Li ni u Hnh hc l mt trong nhng ngnh c lch s lu i nht. Hnh hc bt
ngun t chnh thc tin ca con ngi v tng bc hon thin v pht trin theo
thi gian tr thnh mt mn khoa hc suy din v tru tng. Nh ton hc ngi c
Christian Felix Klein l ngi u tin a ra cch phn loi hnh hc theo nhm
cc bin i trong n. Da vo cch phn loi ny th hnh hc cao cp gm c hnh hc
x nh, hnh hc affine, hnh hc Euclid. S lng khi nim v nh l khng nhiu,
nhng qua hnh hc affine chng ta c th suy ra c cc khi nim v nh l ca
hnh hc s cp v gii bi ton hnh hc phng mt cch tng qut.Sau y em xin
gii thiu v php bin i affine. M in hnh l php thu x. MC LCI ) nh
nghaII) Biu thc ta ca php thu xIII)Tnh cht ca php thu xIV) Mnh V)
Phn loi cc php thu xVI) Cc dng nh ca mt im qua php thu xVII) Cc php
thu x trong khng gian affine A 2 v A 3 :VII Cc php bin i affine
sinh ra bi cc php thu x :VII. Bi tp:1BI.)nh ngha: 1.) im bt ngim M
c gi l im bt ng ca php affine f: A A nu f(M) = M.Theo h qu ca nh l
v s xc nh php affine ta suy ra: Php affine l php ng nht khi v ch
khi n c ba im bt ng khng thng hng.Ta thy rng: Nu php affine f c hai
im bt ng phn bit A, B th mi im nm trn ng thng AB u l im bt ng.2)nh
ngha php thu x : Trong khng gian affine A cho phng v khng gian vct
con (khc {0}), ca Atha mn A = . Cho l mt s thc khc 0. Xt nh xf :AA
bin M Athnh f (M) xc nh bi t M1 = v f (M) l im tha mn : M1f(M) =
M1M . Ta gi f l php thu x vi c s c phng v h s .Php thu x vi c s c
phng v h s = -1 c gi l php ixng xin theo phng qua phng .II.Biu thc
ta ca php thu xGi s cho php thu x f c trc l ng thng d v hai im tng
ng l A v A = f(A), (A khng thuc d). Chn trn d hai im phn bit O v B,
t= = , ta c mc tiu affine . Gi s i vi mc tiu vectc ta = (a,b),
B(1,0), A(0,1), B(1,0), A(a,b)Khi : f(O) = O = (0,0),=== (1, 0),===
(a, b)Vy i vi mc tiu , php thu x affine c biu thc ta nh sau:2OA dx
ACh rng b 0, v A khng thuc d. Nu a = 0, b = 1, php thu x l php ng
nhtIII)Tnh cht ca php thu xGi s cho php thu x affine f khng phi l
php ng nht1.Phng thu xNu im M v nh M = f(M) khng trng nhau th ng
thng MM lun c phng c nh.Phng ca cc ng thng gi l phng thu x.Tht vy,
gi s M = (x, y), th M = (x + ay, by).Khi = (ay, (b - 1)y) = y(a, b
- 1). Vy, MM c phng xc nh bi vect (a; b - 1).2. Trc thu xHai ng
thng tng ng m v m = f(m) hoc cng song song vi phng thu x, hoc ct
nhau ti mt im nm trn trc thu x d.Tht vy: Trng hp ng thng m song
song vi trc thu x d th nh m song song vi d, v d trng vi d, nn m cng
song song vi d.Trng hp ng thng m ct trc thu x d ti im I th m ct d
(tc d) ti I = f(I) = I. Vy m v m ct nhau ti im I trn trc thu x d3
.T s thu xNu M khng phi l im bt ng v ng thng MM ct d ti im M0 th
(M, M, M0) = k, trong k l mt s khng i khc 0 v khng ph thuc M. S k
gi l t s thu x.Chng minh:Nu M = (x, y) th chn mc tiu affine nh trn
ta c M = (x + ay, by). im M0 nm trn d nn M0 = (x0, O). Ta c M, M,
M0 thng hng v(M, M, M0)= k, tc= k , hay ta c: x + ay - x0 = k(x -
x0) v by = ky. V M khng thuc d nn y 0 v do k = b khng ph thuc vo im
M.3IV) Mnh : Php thu x vi c s c phng v h s l mt php affinee trn A
lin kt vi nh x tuyn tnhf= p1+ p2 trong p1 v p2 ln lt l cc php chiu
t tng trc tip A = ln thnh phn th nht v thnh phn th hai Chng minh :
Vi mi (M, N A ) ,f(M)f(N) =f(M)M1 + M1N1 + N1f(N) = MM1 + M1N1 +
N1N = (M1N1)+ (MM1 + N1N)Mt khc, ta li c MN = M1N1 + (MM1 +
N1N)VM1N1 , (MM1 +NN1) Nn ta c M1N1 = P1(MN) v (MM1 + N1N) =
P2(MN).Vy f(M)f(N) = (p1+ p2)(MN)V)Phn loi cc php thu x:1)Php thu x
n:a)nh ngha :- Php bin i affine c gi l php thu x n nu c mt siu phng
m mi m ca n u l im bt ng.- Siu phng c gi l c s nn hay nn thu x.-
Nhn xt rng php ng nht l mt trng hp c bit ca php thu x n, lc siu
phng bt k no ca A u l c s nn ca php thu x.b) nh l:N1MNf (N)f
(M)4M1Nu f l mt thu x n khc php ng nht th c duy nht mt im bt
ngOsaochomi ngthngquaOubinthnhchnhn(bt bin) nhng ni chung khng bt
ng ( tc l mt im bt k ca ng thng c th bin thnh mt im khc cng thuc ng
thng ) im gi l tm thu x.Nhn xt: - Nu tm thu x O khng nm trn nn thu
x th php thu x n chnh l php thu x tm O v c s thu x l siu phng .- Nu
tm thu x O thuc siu phng th thu x fc gi l thu x n c bit.Chng minh:-
Gi s f l php thu x n, khc php ng nht v c c s nn l . Do l siu phng
nn c th chn trong h n imc lp x nh l { }1 2, ,....,nAA Av c cc vect
i din l1 2, ,....,nee e. Gi d l ng thngbt kkhngnmtrongvct ti A. Ly0
0vA d A tac '0 0( ) A f A v A0 , A,A0 thng hng. Ta c th tm c cc
vect e0 , a, e0 l i din cho A0 , A,A0 m e0 = e0 + a ( iu ny c th lm
c v gi s x, y, z l ba vect i din cho A0 , A,A0 th do ba im A0 ,
A,A0 thng hng nn c mt vect biu th tuyn tnh qua hai vcto cn li, tc:
z = mx + ny, t e0 = z, e0 = mx, a = ny cng l ba vect i din cho A0,
A0, A m e0 = e0 + a ). - LyElimcvctoi dinle=e0+e1++enth tac { }0 1
2, , ,...., ,nAAA AElmctiuca A. Do A thuc nn vcto i din ca n biu th
tuyn tnh qua h vcto 1 2, ,....,nee e, tc:a = a1.e1+ a2.e2+ + an.en,
vy A = (0 : a1 : a2 : : an ) v5 A0 = A + A0 = ( 1 : a1 : a2 : : an
). Xt E0= A1+A2++An= (0:1:1::1) do cc A1, A2,.,Anbt ng nn E0bt
ng.Gi l php bin i tuyn tnh lin kt vi f .Do 0 0' ( ) A f A v A1,
A2,.,An bt ng, f(E0)=E0 nn ta c:( ) ( )( )( ) ( )( ) ( ) ( ) ( )0 0
0 0 0 1 1 2 21 2 1 21 2 1 2 1 2 1 21 1 2 2 1 2. ' . . . ....
..voi1, 2,...,... . ...Cho1 , ta c :... ... ... .... . .... . ...n
ni i in nn n n nn n ne k e k e ae a e a ee ke i ne e e k e e eke e
e e e e e e e e e eke k e k e e e e + + + + + + + + + ++ + + + + +
+ + + + + + + + + + + + 1, 1,ik i n Ma trn ca f l :00 100 ... ...
0. 1 0 ... 0... 0 ... ... ..... ... ... ... 0. 0 ... 0 1nkk aMk a 1
1 1 1 1 1 1 ]Nhnthy ( )0 0 1 0 2 01: : : .... :nO k k a k a k a lmt
im bt ng ca f. By gi chng minh mt ng thng d bt k qua O bt ng. Ly X
= ( x0 : x1 : x2 :.: xn ) thuc d v X =f(X) = ( k0.x0 : k0.a1.x0 +
x1 : k0.a2.x0 + x2 : . : k0.an.x0 + xn )= X + x0.O tc l f(X) thuc
ng thng ni O vi X ( l d ). Vy d l ng thng bt ng. (pcm)Nhn xt:Vy mt
php thu x no gi bt ng mt siu phng th hoc l thu x tm hoc l thu x c
bit.2) Php thu x cp:a) nh ngha:- Trong A cho m phng () v (n m 1)
phng () b nhau, tc l 6 = A = ta bo v l(m, n m 1) cp.- Cho f l mt
php bin i ca A , ta ni f l thu x cp ( thu x (m,n m 1) cp ) vi ( , )
l cp nn nu f gi bt ng mi im nm trn v . Tc l:( ),X f X X .- Trong
trng hp m = 0 th thu x ( 0, n 1 ) cp c gi l thu x tm vi tm l = O v
c s thu x l siu phng.b) nh l:Cho f l thu x (m,n m 1) cp nn l ( , )
mnPf id khi tn ti duy nht{ }\ 0,1 k K sao cho ( ) thX' X f X X th
ng thng ni X v X ct ti A v ct ti B u c t s kp ca hng 4 im (ABXX) =
k.Chng minh:Gi f l php thu x (m,n m 1) cp nn l ( , ) vi l ci phng m
chiu (m < n) v l ci phng (n m 1) chiu b vi v hai ci phng ny cha
ton nhng im kp ca f. V dim = m nn c th chn trong m+1 im c lp x nh l
0 1, ,....,mA A A . V dim = n m 1 nn ta c th chn trong n m im c lp
x nh l 1 2, ,....,m m nA A A+ +. Chn thm mt im E khng nm trn v ( iu
ny c th lm c v v l hai phng cho nhau ) ta c mt mc tiu x nh trong Pn
l { }0 1 1: , ,...., , ,..., ,m m nRA A A A A E +. Gi { }0 1 1,
,...., , ,...,m m ne e ee e+l c s nn ca mc tiu R. i vi mc tiu trn
th m - phng c phng trnh l 1 2... 0m m nx x x+ +cn (n m 1 ) phng c
phng trnhl0 1... 0mx x x . 7Qua php thu x f cc im thuc v u kp nn ta
suy ra biu thc ta ca f i vi mc tiu chn c dng :''. .voi p 0 v0,1,...
