bro Ml, MN 4- \SnT— W+$L l 0
T @c) The only 2 x 2 matrixA with A2 = I2 is the identity matrix O][O 1].
[\lo +r- {my}!
) .3 3 A: o 4] T a (d) LetA and B be two n x n matrices such that ker(B) g ker(A). Then the matrix AB is not invertible. \
I\jal aqu l“) 4 §iMPL¢ (cu-N Wan/\YOL. A: R: L
M leg/- (2)21'aggmg4jzga3 L331— NC
@“ré‘. AB: 3:; a“) In) “.1; a. ’3 Wk
T (e) If A Is a 5 x 4 matrix of rank 4 then the kernel of A m st e nonzero.
No look a‘t’ : \ooo —-‘
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(6)) g o <'> Wk CM - L1 @ F (f) LetA and B be two matrices such that the productAB 15 deﬁned. If B has a column of
zeros, then AB must also have a column of zeros. rl: B: [as] 1m“ Agzla'oﬁm Aﬂ Question 2 (15 points total) 0 1 0 0
Suppose that A =[ 1 0] and B = [0 I] are matrices for two linear transformations. (a) (4 points) Brieﬂy describe the two transformations given by A and B geometrically
__\ .
A— Sen/)5 5;“ “b '5; / we) Ektn e‘ ' $— ’hrwt§ A- 'IS a: 10° Cka-wl‘s‘ M‘l‘aj‘xcn (- cits—l: A: T: 70° s“; 10‘] > ‘€|"\ 10° 0: TOO B colicqajys Ow, x-mc.x soul-m "h. ‘3‘ 3 - 5- l/UU'M—l LBS 94‘st x~ux.5)
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(b) (4 points) What are the ranks of A and B? Is either A or B invertible? If not, justify your answer, if
so then ﬁnd its inverse. - ”40%: A, $0 mimosa-:1, caulk “Ragga
mi m»: CRY—r1. V '
Ag mites):|<1/ r3 :3 no'l’ Ammo. —)~ l 0—! _ o~l
”l ‘ Mai» 017‘ l: l -._-.,- ._._.A ,. “an--. ._., -.;,«;._.‘,. _. ,
mama , ,- 3—,}.54; (c) (4 points) Do A and B commute? Justify your answer by describing the transformations AB and BA geometrically.
A—B mmm‘l‘g ’hvc, +MAsForMn‘l-CA ‘I’Ld” Ppblgﬁg 04—1-0 M 5~axVSJ Mn rb‘l‘dts ﬁoo CJOCJQMNYE, 'h‘w-ﬁ 11'“— ‘aqjﬁ 5‘: AB (”[3 M x~a>ciS (‘l‘ha‘l'ts 1kg j-«axig «QLF‘ a (10‘ rat‘s‘ll‘crvxj.
HOME KA- i—qarzsen‘l’s a Ol0° FO‘l’tCHorx 41‘: M Plane ‘FDLIwQA
53 Pro‘de'i'on oven 11“ 3mm] 5., m. cm: 55? RA 3 m
yaxk. As ‘l\~<— DMD: (A—B\# WW3: (B A—X we +wr> +mms¥s¢mmg
«we. 3'3me , SD pt an) E Ac .MT' ammud‘e/ ﬁe“ AQ¢ RA (d) (3 points) What is A2002 ? . I $t‘hcn, A— repmsen‘l'g at ‘10” @locicvou‘se> Women Mm A“.
