Linear Algebra and Its Applications Solution

《Linear Algebra and Its Applications Solution》由会员上传分享，免费在线阅读，更多相关内容在学术论文-天天文库。

1、1.1SOLUTIONSNotes:Thekeyexercisesare7(or11or12),19–22,and25.Forbrevity,thesymbolsR1,R2,…,standforrow1(orequation1),row2(orequation2),andsoon.Additionalnotesareattheendofthesection.xx12+=571571.−−=275xx12−−−−275xx12+=57157ReplaceR2byR2+(2)R1andobtain:39x2=039xx12+=57157ScaleR2

2、by1/3:x2=3013x1=−8108−ReplaceR1byR1+(–5)R2:x2=3013Thesolutionis(x1,x2)=(–8,3),orsimply(–8,3).244xx12+=−244−2.571xx12+=15711xx12+=22−122−ScaleR1by1/2andobtain:571xx12+=15711xx12+=22−122−ReplaceR2byR2+(–5)R1:−=32x21032−1xx12+=22−122−ScaleR2by–1/3:x2=−7017−x1

3、=121012ReplaceR1byR1+(–2)R2:x2=−7017−Thesolutionis(x1,x2)=(12,–7),orsimply(12,–7).12CHAPTER1•LinearEquationsinLinearAlgebra3.Thepointofintersectionsatisfiesthesystemoftwolinearequations:xx12+=57157xx12−=22−122−−xx12+=57157ReplaceR2byR2+(–1)R1andobtain:−=79x2−079−−xx12+=57

4、157ScaleR2by–1/7:x2=9/7019/7x1=4/7104/7ReplaceR1byR1+(–5)R2:x2=9/7019/7Thepointofintersectionis(x1,x2)=(4/7,9/7).4.Thepointofintersectionsatisfiesthesystemoftwolinearequations:xx12−=51151−375xx12−=375−xx12−=51151−ReplaceR2byR2+(–3)R1andobtain:82x2=082xx12−=51151−S

5、caleR2by1/8:x2=1/4011/4x1=9/4109/4ReplaceR1byR1+(5)R2:x2=1/4011/4Thepointofintersectionis(x1,x2)=(9/4,1/4).5.Thesystemisalreadyin“triangular”form.Thefourthequationisx4=–5,andtheotherequationsdonotcontainthevariablex4.Thenexttwostepsshouldbetousethevariablex3inthethirdequationtoel

6、iminatethatvariablefromthefirsttwoequations.Inmatrixnotation,thatmeanstoreplaceR2byitssumwith3timesR3,andthenreplaceR1byitssumwith–5timesR3.6.Onemorestepwillputthesystemintriangularform.ReplaceR4byitssumwith–3timesR3,which16401−−02704−produces.Afterthat,thenextstepistoscalethefourthr

7、owby–1/5.00123−00051−57.Ordinarily,thenextstepwouldbetointerchangeR3andR4,toputa1inthethirdrowandthirdcolumn.Butinthiscase,thethirdrowoftheaugmentedmatrixcorrespondstotheequation0x1+0x2+0x3=1,orsimply,0=1.Asystemcontainingthisconditionhasnosolution.Furth