LU分解的矩阵逆运算

算法名称：矩阵求逆（基于LU分解法）

LU分解算法评价：
LU分解大约需要执行N3/3次内层循环（每次包括一次乘法和一次加法）。这是求解一个（或少量几个）右端项时的运算次数，它要比Gauss-Jordan消去法快三倍，比不计算逆矩阵的Gauss-Jordan法快1.5倍。
当要求解逆矩阵时，总的运算次数（包括向前替代和回代部分）为N3，与Gauss-Jordan法相同。

算法描述：
简言之，我们只需对原始矩阵进行一次LU分解，然后变换右端向量b就可以了，即设我们的原始矩阵为4×4阶方阵，那么我们的b依次取



然后重新排列成的矩阵就是逆矩阵了。

运行示例：
Origin matrix:
| 0.0 2.0 0.0 1.0 |
| 2.0 2.0 3.0 2.0 |
| 4.0 -3.0 0.0 1.0 |
| 6.0 1.0 -6.0 -5.0 |
-----------------------------------------------
Its inverse matrix:
| -0.025641025641025623 0.1282051282051282 0.08974358974358977 0.0641025641025641 |
| 0.17948717948717946 0.10256410256410259 -0.12820512820512822 0.05128205128205129 |
| -0.5299145299145299 0.3162393162393163 -0.14529914529914528 -0.00854700854700854 |
| 0.6410256410256411 -0.20512820512820518 0.25641025641025644 -0.10256410256410257 |
-----------------------------------------------

示例程序：


package com.nc4nr.chapter02.matrixinver;




public class MatrixInver {




double[][] a = {


{0.0, 2.0, 0.0, 1.0},


{2.0, 2.0, 3.0, 2.0},


{4.0, -3.0, 0.0, 1.0},


{6.0, 1.0, -6.0, -5.0}


};




double[] b = null;




int anrow = 4;


double[] vv = new double[anrow];


int[] indx = new int[anrow];




private void lucmp() {


int n = anrow, imax = 0;


for (int i = 0; i < n; i++) {


double big = 0.0;


for (int j = 0; j < n; j++) {


double temp = Math.abs(a[i][j]);


if (temp > big) big = temp;


}


vv[i] = 1.0 / big;


}


for (int j = 0; j < n; j++) {


for (int i = 0; i < j; i++) {


double sum = a[i][j];


for (int k = 0; k < i; k++) sum -= a[i][k] * a[k][j];


a[i][j] = sum;


}


double big = 0.0;


for (int i = j; i < n; i++) {


double sum = a[i][j];


for (int k = 0; k < j; k++) sum -= a[i][k] * a[k][j];


a[i][j] = sum;


double dum = vv[i] * Math.abs(sum);


if (dum >= big) {


big = dum;


imax = i;


}


}


if (j != imax) {


for (int i = 0; i < n; i++) {


double mid = a[imax][i];


a[imax][i] = a[j][i];


a[j][i] = mid;


}


double mid = vv[j];


vv[j] = vv[imax];


vv[imax] = mid;


}


indx[j] = imax;


if (j != n - 1) {


double dum = 1.0/a[j][j];


for (int i = j + 1; i < n; i++) a[i][j] *= dum;


}


}


}




private void lubksb(double[] b) {


int n = anrow, ii = 0;


// y


for (int i = 0; i < n; i++) {


int ip = indx[i];


double sum = b[ip];


b[ip] = b[i];


if (ii != 0)


for (int j = ii - 1; j < i; j++) sum -= a[i][j] * b[j];


else


ii = i + 1;


b[i] = sum;


}


// x


for (int i = n - 1; i >= 0; i--) {


double sum = b[i];


for (int j = i + 1; j < n; j++) sum -= a[i][j]*b[j];


b[i] = sum / a[i][i];


}


}




private void output(double a[][], int anrow) {


for (int i = 0; i < anrow; i++) {


System.out.println(" | " + a[i][0] + " " +


a[i][1] + " " +


a[i][2] + " " +


a[i][3] + " | ");


}


System.out.println("-----------------------------------------------");


}




public MatrixInver() {


System.out.println("Origin matrix:");


output(a,4);


lucmp();


double[] b = new double[anrow];


double[][] y = new double[anrow][anrow];


for (int i = 0; i < anrow; i++) {


for (int j = 0; j < anrow; j++) b[j] = 0;


b[i] = 1.0;


lubksb(b);


for (int j = 0; j < anrow; j++) y[j][i] = b[j];


}


System.out.println("Its inverse matrix:");


output(y,4);


}




public static void main(String[] args) {


new MatrixInver();


}




}
http://blog.csdn.net/BoyMgl/archive/2007/12/03/1914288.aspx