.. .voi q 0 v1, 2,... .i ij jkx px i mkx qx j m m n ' + +Ma trn A
ca f c (m + 1) s p v c (n m) s q trn ng cho chnh, cc phn t khc u
bng 0, tc:10 . . . . 00 . . . . .. . . . . . .. . . . . .. . . . .
.. . . . . . 00 . . . . 0nppA pqq+ 1 1 1 1 1 1 1 1 1 1 ]Nu p = q th
f l nh x ng nht.Nu X v X l hai im tng ng ca f ( tc l X = f (X) ) vi
X khng l im kp ca f. Gi s X = (x0 :x1 ::xm :xm+1 :.:xn ) v X khng
thuc (do X khng l im kp ca f ) nn trong cc s x0 ,x1 ,,xm phi c t
nht mt s khc 0 v trongcc s xm+1 ,.,xnphi c t nht mt s khc 0 (do X
khng thuc ). Ta c X = (px0 :px1 ::pxm :qxm+1 :.:qxn). Gi s ng thng
ni X v X ct ti A v ct ti B. im A thuc ng thng XX nn A c ta l:[ ] [
] [ ] . . ' A X X + , mt khc A thuc nn ta ca A tha phng trnh ca .
Gi s[ ] [ ]0 1 1: :...: : :....:m m nA a a a a a+th 1 2.... 0m m na
a a+ + . Ta c: . . . 0 voi 1, 2,...,j j ja x qx j m m n + + +. V c
t nht mtxj 0 nn . 0q + , ta ly 1 vq khi [ ] [ ] [ ] [ ]0 1. ' ( ) :
( ) : .... : ( ) : 0: .... : 0mA q X X q px q px q px Tng t ng thng
XX ct ti B c ta l [ ] [ ] [ ] [ ]1 2. ' 0: .... : 0: ( ) : ( ) :
.... : ( )m m nB p X X p qx p qx p qx+ + Do X = (x0 :x1 ::xm :xm+1
:.:xn ) c xi 0 vi 0 i m v xj 0 vi m +1 j n nn 0 do p qi ji jx xpx
qx 8Ta c:( ) ( )' '( ) 00 ( )1 1: :( ) 00 ( )i i ij i ji i ij j jx
q px xx x p qxpk ABXX XXABpx q px px q p qqx qx p qx V k 0 v p v q
u khc 0; k 1 v p q (do f khc nh x ng nht). Vy t s kp( ) ' ABXX khng
ph thuc vo im X.3) Php thu x trtnh ngha: Nu tn ti im A c nh A =
f(A) sao cho ng thng AA song song vi trc thu x d, th php thu x f c
gi l php thu x trtKhi , vi mi im M v nh M = f(M), ta c MM song song
vi trc thu x d.V d 1: Trn mt phng affine cho tam gic ABC. Xc nh php
affine f v tm im bt ng ca f nu bit f(A) = A, f(B) = C, f(C) =
B.Gii:Theo trn, bit nh ca ba im khng thng A, B, C l A, C, B nn php
affine f c xc nh duy nht Trung im M ca BC l mt im bt ng ca f. Tht
vy, nu M = f(M) th do f bo ton t s n: (BCM) = (CBM) = -1, nn M cng
l trung im ca BC, vy f(M) = M9BCMAng trung tuyn AM c A, M l hai im
bt ng nn l ng thng gm ton cc im bt ng, do fl mt php thu x affine,
trc thu x l trung tuyn AM, phng thu x l BC, t s thu x k = -1V d 2:
Php co v mt ng thngnh ngha: Trong mt phng affine, cho ng thng d v
vectkhng song song vi d v mt s thc k, (k 0). nh x f c gi l php co v
ng thng d theo vectv t s k nu = k. (*), trong M0 l giao im ca ng
thng d v ng thng m ia qua im M, c vect ch phng .Mi php co f v ng
thng d theo vectv t s k, (k 0) l mt php thu x affine. Thc vy, ng
thc (*) chng t f l mt song nh. Ta chn h ta affine xOy, c trc honh
Ox trng vi d, trc Oy c vect ch phng.Ta ca cc im M = (x, y), M = (x,
y), M0 = (x, 0), do t ng thc (*) ta c:
Vi ma trn A =khng suy bin v k 0 nn l biu thc ta ca mt php
affine. Hn na, ng thng d gm ton cc im bt ng v php co l mt php thu x
affine trc d, t s k.c bit:Nu k = 1, ta c php ng nht ca mt phng
affineNu k = -1 th f c gi l mt php i xng xin, trc l d v theo phng
ca vect M M0d10m'MB'dM'AMVI).Cc dng nh ca mt im qua php thu xT cc
tnh cht ca php thu x ta suy ra cch dng nh ca mt im bt k.Gi s cho
php thu x f c trc d v hai im tng ng A v A = f(A). Khi nh M ca im M
dng nh sau:1. Khi AA khng song song vi dNuMkhngnmtrnAAth
quaAdngngthngmnhcang thng m = AM, (m, d v m ng quy hoc song song).
