Not Yet Answered # Transpose a matrix

Narue 5,707 murschech 11 murschech 11 Narue 5,707 aldrinsms -2 Dave Sinkula 2,398 MANDALSHASHI thinkammar Ancient Dragon 5,243 nucleon 114 marvind786 -8 Nit90esh -4 srivanip -3

0

If it's a square matrix then you can do it with something like this:

```
for (int i = 0; i < 4; i++) {
for (int j = i + 1; j < 4; j++) {
int save = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = save;
}
}
```

If it's a non-square matrix then you're SOL.

0

If it's a square matrix then you can do it with something like this:

`for (int i = 0; i < 4; i++) { for (int j = i + 1; j < 4; j++) { int save = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = save; } }`

If it's a non-square matrix then you're SOL.

Your program won't work. The reason is that when you write matrix*[j] = matrix[j] you've lost the contents of matrix[j] forever and you can't get it back to put in position [[j]. This will do it.*

```
for (int i = 0; i < 4; i++)
for (int j = 0; j < 4; j++)
{ save = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = save;
}
```

0

If it's a square matrix then you can do it with something like this:

`for (int i = 0; i < 4; i++) { for (int j = i + 1; j < 4; j++) { int save = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = save; } }`

If it's a non-square matrix then you're SOL.

Sorry. I didn't read your code carfully enough before I replied.

0

>Sorry. I didn't read your code carfully enough before I replied.

Maybe you should run example code to see if it works before trying to correct it with an incorrect solution. Starting the inner loop at 0 instead of i + 1 will result in a lot of shuffling to get the same matrix that you started with, not a transposed matrix as the OP requested. So maybe you should also test your own code before posting it. Especially if you're trying to correct someone and using it as an example.

On a purely stylistic point, I have two issues with your code. First, even though it's valid the way you did it, you should always put braces around a loop or if construct that has more than a one line body. This way you don't have to rely solely on indention to prove that your code is correct. You also avoid certain pitfalls.

Second, starting the statements of a block on the same line as the opening brace is a formatting nightmare. It makes code harder to read and harder to reformat so that it's easier to follow.

Something more like this (following your apparent brace indention style):

```
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++)
{
save = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = save;
}
}
```

You also failed to define save before using it, but I'll let you slide on that. ;)

0

How to make a program yhat make the transpose of a matrix using array and pointers C ofcourse?

0

Wonderful! Be a pompous ass and dredge up an old thread (after reiterating the same demand a 3rd time), offering nothing, and pretend you are being courteous! Well, done. *plonk*

0

// prog to mulltiply matrix

```
#include<stdio.h>
#include<conio.h>
void main()
{
int a[10][10],b[10][10],c[10][10],i,j,k,m,n,m1,n1;
clrscr();
printf("Enter array size ");
scanf("%d%d",&m,&n);
printf("Enter array size 2");
scanf("%d%d",&m1,&n1);
if(n!=m1)
{
printf("Wrong choice entered");
getch();
exit(0);
}
else
{
printf("Enter elements");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
}
}
printf("Enter element 2");
for(i=0;i<m1;i++)
{
for(j=0;j<n1;j++)
{
scanf("%d",&b[i][j]);
}
}
for(i=0;i<m;i++)
{
for(j=0;j<n1;j++)
{
c[i][j]=0;
for(k=0;k<n;k++)
{
c[i][j]=c[i][j]+(a[i][k]*b[k][j]);
}
}
}
printf("REQUIRED MATRIX");
for(i=0;i<m;i++)
{
for(j=0;j<n1;j++)
{
printf("\n%d",c[i][j]);
}
printf("\n");
}
getch();
}
}
```

0

Hi there

I read you method of transposing a matrix

it works

but i have toa sk a question!

that what is the logic behind initializing inner loop variable j=i+1 instead of j=0???

i shalll be thankful to you if you reply to this post

thanx

0

>>what is the logic behind initializing inner loop variable j=i+1 instead of j=0???

which post # are you talking about?

0

You only need to loop through less-than-half of the matrix since in each execution of the inner loop you're swapping two elements (and the main diagonal stays put). If you iterated through the whole matrix, you'd just put it back the way it started. Starting j at i+1 loops through the upper triangle of the square (or lower, depending on how you look at it), swapping elements with the opposite triangle.

-2

i want to find determinant of a matrix in java where the order of the matrix is taken at run time.. i'm implementing hill cipher in java.. please help me out..

-1

```
/*this is easy transpose of matrix*/
void(int mat[],int m,int n)
{
int t[10][10],i,j;
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
t[i][j]=mat[j][i]
}
}
}
```

*Edited 5 Years Ago by Narue*: Added code tags

-1

The above code needs a small change. If you use the above code for transpose, you will end up getting the same matrix as before because, you are replacing rows and columns twice.

You need to do the replacement only once.

```
for ( i=0; i<3; i++)
{
for ( j=0; j<3; j++)
{
if(i>j)
{
temp = inputmatrix[i][j];
inputmatrix[i][j] = inputmatrix[j][i];
inputmatrix[j][i] = temp;
}
}
};
```

*Edited 4 Years Ago by srivanip*

This article has been dead for over six months. Start a new discussion instead.

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