111-2006&2007-2-M20-April2007

111-2006&2007-2-M20-April2007 - Kuwait University...

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Unformatted text preview: Kuwait University Math 111 Date: April 30, 2097 Dept. of Math. 85 Comp. Sci. Second Exam Duration: 90 minutes -Calculators, mobile phones, pagers and all other mobile communication equipments are not allowed Answer the following questions: 1. Anewer each of the following as True or False (Justify your answer): (6 pts.) (3.) If U is a vector in R3, and (2U) - (3U) = 0, then U is the zero vector. (b) The set {(1, O, —2) , (0, 0,0) , (—2, O, 1)} of vectors in R3 is linearly independent. (0) The set W :- {(z,y, z) 6 R313: + y + z z 1} is a subspace of R3. (d) P (7, ——11 —2) is the point of intersection of the two lines L1 and L2 whose para— metric equations are: ——t,z=4—2tandLg:a:=9+s,y=5+3s,z=—3—s. (a) Find adj A. (b) Find det (A) by cofactor expansion. (c) Use the above to find A“. (5 pts.) 3. (a) PrOve that if A is an n X n, nonsingular matrix, then det (adj A) = (det (A))”'1 . ‘ (2 p'ts-l (b) Let Y be a vector parallel to X = (1, —1,2,3,—1) and X - Y = —1. Prove that Z = 4Y is a unit vector in R5. (2 pts.) 4. Let A(1,1,5),B(2,2,1),C(1,—2,2) and D(—2,1,2) be four points in R3. (a) Find an equation of- the plane through B, C and D. (2 pts.) (b) Find parametric equations of the line through A and perpendicular to the plane through B, C and D. ' (2 pts.) 5. Determine whether W = {(a, b, c) : a + b — 2c = 0} is a subspace of R3. (3 pts.) 6. Is X = (1,1,1) 3. linear combination of X1 = (0,1, —1) and X2 = (1,—1,2)? Justify your answer. (3 pts.) .9 99 21) T1110. (2171') - (3U) 2 6 ‘IUH2 2 0 fl HUN 2 O 2» "201 (,1 / K (b) False. (0.11m e OR. (0,070)=(01111022)+(f1jl’0‘m’1m"(w—9 011). ( 1‘1 2 (n/Jf, C’) E W, then, n+7) +1: 2 l ;11'1r‘lr//+/)’#C’ — 1. Huh ((1 + b + + ((17 + b’ + C') 2 2 :.> X + Y 52 W1 (d) False. 72147272 ~122—t&—2 24—2t ‘ lies 011 L11 But, 7 2 9 + s S, s 2 —2, while ~2 2 1321—?) does not belong to L2. 22» ——3~s‘ 22* 52—l :> 2' (a) A“ : 7- z 5-1413 2 411421 = —2.,,422 zm<l A}; 1 2 —13‘ ad} ,4 2 l ‘2 ~15 L ~Ll U 12 (b) (1e1134):3(7)+ 3 (5) + 7 (—4) 2E (‘4 1 "1 722—1? (Cl/l 2-~‘———ncl](A)2— 5 2 ~11) ~4 0 l2 2 2,1423 2 ll. ‘7 ‘lrlw‘qgg :- r1 {J . 7 l O Clo“ (.4) 8 a 3. (51.) 811100 A mlj (A) 2 (loll/l) In => dell/4 ald‘j :> (lotldm:(ll/4:11”) :> (lot: (/41) (101" (mli A) 2 (dot (14))fl'cletflnl 2x (letmrlj A) 2 ((lolxl’x’llln‘l since} def, (A :1 7’: {l and (lot (In) 2 l (W WWLN: llfx'll 1/11)“ (9112—1 (2113+ (r313 lay-~11" 15:11-11:12. and 1/ are paw‘xllr’l‘ then, I 2 l~ll 2 3X ~ Yl 2 [3X1] M’fl ::: MYH 2 :x : ll/llflj 7: r1 jll“"l 2 l :2 Z is a, unit vector .— 217 2 .19’ — 1:31;: equation of 1(1— 2) — 3131/ — '2‘) ~ 15» 1:) = o 2» (b) Equation of the line through A and perpmuliculm‘ to the plane through BC and D is (my, 2‘) 2 (’l‘ l, t (—3, —3, ~15) \ :2) l3211‘;11’1'(:>11"1(1 01111211110115 are: [121—3114212312 :5— 151,. 51 If a, 2 l) 2 c 2 0 22 (O. 0, O) 6 W” :> [V 2’ (15, Let, X 2 l), (f) ., l” 2 (ml, b" 0’) E W’, thus, (I, w b 2 9r 0 1K: (1’ l1" 7 rlrr’ 2: (l (a X+Y :; l1 (:‘1 + (0/, «5’, (3’) : (a, + (1/, l) + (2’, T c’, + d” 53 ll". lmmnso. "(1+ (1’ + 1 l / / 1 / 1 (11+ 5) 2 2 (r: at a") 2 ((1—1— 1') — 2(3) + 01’ —‘- b’ — 260 + 0 2 ()1 (/3) For any 1;: E F; ' , 2 21.704412 20 2 .1: (0) 2 _ 6. l. O) 2 (3, (0. l, 21) + Cg —l) 2) + Cr; (1, 1, l); Tl‘111s,solv111g thislinearsystem1n cl‘ (:2, Cg, we have; 0 1 l 1 1 1 O l 2 l — ‘ l 2 1‘ O l ‘ l :the system has no solution, X cannot be ~1 2 l 1 l L 0 0 1 a. linear combination of X1 and X2. UN 99 ...
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111-2006&amp;amp;2007-2-M20-April2007 - Kuwait University...

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