Difference between revisions of "Pertemuan 3"
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dx + ey + fz = r\ | dx + ey + fz = r\ | ||
gx + hy + iz = s\ | gx + hy + iz = s\ | ||
| − | [[File: | + | [[File:]] |
| + | |||
| + | ---- | ||
| + | |||
| + | == Python == | ||
| + | |||
| + | |||
| + | import numpy as np | ||
| + | from numpy import array | ||
| + | |||
| + | # Mendefinisikan Matrix [A] | ||
| + | |||
| + | A=array([[4, -2, 1], [-2, 4, -2], [1, -2, 4]], float) | ||
| + | |||
| + | # Mendefinisikan Matrix [B] | ||
| + | b = array ([[11], [-16], [ 17]], float) | ||
| + | |||
| + | c=array([[4, -2, 1, 11], [-2, 4, -2, -16], [1, -2, 4, 17]], float) | ||
| + | |||
| + | print ('Matrix dari persamaan adalah : ') | ||
| + | print (c) | ||
| + | #define rows column | ||
| + | |||
| + | n=len(b) #number of rows and colomn, due too the matrix that is square matrix | ||
| + | print ('n adalah ' + str(n)) | ||
| + | |||
| + | for k in range (0,n-1): | ||
| + | for i in range (k+1,n): | ||
| + | if A[i,k] != 0.0 : #perintah jika sebuah baris yang akan di eliminasi, ada bagiannya yang tidak 0 maka akan di substract dengan pivot | ||
| + | lam = A[i,k]/A[k,k] | ||
| + | A[i, k+1:n] = A[i, k+1:n] - lam*A[k, k+1:n] | ||
| + | b[i]=b[i] - lam*b[k] | ||
| + | |||
| + | x=np.zeros((3), float) | ||
| + | for k in range (n-1,-1,-1): | ||
| + | x[k]=(b[k]-np.dot(A[k,k+1:n], x[k+1:n]))/A[k,k] | ||
| + | print (x[k]) | ||
Latest revision as of 09:51, 6 March 2019
Belajar python
Eliminasi Gauss- Jordan
Misalkan ax + by + cz = q\ dx + ey + fz = r\ gx + hy + iz = s\ [[File:]]
Python
import numpy as np from numpy import array
- Mendefinisikan Matrix [A]
A=array([[4, -2, 1], [-2, 4, -2], [1, -2, 4]], float)
- Mendefinisikan Matrix [B]
b = array ([[11], [-16], [ 17]], float)
c=array([[4, -2, 1, 11], [-2, 4, -2, -16], [1, -2, 4, 17]], float)
print ('Matrix dari persamaan adalah : ') print (c)
- define rows column
n=len(b) #number of rows and colomn, due too the matrix that is square matrix print ('n adalah ' + str(n))
for k in range (0,n-1):
for i in range (k+1,n):
if A[i,k] != 0.0 : #perintah jika sebuah baris yang akan di eliminasi, ada bagiannya yang tidak 0 maka akan di substract dengan pivot
lam = A[i,k]/A[k,k]
A[i, k+1:n] = A[i, k+1:n] - lam*A[k, k+1:n]
b[i]=b[i] - lam*b[k]
x=np.zeros((3), float) for k in range (n-1,-1,-1):
x[k]=(b[k]-np.dot(A[k,k+1:n], x[k+1:n]))/A[k,k] print (x[k])
