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();
}
}
