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The code below is designed to take in an Augmented matrix (Concatenating A and B) and produce the linear combination of the system in terms

The code below is designed to take in an Augmented matrix (Concatenating A and B) and produce the linear combination of the system in terms of the free variables. The free variables will be given the variables t1, t2, ..., tn for the nth free variable. However, the code is not working properly. The code needs to be fixed so that it matches the input of the expected outputs below. The code should be able to output the linear combination of the system in terms of the free variables regardless of the size of the matrix, for example, a 1 x 10, 5 x 5, or 7 x 3. Look at the 4 test cases below and make sure that the coed could output the linear combination as a whole with its column matrices and free variables.

Code:

A = [2 3 4 5 6; 1 2 5 4 3; 4 5 2 1 6];

B = [0; 0; 0];

Aug = [A, B]

[basic_cols, non_basic_cols, num_basic, num_non_basic] = augmented_rref(Aug);

function [basic_cols, non_basic_cols, num_basic, num_non_basic] = augmented_rref(Aug)

% Function to find the basic and non-basic columns of an augmented matrix

% Input: Augmented matrix, with last column as B from the left side

% Output: Basic and non-basic columns in short and long form, and their counts

[m, n] = size(Aug);

RREF = rref(Aug);

pivots = [];

% Finding pivot columns

for i = 1:m

for j = 1:n

if RREF(i,j) == 1 && ~ismember(j, pivots)

pivots = [pivots, j];

break

end

end

end

basic_cols = cell(1, length(pivots));

non_basic_cols = cell(1, n-length(pivots));

num_basic = length(pivots);

num_non_basic = n-length(pivots);

% Separating basic and non-basic columns

for i = 1:n

if ismember(i, pivots)

basic_cols{i} = ['X', num2str(i)];

else

non_basic_cols{i-length(pivots)} = ['X', num2str(i)];

end

end

% Finding the solution in terms of free variables

solution = zeros(n-1,1);

free_cols = setdiff(1:n-1,pivots);

if ~isempty(free_cols)

for i = free_cols

solution(i) = 0;

for j = pivots

if RREF(find(pivots==j),i) ~= 0

solution(i) = solution(i) - RREF(find(pivots==j),i) * solution(j);

end

end

solution(i) = solution(i) + RREF(end,i);

end

end

% Printing the solution in printable form

solution_str = [];

for i = 1:length(free_cols)

solution_str = [solution_str, 't', num2str(i), '['];

for j = 1:n-1

if ismember(j,pivots)

solution_str = [solution_str, num2str(RREF(find(pivots==j),free_cols(i))), ';'];

elseif j == n-1

solution_str = [solution_str, num2str(RREF(end,free_cols(i))), ']'];

else

solution_str = [solution_str, '0;'];

end

end

if i ~= length(free_cols)

solution_str = [solution_str, ' + '];

end

end

solution = solution_str;

disp(['Solutions: ', mat2str(solution_str)]);

% Outputting results

disp(['Augmented Matrix: ', mat2str(Aug)]);

disp(['RREF: ', mat2str(RREF)]);

disp(['Basic Columns: ', mat2str(Aug(:, pivots))]);

disp('Short Form:');

disp(['Basic Columns: {', strjoin(basic_cols(~cellfun('isempty',basic_cols)), ', '), '}']);

disp(['Non-Basic Columns: {', strjoin(non_basic_cols(~cellfun('isempty',non_basic_cols)), ', '), '}']);

disp('Long Form:');

disp(['Basic Columns: ', mat2str(Aug(:, pivots))]);

disp(['Non-Basic Columns: ', mat2str(Aug(:, setdiff(1:n, pivots)))]);

disp(['Number of Basic Columns: ', num2str(num_basic)]);

disp(['Number of Non-Basic Columns: ', num2str(num_non_basic)]);

end

Test Cases

1) Input:

A = [4 8; 1 7; 2 4]

B = [12;8;6]

Output:

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2)

Input:

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Code form:

A = [2 3 4 56 6 8;1 2 4 5 6 7;3 6 12 15 18 21;4 6 8 112 12 16]

B = [9;8;24;18]

Output:

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3)

Inputs

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Code form:

A = [3 4 5;1.5 2 2.5]

B = [1;0.5]

image text in transcribed

4)

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4128741286 1) Given the RREF of A is 100010110 this gives the general solution is =x1x2x3=t1t1t1=t1111 213432664412856515112661812872116982418 2) Given the RREF of A is 100001004400974600660056006700 this gives the general solution is = x1x2x3x4x5x6x7=4t197t2+6t3+5t4+6t54t1+46t26t36t47t5t1t2t3t4t5=t14410000+t2974601000+t36600100+t45600010+t56700001 (31.54252.5121) 3)Given the RREF of A is : [10340350310] this gives the general solution is =34t135t231t3t1t2t3=t134100+t235010+t331001 Input: AB=214325452541636=000 Expected Output: Code form: t1[7;-6;1;0;0] +t2[-5;3;0;-1;1] x1x2x3x4x5=7t16t1+t1t2t25t23t2=t176100+t253011

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