本文实例讲述了java实现的求逆矩阵算法。分享给大家供大家参考，具体如下：
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package demo;public class matrixinverse { public static double det(double [][]matrix,int n)//计算n阶行列式（n=n-1） { int t0; int t1; int t2; double num; int cha; double [][] b; if(n>0) { cha=0; b=new double[n][n]; num=0; if(n==1) { return matrix[0][0]*matrix[1][1]-matrix[0][1]*matrix[1][0]; } for (t0=0;t0<=n;t0++)//t0循环 { for (t1=1;t1<=n;t1++)//t1循环 { for (t2=0;t2<=n-1;t2++)//t2循环 { if(t2==t0) { cha=1; } b[t1-1][t2]=matrix[t1][t2+cha]; } //t2循环 cha=0; } //t1循环 num=num+matrix[0][t0]*det(b,n-1)*math.pow((-1),t0); } //t0循环 return num; } else if(n==0) { return matrix[0][0]; } return 0; } public static double inverse(double[][]matrix,int n,double[][]matrixc){ int t0; int t1; int t2; int t3; double [][]b; double num=0; int chay=0; int chax=0; b=new double[n][n]; double add; add=1/det(matrix,n); for ( t0=0;t0<=n;t0++) { for (t3=0;t3<=n;t3++) { for (t1=0;t1<=n-1;t1++) { if(t1<t0) { chax=0; } else { chax=1; } for (t2=0;t2<=n-1;t2++) { if(t2<t3) { chay=0; } else { chay=1; } b[t1][t2]=matrix[t1+chax][t2+chay]; } //t2循环 }//t1循环 det(b,n-1); matrixc[t3][t0]=det(b,n-1)*add*(math.pow(-1, t0+t3)); } } return 0; } public static void main(string[]args)//测试 { double[][] testmatrix = { {1, 22, 34,22}, {1, 11,5,21} , {0,1,5,11}, {7,2,13,19}}; double[][]inmatrix=new double[4][4]; inverse(testmatrix,3,inmatrix); string str=new string(""); for (int i=0;i<4;i++) { for (int j=0;j<4;j++) { string strr=string.valueof(inmatrix[i][j]); str+=strr; str+=" "; } str+="\n"; } system.out.println("服务器之家测试结果："); system.out.println(str); }} |
运行结果：
希望本文所述对大家java程序设计有所帮助。
原文链接：http://blog.csdn.net/u014581901/article/details/50805054
