blä 5 Find the SUM for mean of the matrix *? F F= 1 1 1 1 2 2 2 3 3 4 4 4 4 sum(mean (F)) sum (sum F) mean(mean(F)) bläs 5 If A = [ 1450 26] and the MATLAB :Code A (4) =[], then the result is A = 14 526 A = 15026 A = 4
Q: 3-5 6. Evaluate the determinant for the following matrix: O A. 8 о В. -2 O C.5 O D. -4
A: The Answer is in step-2.
Q: 3-5 Evaluate the determinant for the following matrix: 1 O A. 8 О В.-2 O C.5 O D. -4
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- 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.Consider the matrix A defined as: [1 0 1] A=[1 2 2] [1 2 1] 121 Compute the determinant of A. Detail each of the computation steps. Compute the inverse of A: A−1 (refer to ”Wiki: adjoint matrix” for the computation of the adjoint matrix A∗).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.
- #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]…Pls Use Python Using NumPy, write the program that determines whether the A=({{1, 5, -2}, {1, 2, -1}, {3, 6, -3}}) matrix is nilpotent. Itro: Nilpotent Matrix: A square matrix A is called nilpotent matrix of order k provided it satisfies the relation, Ak = O and Ak-1≠O where k is a positive integer & O is a null matrix of order k and k is the order of the nilpotent matrix A . The following picture is an example of the intro: Ps: Please also explain step by step with " # "## 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]…
- #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]…Matlab screenshot - If x=[ 2 6 12; 15 6 3; 10 11 1], then a) replace the first row elements of matrix x with its average value. b) reshape this matrix into row vector.Write a MATLAB program to solve(i) an arbitrary matrix problem A= M*N where M and N arematrices in which you need to input the numbers onyour screen. Note M and N has a size mxm.(ii) Finding an inverse of the matrix
- Using 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][0Construct 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.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;…