因为专业实习的工作需要，要用Java编程实现对已知矩阵求逆矩阵的过程，折腾了3天，终于编出来了。网上通行的矩阵求逆的主要方法是高斯－约旦法，感觉不是很清楚，也不是很好懂。我采用的是线性代数中最传统的方法，为矩阵右加一个单位矩阵后进行初等行变换，当左边变成单位矩阵时，右边就是求得的逆矩阵。

cMatrix.java代码：

/**

* Classname:@(#)cMatrix.java

*

* Version

information:V1.0

*

* Date:2005-07-21

*

* Copyright

notice:Copyright (c) 2005 Lichee, org.

*/

import java.lang.*;

import java.io.*;

import java.math.*;

import

java.util.*;

/**

* 求逆矩阵

*

*

* @version  1.0 21 July 2005

*

@author  farewell

*/

public class cMatrix{

public

cMatrix(){

double Vp[][] = new double [10][10];

int i,j;

i = j =

3;

for(int k=1;k<10;k++)

{

for(int t=1;t<10;t++)

{

Vp[k][t] = 0.0;

}

}

//input the data of the matrix

Vp[1][1] = 1;

Vp[1][2] = 2;

Vp[1][3] = 3;

Vp[2][1] = 2;

Vp[2][2] = 2;

Vp[2][3] = 1;

Vp[3][1] = 3;

Vp[3][2] = 4;

Vp[3][3] = 3;

System.out.println("Print out the array first.");

for(int

x=1;x<=i;x++)

{

for(int y=1;y<=j;y++)

{

System.out.print(Vp[x][y]+"  ");

}

System.out.println();

}

System.out.println("\nThen Try to change it.Use identity matrix.");

//add the

identity array at the end

for(int k=1;k<=i;k++)

{

for(int

t=j+1;t<=j*2;t++)

{

if((t-k)==j)

{

Vp[k][t]=1.0;

}

else

{

Vp[k][t]=0;

}

}

}

for(int x=1;x<=i;x++)

{

for(int

y=1;y<=j*2;y++)

{

System.out.print(Vp[x][y]+"  ");

}

System.out.println();

}

//get the change of the array

for(int k=1;k<=i;k++)

{

if(Vp[k][k]!=1)

{

double

bs = Vp[k][k];

Vp[k][k]=1;

for(int p=k+1;p<=j*2;p++)

{

Vp[k][p]/=bs;

//

System.out.println("Vp["+k+","+p+"]="+Vp[k][p]);

}

}

for(int q=1;q<=i;q++)

{

if(q!=k)

{

double bs = Vp[q][k];

for(int p=1;p<=j*2;p++)

{

//

System.out.println(bs);

Vp[q][p]-=bs*Vp[k][p];

//

System.out.println("Vp["+q+","+p+"]="+Vp[q][p]);

}

}

else

{

continue;

}

}

}

//print out the

result of the change

System.out.println("\nThe result is:");

for(int

x=1;x<=i;x++)

{

for(int y=j+1;y<=j*2;y++)

{

System.out.print(Vp[x][y]+"  ");

}

System.out.println();

}

}

public static void main(String args[])

{

cMatrix cm = new

cMatrix();

}

}