C program to find Inverse of Matrix : For a square matrix A, the inverse matrix is written A-1. When A is multiplied by A-1 the result is the identity matrix I. Non-square matrices do not have inverses. There are several different methods for calculating inverse of Matrix like Augmented Matrix Method, Adjoint Method and few shortcut methods. Today i have written a C program to find Inverse of Square or say NxN Matrix using Adjoint Method.
Note: Not all square matrices have inverses. A square matrix which has an inverse is called invertible or non-singular, and a square matrix without an inverse is called non-invertible or singular.
Algorithm to calculate Inverse of Matrix using Adjoint Method :
1) calculating the Matrix of Minors
2) then turn that into the Matrix of Cofactors
3) then the Adjugate
4) multiply that by 1/Determinant.
But it is best explained by working through an example!
Example: find the Inverse of A:
It needs 4 steps. It is all simple arithmetic but there is a lot of it, so try not to make a mistake!
Step 1: Matrix of Minors
The first step is to create a “Matrix of Minors”:
For each element of the matrix:
- ignore the values on the current row and column
- calculate the determinant of the remaining values
Put those determinants into a matrix (the “Matrix of Minors”)
Determinants
For a 2×2 matrix (2 rows and 2 columns) the determinant is easy: ad-bc
Think of a cross:
|
 |
The Calculations
Here are the first two, and last two, calculations of the “Matrix of Minors” (notice how I ignore the values in the current row and columns, and calculate the determinant using the remaining values):
And here is the calculation for the whole matrix:
Step 2: Matrix of Cofactors
This is easy! Just apply a “checkerboard” of minuses to the “Matrix of Minors”. In other words, you need to change the sign of alternate cells, like this:
Step 3: Adjugate (also called Adjoint)
Now “Transpose” all elements of the previous matrix… in other words swap their positions over the diagonal (the diagonal stays the same):
Step 4: Multiply by 1/Determinant
Now find the determinant of the original matrix. This isn’t too hard, because we already calculated the determinants of the smaller parts when we did “Matrix of Minors”.
So: multiply the top row elements by their matching “minor” determinants:
Determinant = 3×2 – 0×2 + 2×2 = 10
And now multiply the Adjugate by 1/Determinant:
And we are done!
Now let us see how to implement above explained method using C program which takes order of Square Matrix as input and then Matrix and displays Determinant and Inverse Matrix as output.
C Program to find Inverse of Matrix :
#include<stdio.h>
#include<math.h>
#include<conio.h>
float determinant(float[][],float);
void cofactor(float[][],float);
void transpose(float[][],float[][],float);
int main()
{
float a[25][25],k,d;
int i,j;
printf("-------------------------------------------------------------\n");
printf("----------------made by C code champ ------------------------\n");
printf("-------------------------------------------------------------\n");
printf("\n C Program to find inverse of Matrix\n\n");
printf("Enter the order of the Matrix : ");
scanf("%f",&k);
printf("Enter the elements of %.0fX%.0f Matrix : \n",k,k);
for (i=0;i<k;i++)
{
for (j=0;j<k;j++)
{
scanf("%f",&a[i][j]);
}
}
d=determinant(a,k);
printf("Determinant of the Matrix = %f",d);
if (d==0)
printf("\nInverse of Entered Matrix is not possible\n");
else
cofactor(a,k);
printf("\n\n**** Thanks for using the program!!! ****");
getch();
}
/*For calculating Determinant of the Matrix */
float determinant(float a[25][25],float k)
{
float s=1,det=0,b[25][25];
int i,j,m,n,c;
if (k==1)
{
return (a[0][0]);
}
else
{
det=0;
for (c=0;c<k;c++)
{
m=0;
n=0;
for (i=0;i<k;i++)
{
for (j=0;j<k;j++)
{
b[i][j]=0;
if (i != 0 && j != c)
{
b[m][n]=a[i][j];
if (n<(k-2))
n++;
else
{
n=0;
m++;
}
}
}
}
det=det + s * (a[0][c] * determinant(b,k-1));
s=-1 * s;
}
}
return (det);
}
void cofactor(float num[25][25],float f)
{
float b[25][25],fac[25][25];
int p,q,m,n,i,j;
for (q=0;q<f;q++)
{
for (p=0;p<f;p++)
{
m=0;
n=0;
for (i=0;i<f;i++)
{
for (j=0;j<f;j++)
{
if (i != q && j != p)
{
b[m][n]=num[i][j];
if (n<(f-2))
n++;
else
{
n=0;
m++;
}
}
}
}
fac[q][p]=pow(-1,q + p) * determinant(b,f-1);
}
}
transpose(num,fac,f);
}
/*Finding transpose of matrix*/
void transpose(float num[25][25],float fac[25][25],float r)
{
int i,j;
float b[25][25],inverse[25][25],d;
for (i=0;i<r;i++)
{
for (j=0;j<r;j++)
{
b[i][j]=fac[j][i];
}
}
d=determinant(num,r);
for (i=0;i<r;i++)
{
for (j=0;j<r;j++)
{
inverse[i][j]=b[i][j] / d;
}
}
printf("\n\n\nThe inverse of matrix is : \n");
for (i=0;i<r;i++)
{
for (j=0;j<r;j++)
{
printf("\t%f",inverse[i][j]);
}
printf("\n");
}
}
We hope you all have enjoyed the article on C program to find Inverse of NxN Matrix. If you have any issues ask me in form of comments.
References:
1. Mathwords
2. Mathisfun(Example)
4. Wikipedia
Well written! It really helped out with my assignment! Many thanks and bravo!
Well written! It really helped out with my assignment! Many thanks and bravo!
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Thank you so much
I need it to Pascal
Please help
sir i like the code since its nt too long nd works… but i will feel more comfortable if u add comments to make understand wht actually is happening in each n every line….
regards
full of error n ur program.please correct an post it again with output.
anyway thank you
Its perfectly correct and working fine. Please copy the code correctly and execute it with the help of Dev C++ compiler or Code blocks.
The code is very difficult and solving of determinant of matrix can by written with recursion techniques (except other parts of code). Its looks like beginners programming trying.
this program 100% correct .
please fill up blank [] with 25 .
as like ……………….
#include
#include
#include
float determinant(float[25][25],float);
void cofactor(float[25][25],float);
void transpose(float[25][25],float[25][25],float);
int main()
{………………………………………………………………………………..
Tnx bro ,
solve this problem . I am new in c . A lot of good wish for U .
The programme in c about inverse of nxn matrix is not compiled when tested what could be the reason ?
bmsrao