Giao im ca m vi ng thng qua M v song song vi AA chnh l nh M ca M cn
dngCnnuMnmtrnAA, th ta dng nh N ca im N khng nm trn AA theo cch
trn, ri dng nh M ca im M (thay hai im A v A bng hai im tng ng N v
N)2. Khi AA // d (tc l f thu x trt)- Nu M khng nm trn AA, gi I l
giao im ca trc d vi ng thng AM, th nhM ca Mchnhl giaoimca ng thng
IA vi ng thng qua M v song song vi AA.11dAA'M'M - Nu M nm trn AA th
d thy rng M l im sao cho= IJAMAMN 'NVII) Cc php thu x trong khng
gian a ff in A2v A3: 1. Trong khng gian A2: - Thu x ( 0, 1) cp nn l
( O ,d ) vi O l mt im v d l ng thng khng qua O. Vi mi im vM d M O
ng thng OM ct d ti A v nu M = f(M) th (OAMM) = k ( vi k l mt s cho
trc ). 12- Thu x n c bit c tm O v c c s nn l ng thng i qua O. Nu ta
bit mt cp im tng ng M v M =f(M) th nh N ca im N c xc nh :+) O, N, N
thng hng.+) ng thng MN ct ng thng MN ti mt im nm trn d.2. Trong
khng gian A 3 : - Thu x ( 0, 2 ) cp nn l ( O, ) vi O l mt im, l mt
phng khng qua O. VivM P M O ng thng OM ct ti A v M = f(M) c xc nh:
+) M, M, O, A thng hng.+)( OAMM ) = k ( vi k l mt s cho trc ).13-
Thu x ( 1, 1 ) cp nn l ( d, d ) vi d v d l 2 ng thng cho nhau. Php
thu x trn c gi l php thu x song trc vi trc l d v d . nh M ca im M
khng thuc d v d c xc nh:+) ng thng MM ct d v d ti hai im A v
B.+)(ABMM) = k ( vi k l mt s cho trc ).- Thu x n c bit tm O v c nn
l mt phng P cha im O. Nu bit mt cp im tng ng M v M =f(M) th nh N ca
im N c xc nh :+) O, N, N thng hng.+) ng thng MN ct MN ti mt im nm
trn P.14VII. Cc php bin i a ff ine sinh ra bi cc php thu x : Ta bit
rng mi php bin i x nh bo tn siu phng v tn W ca Pn v u sinh ra mt
php bin i affine trong khng gian affine An = Pn \W. Sau y, ta xt mt
vi trng hp khi f l mt php thu x no .1. Gi s f l php thu x ( 0, n 1
) cp vi nn l ( O, ) v h s thu x l k. Vi mi im M khng l im bt ng
(i,e) vM M O ) nh ca n l M = f(M) c xc nh sao cho (OAMM) = k ( k0 v
k1 ) trong A l giao im ca ng thng OM vi siu phng .+) Nu chn l siu
phng v tn v xt khng gian affine An = Pn \ th A l im v tn nn ta c t
s n ( ) ( ) ( ) ' ' ' MMO MMOA OAMM k . Nh vy: 1. ' ' . OM k OM OM
OMk . Vy f sinh ra php v t tm O t s 1k.15+) Nu chn siu phng W no i
qua O lm siu phng v tn th O l im v tn nn: ( ) ( ) ' ' AMM OAMM k .
Ngoi ra cc ng thng MM lun song song vi nhau ( phng l ca chng c xc
nh bng phng ca im v tn O ). Vy f sinh ra trn A n = Pn \W mt php thu
x affine c c s l W, phng thu x l l, t s thu x l k.2. Gi s f l php
thu x n c bit c tm O nm trn nn . Nu ly hai cp im M, M = f(M) v N, N
= f(N) th MM v NN u qua O v MN giao vi MN ti mt im thuc . Nu ly l
siu phng v tn th trong A n = Pn\W ta c: MN song song MN v MM song
song vi NN, suy ra: MM = NN.Vy f sinh ra trong A n mt php tnh tin.
