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.

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<n;j++){

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

for (int i=j+1;i<n;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<n;i++){

double alpha= A[i][j]/A[j][j];

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

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

StdOut.println(x[i]);

}

}

}

the answer is simply just 6n add the two together as they have an n

what is the answer to this