Introduction-to-Linear-Algebra练习题答案

INTRODUCTI[e-mail]eofTechnology://math.mit.edu/˜gs(1,1,1);plane;sameplane!3v=(2,2)andw=(1,−1).43v+w=(7,5)andv−3w=(−1,−5)andcv+dw=(2c+d,c+2d).5u+v=(−2,3,1)andu+v+w=(0,0,0)and2u+2v+w=(addfirstanswers)=(−2,3,1).6Thecomponentsofeverycv+dwaddtozero.Choosec=4andd=10toget(4,2,−6).8Theotherdiagonalisv−w(orelsew−v).Addingdiagonalsgives2v(or2w).9Thefourthcornercanbe(4,4)or(4,0)or(−2,2).10i+jisthediagonalofthebase.11Fivemorecorners(0,0,1),(1,1,0),(1,0,1),(0,1,1),(1,1,1).Thecenterpointis(12,12,12).Thecentersofthesixfacesare(12,12,0),(12,12,1)and(0,12,12),(1,12,12)and(12,0,12),(12,1,12).12Afour-dimensionalcubehas24=16cornersand2·4=8three-dimensionalsidesand24two-dimensionalfacesand32one-dimensionaledges.SeeWorkedExample2.4A.13sum=zerovector;sum=−4:00vector;1:00is60◦fromhorizontal=(cosπ3,sinπ3)=(12,√32).14Sum=12jsincej=(0,1)isaddedtoeveryvector.15Thepoint34v+14wisthree-fourthsofthewaytovstartingfromw.Thevector14v+14wishalfwaytou=12v+12w,andthevectorv+wis2u(thefarcorneroftheparallelogram).16Allcombinationswithc+d=1areonthelinethroughvandw.ThepointV=−v+2wisonthatlinebeyondw.17Thevectorscv+cwfilloutthelinepassingthrough(0,0)andu=12v+12w.Itcontinuesbeyondv+wand(0,0).Withc≥0,halfthislineisremovedandthe“ray”startsat(0,0).18Thecombinationswith0≤c≤1and0≤d≤1filltheparallelogramwithsidesvandw.19Withc≥0andd≥0wegetthe“cone”or“wedge”betweenvandw.20(a)13u+13v+13wisthecenterofthetrianglebetweenu,vandw;12u+12wisthecenteroftheedgebetweenuandw(b)Tofillinthetrianglekeepc≥0,d≥0,e≥0,andc+d+e=1.3421Thesumis(v−u)+(w−v)+(u−w)=zerovector.22Thevector12(u+v+w)isoutsidethepyramidbecausec+d+e=12+12+121.23Allvectorsarecombinationsofu,v,andw.24Vectorscvareinbothplanes.25(a)Chooseu=v=w=anynonzerovector(b)Chooseuandvindifferentdirections,andwtobeacombinationlikeu+v.26Thesolutionisc=2andd=4.Then2(1,2)+4(3,1)=(14,8).27Thecombinationsof(1,0,0)and(0,1,0)fillthexyplaneinxyzspace.28Anexampleis(a,b)=(3,6)and(c,d)=(1,2).Theratiosa/candb/dareequal.Thenad=bc.Then(dividebybd)theratiosa/bandc/dareequal!ProblemSet1.2,page171u·v=1.4,u·w=0,v·w=24=w·v.2kuk=1andkvk=5=kwk.Then1.4(1)(5)and24(5)(5).3Unitvectorsv/kvk=(35,45)=(.6,.8)andw/kwk=(45,35)=(.8,.6).Thecosineofθisvkvk·wkwk=2425.Thevectorsw,u,−wmake0◦,90◦,180◦angleswithw.4u1=v/kvk=1√10(3,1)andu2=w/kwk=13(2,1,2).U1=1√10(1,−3)or1√10(−1,3).U2couldbe1√5(1,−2,0).5(a)v·(−v)=−1(b)(v+w)·(v−w)=v·v+w·v−v·w−w·w=1+()−()−1=0soθ=90◦