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Unformatted text preview: Bhu, mun NAME: SECTION: A (10:3011:20);B(11:3012:20) MATH 21241, 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. (203;. [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} Lem5 #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§="‘CZ C3+CLE= :) Cat(4:11 "CLEtarwl
0H 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 “019% WW
have” can be. M'Hm 0L; Q 13403? 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 Rinac.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}: 53 :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 TQSlc’bx) — x+
Ll : 4+af+ux+a
1 1 ~\ +Hx+2xl~ /, ...
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 Algebra

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