{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

math-51-f09-hw1-sol

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

This preview shows pages 1–6. Sign up to view the full content.

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

View Full Document

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

View Full Document

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: 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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern