Basic matrix operations on Matlab

Matlab - basic matrix operations

%Question 1..

clear all
close all
clc

A = [20 -10 0 0 0 0 0 0
    -10 20 -10 0 0 0 0 0
    0 -10 15 -5 0 0 0 0
    0 0 -5 10 -5 0 0 0
    0 0 0 -5 20 -15 0 0
    0 0 0 0 -15 30 -15 0
    0 0 0 0 0 -15 30 -15
    0 0 0 0 0 0 -15 15];

B = [100
    75
    100
    50
    75
    50
    100
    75];

n = length(B); %number of equations and unknowns in this script (q1)

%Calculation of the solution

%  LU Decomposition method 
%(Initialising U and L....)
U = A;
L = eye(n);

% > LU Decomposition method <
%(step 1: Forward Elimination)

for k = 1 : n-1
    for i = k+1 : n
        L(i,k) = U(i,k) / U(k,k);
        U(i,:) = U(i,:) - L(i,k) * U(k,:);
    end
end

% >> LU Decomposition method <<
%(step 2: Forward Substitution)

D = zeros(n,1); % initialise solution vector
D(1) = B(1);
for i = 2 : n
    D(i) = ( B(i) - L(i,1:i-1) * D(1:i-1) );
end

% >>> LU Decomposition method <<<
%(step 3: Back Substitution)

X = zeros(n,1); % initialise solution vector
X(n) = D(n) / U(n,n);
for i = n-1 : -1 : 1
    X(i) = ( D(i) - U(i,i+1:n) * X(i+1:n) ) / U(i,i);
end

%Print solution

fprintf('The calculated displacements of the %d masses are: \n', n)
for i = 1:n
    fprintf('X%d = %6.2f mm\n', i, X(i))
end



%question2
%%  Read or Input any square Matrix
A = [20 -10 0 0 0 0 0 0
    -10 20 -10 0 0 0 0 0
    0 -10 15 -5 0 0 0 0
    0 0 -5 10 -5 0 0 0
    0 0 0 -5 20 -15 0 0
    0 0 0 0 -15 30 -15 0
    0 0 0 0 0 -15 30 -15
    0 0 0 0 0 0 -15 15];% coefficients matrices

C = [100
    75
    100
    50
    75
    50
    100
    75];% constants vectors
n = length(C);
X = zeros(n,1);
Error_eval = ones(n,1);

%% Checking if the matrix A is diagonally dominant
for i = 1:n
    j = 1:n;
    j(i) = [];
    B = abs(A(i,j));
    Check(i) = abs(A(i,i)) - sum(B); % Is the diagonal value greater than the remaining row values combined?
    if Check(i) < 0
        fprintf('The matrix is not strictly diagonally dominant at row %2i\n\n',i)
    end
end

%% Starts the Iterative method

iteration = 0;
while max(Error_eval) > 0.00001
    iteration = iteration + 1;
    Z = X;  % save current values to calculate error later
    for i = 1:n
        j = 1:n; % define the array of the coefficients' elements
        j(i) = [];  % eliminate the unknow's coefficient from the remaining coefficients
        Xmass = X;  % copy the unknows to a new variable
        Xmass(i) = [];  % eliminate unknowns under question from the set of values
        X(i) = (C(i) - sum(A(i,j) * Xmass)) / A(i,i);
    end
    Xsolution(:,iteration) = X;
    Error_eval = sqrt((X - Z).^2);
end
%Print solution

fprintf('The calculated displacements of the %d masses are: \n', n)
for i = 1:n
    fprintf('X%d = %6.2f mm\n', i, X(i))
end

%question 3
%Coefficent of matrix A 
A = [20 -10 0 0 0 0 0 0
    -10 20 -10 0 0 0 0 0
    0 -10 15 -5 0 0 0 0
    0 0 -5 10 -5 0 0 0
    0 0 0 -5 20 -15 0 0
    0 0 0 0 -15 30 -15 0
    0 0 0 0 0 -15 30 -15
    0 0 0 0 0 0 -15 15];

n = length(A);

%Factorisuing A...

% >>> Using the LU Decomposition method <<<
%(Initialise U and L)
U = A;
L = eye(n);

% >>> Using the LU Decomposition method <<<
%(step 1: Forward Elimination)

for k = 1 : n-1
    for i = k+1 : n
        L(i,k) = U(i,k) / U(k,k);
        U(i,:) = U(i,:) - L(i,k) * U(k,:);
    end
end

% The solution is calculated with B

B =[100
    75
    100
    50
    75
    50
    100
    75] ;

% >>> Using the LU Decomposition method <<<
%(step 2.1: Forward Substitution)

D = zeros(n,1); % initialise the solution vector
D(1) = B(1);
for i = 2 : n
    D(i) = ( B(i) - L(i,1:i-1) * D(1:i-1) );
end

% >>> LU Decomposition method <<<
%(step 2.2: Back Substitution)

X_B = zeros(n,1); % initialise solution vector
X_B(n) = D(n) / U(n,n);
for i = n-1 : -1 : 1
    X_B(i) = ( D(i) - U(i,i+1:n) * X_B(i+1:n) ) / U(i,i);
end

%The solution is calcuated with C

for W_6 = linspace(70,130,15);
    C = [100
    75
    100
    50
    75
    W_6
    100
    75];

% >>> Using the LU Decomposition method <<<
%(step 2.1: Forward Substitution)

D = zeros(n,1); % initialise solution vector
D(1) = C(1);
for i = 2 : n
    D(i) = ( C(i) - L(i,1:i-1) * D(1:i-1) );
end

% >>> Using the LU Decomposition method <<<
%(step 2.2: Back Substitution)

X_C = zeros(n,1); % initialise solution vector
X_C(n) = D(n) / U(n,n);
for i = n-1 : -1 : 1
    X_C(i) = ( D(i) - U(i,i+1:n) * X_C(i+1:n) ) / U(i,i);
end

%Printing solutions
fprintf('The calculated displacements of the %d masses are: \n', n)
for i = 1:n
    fprintf('X%d = %6.2f mm (with B), and X%d = %6.2f mm (with C)\n', i, X_B(i), i, X_C(i))
end
end