Write a program in c to find transpose of a matrix using functions?

Answer 1

Sorry to delete the previous version of this program, but this one is much more concise and sophisticated. I have written this program for a 3*3 matrix, but if you want to change it, just enter additional variables to store the order of the matrix and then replace the "3"s of the program with them as needed. It's easy. So here we go:

#include

#include

void swap(int *x,int *y);

main()

{

int i,j,k,m,n;

int matrix[3][3];

printf("Enter the matrix: \n\n");

for (i=0;i<3;++i)

{

printf("\nEnter #%d row: ",(i+1));

for (j=0;j<3;++j)

scanf("%d",&matrix[i][j]);

}

printf("\n\n");

for (j=0;j<3;++j)

{

for (i=0;i<3;++i)

if (i>j)

swap(&matrix[i][j],&matrix[j][i]);

}

printf("The transposed matrix is: \n\n");

for (i=0;i<3;++i)

{

for (j=0;j<3;++j)

printf("%4d",matrix[i][j]);

printf("\n\n");

}

}

void swap(int *x,int *y)

{

int t=*x;

*x=*y;

*y=t;

return;

}

Answer 2

#include<iostream.h>

#include<conio.h>

class A

{

int a[2][2],b[2][2],

x[2][2],y[2][2],

first[2][2],second[2][2],

c,d,e;

public:

void read()

{

y[0][0]=0;

y[1][0]=0;

y[0][1]=0;

y[1][1]=0;

x[0][0]=0;

x[0][1]=0;

x[1][0]=0;

x[1][1]=0;

first[0][0]=0;

first[0][1]=0;

first[1][0]=0;

first[1][1]=0;

second[0][0]=0;

second[0][1]=0;

second[1][0]=0;

second[1][1]=0; cout<<"Enter the value for first matrix : "<<endl;

for(c=0;c<2;c++)

{

for(d=0;d<2;d++)

{

cin>>a[c][d];

}

} cout<<"Enter the value for second matrix : "<<endl;

for(c=0;c<2;c++)

for(d=0;d<2;d++)

cin>>b[c][d]; for(c=0;c<2;c++)

for(d=0;d<2;d++)

x[c][d]=a[c][d]+b[c][d]; for(c=0;c<2;c++)

for(d=0;d<2;d++)

for(e=0;e<2;e++)

y[c][e]+=a[c][d]*b[d][e]; for(c=0;c<2;c++)

for(d=0;d<2;d++)

first[c][d]=a[d][c];

first[c][d]=a[c][d]; for(c=0;c<2;c++)

for(d=0;d<2;d++)

second[c][d]=b[d][c];

second[c][d]=b[c][d];

}

void addition()

{

cout<<"Addition of matrix : "<<endl;

for(c=0;c<2;c++)

{

for(d=0;d<2;d++)

{

cout<<" "<<x[c][d];

}

cout<<endl;

}

} void multiplication()

{

cout<<"Multiplication of matrix : "<<endl;

for(c=0;c<2;c++)

{

for(d=0;d<2;d++)

{

cout<<" "<<y[c][d];

}

cout<<endl;

}

} void transpose()

{

cout<<"Transpose for first matrix : "<<endl;

for(c=0;c<2;c++)

{

for(d=0;d<2;d++)

{

cout<<" "<<first[c][d];

}

cout<<endl;

}

cout<<"Transpose for second matrix : "<<endl;

for(c=0;c<2;c++)

{

for(d=0;d<2;d++)

{

cout<<" "<<second[c][d];

}

cout<<endl;

}

}

}; void main()

{

clrscr();

A a;

a.read();

a.addition();

a.multiplication();

a.transpose();

getch();

