The statement that extracts the element (8) from the following matrix -:is 5 1 4 A = 9 3 0 L6 8 2. x = A(2,3); O x = A(3, 2); O x = A(2); O x = A ( 3 ); O x = A(end); O
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![3 نقاط
*
The statement that
extracts the element (8)
from the following matrix
-:is
[5 1 4]
930
8 2
x = A(2,3); O
x = A(3, 2); O
x = A(2); O
x = A ( 3 ); O
x = A(end); O
A =
L6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8f1c690d-a2fa-4776-bd49-4f527b1cb482%2F97dfae7f-da44-45e0-ac80-4a38c0e98b95%2F3owp5ue_processed.jpeg&w=3840&q=75)
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- Q3 // (A) Using nested for-loops to generate a matrix that has elements shown below (without typing the numbers explicitly): 12 8 4 0 -4 14 10 6 2 -2 16 12 8 4 0Java Objective:Design and implement simple matrix manipulation techniques.Details:Your java program should use 2D arrays to implement simple matrix operations.Your program should do the following:• Read the number of rows and columns of a matrix M1 from the user. Use an input validation loop to make sure the values are greater than 0. • Read the elements of M1 in row major order• Print M1 to the console; make sure you format as a matirx• Repeat the previous steps for a second matrix M2• Create a matrix M3 that is the transpose of M1 and print it to the console• Check if M1 and M2 can be added (should have the same dimensions). If possible, add M1 and M2 and print the result to the console. Otherwise print an error message.• Multiply M1 and M2 if possible and print to the console. If the matrices cannot be multiplied, print an error message. Implementation requirements:• Use a helper method for reading a positive integer using an input validation loop.• Use a helper method for printing…Function Name: odd_even_diagParameters: a 2D list (list of lists)Returns: list of lists Description: Given a 2-dimensional matrix (list of lists) with any size (n*n), modify it according to the following rules: Find the sum of the main diagonal. If the sum is an odd number, change all the values of the given matrix (except the main diagonal) to 0. If the sum is an even number, change all the values of the given matrix (except the main diagonal) to 1. Return the resulting matrix. Example 1:If argument is: [[1, 2], [4, 3]] odd_even_diag should return: [[1, 1], [1, 3]] because the sum 1 + 3 is even. Example 2:If argument is: [[1, 2, 3], [4, 5, 6], [7, 8, 9]] odd_even_diag should return: [[1, 0, 0], [0, 5, 0], [0, 0, 9]] because the sum 1 + 5 + 9 is odd.
- def large_position(matrix: list[list[int]], row: int, col: int) -> int: """ Returns the area of the largest rectangle whose top left corner is at position <row>, <col> in <matrix>. You MUST make use of the helper <longest_chain> here as you loop through each row of the matrix. Do not modify (i.e., mutate) the input matrix. >>> case100 = [[1, 0, 1, 0, 0], ... [1, 0, 1, 1, 1], ... [1, 1, 1, 1, 1], ... [1, 0, 0, 1, 0]] >>> large_position(case1, 0, 0) 4 """ Helper: def longest_chain(lst: List[int]) -> int: count = 0 # defining count variable as 0. Using it for counting the length of the sequence.i = 0 # defining i variable as 0. Using it for index positionwhile (i < len(lst) and lst[i] != 0): # iterating through out the list while i is lesser than the length of the list and until we encounter any 0i += 1 # increasing the index valuecount += 1 # increasing the countreturn…def large_position(matrix: list[list[int]], row: int, col: int) -> int: """ Returns the area of the largest rectangle whose top left corner is at position <row>, <col> in <matrix>. You MUST make use of the helper <longest_chain> here as you loop through each row of the matrix. Do not modify (i.e., mutate) the input matrix. >>> case100 = [[1, 0, 1, 0, 0], ... [1, 0, 1, 1, 1], ... [1, 1, 1, 1, 1], ... [1, 0, 0, 1, 0]] >>> large_position(case1, 0, 0) 4 """ Helper: def longest_chain(lst: List[int]) -> int: count = 0 # defining count variable as 0. Using it for counting the length of the sequence.i = 0 # defining i variable as 0. Using it for index positionwhile (i < len(lst) and lst[i] != 0): # iterating through out the list while i is lesser than the length of the list and until we encounter any 0i += 1 # increasing the index valuecount += 1 # increasing the countreturn…java programming language *matrices refers to 2d arrays note: In the question, it is asked to rotate the points(counterclockwise) inside of the matrix, obviously not the matrix itself. For example,|[1, 2]| -> |[1.866, 1.232]||[3, 4]| (30deg) => |[4.598, 1.964]||[5, 6]| -> |[7.330, 2.696]| it should work in any Nx2 matrice like 2x2, 3x2,4x2. N and the angle (in degrees) should be determined by the user
- please answer all questions You need to generate a (5x12) rain matrix (Rain[5][12]) that contains the average rainfall over 12 months for a 5 year period. You program should do the following steps: 1) Create a 5x12 integer rainfall matrix 2) Create a 5 element array of int pointers(int * Rain[5]) and assign each pointer to the address of the first element of each row 3) Populate the elements of the matrix with pseudo-random numbers having a range from 0-100 inches. Use the rand( ) function to generate your temperatures. The output of rand() should be mod with 100 (rand()%100) in order to get reasonable rainfall values. Init the random number generator with long seed = srand(67) This task should be performed by a function, void RainGen(int* Rain[ ], int R, int C), that populates all the matrix entries. 4) After generating all rainfall matrix entries, print the entire matrix using the function, void Print(int* Rain[ ], int R, int C); 5) Calculate the…C++ Chapter 8 - 2D Array -Find the sum above or below the main diagonal in any n by n matrix, please solve the question with new solution, not the same as in questions bankIn this c langauge program. please explain everyline of this code. Source Code: #include<stdio.h> #include<stdlib.h> int main() { int a[10][10],b[10][10],mul[10][10],r,c,i,j,k; system("cls"); printf("Enter rows of the matrix: "); scanf("%d",&r); printf("Enter Column of the matrix: "); scanf("%d",&c); printf("Enter the first matrix elements\n"); for(i=0;i<r;i++) { for(j=0;j<c;j++) { scanf("%d",&a[i][j]); } } printf("Enter the second matrix elements\n"); for(i=0;i<r;i++) { for(j=0;j<c;j++) { scanf("%d",&b[i][j]); } } printf("multiplication of matrix is \n"); for(i=0;i<r;i++) { for(j=0;j<c;j++) { mul[i][j]=0; for(k=0;k<c;k++) { mul[i][j]+=a[i][k]*b[k][j]; } } } for(i=0;i<r;i++) { for(j=0;j<c;j++) { printf("%d\t",mul[i][j]); } printf("\n"); } return 0; }…
- def large_position(matrix: list[list[int]], row: int, col: int) -> int: """ Returns the area of the largest rectangle whose top left corner is at position <row>, <col> in <matrix>. You MUST make use of the helper <longest_chain> here as you loop through each row of the matrix. Do not modify (i.e., mutate) the input matrix. >>> case = [[1, 0, 1, 0, 0], ... [1, 0, 1, 1, 1], ... [1, 1, 1, 1, 1], ... [1, 0, 0, 1, 0]] >>> large_position(case1, 0, 0) 4 """ You must refer to this while writing for the code above Helper: def longest_chain(lst: List[int]) -> int: count = 0 i = 0while (i < len(lst) and lst[i] != 0): i += 1 count += 1 return count """ Please do this on python you should not use any of the following: dictionaries or dictionary methods try-except break and continue statements recursion map / filter import ""'Array TypesObjective: Based on the given values and initializations, give what is being required of each statement.1. Given A[10], α=2000, esize=4 bytes:a) Find the number of elements.b) Find the address of the 6th element.c) Find the index no. of the 8th element.2. Given E[3][4], α=2020, esize=4 bytes:a) Find the total no. of elements.b) Find the address of the last element.c) Find the address of the 10th element.in c++. Write a function that returns the sum of all the elements in a specified column in a matrix using the following header: const int SIZE = 4; double sumColumn(double m[][SIZE], int rowSize, int columnIndex); Write a test program that reads a 3-by-4 matrix and displays the sum of each column. please leave comments on each step.