# Numerical Methods: Inverse of nxn matrix using C

**Source Code:**

#include<stdio.h> int main(){ float matrix[10][10], ratio,a; int i, j, k, n; printf("Enter order of matrix: "); scanf("%d", &n); printf("Enter the matrix: \n"); for(i = 0; i < n; i++){ for(j = 0; j < n; j++){ scanf("%f", &matrix[i][j]); } } for(i = 0; i < n; i++){ for(j = n; j < 2*n; j++){ if(i==(j-n)) matrix[i][j] = 1.0; else matrix[i][j] = 0.0; } } for(i = 0; i < n; i++){ for(j = 0; j < n; j++){ if(i!=j){ ratio = matrix[j][i]/matrix[i][i]; for(k = 0; k < 2*n; k++){ matrix[j][k] -= ratio * matrix[i][k]; } } } } for(i = 0; i < n; i++){ a = matrix[i][i]; for(j = 0; j < 2*n; j++){ matrix[i][j] /= a; } } printf("The inverse matrix is: \n"); for(i = 0; i < n; i++){ for(j = n; j < 2*n; j++){ printf("%.2f", matrix[i][j]); printf("\t"); } printf("\n"); } return 0; }

it is very useful program with nice style

Guy you mix I/O with business logic.

Nice and useful code. Thank you.

what when matrix[i][i]=0. You have not taken care of such situation. When u are going to deal with n*2n matrix, why create a matrix of size 10*10. You should have created a matrix of 5*10 or 10*20 or …

I just want to ask this type similar question . Did you solve if matrix[0][0] = 0; ?

How amazing work. You help me out. Thanks very much.

Is this code work properly with a 6×6 matrix of complex numbers?

for solving divide by 0 when matrix[0][0] == 0 use pivoting

example

http://ganeshtiwaridotcomdotnp.blogspot.in/2009/12/c-c-code-gauss-jordan-method-for.html

Thanks.

you must check if the matrix is invertible or not by calculating the determinant of the matrix ,that is,if the determinant is zero ,the matrix has no inverse

values getting replaced…..so no problem if 0 present as diagonal element

inverse of such a matrix will never exist

??

what approach did u use? I don't completely understand your program.

is this code working when i runthiscode it is giving nan

logic seems incorrect as matrix[i][i] is 0.

doesnt work for huge matrix.

Very smooth code, saved me a lot of work 🙂