L3 Alg effective TP 3

'm'+str(1)

p=3;q=4

m=var(['m'+str(i)+str(j) for i in [1..p] for j in [1..q]])

m[4*(2-1)+(1-1)]

A=matrix(p,m)

A

2*A

def E(n,i,j):    return(matrix(ZZ,n,n,{(i-1,j-1):1}))

E(3,1,2)

E(3,1,2)*M

def I(n):    return(identity_matrix(n))

(I(3)+2*E(3,1,2))*A

A*E(4,2,3)

A*(I(4)+2*E(4,2,3))

(I(3)+2*E(3,1,2))*(I(3)-2*E(3,1,2))

def M(n,i,j,x):    return(I(n)+x*E(n,i,j))

M(2,1,2,1)

def D(l):    n=len(l)    D=matrix(ZZ,n,n,0)    for i in [0..n-1]:        D[i,i]=l[i]    return(D)

D([1,-1])

D([1,-1])*M(2,1,2,1)*M(2,2,1,-1)*M(2,1,2,1)

def P(n,i,j):    P=I(n)    P[i-1,i-1]=0    P[i-1,j-1]=1    P[j-1,i-1]=1    P[j-1,j-1]=0    return(P)

P(3,1,3)

P(3,1,2)*A

A*P(4,2,4)

B=matrix(ZZ,1,3,[30,42,105]);B #Cf Feuille de TD 4 ex 2.

gcd(gcd(30,42),105)

BB=1/3*B;BB

BB*M(3,1,2,-1) # col2 remplacée par col2 - col1

BB*M(3,1,2,-1)*M(3,2,1,-2) #col1 de la nouvelle matrice remplacée par col1 -2col2

BB*M(3,1,2,-1)*M(3,2,1,-2)*M(3,1,2,-2) #etc.

BB*M(3,1,2,-1)*M(3,2,1,-2)*M(3,1,2,-2)*M(3,1,3,-17)

BB*M(3,1,2,-1)*M(3,2,1,-2)*M(3,1,2,-2)*M(3,1,3,-17)*M(3,3,1,-2)

BB*M(3,1,2,-1)*M(3,2,1,-2)*M(3,1,2,-2)*M(3,1,3,-17)*M(3,3,1,-2)*P(3,1,3) #échange de la col3 avec la col1

PP=M(3,1,2,-1)*M(3,2,1,-2)*M(3,1,2,-2)*M(3,1,3,-17)*M(3,3,1,-2)*P(3,1,3);PP #matrice de changement de base

PP*vector([1,0,0])

B*PP*vector([1,0,0]) # on a bien obtenu une relation de Bezout !

B*PP*vector([0,1,0]) # le 1er vecteur d'une base de solutions de l'équation homogène

B*PP*vector([0,0,1]) # le 2ème vecteur

a=[-1,2,0,0,3];a

min(abs(x) for x in a if x!=0 )

[x for x in a if x!=0]

A

[A[i][j] for i in [0..2] for j in [0..3]]

A.nrows()

A.ncols()

A[:,1]=A[:,1]-2*A[:,3];A

A[0,:]=A[0,:]+A[2,:]

show(A)