C program to multiply two matrices has been shown here. Matrix $[A]_{m \times n}$ can be multiplied with matrix $[B]_{p \times q}$, iff $n = p$ i.e. the number of columns of the first matrix should to be equal to the number of rows in the second matrix. The order of the resultant matrix $C$ would be $m \times q$.

Algorithm, pseudocode and time complexity of the program have also been shown below.

#### Page content(s):

1. Algorithm

2. Pseudocode

3. Time Complexity

4. Program & Output

## 1. Algorithm for matrix multiplication

1. Take two matrice $A_{m\times n}$ and $B_{p\times q}$ as input.

2. Declare another matrix $C_{m\times q}$ and initailize it with 0.

3. Check if $n = p$

4. If step [3] is false then display "Matrices can not be multiplied!" and go to step [8]

5. If step [3] is true, then

6. Repeat for each $i \in [0, m - 1]$

6.1. Repeat for each $j \in [0, q - 1]$

6.1.1. Repeat for each $k \in [0, p - 1]$

6.1.1.1. Perform $C[i][j] = C[i][j] + A[i][k] * B[k][j]$

7. Display $C_{m\times q}$ as the resultant matrix.

8. Exit program.

## 2. Pseudocode for matrix multiplication

Input: Two matrices $A_{m\times n}$ and $B_{p\times q}$

Output: $A * B$

1. Procedure matrixMultiplication($A_{m\times n}$, $B_{p\times q}$):

2. Initialize $C_{m\times q} = 0$

3. If $n == p$:

4. Repeat for each $i \in [0, m-1]$

5. Repeat for each $i \in [0, q-1]$

6. Repeat for each $k \in [0, p-1]$

7. $C_{ij}\leftarrow C_{ij} + A_{ik} * B_{kj}$

8. Return $C_{m\times q}$

9. Else:

10. Return Matrices can not be multiplied!

11. End Procedure

## 3. Time Complexity for matrix multiplication

Time Complexity: O($n^3$)

Where $n$ is the row and column size of two matrices.

## 4. C Program for matrix multiplication

Code has been copied
/****************************************
alphabetacoder.com
C Program for Matrix Multiplication
*****************************************/
#include <stdio.h>

int main() {
// declare variables
int m, n, p, q, i, j, k;
int A[10][10] = {0}, B[10][10] = {0}, C[10][10] = {0};

//take input of the order of first matrix
printf("Enter the number of row and column of first matrix: ");
scanf("%d%d", & m, & n);

//take input of the first matrix
printf("Enter the first matrix of order %d x %d =\n", m, n);
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
scanf("%d", & A[i][j]);

//take input of the order of second matrix
printf("Enter the number of row and column of second matrix: ");
scanf("%d%d", & p, & q);

//take input of the first matrix
printf("Enter the second matrix of order %d x %d =\n", p, q);
for (i = 0; i < p; i++)
for (j = 0; j < q; j++)
scanf("%d", & B[i][j]);

// check if number of columns in first matrix
// is same as number of rows in second matrix.
// If not, then matrices can not be multiplied
if (n != p)
printf("\nMatrices can not be multiplied!");
else {
// do matrix multiplication
for (i = 0; i < m; i++) {
for (j = 0; j < q; j++) {
for (k = 0; k < p; k++)
C[i][j] += A[i][k] * B[k][j];
}
}

printf("The resultant matrix after multiplication:\n");
//display the result
for (i = 0; i < m; i++) {
for (j = 0; j < q; j++) {
printf("%d ", C[i][j]);
}
//new  line
printf("\n");
}
}

return 0;
}


Output

#### Case 1:

Enter the number of row and column of first matrix: 3 3

Enter the first matrix of order 3 x 3 =

1 3 4

5 6 1

0 5 2

Enter the number of row and column of second matrix: 3 2

Enter the second matrix of order 3 x 2 =

2 3

1 0

7 1

The resultant matrix after multiplication:

33 7

23 16

19 2

#### Case 2:

Enter the number of row and column of first matrix: 3 3

Enter the first matrix of order 3 x 3 =

3 5 7

4 5 9

0 1 1

Enter the number of row and column of second matrix: 2 3

Enter the second matrix of order 2 x 3 =

9 4 5

3 7 8

Matrices can not be multiplied!