C program to find transpose of a matrix

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C program to find transpose of a matrix

Write a C program to read elements in a matrix and find transpose of the given matrix. How to find transpose of a given matrix in C. Logic to find transpose of a matrix in C programming.

Required knowledge

Basic C programming, C for loop, ArrayS

Transpose of a matrix

Transpose of a matrix A is defined as converting all rows into columns and columns into rows. Transpose of matrix A is written as AT.

Transpose of a matrix A is defined as -

AT_ij = A_ji ; Where 1 ≤ i ≤ m and 1 ≤ j ≤ n

Logic to find transpose of a matrix

Below is the step by step descriptive logic to find transpose of a matrix.

1. Input elements in matrix A from user.
2. Declare another matrix of same size as of A, to store transpose of matrix say B.
3. To iterate through each element of matrix run two loops. Run an outer loop from 0 to MAX_ROWS to iterate through rows. The loop structure should look like for(row=0; row<MAX_ROWS; row++).
4. To iterate through each column of the matrix, run an inner loop from 0 to MAX_COLS. The loop structure must look like for(col=0; col<MAX_COLS; col++).
5. Inside inner loop we will perform actual transpose of the matrix. As per definition column of transpose matrix is equal to row of original matrix and vice versa. Means assign B[col][row] = A[row][col].

Program to find transpose of a matrix

/**
 * C program to find transpose of a matrix
 */

#include <stdio.h>
#define MAX_ROWS 3
#define MAX_COLS 3

int main()
{
    int A[MAX_ROWS][MAX_COLS];  // Original matrix
    int B[MAX_COLS][MAX_ROWS];  // Transpose matrix

    int row, col;

    /* Input elements in matrix A from user */
    printf("Enter elements in matrix of size %dx%d: \n", MAX_ROWS, MAX_COLS);
    for(row=0; row<MAX_ROWS; row++)
    {
        for(col=0; col<MAX_COLS; col++)
        {
            scanf("%d", &A[row][col]);
        }
    }

    /*
     * Find transpose of matrix A
     */
    for(row=0; row<MAX_ROWS; row++)
    {
        for(col=0; col<MAX_COLS; col++)
        {
            /* Store each row of matrix A to each column of matrix B */
            B[col][row] = A[row][col];
        }
    }
    
    /* Print the original matrix A */ 
    printf("\nOriginal matrix: \n");
    for(row=0; row<MAX_ROWS; row++)
    {
        for(col=0; col<MAX_COLS; col++)
        {
            printf("%d ", A[row][col]);
        }

        printf("\n");
    }
    
    /* Print the transpose of matrix A */
    printf("Transpose of matrix A: \n");
    for(row=0; row<MAX_COLS; row++)
    {
        for(col=0; col<MAX_ROWS; col++)
        {
            printf("%d ", B[row][col]);
        }

        printf("\n");
    }

    return 0;
}

Output:

Enter elements in matrix of size 3x3:

10 20 30
40 50 60
70 80 90

Original matrix:

10 20 30
40 50 60
70 80 90

Transpose of matrix A:

10 40 70
20 50 80
30 60 90

Process returned 0 (0x0) execution time : 27.624 s

Program to find transpose of a matrix

/**
 * C program to find transpose of a matrix
 */

#include 


int main()
{
    int A[10][10];  // Original matrix
    int B[10][10];  // Transpose matrix

    int row, col,m,n;
    printf("Enter number of rows:");
    scanf("%d",&m);
    printf("Enter number of columns:");
    scanf("%d",&n);

    /* Input elements in matrix A from user */
    printf("Enter elements in matrix of size %dx%d: \n", m, n);
    for(row=0; row<m; row++)
    {
        for(col=0; col<n; col++)
        {
            scanf("%d", &A[row][col]);
        }
    }

    /*
     * Find transpose of matrix A
     */
    for(row=0; row<m; row++)
    {
        for(col=0; col<n; col++)
        {
            /* Store each row of matrix A to each column of matrix B */
            B[col][row] = A[row][col];
        }
    }

    /* Print the original matrix A */
    printf("\nOriginal matrix: \n");
    for(row=0; row<m; row++)
    {
        for(col=0; col<n; col++)
        {
            printf("%d ", A[row][col]);
        }

        printf("\n");
    }

    /* Print the transpose of matrix A */
    printf("Transpose of matrix A: \n");
    for(row=0; row<m; row++)
    {
        for(col=0; col<n; col++)
        {
            printf("%d ", B[row][col]);
        }

        printf("\n");
    }

    return 0;
}

Output:

Enter number of rows:3
Enter number of columns:3
Enter elements in matrix of size 3x3:

1 2 3
4 5 6
7 8 9

Original matrix:

1 2 3
4 5 6
7 8 9

Transpose of matrix A:

1 4 7
2 5 8
3 6 9

Process returned 0 (0x0) execution time : 14.592 s

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