Python同情 | Matrix.rref() 方法

借助sympy.Matrix().rref()方法，我们可以将矩阵简化为 Row 梯形形式。 Matrix().rref()返回两个元素的元组。第一个是简化的行梯形形式，第二个是枢轴列的索引元组。

Syntax: Matrix().rref()

Returns: Returns a tuple of which first element is of type Matrix and second one is of type tuple.

示例 #1：

# import sympy 
from sympy import * 
  
M = Matrix([[1, 0, 1, 3], [2, 3, 4, 7], [-1, -3, -3, -4]])
print("Matrix : {} ".format(M))
   
# Use sympy.rref() method 
M_rref = M.rref()  
      
print("The Row echelon form of matrix M and the pivot columns : {}".format(M_rref))  

输出：

Matrix : Matrix([[1, 0, 1, 3], [2, 3, 4, 7], [-1, -3, -3, -4]]) 
The Row echelon form of matrix M and the pivot columns : (Matrix([
[1, 0,   1,   3],
[0, 1, 2/3, 1/3],
[0, 0,   0,   0]]), (0, 1))

示例 #2：

# import sympy 
from sympy import * 
  
M = Matrix([[14, 0, 11, 3], [22, 23, 4, 7], [-12, -34, -3, -4]])
print("Matrix : {} ".format(M))
   
# Use sympy.rref() method 
M_rref = M.rref()  
      
print("The Row echelon form of matrix M and the pivot columns : {}".format(M_rref))  

输出：

Matrix : Matrix([[14, 0, 11, 3], [22, 23, 4, 7], [-12, -34, -3, -4]]) 
The Row echelon form of matrix M and the pivot columns : (Matrix([
[1, 0, 0, 1405/4254],
[0, 1, 0,    10/709],
[0, 0, 1, -314/2127]]), (0, 1, 2))