programming-examples/java/Numerical_Problems/Java Program to Find Inverse of a Matrix.java

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2019-11-15 12:59:38 +01:00
/*
This is the java program to find the inverse of square invertible matrix. The matrix is invertible if its determinant is non zero.
*/
//This is sample program to find the inverse of a matrix
import java.util.Scanner;
public class Inverse
{
public static void main(String argv[])
{
Scanner input = new Scanner(System.in);
System.out.println("Enter the dimension of square matrix: ");
int n = input.nextInt();
double a[][]= new double[n][n];
System.out.println("Enter the elements of matrix: ");
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
a[i][j] = input.nextDouble();
double d[][] = invert(a);
System.out.println("The inverse is: ");
for (int i=0; i<n; ++i)
{
for (int j=0; j<n; ++j)
{
System.out.print(d[i][j]+" ");
}
System.out.println();
}
input.close();
}
public static double[][] invert(double a[][])
{
int n = a.length;
double x[][] = new double[n][n];
double b[][] = new double[n][n];
int index[] = new int[n];
for (int i=0; i<n; ++i)
b[i][i] = 1;
// Transform the matrix into an upper triangle
gaussian(a, index);
// Update the matrix b[i][j] with the ratios stored
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];
// Perform backward substitutions
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;
}
// Method to carry out the partial-pivoting Gaussian
// elimination. Here index[] stores pivoting order.
public static void gaussian(double a[][], int index[])
{
int n = index.length;
double c[] = new double[n];
// Initialize the index
for (int i=0; i<n; ++i)
index[i] = i;
// Find the rescaling factors, one from each row
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;
}
// Search the pivoting element from each column
int k = 0;
for (int j=0; j<n-1; ++j)
{
double pi1 = 0;
for (int i=j; i<n; ++i)
{
double pi0 = Math.abs(a[index[i]][j]);
pi0 /= c[index[i]];
if (pi0 > pi1)
{
pi1 = pi0;
k = i;
}
}
// Interchange rows according to the pivoting order
int itmp = index[j];
index[j] = index[k];
index[k] = itmp;
for (int i=j+1; i<n; ++i)
{
double pj = a[index[i]][j]/a[index[j]][j];
// Record pivoting ratios below the diagonal
a[index[i]][j] = pj;
// Modify other elements accordingly
for (int l=j+1; l<n; ++l)
a[index[i]][l] -= pj*a[index[j]][l];
}
}
}
}
/*
Enter the dimension of square matrix:
2
Enter the elements of matrix:
1 2
3 4
The Inverse is:
-1.9999999999999998 1.0
1.4999999999999998 -0.49999999999999994