KAREN TRANSFORMACIONES DIA 6

168 days ago by karencepeda@tecpabellon

x1=8 x2=4 w1 = (2*x1)-x2 w2 = (5*x1)+(3*x2) w3 = (6*x1)-(2*x2)+3 w4 = (10*x2)+5 print "w1= ",w1 print "w2= ",w2 print "w3= ",w3 print "w4= ",w4 
        
w1=  12
w2=  52
w3=  43
w4=  45
w1=  12
w2=  52
w3=  43
w4=  45
#1 x=0 y=0 z = matrix([(2*x-x,5*x+3*x,6*x-2*x+3,10*x+5)]) print z 
        
[0 0 3 5]
[0 0 3 5]
No se obtienen ceros puesto que en las ecuacion w3 y w4 se tiene al final una suma que es de +3 y +5 
       
#2 var('u1,u2,v1,v2,c') u = matrix([u1,u2]) v = matrix([v1,v2]) print "u+v = ",u+v print "cu = ",c*u 
        
u+v =  [u1 + v1 u2 + v2]
cu =  [c*u1 c*u2]
u+v =  [u1 + v1 u2 + v2]
cu =  [c*u1 c*u2]
#3 x=u1+v1 y=u2+v2 z = matrix([2*x-x,5*x+3*x,6*x-2*x+3,10*x+5]) print z 
        
[          u1 + v1       8*u1 + 8*v1   4*u1 + 4*v1 + 3 10*u1 + 10*v1 +
5]
[          u1 + v1       8*u1 + 8*v1   4*u1 + 4*v1 + 3 10*u1 + 10*v1 + 5]
#4 x=u1 y=u2 x1=v1 y1=v2 zu = matrix([2*x-x,5*x+3*x,6*x-2*x+3,10*x+5]) zv = matrix([2*x1-x1,5*x1+3*x1,6*x1-2*x1+3,10*x1+5]) print zu print zv print zu+zv 
        
[       u1      8*u1  4*u1 + 3 10*u1 + 5]
[       v1      8*v1  4*v1 + 3 10*v1 + 5]
[           u1 + v1        8*u1 + 8*v1    4*u1 + 4*v1 + 6 10*u1 + 10*v1
+ 10]
[       u1      8*u1  4*u1 + 3 10*u1 + 5]
[       v1      8*v1  4*v1 + 3 10*v1 + 5]
[           u1 + v1        8*u1 + 8*v1    4*u1 + 4*v1 + 6 10*u1 + 10*v1 + 10]
#5 var('cu1,cu2') x=cu1 y=cu2 T(cu) = matrix([2*x-x,5*x+3*x,6*x-2*x+3,10*x+5]) print "T(cu)= ",T(cu) x=u1 y=u2 t= matrix([2*x-x,5*x+3*x,6*x-2*x+3,10*x+5]) print "c*T(u)= ",c*t 
        
T(cu)=  [       cu1      8*cu1  4*cu1 + 3 10*cu1 + 5]
c*T(u)=  [          c*u1         8*c*u1   (4*u1 + 3)*c 5*(2*u1 + 1)*c]
T(cu)=  [       cu1      8*cu1  4*cu1 + 3 10*cu1 + 5]
c*T(u)=  [          c*u1         8*c*u1   (4*u1 + 3)*c 5*(2*u1 + 1)*c]
NO ES UNA TRANSFORMACION LINEAL, PUESTO QUE AL FALLAR EN UNA SOLA PRUEBA ES DE CONCLUIR QUE FALLAR EN TODAS LAS DEMAS