(c)(v−2w)·(v+2w)=v·v−4w·w=−36(a)cosθ=1(2)(1)soθ=60◦orπ3radians(b)cosθ=0soθ=90◦orπ2radians(c)cosθ=−1+3(2)(2)=12soθ=60◦orπ3(d)cosθ=−1/√2soθ=135◦or3π4.7Allvectorsw=(c,2c);allvectors(x,y,z)withx+y+z=0lieonaplane;allvectorsperpendicularto(1,1,1)and(1,2,3)lieonaline.8(a)False(b)True:u·(cv+dw)=cu·v+du·w=0(c)True9Ifv2w2/v1w1=−1thenv2w2=−v1w1orv1w1+v2w2=0.10Slopes21and−12multiplytogive−1:perpendicular.11v·w0meansangle90◦;thisishalfoftheplane.12(1,1)perpendicularto(1,5)−c(1,1)if6−2c=0orc=3;v·(w−cv)=0ifc=v·w/v·v.13v=(1,0,−1),w=(0,1,0).14u=(1,−1,0,0),v=(0,0,1,−1),w=(1,1,−1,−1).1512(x+y)=5;cosθ=2√16/√10√10=.8.16kvk2=9sokvk=3;u=13v;w=(1,−1,0,...,0).17cosα=1/√2,cosβ=0,cosγ=−1/√2,cos2α+cos2β+cos2γ=(v21+v22+v23)/kvk2=1.518kvk2=42+22=20,kwk2=(−1)2+22=5,k(3,4)k2=25=20+5.19v−w=(5,0)alsohas(length)2=25.Choosev=(1,1)andw=(0,1)whicharenotperpendicular;(lengthofv)2+(lengthofw)2=12+12+12but(lengthofv−w)2=1.20(v+w)·(v+w)=(v+w)·v+(v+w)·w=v·(v+w)+w·(v+w)=v·v+v·w+w·v+w·w=v·v+2v·w+w·w.Noticev·w=w·v!212v·w≤2kvkkwkleadstokv+wk2=v·v+2v·w+w·w≤kvk2+2kvkkwk+kwk2=(kvk+kwk)2.22Comparev·v+w·wwith(v−w)·(v−w)tofindthat−2v·w=0.Divideby−2.23cosβ=w1/kwkandsinβ=w2/kwk.Thencos(β−a)=cosβcosα+sinβsinα=v1w1/kvkkwk+v2w2/kvkkwk=v·w/kvkkwk.24Weknowthat(v−w)·(v−w)=v·v−2v·w+w·w.TheLawofCosineswriteskvkkwkcosθforv·w.Whenθ90◦thisispositiveandv·v+w·wislargerthankv−wk2.25(a)v21w21+2v1w1v2w2+v22w22≤v21w21+v21w22+v22w21+v22w22istruebecausethedifferenceisv21w22+v22w21−2v1w1v2w2whichis(v1w2−v2w1)2≥0.26Example6gives|u1||U1|≤12(u21+U21)and|u2||U2|≤12(u22+U22).Thewholelinebecomes.96≤(.6)(.8)+(.8)(.6)≤12(.62+.82)+12(.82+.62)=1.27Thecosineofθisx/px2+y2,nearsideoverhypotenuse.Then|cosθ|2=x2/(x2+y2)≤1.28Tryv=(1,2,−3)andw=(−3,1,2)withcosθ=−714andθ=120◦.Writev·w=xz+yz+xyas12(x+y+z)2−12(x2+y2+z2).Ifx+y+z=0thisis−12(x2+y2+z2),sov·w/kvkkwk=−12.29Thelengthkv−wkisbetween2and8.Thedotproductv·wisbetween−15and15.30Thevectorsw=(x,y)withv·w=x+2y=5lieonalineinthexyplane.Theshortestwis(1,2)inthedirectionofv.31Threevectorsintheplanecouldmakeangles90◦witheachother:(1,0),(−1,4),(−1,−4).Fourvectorscouldnotdothis(360◦totalangle).HowmanycandothisinR3orRn?ProblemSet1.31(x,y,z)=(2,0,0)and(0,6,0);n=(3,1,−1);dotproduct(3,1,−1)·(2,−6,0)=0.24x−y−2z=1isparalleltoeveryplane4x−y−2z=dandperpendicularton=(4,−1,−2).3(a)True(assumingn6=0)(b)False(c)True.4(a)x+5y+2z=14(b)x+5y+2z=30(c)y=0.5Theplanechangestothesymmetricplaneontheothersideoftheori