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midterm2_blue - Bhu mun NAME SECTION...

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Unformatted text preview: Bhu, mun NAME: SECTION: A (10:30-11:20);B(11:30-12:20) MATH 21-241, MATRIX ALGEBRA Spring 2011 Midterm 2 Total: 100 points 0 You must Show ALL work for full credit. 0 If you have any questions during the exam, please raise your hand. -- #2 (20 pts) - -- 1. (15 pts) a) (5 pts) Which 3 X 3 matrix represents the transformation that rotates the :r—y plane through 90“, leaving the z—axis alone? on 010° emote” o O —1 o 31h 010° CGEDUJO 0 W l O 0 0 O I O O l b) (5 pts) Let A be an m X n matrix, True or false: The null space of A is a subspace of R“. Justify your answer. iFfiLKi NUS: {Mm—‘03 men)lhht) =v xe Rn 1not film- c) (5 pts) True or false: If the columns of A are linearly independent, then Ax = b has exactly one solution for every 5. Justify your answer. [Panel Ii?- Ais mm willie wwn ) We wagmm 04: A don‘t- lom alrpawm‘ng. MfOr at“, mm lo maid not 194 I‘m con. (2-03;. [1 a] 3 (“my andLlflH) cm hmdm bur m3 1 O 3 ‘L mikj won a plane.1 noi R3) 2. (20 pts) a) (10 pts) Find a basis for the subspace of 1R3 determined by the vectors a: = (m1,$2,$3) for which 21:2 = $1 + $3 + -1. What is the dimension of the subspace? ' ' ‘9 I This Corticdfian isn‘t :‘0 .Cad- a 5009;,on . CUOMWYNOOG‘ error). 140dequ V1. =01? .5) and Vz=(0."t.3) .,~_>VE+U;=(MH:8) kit?" Lt=0+?>-H w u ems“. ° However, 5‘19 561 must HM: 5mm Wind: 0 K: 1‘ l O X1+K5+1 = K1 [1 .1— X3 ‘11“ l I a) {({.110)l(0tl‘lj.[0,E:0)3 X3 ' 0 ' ‘ O {s a [00.53, ' dfimg =5. ‘3“ vi“ not be pancdj'ecd. TM mac to; flee vofiaw. M 8‘0an OJ} Como} Lem-5 #19- 8 «50+ bow?- at (Schemes ' b) (10 pts) 'Let Mn be the vector space of n X n matrices. Determine if Sis a subspace of Mn, where S = {A E M'n : AT : ~A}. Lest A,B€Mn, cascaiqr_' (M3? = 9512,? = —-Ar+(—s) = wows) me e; (9,sz ELM": cbpr) tea/3: mates. S is a 5ubspau'OL Mn. 3. (15 pts) Find the complete solution a: = 533, +mn t0 the linear system As: : b. Clearly Show how you obtain xp and mm. 1 3 —4 —2 A = 1 5 2 ; b = 4 . ——3 —7 6 12 _.L} —o. l'3—Ll' —2 —R‘7_+Q3 0 l2] 6 6 ) (”Ll-‘3. O 0 Vin—’23 0 =3 dim UUQ = 5—3 =Ol => mmzi‘og :3 Km = (9,0,0) :7 X=BLP. lkaVhLEW= ~51. Uk+cl =-'2 a) u:—4i 1V +6w :6 31:6 a.)\f:3 filw :O :3 LU :0 4. (15 pts) Determine whether or not the following vectors form: a.) (8 pts) an independent set b) (7 pts) a spanning set for R4 'Ul I (1,1,0,0) , ’02 = (1,0,1,0) , ’U3 = (0,0,1,1) , v4 = (0,1,0,1) . (Hint: for part b), try to find constants cl, e2, (:3, C4 such that clvl+02v2+63vg+04v4= (0, 0, 0, 1). Why does this tell you whether the given set is a spanning set or not?) sq our, +CLVL +C5V3 +CqVL1 —._ (0.0,0403 Q‘Ciilgottfl 14:10“)! to) {C3 (09.44),» Ct, (0,1,0, [3 a (0.0,0.0). Q4 {'91 2:0 =7 C4:-C?_ C—l +C({ =0 :) —CZ l—CL‘ :20 :7 CH'JCZ C1 1,— C3 :0 :3 C’B :d.Q2 C5 + Ca : 0- :3 fl [2 HQ, =0 :3 Ci, :»C2,.(SQ,WLL 0.1 (:23) :> not Q mfgm SW ' tr) no+ {WWW vb) Q4+C1=O =>IC4:~C; CLthgb :>C§="‘C-Z C3+CLE= :) Cat-(4:11 "CL-Etarwl 0-H X 2) (Good) Cat/mai- 1oe, mum cu a fire. 01: V..u?.,\r5,v,1_ 3? JAM/Lg, cannot“ 10%. CE. mummy? M" ‘12)” EL:- I (”hm “01-9% WW have” can be. M'Hm 0L; Q 134-03? W. 4 5. (15 pts) What is the echelon form U of A? Find the rank of A; WITHOUT finding the bases, determine the dimen— sions of its four fundamental spaces. ' l 2 0 ’L t i 2. o P. I m? —i*?_\‘0 Rina-c.0013! i 2 ’5 :3, 2 —Q+Q3 0 O 3 "01 1 D 2 o 2 i e o m 3 I O o 0 —\3 *1 (unites? dim CCQ§==dim RUB: 3 dim Muir) = 5m}: r» 2 din, MAT}: 5-3 :0. 6. (20 pts) a) {8 pts) Is the following transformation a linear transfor— mation? Justify your answer. T : R3 —> R, T(v) = the largest component of e. so, = Low: new); my” LUCUUICM713\= Thu} :3" \rwa : LOAD) 3 1%wa : (010103 = o. rmwm 2. 2_ W Tva) + no +Ttw)_ b) (12 pts) Let T : P2 —) P2 be a linear transformation given by the formula T(p(m)) = p(2: + 1) (the right side is p evaluated at a: + 1). Find the matrix of T in the standard basis for P2 and use it to determine T(p(:t)) for p(:r:) : ——3 + 2x2. E73), 3 ileKtKfl) Tm = L'. THY-«Hi; TU?) =Q<+QL= l+M+¥L : l ' i —3 o 1 2 ’ l“ o e o 1 l ‘l '1' ‘ 1 — 43’ch :31, A? 2 = -\+L{x+?/x D TQ-Slc’bx) — x+ Ll : 4+af+ux+a 1 1 ~\ +Hx+2xl~ /, ...
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