2. Matrices In this exercise you will write several functions that operate on ma

Question

2. Matrices In this exercise you will write several functions that operate on matrices or two-dimensional arrays. For this exercise, we will restrict our attention to square matrices collections of numbers with the same number of columns and rows. You will place all the function prototypes in a header file named matrix_utils.h with their definitions in a source file named matrix_utils.c. You should test your functions by writing a driver program, but you need not handitn (a) Write a function that constructs a new n × n identity matrix. An identity matrix is a square matrix with ones on the diagonal and zeros everywhere else. int **getIdentityMatrix(int n); b) Write a function that takes two matrices and determines if they are equal (all of their elements are the same). int isEqual(int **A, int **B, int n); c) Write a function that takes a matrix and an index i and returns a new array that

Explanation / Answer

If you post more than 4 parts in a question, as per chegg guidelines I have to solve only 4 parts.

(A)

int **getIdentityMatrix(int n)

{

    // dynamically allocate memory for n x n matrix

    int **arr = (int **)malloc( n * sizeof(int *) );

   

    int i, j;

   

    for( i = 0 ; i < n ; i++ )

        arr[i] = (int *)malloc( n * sizeof(int) );

   

    // initialize array with 0

    for( i = 0 ; i < n ; i++ )

        for( j = 0 ; j < n ; j++ )

            arr[i][j] = 0;

       

    // set the disgonal values to 1

    for( i = 0 ; i < n ; i++ )

        arr[i][i] = 1;

   

    return arr;

}

(b)

int isEqual( int **A, int **B, int n )

{

    int i, j;

   

    // traverse the row

    for( i = 0 ; i < n ; i++ )

        // traverse the column

        for( j = 0 ; j < n ; j++ )

            // if the current (i, j) element of the two matrix are not equal

           if( A[i][j] != B[i][j] )

                return 0;

   

    // if the matrix are equal

    return 1;

}

(c)

int *getRow(int **A, int n, int i)

{

    int *arr = (int *)malloc( n * sizeof(int) );

   

    int j;

   

    for( j = 0 ; j < n ; j++ )

        arr[j] = A[i][j];

   

    return arr;

}

(d)

int *getCol(int **A, int n, int j)

{

    int *arr = (int *)malloc( n * sizeof(int) );

   

    int i;

   

    for( i = 0 ; i < n ; i++ )

        arr[i] = A[i][j];

   

    return arr;

}

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