>	restart;with(linalg);

Warning, the protected names norm and trace have been redefined and unprotected

>	Exercice 2

question a)

>	j:=exp(I*2*Pi/3):j:='j':m:='m':x:='x':y:='y':z:='z':

>	S:=matrix(3,4,[[x,j*y,z,3],[j*x,j^2*y,z,m],[j^2*x,y,z,2*m]]);

>	S:=Transvect_ligne(2,-j,1,S);

>	S:=Transvect_ligne(3,-j^2,1,S);

>	S:=Permut_colonne(2,3,S);

>

evalm(S);

>	S:=Dilat_ligne(1/(1-j),2,S);

>	S:=Transvect_ligne(3,j^2-1,2,S);

>	S[3,2]:=0:S[3,3]:=0:S[2,2]:=z:

>

evalm(S);

>

Equation de compatibilité:

>	Eq_comp:=S[3,4]=0;

>	m:=solve(Eq_comp,m);

y est un paramètre libre et

>	z:=simplify(S[2,4]);

>	x:=3-z-j*y;

>

>

question b)

>	l:='l':lambda:='lambda':l:=lambda:a:='a':x:='x':y:='y':z:='z':

>	S0:=matrix(4,5,[[l*x,y,z,t,a^3],[x,l*y,z,t,a^2],[x,y,l*z,t,a],[x,y,z,l*t,1]]);

>	A0:=matrix(4,4,[[l,1,1,1],[1,l,1,1],[1,1,l,1],[1,1,1,l]]);

>

1er Cas

>	On suppose l, "différent de ", 0,1,-1,-2,-3;

>	S:=evalm(S0);

>	S:=Dilat_ligne(1/l,1,S);

>	S:=Transvect_ligne(2,-1,1,S):S:=Transvect_ligne(3,-1,1,S):S:=Transvect_ligne(4,-1,1,S);

>	S:=Dilat_ligne(l/(l^2-1),2,S);

>

>	S:=Transvect_ligne(3,1/l-1,2,S):S:=Transvect_ligne(4,1/l-1,2,S);

>	S:=Dilat_ligne(z/S[3,3],3,S);

>	S:=Transvect_ligne(4,-(l-1)/(l+1),3,S);

>	S:=Dilat_ligne(t/S[4,4],4,S);

>	S:=Transvect_ligne(3,-S[3,4]/t,4,S):S:=Transvect_ligne(2,-S[2,4]/t,4,S):S:=Transvect_ligne(1,-S[1,4]/t,4,S);

>	S:=Transvect_ligne(2,-S[2,3]/z,3,S):S:=Transvect_ligne(1,-S[1,3]/z,3,S);

>	S:=Transvect_ligne(1,-S[1,2]/y,2,S);

Voilà la solution du système de CRAMER

>

2eme Cas

>	On suppose lambda=0;

>	A:=subs(lambda=0,evalm(A0));

>	Gsyst_cramer(A,[x,y,z,t],[a^3,a^2,a,1]);

", L[1], matrix([[x+z+t], [y+z+t], [x+y+t], [x+y+z]]), "=", matrix([[a^2], [a^3], [a], [1]])" align="center">

>

3eme Cas

>	On suppose lambda=1;

>	A:=subs(lambda=1,evalm(A0));

>	Gsyst(A,[x,y,z,t],[a^3,a^2,a,1]):

", C[3], matrix([[x, z, y, t], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]), "=", matrix([[a^3], [a^2-a^3], [a-a^3], [1-a^3]])" align="center">

", C[4], matrix([[x, t, y, z], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]), "=", matrix([[a^3], [a^2-a^3], [a-a^3], [1-a^3]])" align="center">

", C[4], matrix([[x, t, z, y], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]), "=", matrix([[a^3], [a^2-a^3], [a-a^3], [1-a^3]])" align="center">

>	D'où nécessairement a=1;

>	Et on trouve l'hyperplan x+y+z+t=1;

>

4eme Cas

>	On suppose lambda=-1;

>	A:=subs(lambda=-1,evalm(A0));

>	Gsyst_cramer(A,[x,y,z,t],[a^3,a^2,a,1]):

", L[2], matrix([[x-y-z-t], [2*y+2*t], [2*z+2*t], [2*y+2*z]]), "=", matrix([[-a^3], [a+a^3], [a^2+a^3], [1+a^3]])" align="center">

>

>

5eme Cas

>	On suppose lambda=-2;

>	A:=subs(lambda=-2,evalm(A0));

>	Gsyst_cramer(A,[x,y,z,t],[a^3,a^2,a,1]):

", L[3], matrix([[x-1/2*y-1/2*z-1/2*t], [y-z-t], [3*z], [3*t]]), "=", matrix([[-1/2*a^3], [-2/3*a^2-1/3*a^3], [1+a^3+a^2], [a+a^3+a^2]])" align="center">

>

6eme Cas

>	On suppose lambda=-3;

>	A:=subs(lambda=-3,evalm(A0));

>	Gsyst(A,[x,y,z,t],[a^3,a^2,a,1]):

>	last:=matrix([[-1/2*a^3+t-1/4*a-1/4*a^2], [-1/2*a^2-1/4*a^3+t-1/4*a], [-1/2*a-1/4*a^3-1/4*a^2+t], [t]]);

Equation de compatibilité:

>	Eq_comp:=1+a+a^2+a^3=0;

>	a_possible:=solve(Eq_comp,a);

D'où trois espaces affines possibles:

>	matrix([[x], [y], [z], [t]])=subs(a=a_possible[1],evalm(last));

>	matrix([[x], [y], [z], [t]])=subs(a=a_possible[2],evalm(last));

>	matrix([[x], [y], [z], [t]])=subs(a=a_possible[3],evalm(last));

Seule la solution particulière change la direction vectorielle est dirigé par le vecteur

>	u:=[1,1,1,1];

Autre Méthode:

1er Cas

>	On suppose l, "différent de ", 1,-3;

>	S:=evalm(S0);

>	S:=Permut_ligne(1,2,S);

>	S:=Transvect_ligne(2,-l,1,S):S:=Transvect_ligne(3,-1,1,S):S:=Transvect_ligne(4,-1,1,S);

>	S:=Permut_ligne(2,3,S);

>	S:=Dilat_ligne(1/(1-l),2,S);

>	S:=Transvect_ligne(3,l^2-1,2,S):S:=Transvect_ligne(4,l-1,2,S);

>	S:=Permut_ligne(3,4,S);

>	S:=Dilat_ligne(1/(1-l),3,S);

>	S:=Transvect_ligne(4,-(2-l-l^2),3,S);

>	S:=Dilat_ligne(1/(3-2*l-l^2),4,S);

>	S:=Transvect_ligne(3,1,4,S):S:=Transvect_ligne(1,-1,4,S);

>	S:=Transvect_ligne(2,1,3,S):S:=Transvect_ligne(1,-1,3,S);

>	S:=Transvect_ligne(1,-l,2,S);

Voilà la solution du système de CRAMER dans le cas lambda différent de 1 et -3

>