Question

我只是处理新手而且比Java或Javascript更让我困惑！我必须解决大学任务的联立方程。（这是一个他们没有真正向我们解释他们在代码中做了什么的课程。）我知道他们如何在下面的代码中用两个方程式来计算代码，但他们现在希望我们用3个方程来做。有谁知道我会怎么做？我想我只需要在每个矩阵中添加额外的位，但显然比这更复杂。我有3个方程式：

X + Y + Z = 9 X + 2Y + 3Z = 23 X + 5Y-3Z = -7

两个方程的代码如下：

// import Jama.*;
// Solve 3x+2y=3
//      -2x-y=-1
// AX=B
// X=InvA B

import java.io.StringWriter;

void setup() 
{
   size(150,110);
   fill(0,0,0);

   double [][] Aline12={{ 3, 2},      // Create a 2D array to store A
                       {-2,-1}};

   Matrix A = new Matrix(Aline12);    // Copy array to A Matrix data structure

   double [][]  B1ine12 = {{3},       // Create a 2D array to store B
                          {-1}};


   Matrix B = new Matrix(B1ine12);    // Copy array to B Matrix data structure

   Matrix X=(A.inverse()).times(B);   // Solve for X

   text("A",10,12);
   app_print(A,0,16);

   text("B",110,12);
   app_print(B,100,16);

   text("X",10,65);
   app_print(X,0,70);
}

// Method added to allow printing on applet screen at (x,y) 
void app_print(Matrix P, int x, int y)
{ 
  StringWriter stringWriter = new StringWriter();
  PrintWriter writer = new PrintWriter(stringWriter);
  P.print(writer,5,2);  
  text(stringWriter.toString(),x,y); 
}

Answer 1

你将以解决2个方程组的方式解决它，只需添加第三个变量。同样在实践中，你几乎从不想采用矩阵的逆，有更好的方法，如LU分解来解决Ax=B。由于您使用的是Jama，您可以尝试以下代码段

   double [][] Aline12={{1.0, 1.0, 1.0},      // Create a 2D array to store A
                        {1.0, 2.0. 3.0},
                        {1.0, 5.0, -3.0}};

   Matrix A = new Matrix(Aline12);    // Copy array to A Matrix data structure

   double [][]  B1ine12 = {{9},       // Create a 2D array to store B
                          {23},
                          {-7}};


   Matrix B = new Matrix(B1ine12);    // Copy array to B Matrix data structure

   Matrix X = A.solve(B) // Solve for X. See Jama documentation on solve

Jama docs
http://math.nist.gov/javanumerics/jama/doc/

如何通过处理

1 个答案: