111-2008&amp;2009-1-M20-December2008

# 111-2008&amp;2009-1-M20-December2008 - Kuwait...

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Unformatted text preview: Kuwait University Math 111 Date: December 17, 2008 Dept. of Math. & Comp. Sci. Second Exam Duration: 90 minutes Calculators, mobile phones, pagers and all other mobile communication equipments are not allowed Answer the following questions: 3 0 O —1 U 1 1 0 1. LetA— 0 _1 0 2 4 l 1 72 (a) Find the cofactor A41. _ (b) Find det {0) Show that adj A is a nonsingular matrix and ﬁnd its inverse, (adj A)‘1 . (1+2+2 ptS.) ENC} 2a—a:+3u use“ a ande 2b—y+3v y-wv v .Ifdet(A}=2,ﬁnd: 2c—z+3w z—w 'w o b 2.LetA= a: y u 1) (a) det ((2Ar1 AT) , (b) det . (2+3 pts.) 3. Let P(—1,—1,0),Q(1,1,0), and R(2, 1,:1) be three points in R3. (a) Show that for all a 6 R, the points P, QR are not on the same line. (b) Find the area of the triangle PQR1 in term of a. 7 (2+2 pts.) 2 4. If “UH = 1, “V” = 2 and the angle between U and v is 9 e ﬁnd |l2U + v”. (3 pts.) 5. Let the lines L1 : 3:: l-l-t , y: 5—3t , z= 3+t and L2 : as: 3—H , y= 572114 , z: 1+3u (a) Show that L1 and L2 intersect each other. (b) Write an equation for the plane which contains L1 and L2. (2+2 pts.) 6. (a) For nonzero vectors U, V in R3, show that the vector U X V is orthogonal to U. (b) Show that, if det (A) 3% 01 then A is a nonsingular matrix. W\ 99 Kuwait University Math 111 Date: December 17, 2008 Dept. of Math. 8: Comp. Sci. Second Exam '1 Answer Key '0 0 —1 1. (a)A41=— 11 0 =. 710 2 1 l 0 (b)det(A)=3A11+4A41=3 —1 0 2 +4. 1 l *2 (c) det (adj A) = [det (A)]3 = —8 512 0 => A is nonsingnlar. ,_. (adj “do—1' = det1(A)A : _%'A = bah- Grab—I c r—‘ b—‘ ONI l [\3 O CNN-B i MIP—‘MIHIQIH Q to F‘H 9: ) det({2A}’1AT) = det ((214)?) det (AT) = m det (AT) = 2a—\$+3u men it (b) det(B): 2b—y+3v yet) '1} 2c—z+3w .2711; w Ceqc'z-l-Cs 2a—5L‘+3u :1: u 251 :5 u a a: u = 2bey+3u y 1) = 2!) y v :2 b y e : 2c—z+3w z w Cldcﬁcraca 20 z w c z w 2 det (A) . 3. (a) Solution{1):The vector m m (21 2,0) and the parametric equations of the line PQ are: a: = 1 + 2t,y = 1 + 212,2 2 0. The point R(2,1,a) does not satisfy these equations for any a E R, since the system: 2 = 1 + 2t,1 = 1 + 215,11. : 0 is inconsistent. (2a, w2c1, —2) % (0,0,0) for all a 6 IR. So, the two lines PQ and PR cannot be parallel or coincides. Hence, the points P, Q, R are not on the same fine. (b) Area = 1 “iv—d x 1711’” = %[|[(2a, —2a, —2)||] = W 2 «2:12 + 1. came: MMHJ. S) 0??“ || Solutionﬂ): P—Q' : (2,2, 0) , P—Ii n (3,2,(1) and ﬁaxﬁi = det ([ 4. ||2U+ V||2 = (2U+ V) . (2U + V) 2 4||U||2 + 4w . V) + M? = 41ng +4HUII nvncos + llVll” = 4(1)? + 411m} (—é) +(212 => H2U + VII - 5. (a) The system: 1 + t 3 —— 1.1,6—3t = 5 — 211,3 +13 = 1+ Bu or equivalently, 1 1 2 t + u = 2,3t — 21; = Lt i 311 = —2 is consistent since: 3 e2 1 E —2 => t: landu: 1. So, thetwolines 1 —3 l 1 _ 2 1 0 —5 —5 E 0 0 -4 —4 0 intersect at P (2,3, 4). 1' j k l —3 1 —1 —2 3 17(x_2)—4(y—3)—5(z_4)=0»:_ n OD—‘O Ob—kH (b) N = = —7i — 4j — 516. An equation of the plane is: ...
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111-2008&amp;2009-1-M20-December2008 - Kuwait...

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