资源描述:
《高等代数与解析几何 习题解答10.pdf》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、1ÙIC:C§1²¡IC1.üIX[O;η,η][O;η′,η′]kú:.3IX[O;η,η]121212e,#IXÄþ:√√22−η′=√2,η′=√2.122222(1)ÑICúª;!!10(2)ÑIX¥Äþη1=,η2=3#IXe01I©þ;!−→1′′(3)®þv3[O;η1,η2]©þ,¦§3#IX[O;η1,η2]−1e©þ.√√22−):(1)Ï(η′,η′)=(η,η)T,Ù¥T=√2√2,¤±I12122222Cúª√√!!22!xx′−x′=T=
2、√2√2.yy′22y′22√√22(2)d(1):(η,η)=(η′,η′)T−1,Ù¥T−1=√2√2,¤121222−22·106·1ÙIC:C√√√√√2±η=2η′−2η′,η=2η′+2η′,=:η=√21122121,η2=22222−√22√2.22!!!!xx′x′x−1−→(3)l=Tí=T.y3v=η1−η2,yy′y′y¤±√√!22!!x′10=√2√2=√,y′22−1−2−22−→ùÒ´v3#IXe©þ.2.3²¡IX[O;η,η]¥,®#
3、IX[O′;η′,η′]1212:O′I(3,2),:M(5,3)3#IXx′¶þ,:M#Ix′>0.Á^Ý/ªÑl[O;η,η][O′;η′,η′]ICúª.!1212!!!3−−−→532):ÏX=,dK¿O′M=−=,02321√25√√−−−→′√5255OMüþ´,=cosθ=,sinθ=.¤±T=555√√5255−√5√5.ÏdCúª52555√√255x−3x′√5√5′y=5252y.15510013.gC3IX[O
4、;η1,η2]¥§´:22x−xy+y+2x−4y+1=0.§1²¡IC·107·(1)#IX[O′;η′,η′],¦O′3ÎIXeI(0,2),12k√√′22η1=η1+η22√2√22η′=−η+η,21222Á^Ý/ªÑICúª;(2)¦C3#IXe§.√√!220√2−2):(1)âK,X0=,T=√,1ª
5、T
6、=1.22222¤±ICúª:√√22x−0x′√2√2′y=222y.1221001x′2y′2(2)C§+=1.624
7、.k²¡IX[O;η,η],e#IX[O′;η′,η′]÷v:x′1212¶Úy′¶3ÎIX¥§©O´x−2y+2=0Ú2x+y+4=0.(1)¦lÎIX#IXCúª;(2)¦x−y+2=03#IX¥§;(3)¦3x′+y′+1=03ÎIX¥§.x−2y+2=0):(1)ÏO′:I(x,y)´§
8、),=002x+y+4=0!!x0−2X0==.y00w,x−2y+2=0√X√ê2:1,2√x+y+4=0√Xê255525−1:2.¤±η′=η+η,η′=−η+η,=(η′,η′)=151522515
9、212√√√√255255−−(η,η)√5√5.du√5√5=1,η′,η′¤mÃX.¤12125255255555±ICúª·108·1ÙIC:C√√255x−−2x′√5√5′y=5250y.1551001(2)x−y+2=03#IX¥§:√√!√√!255525′′′′x−y−2−x+y+2=0,5555=x′−3y′=0.(3)d(1)ICúª±√√√25545x′x′√5√55√y=−525−25y,15
10、5510013x′+y′+1=03ÎIXe§:√√√!√√√!25545525253x+y++−x+y−+1=0,555555√=5x+5y+10+5=0.§2g§z{1.z{g§225x+4xy+2y−24x−12y+12=0,¿xѧã/±9#I!¶.!52−12):ÝA=,B=,ÏdI1=Tr(A)=7>0,22−652−12I2=
11、A
12、=6>0,I3=22−6=−108<0.d´ý.A−12−612A´§λ2−7λ+6=0,)λ=6,λ=1.{z §12108x′2y′26x′2+y′2−=0
13、,=+=1.6318§2g§z{·109·xÑÙã/,I¦ÑICúª.éAuA61üA√√√√!√√!255255525√5−5þ©O´,−,,¤±T=√,555552555
14、T
15、=1.2¦¥%(=#IX:)O′(x,y