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math-51-f09-hw1-sol

math-51-f09-hw1-sol - SOLUTION TO HOMEWORK ASSIGNMENT 1 1...

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Unformatted text preview: SOLUTION TO HOMEWORK ASSIGNMENT 1 1 2H—1H3W—3 9 [2 HO 3 2 9 “—5 (b)2 (a+b)— 30 —2([ 3 + H4 )—3[ “2] —2 7 — liG __ H20 4 2 4 H6 12 0 1. 5. Solution. % fiW 1..8 Solution: 11: [31% ,v [g] b)3u+2v= [:1]. 2 SOLUTION TO HOMEWORK ASSIGNMENT 1 2.1. Solution: (21) Suppose 3 = clv + ng = 01 [i] + 02 E], we get cl + 262 = 1 201 + 02 = 0. Solving the equations, we obtain c1 = —%, C2 = g Hence [a] _ (b) Suppose [(1)] = 61V + 62W 2 c1 E] + 02 [i], we get c1 + 202 = 0 201 + 02 = 1. Solving the equations, we obtain c1 = g, c2 = —%. Hence [(1)] = MH-miaimm- 1 Z ~—§v + 3W SOLUTION TO HOMEWORK ASSIGNMENT 1 d Suppose $1 =61V+62W=Cl 1 +62 2 ,We get (112 2 1 01 +202 2 51:1 261 + 62 = 1132. . . . _ _ ' x Solv1ng the equatlons, we obtain c1 = 29‘2—3zl,02 = 313—12. Hence [ 1] 211.102“, 3 . 2.14. Solution: 0 We let x0 = 4 . The direction vector is 1 2 0 2 v = -2 — 4 = —6 4 1 3 Hence the parametric representation of the line is 0 2 { 4 +25 —6 t6 R}. 1 3 2.16. Solution: 1 We let x0 = 1 . The direction vectors are 1 2 1 1 V1 — —3 — 1 = —4 1 1 0 4 1 3 V2 — 5 ~" 1 = 4 2 1 1 Hence the parametric representation of the line is 1 1 3 {1 +3 —-4 +15 4 s,tER}. 1 0 1 3.2. Solution: First of all, E] and [g] are linearly independent. Next, suppose [ C2 [3] , then we get 261 + 302 2 1 cl + 202 = 2. Solving the equations, we obtain 01 = —4,62 = 3. Hence [I] = —4 [ 3.6. Solution: 4 SOLUTION TO HOMEWORK ASSIGNMENT 1 1 2 3 It is obvious that ——1 and 1 are linearly independent. Next, suppose 3 = 3 3 3 1 2 cl ——1 + 02 1 , then we get 3 3 61 + 2C2 = 3 —01 + 02 = 3 361 + 302 = 3. . 3 1 2 Solving the equations, we obtain 01 = —1,02 = 2. Hence 3 = —1 ——1 + 2 1 . 3 3 3 3.7. Solution: Suppose c1-(u + v) + 02(u + W) + 63(V + w) = 0, then (01 + 02)u + (c1 + 03)v + (02 + 63)W = 0. Since u,v, w are linearly independent, we obtain C1 + 62 = 0 61 + 03 = 0 62 + 03 = 0. It implies that c1 = 02 = 03 : 0. Therefore, 11 + v,u + w,v + w are linearly independent. 3.12. Solution: It is true. Suppose not, then one of v1,v2,v3 is a linear combination of others. Without loss of generality, let V3 = c1v1 + 02V2 for some scalars 01,02. It implies that span(v1,V2,v3) = span(v1,v2), which is a plane or a line. It contradicts to span(v1,V2,V3) = R3. 4.2. Solution 1 1 —2 1 3 2 6 0 2 6 (a)u-(W—V)— _1( 1 ~ 4 — _1 - _3 _30. —2 —3 3 —2 —6 (b)u-w— u-v=u (—w v)=30. <c>uu+vu¢(—1T+—22TW—m (d) Hun+nvn= W+W= ym (e) v - w =(— 2)(1) + 0(6) +4(1) + 3(—-3) = —7. (f) (2v) -W = 2(v-w) = —14. (g) v- (2w) 2 2(v ~ w) = —14. 4.7 Soluti n Let W = Egg be the velocity vector of wind. The velocity of the plane relative to Wind is V = [—200]. Hence the velocity of the plane relative to the ground is w + v : [10%;fi500} The speed is (10x/2— — 500)2 + hex/5)? m 486mph. 4.9. Solution: (80 ||X+yll2 - llXIIZ— llyllz= (X+y) (X+Y) —X-X-y-y= ZX-y. SOLUTION TO HOMEWORK ASSIGNMENT 1 5 (b) From [IWH = ”Z”, we get ”W”2 = “2H2, i.e. W-W = z-z. Now (w+Z)-(W—z) = w - w — z - z = 0, hence they are orthogonal. 4.14. Solution: Let 6 be the angle between 11 and v. Then c039 = “fin-“Q” = W = 1125 . 4.20. Solution: a l ’01 7.01 112 102 (a) Let v = , and W = _ . Then run wn v-W=v1w1+vzw2+---+vnwn and w-v=w1v1+wzvz+---+'wnvn. Hencev-w=w-v. $1 332 (b) Let x = : . Then an (v+w) -x= (v1 +w1):c1 +(122 +1122)a:2+---+(vn+wn)acn and v-x+W-x=v1x1+02m2+---+vnwn+w1x1+w2x2+---wnxn. Hence (v+w) -x=v-x+w‘x. (c) (cv) -W = (cv1)w1 +(c112)'w2 + - - - + (0117mm, = C(v1w1 + @2102 + - - ~ +vnwn), C(V ~ w) = C(v1w1 + 021112 + - - - + onwn). Hence (cv) - W = C(v - W). 5.5. Solution: Switch these two equations, we get —x1 +4102 + 6123 = 2 22:1 + 311:2 — x3 = 0. Using the first equation to eliminate .271 form the second equation, ——w1 + 4% + 6333 = 2 11.732 + 11153 = 4. Dividing the second equation by 11, —:E1 +4332 + 61133 = 2 4 11 ' Using the second equation to eliminate :52 from the first equation, $2+$3= _ 2 =_._ (131+ {173 11 16 +51: — 4 2 3—11~ Nowweobtain 6 $1—~—-+2(l73 6 SOLUTION TO HOMEWORK ASSIGNMENT 1 a: — 4 x 2 — 11 37 Where 133 E R. 2 5 1 5.10. Solution: Suppose v = c1 1 + 02 8 + C3 6 , then we get 1 0 —2 201 +502+C3 =8 01 +802 +603 = —5 c1 — 2C3 = ——11 Switch the equations, 01 — 263 = —11 Cl + 862 + 663 = —5 2C1+5Cz+63=8 Using the first equation to eliminate c1 from the other equations, cl — 203 = —11 8C2 + 863 = 6 562 + 503 = 30 the second and third equation are inconsistent. Hence there is no solution to the original equations. That means v is not in the span of the set. 1'1 2 5.18. Solution: Let x = 232 be a vector which is orthogonal to —1 and $3 5 —1 0 . Then 3 2x1 —m2+5:t:3=O —.’121 +0332 +3$3 = 0 Using the second equation to eliminate $1 from the first equation, —ac2 + 11x3 2 0 —1121 + 3:113 = 0 Hence we get 11:1 = 313 and x2 = 113:3. The set of vectors is (171 3 $2 = $3 11 , $3 1 where :33 E R. Extra problem solution: The total value is S » P. DEPARTMENT OF MATHEMATICS, STANFORD UNIVERSITY, STANFORD, CA 94305 ...
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