(011413 3600 , Le. AV, -‘-'— IL. mm Auoésﬂw AK “A Anew : 6%)500: @53 5'00: Ix) s: Aka». z A}. ‘-IO
‘0" Question 3. (18 points total) (a) (10 points) Find all solutions to the following homogeneous linear system (if there is more than
one solution, express the solutions in vector form, i.e. as a linear combination of column vectors). 2x1——x2 +3x3 +3x4 :0 —2x1 +x3+4x4=0
10xl —3x2 +7x3 +x4 =0
Cmnk out We. Fﬁ! 5'17”"? Nth“ D. ’l 3 3 ( “‘7' MA) +1, butts weTlrkj ’h‘e. —>‘ O ( ‘7’ +9)
Qx-h'“ @[qux a? 0'5 4'1" ‘h‘t. PERT '0 -3 :1— , ’5—(I)
Alga~ weight ‘i‘nlce. 4 RN students +. «vow enemas wlwg poSQlloL . . 1-133 ~l33 +033) lo’l‘V‘
o 4 L/ 4 an 0 “~14 ’9 O ‘ ""7“
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‘ . . \ O 4/; " 1
Okaj/ EM‘Q T 'Dl‘fi— ”‘4 L3 3‘ ’ [O ) —|—I ~¥1 SANQ
o o 0 O
s. X —i-
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X: “ ﬁx} +§X~1 SGLJEOAS'. S L‘ + t _+
XR'ES ‘ O
X‘l’ : O l
1
+t 7‘ :4: 30“ Jon“? \\\'\_e, #95067 éi‘ S I
1 . \
A/OJ‘: chk‘ lg) ‘mJ a) I: be, ariij~l quﬂ‘l'yo n3 ‘ib Viki." 557‘»? Th‘y‘
6 ° 0
(b) (3 points) Is it possibl to give a ' trivial solution to the linear system in part (a) with o m “S
x3 = x 4 = O ? If so, write down one such solution, if not justify your answer OUOQ" \ §> thgj in it; SolvV/‘téa m PM” 63 .-l‘— X5 * *1 J
Then .‘N 1"“: solwﬁm 6: wrl‘l’l’m mg COPPCSPOAAS. ta
*3 2"; =0 1 so, 2 t=o/ ’ Jo 75:. 04:) 301mm calm
X3 3X7 =0 Is 0 [5:]4— 0 [78X 2: (—5“ ——> i.e. 7%“: NT
PaSS‘U‘e 4N 31% a AaZh‘iUKOA I l \ ‘ Sad-.0” m '1'th Casz.
[or ‘ raft x3=xv=o «MP/Id X.R’3 PM,“ ~07Lw‘hdﬁ( m At
”owl Sgs‘lem/ 11;: 7x“ .Zupii'es' “=0 4.1m cﬁhkv r)[email protected]) Question 3 continued (c) (5 points) Find all values for a for which the following homogeneous linear system has nontrivial
solutions. 2x1 4x3 =
—-00c2 +x3 =0
(l—az)x3 :0
A3 arm J +“j "l‘o won‘t 0:!— (th. PFCIP 0'?
5. o I 4: 3. o '/
0 (>1 . -—> I I\ \ b v";
" l -. got) 0 l “/4 ‘9 O l “/04
O O (I ~03) O 0 (1-? —‘: (I ﬁx") l
/ ’\ O
A’+1/Cao‘7— o 7);: $1: / ._3 o l O)
(X: / 5.; Inca/J +0 CM C‘A‘T— «50 TM: 0 Q ‘
t A __
oi =0 :Q'P‘P‘HJ ‘ ('"D‘ >9 So CLLCLQ chvx “IS mars, TL: omb'nmx) M‘Nﬂ'r-IIK wt“ LVV‘C. 00(3 11‘; T‘NLH'Gf (AIM/Lisa
who M PF—Cp 313/ “Lida lMpPE’AS wer ”(VAC “16‘ Mow M ‘y \A W D( 2 O . 119. 5: 51km 19‘ng Q Q l b. o o . . O O \ —'5 O O ‘1 S 0 S blmhho I\ IS CA 31:|N:JO o o \ o o a ,4: 114.00)
salts; "J Ann ‘l‘wiUt‘c' S¢[u1?;m mun ix: 671'ch +\ or - l we. 39:? [l 6 ‘3 “‘90" Sat“. MAS ell—W XI = - ix}.
0 t\ \ X = 2‘: >(
O o o z 3 . -l/ X; -- S
‘D \ 1 'l
5 aw SR \41 01‘ S [-6]
l l
QM Na) —% hoAﬂUI‘KK midfms
17»; 53313;“ lrx<§ no’\+ﬁ-\I|W( Saba! tons on!) poke-x “to/'1 “vi-l So Question 4. (12 points total) u know that the inverses of two 3 X 3 matrices, A and B, are given as follows: Suppose yo
1 0 2 1 2 3
A"=0—31 andB"=04o
2 —2 0 0 O 6
0
(a) (4 points) Find all solutions to BE = — 2
. 0. o _ ~ -
Si‘hCQ, E.) : ['1‘] Own R \ E32 = R ) Ea] : 1:;
o q o
L, z X o .5 ‘7 ‘ i .
