Write a function that solves the matrix equation Ax = b using Gaussian Elimination (book section 6.2). Your function should accept as input a n-by-n matrix A and an n-by-1 vector b, and it should produce a n-by-1 vector x that satisfies Ax = b.

Question

Mathematics

Adolfo Moss

20 April, 05:14

Write a function that solves the matrix equation Ax = b using Gaussian Elimination (book section 6.2). Your function should accept as input a n-by-n matrix A and an n-by-1 vector b, and it should produce a n-by-1 vector x that satisfies Ax = b.

+5

Answers (1)

Know the Answer?

Hugh Byrd · Answer 1

See explaination

Step-by-step explanation:

public class GaussElim{

private static final double eps = 1e-10; % set epsilon value

public static doublic[] fun (double[][] A, double[] b) {

int n=b. length; %calculate length of vector b.

for (int j=0; j
int max=j; %find and swap pivot row.

for (int i=j+1; i
if (Math. abs (A[i][j]) >Math. abs (A[max][j])) {

max=i;

}

}

double[] t1 = A[j]; %swap

A[j]=A[max];

A[max]=t1;

double t = b[j]; %swap

b[j]=b[max];

b[max]=t;

if (Math. abs (A[j][j]) <=eps) {

throw new ArithmeticException ("Matrix is singular."); % if matrix A is a singular matrix then throw error.

}

for (int i=j+1; i
double alpha = A[i][j]/A[j][j];

b[i]=b[i]-alpha*b[j];

for (int k=j; k
A[i][k]=A[i][k]-alpha*A[j][k];

}

}

}

double[] x=new double[n]; % back substitution starts here

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

double sum=0.0;

for (int j=i+1; j
sum=sum+A[i][j]*x[j];

}

x[i] = (b[i]-sum) / A[i][i];

}

return x;

}

public static void main (String[] args) {

int n=3;

double[][] A={{1,2,1},{4,2,0},{-1,5,-3}};

double[] b={5,3,21};

double[] x=fun (A, b);

for (int i=0; i
StdOut. println (x[i]);

}

}

}