Uploaded by Hung Duy Ha (Hiro)

MATLAB Code: Linear Independence of Vectors

advertisement
Q1:
function homework1()
m = 10;
n = 10;
for i = 1:5
vectors = generate_vectors(m, n);
result = check_linear_independence(vectors);
disp(['Case I: ', result]);
end
m = 10;
n = 9;
for i = 1:5
vectors = generate_vectors(m, n);
result = check_linear_independence(vectors);
disp(['Case II: ', result]);
end
m = 10;
n = 15;
for i = 1:5
vectors = generate_vectors(m, n);
result = check_linear_independence(vectors);
disp(['Case III: ', result]);
end
end
function vectors = generate_vectors(m, n)
vectors = cell(1, n);
for i = 1:n
vectors{i} = rand(m, 1);
end
end
function result = check_linear_independence(vectors)
matrix = cell2mat(vectors);
if rank(matrix) == length(vectors)
result = 'Linearly independent';
else
result = 'Linearly dependent';
end
end
Q2:
1.
w1 = [1; 2; -1; 3];
w2 = [4; 1; 1; 8];
w3 = [1; 0; 2; 2];
w4 = [-1; 1; 2; -1];
A = [w1, w2, w3, w4];
[~, R] = rref(A);
maximal_li_set = A(:, any(R, 1));
dimension_S1 = rank(A);
n = size(A, 1);
is_hyperplane = (dimension_S1 == n-1);
is_plane = (dimension_S1 == n-2);
disp('Maximal linearly independent set:');
disp(maximal_li_set);
disp(['S1 is a hyperplane: ', num2str(is_hyperplane)]);
disp(['S1 is a plane: ', num2str(is_plane)]);
2.
A = [1 4 1 -1; 2 1 0 1; -1 1 2 2; 3 8 2 -1];
augmented_matrix = [A, [3; 4; 0; 8]];
rref_matrix = rref(augmented_matrix);
disp(rref_matrix)
3.
% Define the vectors
z1 = [2; 4; -2; 6];
z2 = [1; 0; 2; 2];
z3 = [3; 4; 0; 8];
% Form the matrix A
A = [z1, z2, z3];
% Compute the reduced row echelon form (RREF)
rref_matrix = rref(A);
disp(rref_matrix)
4.
Download