a) Enter the matrix in MATLAB 1234 2000 3000 4000 b) Q5:Let B =[16 2 3 13;5 11 10 8;9 7 6 12;4 14 15 1] find: a. A = B(:,[1 3 2 4]) b. sum(A(:,end)) c. A(1,4) + A(2,4) + A(3,4) + A(4,4)
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![a) Enter the matrix in MATLAB
1234
2000
3000
4000
b) Q5:Let B =[16 2 3 13;5 11 10 8;9 7 6 12;4 14 15 1] find:
a. A = B(:,[1 3 2 4])
b. sum(A(:,end))
c. A(1,4) + A(2,4) + A(3,4) + A(4,4)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F26fb6d36-22ff-438a-87c4-4d47d80b733a%2F8e11d797-fd16-4079-bece-42814f8c9582%2Fwkqgxwf_processed.jpeg&w=3840&q=75)
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- Write a report in which you discuss and compare your Gauss elimination and Gauss-Jordan programs. Gauss elimination #include<stdio.h> int main ( ) { int i, j, k, n; float A[20] [20], c, x [10], sum=0.0; printf("\Enter the order of matrix: "); Sacnf("%d,&n); prinf("\n Enter the elements of augmented matrix row-wise: \n\n"); for(i =1; i<=n; i++) { for(j=1; j<(n+1); j++) { printf("A[%d][%d] : ", i, j); scanf("%f", [i][j]); } } for(j=1; j<n; j++) { for(i=1; j<n; i++) { if(i>j) { c=A[i][j]/A[j][j]; for(k=1; k<=n+!; k++) { A[i][k] = A[i][k]-c*A[j][k]; } } } } x[n]= A[n] [n+1]/A[n][n]; for(i=n-1; i>=1; i--) { sum=0;…%Encode A1, b1 and x1 as the vector of unknowns. A1 = b1 = syms xv1 = %Check the size of A, set it as m1 and n1 [m1,n1] = %Augment A and b to form AM1 AM1 = %Solve the Reduced Rwo Echelon of AM1. RREFA1 = %Collect the last column and set as bnew1, set the remaining elements as Anew1 bnew1= Anew1 = %Check if Anew is an identity matrix, if it is, bnew is the solution if Anew1 =eye(m1,n1) Root1 = bnew1 else display("No Solution") end %Augment the matrix A1 with the identity Matrix of the same size, set the result as AMI1 AMI1 = %Find the reduced row echelon form of AMI1, set the result as RREFAI1 RREFAI1= %Collect the second half of the matrix as AInew1, set the remaining elements as AIold1 AInew1= AIold1 = %Check if AIold1 is an identity matrix, if it is, AInew1 is the inverse %Encode A2, b2 and xv2 as the vector of unknowns. A2 = b2 = syms xv2 = %Check the size of A2, set it as m2 and n2 [m2,n2] = %Augment A2 and b2 to form AM2 AM2 = %Solve the…Construct a square matrix with NN rows and NN columns consisting of nonnegative integers from 00 to 10^{18}1018, such that its determinant is equal to 11, and there are exactly A_iAi odd numbers in the ii-th row for each ii from 11 to NN, or report there isn't such a matrix. Standard input The first line contains a single integer NN. Each of the next NN lines contains a single integer A_iAi. Standard output If there is no solution, output \text{-}1-1. Otherwise, print NN lines, each consisting of NN integers, representing the values of the constructed matrix. If there are multiple solutions, print any. Constraints and notes 2 \le N \le 502≤N≤50 1 \leq A_i \leq N1≤Ai≤N For 40\%40% of the test files, N \le 17N≤17.
- Enter the 5 × 5 Hilbert matrix using MATLAB software 1 1/ 2 1/ 3 1/ 4 1/ 5 1/ 2 1/ 3 1/ 4 1/ 5 1/ 6 1/ 3 1/ 4 1/ 5 1/ 6 1/ 7 1/ 4 1/ 5 1/ 6 1/ 7 1/ 8 1/ 5 1/ 6 1/ 7 1/ 8 1/ 9 H i) Find the determinant of H. ii) Find the transpose and inverse of H iii) Using the commands in the text, find the dimensions of H, the column sums, and the row sums of H.Computer Science Given an N x N matrix M with binary entries i.e every entry is either 1 or 0. You are told that every row and every column is sorted in increasing order. You are required to output a pair (i,j) with 1 <= i and j <= n corresponding to the entry of the matrix satisfying Mij = 1 and Mrs = 0 for all 1 <= r <= i and 1 <= s <= j except for Mij Informally this includes the entry of M = 1 and is closest to the top left corner. for example: M = [ 0 0 0 1 0 0 1 1 0 0 1 1 0 0 1 1] output is (2,3) or (1,4) M = [ 0 1 1 1 1 1 1 1 1] output could be (1,2) or (2,1) Design a divide and conquer algorithm, explain correctness and runtime of the algorithm.Array P = [40, 30, 25, 10, 35, 5, 20] Suppose the dimension of 6 matrices (A1, A2 … A6) are given by array P A1 is a P[0] x P[1] matrix A2 is a P[1] x P[2] matrix . . . A6 is a P[5] x P[6] a) Find the minimum number of scalar multiplications necessary to calculate the product of all the 6 matrices (A1.A2.A3.A4.A5.A6) and show the parenthesization for this multiplication. • Solve the problem manually (you need not to write any code) using bottom-up tabulation approach. Compute and show the ‘m’ matrix and ‘s’ matrix to solve your problem.
- a)Please make an organized output of the species and stoichiometric data. matrix shape = (24, 23) b) Compute the rank of the stoichiometric matrix using your own algorithm. Is the matrix full rank? Is mass conserved for the first 3 reactions and the last 3 reactions? ExplainUsing Java Modify the program 1 in guided activity 3 to compute the multiplication of two vectors with the sizes 1xn and nx1. The result is a number (one element matrix) computed as the sum of the products of the corresponding elements from the first vector with the ones from the second vector: m1[0][0]m2[0][0]+m1[0][1]*m2[1][0]+...+m1[0][n-1]m2[n-1][0#plea# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…
- Let us consider multiplying a 5 by 5 sparse matrix with a 5 by 3 sparse matrix shown in picture. make Python code to implement multiplication of these two sparse matrices#com# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…## Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…
