trabajo 5 dic

170 days ago by karencepeda@tecpabellon

#1 v1 = matrix([4,1,-2]) v2 = matrix([-3,0,1]) v3 = matrix([[1,-2,1]]) print "v1" print v1 print "v2" print v2 print "v3" print v3 
        
v1
[ 4  1 -2]
v2
[-3  0  1]
v3
[ 1 -2  1]
v1
[ 4  1 -2]
v2
[-3  0  1]
v3
[ 1 -2  1]
u = matrix([[4,-3,1],[1,0,-2],[-2,1,1]]) u.echelon_form() 
        
[ 1  0 -2]
[ 0  1 -3]
[ 0  0  0]
[ 1  0 -2]
[ 0  1 -3]
[ 0  0  0]
k1=2 k2=1 u = matrix([[4,1,-2]]) v = matrix([[-3,0,1]]) w=(k1*u)+(k2*v) print w 
        
[ 5  2 -3]
[ 5  2 -3]
w = matrix([[4,-3,1,0],[1,0,-2,0],[-2,1,1,0]]) w.echelon_form() 
        
[ 1  0 -2  0]
[ 0  1 -3  0]
[ 0  0  0  0]
[ 1  0 -2  0]
[ 0  1 -3  0]
[ 0  0  0  0]
var('k1,k2,k3') eqn = [4*k1+k2+4*k3==0, 1*k1+k2-2*k3==0, -2*k1+k2==0] s = solve(eqn, k1,k2,k3); s 
        
[[k1 == 0, k2 == 0, k3 == 0]]
[[k1 == 0, k2 == 0, k3 == 0]]
w = matrix([[4,1,-2],[-3,0,1],[1,-2,1]]) w.determinant() 
        
0
0
SI ES DEPENDIENTE 
       
#2 v1 = matrix([1,0,0,1]) v2 = matrix([0,1,-1,0]) v3 = matrix([[-1,0,-1,0]]) v4 = matrix([[1,1,1,0]]) print "v1" print v1 print "v2" print v2 print "v3" print v3 print "v4" print v4 
        
v1
[1 0 0 1]
v2
[ 0  1 -1  0]
v3
[-1  0 -1  0]
v4
[1 1 1 0]
v1
[1 0 0 1]
v2
[ 0  1 -1  0]
v3
[-1  0 -1  0]
v4
[1 1 1 0]
u = matrix([[1,0,-1,1],[0,1,0,1],[0,-1,-1,1],[1,0,0,0]]) u.echelon_form() 
        
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
k1=6 k2=3 u = matrix([[1,0,0,1]]) v = matrix([[0,1,-1,0]]) w=(k1*u)+(k2*v) print w 
        
[ 6  3 -3  6]
[ 6  3 -3  6]
w = matrix([[1,0,-1,1],[0,1,0,1],[0,-1,-1,1],[1,0,0,0]]) w.echelon_form() 
        
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
var('k1,k2,k3') eqn = [1*k1+k2+4*k3==0, 0*k1+k2-2*k3==0, 0*k1+k2==0] s = solve(eqn, k1,k2,k3); s 
        
[[k1 == 0, k2 == 0, k3 == 0]]
[[k1 == 0, k2 == 0, k3 == 0]]
w = matrix([[1,0,0,1],[0,1,-1,0],[-1,0,-1,0],[1,1,1,0]]) w.determinant() 
        
-1
-1
COMO RESULTA SER DIFERENTE DE CERO POR LO TANTO ES INDEPENDIENTE 
       
#3 v1 = matrix([2,-1,1]) v2 = matrix([3,-4,-2]) v3 = matrix([[5,-10,-8]]) print "v1" print v1 print "v2" print v2 print "v3" print v3 
        
v1
[ 2 -1  1]
v2
[ 3 -4 -2]
v3
[  5 -10  -8]
v1
[ 2 -1  1]
v2
[ 3 -4 -2]
v3
[  5 -10  -8]
u = matrix([[2,3,5],[-1,-4,-10],[1,-2,-8]]) u.echelon_form() 
        
[ 1  0 -2]
[ 0  1  3]
[ 0  0  0]
[ 1  0 -2]
[ 0  1  3]
[ 0  0  0]
k1=6 k2=3 u = matrix([[2,-1,1]]) v = matrix([[ 3,-4,-2]]) w=(k1*u)+(k2*v) print w 
        
[ 21 -18   0]
[ 21 -18   0]
w = matrix([[2,3,5],[-1,-4,-10],[1,-2,-8]]) w.echelon_form() 
        
[ 1  0 -2]
[ 0  1  3]
[ 0  0  0]
[ 1  0 -2]
[ 0  1  3]
[ 0  0  0]
var('k1,k2,k3') eqn = [2*k1+k2+4*k3==0, -1*k1+k2-2*k3==0, 1*k1+k2==0] s = solve(eqn, k1,k2,k3); s 
        
[[k1 == 0, k2 == 0, k3 == 0]]
[[k1 == 0, k2 == 0, k3 == 0]]
w = matrix([[2,3,5],[-1,-4,-10],[1,-2,-8]]) w.determinant() 
        
0
0
si es cero es dependiente