VIII. Bi tp:Bi 1: Trong A 2 cho bin i x nh c biu thc ta :'0 1 2'1 0
2'0 0 1x x xx x xx x x + +' + Chng minh f l php thu x tm. Tm tm, nn
, t s thu x.Gii:Tm cc im kp ca f: - Phng trnh tm cc im kp ca f l: 0
1 2 0 1 21 0 2 0 1 22 0 1 0 1 200 (1)0x x x x x xx x x x x xx x x x
x x + + + + + ' ' + + Xt( ) ( )21 121 1 0 2 . 1 011 1 + + Ta c hai
gi tr ring l = 2 v = - 1(bi 2).16- Vi = 2 thay vo (1) ta c:0 1 2 00
1 2 12 0 1 22 0 12 0 11 2 0x x x xx x x xx x x x + + + ' ' + Vy ta
c im kp l: A = (1: 1: 1).- Vi = -1 thay vo (1) ta c:0 1 20 1 2 0 1
20 1 200 00x x xx x x x x xx x x+ + + + + + '+ + Vy tp hp cc im kp
lp thnh mt ng thng c phng trnh 0 1 2( )0 d x x x + + . Nhn xt A
khng thuc d nn f l thu x tm A v c nn l ng thng 0 1 2( )0 d x x x +
+ .Tnh t s thu x k:Ly B = (1: 0 : -1) thuc d. Ly C = A + B = (2: 1
: 0) th f(C) = D = (1: 2: 3) th k = (ABCD) = 1 2 1 11 1 1 2 1 1: :
21 2 1 1 1 20 1 0 2 Bi 2: Trong A2 cho mc tiu { }0 1 2, , , S S S
E. Vit biu thc ta ca php thu x f : A 2 A 2 trong cc trng hp sau:a)
f l thu x tm, c tm l im S0 = (1: 0: 0), trc thu x l ng thng S1S2 ,
t s thu x k.b) flthuxtm, ctmlimE=(1: 1: 1), trclngthng 0 1 2( ) : 0
d x x x + + t s thu x l k = 2.c) f l thu x c bit, c trc l ng thng 0
1 2( ) : 0 x x x + tm l im S = (1: 0: 1), bin im E = (1: 1:1) thnh
im E = (2: 1: 2).Gii:17Gi e0 , e1 , e2 , e ln lt l cc vcto i din
cho S0 , S1 , S2 , S v l php bin i tuyn tnh lin kt vi f. a) Ly G =
S1 + S2 = (0: 1: 1) thuc S1S2 th E = S0 + G. t E = f(E) th E thuc
S0E tc E = a.S0 + b.E = (a+b : b: b) v gi e l vcto i din cho E. Do
f l thu x n nn (S0GEE) = k.Ta c: 1 1 10 1 0 1: :0 1 0 1 ( )1 1 1a
bb b a bka b a b bb++ + +Chn b = 1 th a+b = k. Suy ra: E = (k: 1:
1).Ta c: f(S0)=S0 , f(S1)=S1 , f(S2)=S2 , f(E)=E nn suy ra:( ) ( )
( )( ) ( )( )0 0 0 1 1 1 2 2 20 1 2 0 1 20 0 1 1 2 2 0 1 2.,.,.. '
( ) . .. . . . .e l e e l e e l ee l e e e e l k e e el e l e l e l
k e e e + + + + + + + +Chn l = 1 th l0= k , l1= l2=1. Suy ra ma trn
ca f l:0 00 1 00 0 1kM 1 1 1 1 ]. Biu thc ca f l: 0 01 12 2' .' ,
0'x k xx xx x 'b) Ly A = (1: 0: -1) , B = (1: 1: -2) v D =B - A =
(0: 1: -1) thuc ng thng d. Ly X = E + D = (1: 2: 0) v X = f(X) th X
thuc ng thng ED tc X = a.E+b.D = (a: a+b: a-b). Lc : m=e0e2l vcto i
din cho A; n=e0+e12e2l vcto i din cho B; p=e1e2lvctoi dinchoD;
x=e0+2e1lvctoi dinchoX, x=ae0+(a+b)e1+(a-b)e2 l vcto i din cho
X.18Do f l php thu x tm vi t s k nn:(EDXX) = k.Ta c: 1 1 11 2 1 1:
:0 1 0 11 2 1aa b b aka a ba b+ +Chn b = 1 th a = k , lc : X = (k:
k+1: k-1).Ta c:f(A)=A,f(B)=B,f(E)=E,f(X)=X suy ra( ) ( )( ) ( )( )
( ) ( )( )( )( )0 2 0 0 2 0 2 0 0 20 1 2 1 0 1 2 0 1 2 1 0 1 20 1 2
2 0 1 2 0 1 2 2 0 1 22 1 2 1 2 10 0 0 1. ( ) ( ) .2 . 2 ( ) ( ) 2 (
) . 2. ( ) ( ) ( ) .2( ) . .3 3 3e e l e e e e l e ee e e l e e e e
e e l e e ee e e e l e e e e e e l e e el l l l l le l e e l + + +
+ ' ' + + + + + + + + + _ + + + ,( ) [ ]0 22 1 2 1 2 11 0 0 1 0 22
1 2 1 2 12 0 1 20 1 0 1 22 1 2 1 20 0 1.2 2 4( ) . . .3 3 32( ) . .
