{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

111-2006&amp;2007-2-M20-April2007

# 111-2006&amp;2007-2-M20-April2007 - Kuwait University...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}