}

Answer 3

#include <stdio.h> #include <conio.h> void transpose(int a[10][10],int ,int); /*Function prototype*/

void main() { int i,j,m,n; int a[10][10], b[10][10]; clrscr(); printf("Enter the order of matrix a\n"); scanf("%d %d", &m, &n); printf("Enter the elements of matrix\n"); for(i=0;i<m;i++) { for(j=0;j<n;j++) { scanf("%d",&a[i][j]); } } printf("Matrix a is\n"); for(i=0;i<m;i++) { for(j=0;j<n;j++) { printf("%3d",a[i][j]); } printf("\n"); } transpose(a,i,j)/*Function call*/ getch(); /* Finding Transpose of matrix*/ void transpose(int a[10][10],int m,int n)/*Function definition*/ for(i=0;i<m;i++) { for(j=0;j<n;j++) { b[i][j] = a[j][i]; } } printf("Its Transpose is\n"); for(i=0;i<m;i++) { for(j=0;j<n;j++) { printf("%3d",b[i][j]); } printf("\n"); } } /*End of main()*/

Answer 4

#include<stdio.h>

#include<conio.h>

void main(){

int arr[3][3], arrT[3][3];

int i,j;

clrscr();

printf("Enter the 3x3 matrix:\n");

for(i=0;i<3;i++)

{

for(j=0;j<3;j++)

{

printf("Enter the element arr[%d][%d] : ",i,j);

scanf("%d",&arr[i][j]);

}

}

for(i=0;i<3;i++)

{

for(j=0;j<3;j++)

{

arrT[j][i] = arr[i][j];

}

}

printf("The transpose of the matrix is: \n");

for(i=0;i<3;i++)

{

for(j=0;j<3;j++)

{

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

}

printf("\n");

}

getch();

}

Answer 5

//matrix and its transpose

#include<stdio.h>

#include<conio.h>

void main() {

clrscr();

int m, n, c, d, matrix[10][10],transpose[10][10];

printf("Enter the number of rows and columns of matrix ");

scanf("%d %d",&m, &n);

//storing the elements of matrix

printf("Enter the elements of matrix \n");

for( c = 0 ; c < n ; c++ ) {

for( d = 0 ; d < m ; d++ ) {

scanf("%d",&matrix[c][d]);

}

}

printf("The matrix you have entered is:\n");

for( c = 0 ; c < n ; c++ ) {

for( d = 0 ; d < m ; d++ ) { //m is number of rows

printf("%d\t",matrix[c][d]);

}

printf("\n");

}

//making transpose

for( c = 0 ; c < n ; c++ ) {

for( d = 0 ; d < m ; d++ ) {

transpose[d][c] = matrix[c][d];

}

}

printf("Transpose of entered matrix :-\n");

for( c = 0 ; c < m ; c++ ) {

for( d = 0 ; d < n ; d++ ) { /* after transpose number of rows is equal to no. of columns of original matrix */

printf("%d\t",transpose[c][d]);

}

printf("\n");

}

getch();

}

Output:

Enter the number of rows and columns of matrix 3 2

Enter the elements of matrix

1 2 3 4 5 6

The matrix you have entered is

1 2 3

4 5 6

Transpose of entered matrix:

1 4

2 5

3 6

Answer 6

#include<iostream.h>

int main()

{

int a[3][3];

int tr[3][3];

int i,j;

cout<<"Enter elements of matrix : ";

for(i=0;i<3;i++)

{

for(j=0;j<3;j++)

{

cin>>a[i][j];

}

}

cout<<"original matrix : ";

for(i=0;i<3;i++)

{

for(j=0;j<3;j++)

{

cout<<a[i][j];

}

cout<<"\n";

}

for(i=0;i<3;i++)

{

for(j=0;j<3;j++)

{

tr[i][j]=a[j][i];

}

}

cout<<"\n\ntranspose of matrix : ";

for(i=0;i<3;i++)

{

for(j=0;j<3;j++)

{

cout<<tr[i][j];

}

cout<<"\n";

}

return 0;

}

Answer 7

it's just easy to display transpose of matrix. First store the elements of matrix in a[i][j], then display the value of a[j][i] instead of a[i][j]. For more information just visit the below site.

link -

s4sourcecode.online/2020/03/c-program-to-display-transpose-of-matrix.html?m=1

Answer 8

for(i=0;i<r;i++)

{

for(int j=0;j<c;j++)

{

b[i][j]=a[i][j];

}

}