• Product of matrices can be done when the matrices are compatible with each other.
  • The matrices are said to be compatible with each other when the number of columns of first matrix is equal to number of rows of second matrix.
  • Thus, if the first matrix has n columns and second matrix has q rows ; then n = q is essential.
  • Thus,  the elements can be multiplied using the following formula: Ci,j = Sum of ( Ai,j * Bi,k) ; where k = 1 to k < n.

Approach:

  • First we will take the number of rows and columns of each matrix as our input.
  • Next we validate if multiplication is possible, based upon number of the condition n = q; where n is number of column of first matrix and q is number of rows of second matrix.
  • Next, using the above mentioned formula we calculate the product of the matrices.
  • Here we are passing arrays between function as an input to next function.

Code:

#include <stdio.h>

void multiply(int arr1[5][5], int arr2[5][5], int, int, int);

int display(int result[5][5], int, int);

int main()

{

   int arr1[5][5], arr2[5][5], r1, c1, r2, c2, i, j;



   printf("Enter rows and column for first matrix: ");

   scanf("%d %d", &r1, &c1);



   printf("Enter rows and column for second matrix: ");

   scanf("%d %d",&r2, &c2);



   // Column of first matrix should be equal to row of second matrix

   while (c1 != r2)

   {

       printf("Error! Column of first matrix is not equal to row of second matrix.\n\n");

       printf("Enter rows and column for first matrix: ");

       scanf("%d %d", &r1, &c1);

       printf("Enter rows and column for second matrix: ");

       scanf("%d %d",&r2, &c2);

   }



   // Storing elements of first matrix.

   printf("\nEnter elements of first matrix row wise:\n");

   for(i=0; i<r1; i++)

       for(j=0; j<c1; j++)

       {

           scanf("%d", &arr1[i][j]);

       }



   // Storing elements of second matrix.

   printf("\nEnter elements of second matrix row wise:\n");

   for(i=0; i<r2; i++)

       for(j=0; j<c2; j++)

       {

           scanf("%d",&arr2[i][j]);

       }



   multiply(arr1, arr2, r1, c2, c1);

       

   return 0;

}

void multiply(int arr1[5][5], int arr2[5][5], int r, int c, int c1)

{

   int i, j, k, result[5][5];

   

   // Initializing all the elements of result matrix to 0

   for(i=0; i<r; ++i)

       for(j=0; j<c; ++j)

       {

           result[i][j] = 0;

       }



   // Multiplying matrices arr1 and arr2

   for(i=0; i<r; i++)

       for(j=0; j<c; j++)

           for(k=0; k<c1; k++)

           {

               result[i][j]+=arr1[i][k]*arr2[k][j];

           }

    

  display(result, r, c);

}



int display(int result[5][5], int r, int c)

{

   int i,j;

   

   printf("\nResult of Matrix Multiplication is:\n");

   for(i=0; i<r; i++)

   {

       for(j=0; j<c; j++)

       {

           printf("%d\t", result[i][j]);

       }

       

       printf("\n");

    }

    return 0;

}

Output:

Enter rows and column for first matrix: 1

5

Enter rows and column for second matrix: 2

3

Error! Column of first matrix is not equal to row of second matrix.



Enter rows and column for first matrix: 1

2

Enter rows and column for second matrix: 2

3



Enter elements of first matrix row wise:

1

2



Enter elements of second matrix row wise:

0

1

3

2

1

4



Result of Matrix Multiplication is:

4 3 11

Report Error/ Suggestion

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CopyRight © 2019