线性方程组解算结果(MATLAB vs Math.NET equation )

Question

1 C#：用Math.NET库求解方程组

// test solver in Math.NET
var A = Matrix<double>.Build.DenseOfArray(new double[,] {
                                {1, 1, 3},
                                {2, 0, 4},
                                {-1, 6, -1}
                            });
Console.WriteLine(A);
var b = Vector<double>.Build.Dense(new double[] { 2, 19, 8 });
Console.WriteLine(b);
var x = A.Solve(b);//Math.NET

Console.WriteLine("Test Solver in Math.NET: " + x);
>> Test Solver in Math.NET: DenseVector 3-Double
 34.5
    5
-12.5

Press any key to continue . . .

2 MATLAB中相同输入的结果：

A = [1 1 3; 2 0 4; -1 6 -1]
B = [2 19 8]
x = B/A
A =

     1     1     3
     2     0     4
    -1     6    -1


B =

     2    19     8


x =

   1.0000e+00   2.0000e+00   3.0000e+00

3在Python中用于相同的输入，并在numpy.linalg的帮助下：

In[10]: 
import numpy as np

# matrix A
A = np.matrix ([[1, 1, 3],[2, 0, 4],[-1, 6, -1]])

# vector b
b = np.array([2, 19, 8])
b.shape = (3,1)
# attempt to solve Ax=b
z = np.linalg.solve(A,b)
z
Out[10]: 
array([[ 34.5],
       [  5. ],
       [-12.5]])

4对于C#(Math.NET)和Math.NET，结果似乎是一样的，而对于MATLAB来说，两者有很大的不同，为什么会这样呢？

Answer 1

C#和Python示例求解方程A*x=b，而MATLAB示例求解x*A=b。

通过转换B，用\代替/，可以改变MATLAB实例来求解/。

Math.NET (和Python)示例可以通过转换A (即A.Transpose().Solve(b)而不是A.Solve(b) )来解决x*A=b问题。