.3 3 3( ) . ' 2 . . ( 1). ( 1).Cho l =1, ta c :. .3 3el l l l l le
l e e l el l l l l le e e ex l x e e l k e k e k el l l l ll e e _
, + + _ _ + + +' , , + + + + + + + _+ + + ,( ) ( ) ( )1 2 1 2 1 2
10 2 0 0 1 0 20 1 22 1 0 0 2 1 1 2 1 0 2 0 1 22 2 2 4. 2. . . .3 3
3 3. ( 1). ( 1).. . 2 . . ( 1). ( 1).l l l l l l ll e l e e l ek e
k e k el l l e l l e l l l e k e k e k e+ + + _ _ _ _ + + + + , , ,
, + + + + + + + + + + + Ta c: 2 1 0 02 1 12 1 0 211 12 1l l l k ll
l k ll l l k l k+ + + ' ' + 19Lc : 0 0 1 21 0 1 22 0 1 22 1 1( ) .
. .3 3 31 2 1( ) . . .3 3 31 1 2( ) . . .3 3 3k k ke e e ek k ke e
e ek k ke e e e + _ _ + + , , + _ _ + +' , , + + +Biu thc ca f l:0
0 1 21 0 1 22 0 1 22 1 1' . . .3 3 31 2 1' . . .3 3 31 1 2' . . .3
3 3k k kx x x xk k kx x x xk k kx x x x+ + + + + +' + + + Thay k=2
ta c: 0 0 1 21 0 1 22 0 1 2' 4.' 4 , 0' 4x x x xx x x xx x x x + +
+ + ' + +c) Ta c: E = (1: 1: 1) v E = f(E) = (2: 1: 2), tm l S =
(1: 0: 1). Xt S0 = (1: 0: 0) th S0 khng thuc 0 1 2( ) : 0 x x x + v
S0 khng thuc ng thng EE v ta ca ng thng EE l: 20( )2 0 0 1 1 22 0 0
1 1 21 1 1 1 1 1: : : : 1: 0: 11 2 2 2 2 1a a a a a ab b b b b b _
_ , ,. Ta cang thng ES0 l( ) 0:1: 1 , gi A l giao im ca ES0 vi suy
ra: A = (0: 1: 1). TacangthngSS0l: ( ) 0:1: 0. TacangthngAE l: ( )
1: 2: 2 , Gi S0 = f(S0) th S0 l giao im ca SS0 vi AE suy ra S0 =
(2: 0: 1). Ly B = (1: 1: 2) v C = (1: -1: 0) thuc , ta c:f(B) = B,
f(C) = C, f(E) = E, f(S0) = S0Tac: ( ) ( )( ) ( )( ) ( )[ ][ ]0 0 1
2 0 0 1 2 1 2 1 20 1 2 1 0 1 20 1 0 0 1 1 0 1 2 0 0 1 2 1 2 1 20 1
2 2 0 1 22 1 2 0 1 2 11( ) ( 4 ). ( 2 ). (4 2 ).22 . 21. ( ) ( 4 ).
