澳洲IT编程assignment代写 计算机代写

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  •  澳洲IT编程assignment代写 计算机代写
    5.  X(end)
    7. size (X)
    10. They are linearly dependent
    11. length (u)
    12.
    12-a)
    u'     % u transpose
    12-b)
    u * v  % ERROR: u and v are both column vectors
    12-e)
    u .* v % element by element multiplication
    13. sum(X) % summation of each column vector of X
    14. Read help for function 'sum'
     澳洲IT编程assignment代写 计算机代写
    sum(X') 
    sum(X,1) % summation of first dimmension, row vector
    sum(X,2) % summation of second dimmension, column vector
    15. mean(X) % mean of each column vector
    16. log(X)  % logarithm of each element of X
    17. abs(X)  % absolute value of each element of X
    19. X(:,2) + v'
    20. [1 1] * X % summation of each column of X
    22. From the table we can construct a matrix (vitamin contents in different food types, observe the link between traditional methods and matrix methods.
    C = [0.5 0.5 0 0; 0.3 0 0.2 0.1; 0.1 0.1 0.2 0.5];
    Consumption of each food, a vector q = [5 10 8];
    Answer for the first question: q * C
    Prices for each vitamin: p = [10 20 25 50];
    Answer for the second question: p * C'
    23. Similar to 22
    Resources matrix of each style
    R = [5 20 16 7 17; 7 18 12 9 21; 6 25 8 5 13];
    Orders
    v = [5 7 12]
    Answer to the first question: v * R
    Costs matrix for purchase and transportation: C = [15 4.5; 8 2; 5 3; 1 0.5; 10 0];
    % Answer to the second question: Quantities * Costs
    24. D^k becomes a zero matrix if k goes to infinity
    25. the power of a diagonal matrix is a diagonal matrix composed by the power of it's diagonal elements
    26. [u v] % horizontal concatenation, u and v must have same number of rows
    27. U = diag(u) elements of vector u become the diagonal elements of matrix U
    28. Read the help for function diag
    diag(u, -1): puts elements of u on the first diagonal above the main diagonal
    diag(u, 1): puts elements of u on the first diagonal below the main diagonal
    29.  
    a) [x2 x3 x4 0]
    b) [x3 x4 0 0]
    c) [x4 0 0 0]
    d) [0 0 0 0]
    e) [-1 -1 -1 x4]
    30.
    a) [x1+x2+x3+x4    x2+x3+x4     x3+x4     x4]
    b) [x1+2x2+3x3+4x4    x2+2x3+3x4     x3+2x4     x4]
    c) [x1+3x2+6x3+10x4    x2+3x3+6x4     x3+3x4     x4]