Exercices sur les Applications Affines - Classes Préparatoires

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fp= => f
f:E → E p∈N∗fp=Ef
1⇒
Ef:E → E f
1−→ f
P: 5x−2y−11z+3 = 0 D:R(2,−1,1)
A=R2R2, f(x, y)=(ax, bx +cy +d)a6= 0 c6= 0
A
(A, ◦)
E3
x0= 3x+ 4y+ 2z−4
y0=−2x−3y−2z+ 4
z0= 4x+ 8y+ 5z−8
fm:R2−→ R2
M(x, y)7−→ M0(x0, y0)
x0= (m+ 1)x+ 4my −m
y0= (m−2)x+ (m−3)y+ 2(m−1)
fm
m fm
f1◦f1=
f0
D
x+ 2y= 1 D03y−x= 0 2
3
f
x0=λ+1
3x−2 3y+2 3z
y0= 2(1 −λ)−2
3x+1 3y+2 3z
z0=−λ+2
3x+2 3y+1 3z
f

X E RX
X
R= (O, −→ e1,−→ e2,−→ e3)
x+ 2y+z= 1 (−→ e1+−→ e2+−→ e3)
(x+y+ 1 = 0
2y+z+ 2 = 0,
3x+ 3y−2z= 0
R= (O, −→ e1,−→ e2,−→ e3)
x0= 3x+ 4y+ 2z−4
y0=−2x−3y−2z+ 4
z0= 4x+ 8y+ 5z−8.
f−−−→ MM0
EA, B, C, D
f f(A) = B f(B) = C f(C) = D f(D) = A
f3=
Pf:P → P f3=f6=
A6=f(A)A, f(A), f2(A)
f
P D P f, g D−→ ∆−→ ∆0
λ µ
f◦g
π, π0E−→ F
λ∈RπλM7−→ (π(M) : λ, π0(M):1−λ)
πλ
−→
F
f:E → E f s
t f =s◦t=t◦s
sB−→ Ft−→ u
s◦t=t◦s⇐⇒ −→ u∈−→ B
f(s, t)f=s◦t=t◦s
f f f ◦f

fR= (O, −→ e1,−→ e2,−→ e3)
x0= (x−2y−2z+ 1)/3
y0= (−2x+y−2z+ 2)/3
z0= (−2x−2y+z−1)/3.
EP, Q, P 0, Q0P6=Q
f f(P) = P0f(Q) = Q0
P P0A, B, C ∈ P O /∈ P
A0, B0, C0P0(OA) (OB) (OC)
α, β, γ [B, C] [C, A] [A, B]
(A0α) (B0β) (C0γ)
A1, . . . , AnnE
B1, . . . , BnAi= (Bi, Bi+1)An= (Bn, B1))
OPO
f M 7−→ M0M0P(OM)P
O
f
f
Ef:E → E
fdet(−→ f) = ±1

s(M)=2P(M)−M−−−−−→ Mp(M) = α−→ u p(M)=(x1, y1, z1)M
Px1=x+ 2α y1=y−α z1=z+α α =−5x+ 2y+ 11z−3
(x1=−9x+ 4y+ 22z−6
y1= 5x−y−11z+ 3
z1=−5x+ 2y+ 12z−3
−−−−−−−→
p(M)s(M) = −−−−−→ Mp(M)
x0=−19x+ 8y+ 44z−12
y0= 10x−3y−22z+ 6
z0=−10x+ 4y+ 23z−6
fa b
0cf
f(0, y) = (0, cy +d)f
(0, e) (0,e−d
c)
g: (x, y)7−→ (ax+a0y+a00 , bx +b0y+b00
g(0, y) = (0, b0y+b00)a0y+a00 = 0 ⇔a0=a00 =
0
g(0, e) (0,−b00 +e
b0)b06= 0
a6= 0 a6= 0 g(1, y)
x+ 2y+z−2=0
−−−→
MM0= (2x+ 4y+ 2z−4)(−→ i−−→ j+ 2−→ k)−−−→ MM0
(1,−1,2)
λ f(−→ v) = λ−→ v
x0=λx = 3x+ 4y+ 2z−4
y0=λy =−2x−3y−2z+ 4
z0=λz = 4x+ 8y+ 5z−8
⇔
(3 −λ)x+ 4y+ 2z−4 = 0
−2x−(3 + λ)y−2z+ 4 = 0
4x+ 8y+ (5 −λ)z−8 = 0
3−λ4 2
−2−3−λ−2
4 8 5 −λ
= 0 ⇔(3 −λ)(λ−1)2= 0
k= 3 k= 1
M f(M) = M mx + 4my −m= 0 (D1) (m−2)x+
(m−4)y+ 2m−1 = 0 (D2)m= 0 2x+ 4y+ 1 = 0
m6= 0 m=4
3
m6=4
3A(−3,1)
fmAm=m+ 1 4m
m−2m−3−3(m−1)2
m6= 1 fm

A1=2 4
−1−2A2
1= 0 −→
f1◦−→ f1= 0Ef1◦f1(M) =
f1◦f1(A) + −→ f1◦−→ f1(−−→ AM)
| {z }
=OE
=A
−−−−−−→
Mf0(M) = (−2x−4y−1)−→ j−→ j= (0,1) ∆
−2x−4y−1 = 0 p−→ j∆p(M)∈∆
p(M)∈M+−→ j p(M) = M+λ−→ j−2x−4λ−4y−1=0 ⇔λ=2x+ 4y+ 1
4
−−−−−−−−→
p(M)f0(M) = 5−−−−−→ p(M)M f0∆−→ j
5
x0=1
5(4x−2y+ 1)
y0=1
15 (−x+ 13y+ 1)
A=
1
3
−2
3
2
3
−2
3
1
3
2
3
2
3
2
3
1
3
(ϕ(−→
i), ϕ(−→ j), ϕ(−→ k))
det(A) = −1ϕ
x+y−z= 0 f
x=1
3x−2 3y+2 3z+λ
y=−2
3x+1 3y+2 3z+ 2(1 −λ)
z=2
3x+2 3y+1 3z−λ
⇔x+y−z−3
2λ= 0
x+y−z+ 3(1 −λ) = 0
t−→ u f t ◦f=f◦t A
M−−→ MA =−→ u−−−−−−−→ f(M)f(A) = −−→ MA f
sAsBX(A, B)∈X2X sA
sBX sB◦sA=t2−−→
AB X(sB◦sA)n=t2n−−→ AB
f(M) = t2n−−→
AB ◦sA(M) = A−−−→
AM + 2n−−→ AB =A+n−−→ AB −(M−(A+n−−→ AB)) Cn
f(M) = C−−−→ CM =sCn(M)A6=B Ck6=Cnn6=k
2x0=x−2y−z+ 1
2y0=−x−z+ 1
2z0=−x−2y+z+ 1
2x0=−5x−3y+ 2z−3
2y0= 3x+y−2z−1
z0=−3x−3y+z−3
P:x+ 2y+z= 2 (−→ e1−−→ e2+ 2−→ e3)