( 2 ). (4 2 ).2. 2 2( ) ( 2 ). ( ). (2e l l l e l l l e l l ee e e
l e e ee e l e e e l l l e l l l e l l ee e e l e e ee l l e l l e
+ + + + + + + + + + + + '+ + + + + +( ) ( )[ ]1 2 20 0 20 1 2 0 0 1
2 1 2 1 2 0 20 1 2 00 1 2 12 2 12 ).. 2cho1 ta c :1( 4 ). ( 2 ). (4
2 ). 224 4 12 0 11 4 2 2l l ee l e e ll l l e l l l e l l e e el l
l ll l l ll l l' + + + + + + + + ' ' Ta c: 0 0 21 0 1 22 0( ) 2.(
)( )e e ee e e ee e + + +' Vy biu thc ca f l: 0 0 1 21 12 0 1' 2',
0' x x x xx xx x x + + ' +21Bi 3: Chng minh rng trong A 2 mt php
bin i x nh f c 3 im bt ng thng hng l mt php thu x tm hay thu x c
bit.Gii.Gi d l ng thng i qua 3 im bt ng thng hng ca f.Do f bo ton t
s kp ca hng 4 im; nn ly im M bt k thuc d th f (M) = M.Do , d l ng
thng ( ng vai tr siu phng trong A 2) bt ng i vi f.Ta xt 3 trng
hp:TH 1: Nu khng c im bt ng no ca f nm ngoi d th f l php thu x c
bit vi nn l d v tm l mt im nm trn d.TH 2: Nu c 1 im bt ng ca f nm
ngoi d th f l php thu x tm.TH 3: Nu c hn 1 im bt ng ca f nm ngoi d
th f l php ng nht. V ly X bt k khng thuc d v nhng im bt ng ngoi d
th mi ng thng qua X u bt ng nn X l im kp. Khi c th xem f l php thu
x c bit hay php thu x tm bt k.Bi 4: Trong A 2 cho php bin i x nh f
c phng trnh:
1 2 32 1 33 1 2 3''' 2 2 3kx x xkx x xkx x x x +' +Chng minh rng
f l mt php thu x . Xc nh tm v nn ca php thu x.Gii:Phng trnh tm im
kp ca f:
1 2 3 1 2 32 1 3 1 2 33 1 2 3 1 2 3002 2 3 2 2 (3 ) 0kx x x kx x
xkx x x x kx xkx x x x x x kx + + + ' ' + + (*)22Xt 3 231 11 1 0 3
3 1 02 2 3( 1) 0kk k k kkk + + Do k = 1 l gi tr ring duy nht bi ba
.Thay k = 1 vo (*) v gii h tng ng ta c: 1 2 31 2 3 1 2 31 2 300 02
2 2 0x x xx x x x x xx x x + + + ' + Nh vy, d: 1 2 30 x x x + l ng
thng cha tt c cc im kp ca f, tc lf (X) = X ,x d .Do , f l php thu x
vi nn thu x l d (pcm). Hn na, tm O ca f phi thuc d. Nn f l thu x c
bit.Ta i tm ta ca tm O.Ly M A\d (ty ). Khi OM l ng thng bt bin. ( )
' . O = MM ' d f M M OMSuy ra Chn M = ( 0: 0: 1 ). Thay vo phng
trnh ca f ta c :f(M) = M = ( 1: -1 : -3 )Phngtrnh ng thng MM: x1 +
x2 = 0Khi :O = MM ' d = 1 1 1 1 1 1: : ( 1:1: 2)1 0 0 1 1 1 _ ,Bi
5: Trong A 2 vi mc tiu {Ai;E} cho php bin i x nh f xc nh bi:f(A1) =
A3; f(A3) = A1; f(A2) = E; f(E) = A2a) Vit phng trnh ca f i vi mc
tiu chnb) Tm cc im kp ca fc) Chng minh rng f l php thu x i hpGii.a)
Vit phng trnhf:Ta c : 231 323 12(1: 0: 0) (0: 0:1)(0:1: 0)
(1:1:1)(0: 0:1) (1: 0: 0)(1:1:1) (0:1: 0)A AA EA AE AGifur l AXTT
lin kt ca fGi {e1, e2, e3}l c s nn ca mc tiu chnV e = e1+ e2,+e3.
Khi :fur(e1),fur(e2),fur(e3) ln lt l vect i din ca A3, E, A1, A2Ta
c 1 32 1 2 33 1( ) (0, 0,1)( ) (1,1,1)( ) (1, 0, 0)f e k kef e g ge
ge gef e h he + +' ururur (*)Ta tm k, g, hTa cfur(e) =fur(e1)
+fur(e2) +fur(e3)0 0 1 11 0 1 00 1 1 0111k g hkgh 1111 1111 + +
1111 1111 ] ] ] ] ' T (*) suy ra 1 32 1 2 33 1( )( )( )f e ef e e e
ef e e + +' urururMa trn ca f trong mc tiu chn l:0 1 10 1 01 1 0M1
1 1 1 ]Phng trnh ca f :241 12 23 31 2 32 23 1 2' 0 1 1' 0 1 0' 1 1
0' ' 'x xx xx xx x xx xx x x111 111 111 111 ] ] ] ' +b) Tm im kp
caf Phng trnh tm im kp ca f :1 2 3 1 2 32 2 23 1 2 1 2 3 +0 (1 ) =
00x x x x x xx x xx x x x x x ' ' + + (**)Xt 1 10 1 01 1 =2 (1- ) -
(1- )Do cc gi tr ring l = 1( bi 2 ) v = -1 ( bi 1 )Thay= 1 vo (**)
v gii h tng ng ta c :1 2 32 1 2 31 2 300 0 00x x xx x x xx x x + +