5:: X 2 Z. 8‘] IS hat.- GALA Sela-hon . n (b) (4 points) Find all solutions to ASE = 0
A .4 «I “l
a w >4=AA2‘)=A @zg (c) (4 points) Consider the transformation T(SE) = (BA)5c', where A and B are the same two matrices as (a) and (b). Find the matrix for the inverse transformation, T'1(5c') we, kmoes (BAY! 2 AME" in parts Question 5. (15 points total) Find examples of linear transformations that have the following features (i.e. write down a possible
matrix for each of the transformations): (a) (3 points) Find an example of a transformation T : 2R3 —) ER“ that has a nonzero kernel.
shag q ; m m, 5&pr s uLL '7‘ Focal AL)
w:— WAD“ A—t 5/ X 3 ms+p‘lx ~ ;F A. ': ”(kc-“x all .t 1P3 1: a ‘11:. keml as
A: :0 e.— «u 2‘ .zt 1193 (b) (3 points) Find an example of a transformation T : 1R3 —> ‘R‘ that has image equal to the line
1 00 oo
oo
at) 0000 2 Sin“. Swag; (T3 =‘ s‘oam JD cglwvmj up md'mx A)
i: chm wL jvts‘l' nu) "hus— Oélvmné a? A— 1-1, lax “WW-M26 a? E) Sat .5 A= ‘ I 1
3 I
I1 spanned by i5 = |l R».
3‘s
‘1‘! (c) (3 points) Find an example of a transformation T : 933 —) m4 that has a kernel equal to its domain. 11:”; i5 « 3+ {“5 ——5 VO‘CU-L CAB-£1633 ‘vawx; «- im'wr‘a‘l CXaMP‘e. phls a." 945‘, (43/ 1.9., A O
:— Q
0
O lad/432R: = Jon/wk (T) (d) (3 points) Is it possible to ﬁnd an example of a transformation T : 913 —) SR" that has an image
equal to its codomain? If so, give an example, if not, explain why not. No, 45 GMT, (T3: 59m A: dun/ms up A; “1“3" \“5 J)“‘T
3 columns, a“; [El / bf awawl Spa-1V a~'+ 5‘ ‘P‘MM‘J
53 yd 3 waders (on JIM (EM-5): rank (4) <3} 50 Ex(A’\ c9a'i’ ’06 RV) (e) (3 points) Is it possible to find an example of a transformation T : 2R3 —-> ER“ so that T 2 :13. If so,
give an example, if not, explain why not. No) ”bk J‘IMQNS‘WAS apt, ‘. 7—; [RE *5 )Rb/ 00
00
be
<30 I 50 TO?) elpﬁ/ now we’— C‘A‘T’ “PPS ) “3"}; “5 T" \ 3 .. .
.waT' isz “h: b!— ‘V‘ R . 01‘ L3 AIMW Mn 9 MVTNULQ
MK, bow“) T&)=A§” whr‘ A’ﬁ L’xg/ 5,) A). Jam; “#EX“. Question 6. (12 points total) 1 1 0
(a) (8 points) Find all values of a for which the inverse of the matrix A = l 0 0 exists and write
‘ 1 2 a
down the matrix for A'l.
l l O I o o I I o I o 0
F7") ”‘3: I o o o I o —(:\ ——5 0—: o ——I I 0 +9)
119k got '(I) o|o< -—IoI +(11)
l ' O l 0 ° —- o o o I
__> o \ o ‘ _‘ 0 $5) —__> ‘ 0
o o u 1 I l ‘ Q 0 l O \ —l o
" '7 0L 3 l
j 0 O ) _J/°‘ At AL I "
3/“ A 4‘ 9r all etc-JR ucq‘l‘ a=o / (67-min Ma oLLL/lc TL-d' AA—‘2I3) (b) (4 points) Suppose a is chosen to be equal to 1 (so that the matrixA'l does in fact exist) then 2 o o 2
calculate the matrix product A’1 0 2 0 A Question 7. (10 points total) Let A be an n x n invertible matrix and let B be an n x r matrix. Show that ker(AB) = ker(B)
Hint: First Show that ker(B) g ker(AB) by taking a vector in ker(B) and Show that this vector must
also be in ker(AB) .1 Then Show that ker(AB) g ker(B) F; ”0 win: ”I'LL Lx Sui".
SuPPoSQ 7C é icer- (E>/ hen ‘hw‘S ' mt-MS B: :8
3;, (Ag)? 2 A (B§>= A (§§=8) SO 2 is aisa in kef‘(AB>/ mg MW @433 g; LwcAB) mud) SAPPOSQ. if": ker (AR>/ TKOA AB? = 5_ Mm m A mm M w m
m—m ' ‘4 =— 50 ,7 BS: = AV” (AB?) , fwm: Mel P‘ A ...

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