C Program to find Sum of Opposite Diagonal Elements of a Matrix

In this article, we will show you, How to write a C Program to find Sum of Opposite Diagonal Elements of a Matrix. Or How to write a C program to find Sum of Opposite Diagonal Elements of a Multi-Dimensional Array with example.

C Program to find Sum of Opposite Diagonal Elements of a Matrix Example 1

This program allows the user to enter the number of rows and columns of a Matrix. Next, we are going to calculate the sum of opposite diagonal elements in this matrix using For Loop.

/* C Program to find Sum of Opposite Diagonal Elements of a Matrix */

#include<stdio.h>
 
int main()
{
 	int i, j, rows, columns, a[10][10], Sum = 0;
  
 	printf("\n Please Enter Number of rows and columns  :  ");
 	scanf("%d %d", &i, &j);
 
 	printf("\n Please Enter the Matrix Elements \n");
 	for(rows = 0; rows < i; rows++)
  	{
   		for(columns = 0;columns < j;columns++)
    	{
      		scanf("%d", &a[rows][columns]);
    	}
  	}
   	  
 	for(rows = 0; rows < i; rows++)
  	{
   		Sum = Sum + a[rows][i - rows  - 1];
  	}
 
 	printf("\n The Sum of Opposite Diagonal Elements of a Matrix =  %d", Sum );

 	return 0;
}

OUTPUT

ANALYSIS
In this Program, We declared single Two dimensional arrays Multiplication of size of 10 * 10.
Below statements asks the User to enter the Matrix size (Number of rows and columns. For instance 2 Rows, 3 Columns = a[2][3] )

printf("\n Please Enter Number of rows and columns  :  ");
scanf("%d %d", &i, &j);

Next, we used for loop to iterate each and every cell present in a[3][3] matrix. Conditions inside the for loops ((rows < i) and (columns < j)) will ensure the compiler, not to exceed the matrix limit. Otherwise the matrix will overflow

scanf statement inside the for loop will store the user entered values in every individual array element such as a[0][0], a[0][1], …..

for(rows = 0; rows < i; rows++).
{
  for(columns = 0; columns < j; columns++)
   {
     scanf("%d", &a[rows][columns]);
   }
}

In the next line, We have one more for loop to find Sum of Diagonal Elements of a Matrix

for(rows = 0; rows < i; rows++)
{
   		Sum = Sum + a[rows][i - rows  - 1];
}

User inserted values are: a[3][3] = {{10, 20, 30}, { 40, 50, 60}, {70, 80, 90}}

Row First Iteration: for(rows = 0; rows < 3; 0++)
The condition (0 < 3) is True.
Sum  = Sum + a[rows][i – rows – 1]
Sum  = Sum + a[0][3 – 0 – 1] => Sum + a[0][2]
Sum  = 0 + 30 = 30

Row Second Iteration: for(rows = 1; rows < 3; 1++)
The condition (1 < 3) is True.
Sum  = Sum + a[1][3 – 1 – 1] => Sum + a[1][1]
Sum  = 30 + 55 = 85

Row Second Iteration: for(rows = 2; rows < 3; 2++)
The condition (2 < 3) is True.
Sum  = Sum + a[2][3 – 2 – 1] => Sum + a[2][0]
Sum  = 85 + 95 = 180

Next, rows value will be incremented. After the increment the condition inside the for loop (rows < 3) will fail. So it will exit from the loop.

At last we used printf statement to print the total Sum as output

C Program to find Sum of Opposite Diagonal Elements of a Matrix Example 2

This program is same as above, but this time we changed the algorithm a bit.

/* C Program to find Sum of Opposite Diagonal Elements of a Matrix */

#include<stdio.h>
 
int main()
{
 	int i, j, rows, columns, a[10][10], Sum = 0;
  
 	printf("\n Please Enter Number of rows and columns : ");
 	scanf("%d %d", &i, &j);
 
 	printf("\n Please Enter the Matrix Elements \n");
 	for(rows = 0; rows < i; rows++)
  	{
   		for(columns = 0;columns < j;columns++)
    	{
      		scanf("%d", &a[rows][columns]);
    	}
  	}
   	  
 	for(rows = 0; rows < i; rows++)
  	{
  		for(columns = 0;columns < j; columns++)
  		{
  			if(rows + columns == ((i + 1) - 2))
  			       Sum = Sum + a[rows][columns];
		}
  	}
 
 	printf("\n The Sum of Opposite Diagonal Elements of a Matrix =  %d", Sum );

 	return 0;
}

OUTPUT

User inserted values are: a[3][3] = {{1, 2, 3}, { 4, 5, 6}, {7, 8, 15}}

Row First Iteration: for(rows = 0; rows < 3; 0++)
The condition (0 < 3) is True. So, it will enter into second for loop

Column First Iteration: for(columns = 0; 0 < 3; 0++)
The condition (columns < 3) is True. So, it will start executing the If Statement
if(rows + columns == ((i + 1) – 2))
=> if(0 + 0 == ((3 + 1) – 2)) – Condition is False

Column Second Iteration: for(columns = 1; 1 < 3; 1++)
The condition (1 < 3) is True.
if(rows + columns == ((i + 1) – 2))
=> if(0 + 1 == ((3 + 1) – 2)) – Condition is False

Column Second Iteration: for(columns = 2; 2 < 3; 2++)
The condition (1 < 3) is True.

if(rows + columns == ((i + 1) – 2))
=> if(0 + 2 == ((3 + 1) – 2)) – Condition is True
Sum = Sum + a[rows][columns] => Sum + a[0][2]
Sum = 0 + 3 = 3

Do the same for remaining iteration where rows = 1 and rows = 2

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