' + Tp hp cc im kp ca f l ng thng d c phng trnh
1 2 30 x x x + Thay=- 1 vo (**) v gii h tng ng ta c :1 2 3221 31
2 3002 00x x xxxx xx x x+ ' ' + + Ta c im kp ca f l I (1:0:1)c) Cch
1:Cho M l mt im ty trong A 2, gi N l nh ca MGi [x] l ma trn ta ca M
[x] l ma trn ta ca N25Theo bi[x] = M [ x]
[x] = M-1 [ x] = M [x] ( V M-1 = M )Suy ra M l nh ca NDo f l php
thu x i hp.Cch 2:Ta thy f l php thu x tm vi tm thu x l im I (1:0:1)
v trc thu x l ng thng d : 1 2 30 x x x + Ly A (1:2:0 ) thA I vA d A
= f (A) = (2:2:1)2 0 0 1 1 2' : : [2: 1: 2]2 1 1 2 2 2AA 1 1 ]( )1
2 2 2 2 1' : : 3: 4: 11 1 1 1 1 1B AA d _ ,Khi ' I AA Ta c1 34 00 4
nn1 1 1 20 2 0 2 2 2( ') : 13 1 3 2 2 24 2 4 2IBAA Do f c h s thu x
l -1. Nn f l php thu x i hp.Bi 6: Trong A 2 cho ng thng d v hai im
phn bit M, M khng thuc d.Chng minh rng:a) C mt php thu x c bit duy
nht f ca A 2 nhn d lm nn thu x v bin M thnh M. Hy dng nh ca mt im
bt k N qua f.b) Cho ty mt s k khc 0. CMR c mt v ch mt php thu x tm
fca A 2nhn d lm nn thu x, k l h s thu x v bin thnh M thnh M. Hy dng
nh ca mt im bt k N qua f.Gii :26a) Gi' S MM d . Ta xy dng mt php
bin i x nh nh sau :LyXA2, Xd,X d{M,M}X0= MX dXt: f : A2 A2
M M = f(M) df(d)=d Xf(X) = SX X0M+ f l php bin i x nh gi bt ng
ng thng d.+ Mt khc,XA2, Xd : X, f(X), S thng hng hay f gi bt bin mi
ng thng qua S.Suy ra f l php thu x c bit tm S, nn l ng thng d..* S
duy nht :F c xy dng nh trn l duy nht v khng ph thuc vo im X; tc l
vi im X bt k u cho ta cng mt kt qu xy dng .im' S MM d l xc nh duy
nht.Vy tn ti duy nht php thu x c bit f nhn d lm nn v bin M thnh M.*
Dng nh ca N qua f: Cch dng: ' S MM d 0X MN d 0( ) ' ' f N N SN XM
TA c N l im cn dng.27X0 Chng minh:Ta c: 0 0 0( ) X d f X X Do M, N,
X0 thng hang nn M, f (N), X0 thng hng ( Do f bo ton 3 im thng hng
).Do f gi bt bin mi ng thng qua S nn N, f (N), S thng hng.Vy 0( ) '
' f N N SN XM l im cn dng.b) Ta xy dng tng t cu a.* Dng nh ca N qua
f:' S MM d Dng A xc nh duy nht sao cho (ASMM) = k0X MN d 0( ) ' ' f
N N AN XM 28 X0Bi 7: Trong A 1 cho thu x cp f vi c s (P,Q) c ma trn
biu din trong mt mc tiu no l a bc d _ ,. Hy tnh h s thu x k.Gii:Gi
{A0,A1,E} l mc tiu ca P1 m trong ma trn ca f l a bc d _ ,.Gi( ) (
): , :p p q qPx y Qx y ta c: ( ) ( ) ( ) ( ): , :p p p p q q q qf P
ax by cx dy f Q ax by cx dy + + + + do P , Q bt ng nn ta c:;p p p p
q q q qp p q qax by cx dy ax by cx dyx y x y+ + + + . t ;p qp qx xy
y ta c:29221. ( ). 01. ( ). 0a b c d c d a ba b c d c d a b + + + +
+ + . Do P v Q phn bit nn . Vy v l hai nghim ca phng trnh 2. ( ). 0
c x d a x b + .Gi A0 = f(A0) = (a:c)v do P v Q phn bit nn 0p pq qx
yx y _ ,th h s thu x l: ( )0 010' : :10p pp p p p pq q q q qq qx x
aay y c y cx ayc ack PQA Aa x x a y cx ay c acy y c t ;a a NM N kc
c M Ta c: ( )( )2 22 2 22 2a da a d aM Nc c c ca a b a a d a ad
bcMN M Nc c c c c c c ++ + ' _ + + + , Ta c: ( )22 2 2 2 221. 11 (
) 22 2kka dcN M N M M N a bc dkad bck M N MN MN ad bcc + _' + + + +
,+ + 30Vy k v 1k l nghim ca phng trnh 2 2221 0a bc dX Xad bc _ + +
+ , Bi 8: Vit phng trnh ca php thu x affine (bng cnh chn mc tiu sao
cho n gin nht)Gii:Cho mc tiu affine (0,e1,e2,.en_), trong = = <
em+1 , .. en>V O (f l php thu x nn ca phng t s ) F(M)=M, M1 = ,
l phng c phng i qua M.Ta c OM= OM1 + M1MOM =OM1 + M1M M(x1,x2,
..,xn) v M(x1,x2, ..,xn) 31M1MMx Ox' 1=x1 .x' m=xmx' m+1=xm+1x'
n=xn
