Question: net beans

dodo zayat
net beans
1 Answer(s)      6 years and 5 months ago
Posted in : Development process

Write a JAVA program to read the values of an NxN matrix and print its inverse.

December 20, 2010 at 3:51 PM

Hi Friend,

Try the following code:

```import java.util.*;
public class InverseMatrix {
public static void main(String argv[]) {
int matrix[][]= new int[3][3];
Scanner input=new Scanner(System.in);
System.out.println("Enter elements of matrix: ");
for (int i=0 ; i < matrix.length ; i++)
for  (int j=0 ; j < matrix[i].length ; j++){
matrix[i][j] = input.nextInt();
}
int len = matrix.length;
int inverseMatrix[][] = invert(matrix);
System.out.println("Inverse of Matrix is: ");
for (int i=0; i<len; ++i) {
for (int j=0; j<len; ++j){
System.out.print(inverseMatrix[i][j]+" ");
}
System.out.println();
}
}

public static int[][] invert(int a[][]) {
int n = a.length;
int x[][] = new int[n][n];
int b[][] = new int[n][n];
int index[] = new int[n];
for (int i=0; i<n; ++i) b[i][i] = 1;

findInverse(a, index);

for (int i=0; i<n-1; ++i)
for (int j=i+1; j<n; ++j)
for (int k=0; k<n; ++k)
b[index[j]][k]
-= a[index[j]][i]*b[index[i]][k];

for (int i=0; i<n; ++i) {
x[n-1][i] = b[index[n-1]][i]/a[index[n-1]][n-1];
for (int j=n-2; j>=0; --j) {
x[j][i] = b[index[j]][i];
for (int k=j+1; k<n; ++k) {
x[j][i] -= a[index[j]][k]*x[k][i];
}
x[j][i] /= a[index[j]][j];
}
}
return x;
}
public static void findInverse(int a[][],
int index[]) {
int n = index.length;
double c[] = new double[n];

for (int i=0; i<n; ++i) index[i] = i;

for (int i=0; i<n; ++i) {
double c1 = 0;
for (int j=0; j<n; ++j) {
double c0 = Math.abs(a[i][j]);
if (c0 > c1) c1 = c0;
}
c[i] = c1;
}

int k = 0;
for (int j=0; j<n-1; ++j) {
double pi1 = 0;
for (int i=j; i<n; ++i) {
double p = Math.abs(a[index[i]][j]);
p /= c[index[i]];
if (p > pi1) {
pi1 = p;
k = i;
}
}
int temp = index[j];
index[j] = index[k];
index[k] = temp;
for (int i=j+1; i<n; ++i) {
int pj = a[index[i]][j]/a[index[j]][j];

a[index[i]][j] = pj;

for (int l=j+1; l<n; ++l)
a[index[i]][l] -= pj*a[index[j]][l];
}
}
}
}
```

Thanks

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