Listen to each assistant's responses to the following instruction, and give each a score out of 100 and an explanation for the score. Here are some additional guidelines: * Do not award bonus points. * If the instruction includes additional context from which the response should be generated, deduct points if the response includes details not found within the provided context. * The assistants should not be making assumptions about the intent of the question. * If the instruction asks for results in a specific format, e.g. JSON, 50% of the score should be whether or not the assistant produced the correct format. * Award points for effort and problem formulation when the instruction is related to solving mathematical equations or coding. Respond in format "[assistant number]. [score]/100 - [explanation] Instruction: Write a MATLAB function that solves a system of linear equations using Gaussian elimination. Assistant 1's response: Here's a MATLAB function that solves a system of linear equations using Gaussian elimination: ```matlab function x = gaussian_elimination(A, b) % This function solves the system Ax = b using Gaussian elimination. % % Inputs: % A: an n x n matrix of coefficients % b: an n x 1 column vector of constants % % Output: % x: an n x 1 column vector of unknowns % Get the size of the coefficient matrix [n, ~] = size(A); % Augment the matrix with the constant vector Ab = [A, b]; % Perform Gaussian elimination for i = 1:n-1 % Find the pivot row max_row = i; for j = i+1:n if abs(Ab(j, i)) > abs(Ab(max_row, i)) max_row = j; end end % Swap the current row and the pivot row temp = Ab(i, :); Ab(i, :) = Ab(max_row, :); Ab(max_row, :) = temp; % Eliminate the entries below the pivot element for k = i+1:n factor = Ab(k, i) / Ab(i, i); Ab(k, :) = Ab(k, :) - factor * Ab(i, :); end end % Back-substitution to find the solution x = zeros(n, 1); x(n) = Ab(n, n+1) / Ab(n, n); for k = n-1:-1:1 x(k) = (Ab(k, n+1) - sum(Ab(k, k+1:n) .* x(k+1:n))) / Ab(k, k); end end ``` You can use this function to solve a system of linear equations like this: ```matlab A = [2, -1, 5; -1, 2, 2; 5, 2, 3]; b = [8; 7; 9]; x = gaussian_elimination(A, b); disp(x); ``` This will output the solution vector x: ``` 1.0000 -1.0000 3.0000 ``` Assistant 2's response: Here's a MATLAB function that solves a system of linear equations using Gaussian elimination: ``` function [x] = gaussian_elimination(A, b) % Perform Gaussian elimination [n, m] = size(A); for i = 1:(n-1) for j = (i+1):n m = A(j,i)/A(i,i); A(j,:) = A(j,:) - m*A(i,:); b(j) = b(j) - m*b(i); end end % Back-substitution to solve for x x = zeros(n,1); x(n) = b(n)/A(n,n); for i = (n-1):-1:1 x(i) = (b(i) - A(i,(i+1):n)*x((i+1):n))/A(i,i); end end ``` This function takes as input the coefficient matrix `A` and the column vector `b` of the system of linear equations `A*x = b`. It first performs Gaussian elimination to convert `A` into an upper triangular matrix, and simultaneously modifies `b` accordingly. Then, it performs back-substitution to solve for `x`. Finally, it returns the solution vector `x`. Assistant 3's response: Here is a simple MATLAB function that solves a system of linear equations using Gaussian elimination: ```matlab function x = solve_linear_system(A, b) [m, n] = size(A); disp(sprintf('The system has %d equations and %d unknowns.', m, n)); A = [A, repmat(b, 1, n)]; A = A'; pivots = zeros(n, 1); for k = 1:n pivot = 0; for i = 1:k-1 pivot = pivot + A(i, k) * pivots(i); end pivots(k) = pivot; end x = zeros(n, 1); for k = n:1 x(k) = -pivots(k) * x(k-1) / pivots(k-1); x(k) = x(k) + b(k); end end ``` To use the function, simply call it and pass in the coefficient matrix A and the right-hand side b, like this: ```matlab A = [1 2 3; 4 5 6; 7 8 9]; b = [10; 20; 30]; x = solve_linear_system(A, b); ``` The function will then solve the system of linear equations using Gaussian elimination and return the solution x. Assistant 4's response: Here is an example MATLAB function that solves a system of linear equations using Gaussian elimination: ``` function [x] = solve_linear_equations(A, b) m = size(A, 1); n = size(A, 2); if numel(b) ~= n error('Number of RHS vectors must match the number of equations.'); end % Gaussian elimination for i = 1:n for j = i:n if abs(A(i,j)) < abs(A(j,i)) temp = A(i,j); A(i,j) = A(j,i); A(j,i) = temp; temp = b(i); b(i) = b(j); b(j) = temp; end end end % Back substitution for j = n:-1:1 for i = n:-1:j+1 if abs(A(i,j)) < abs(A(j,i)) temp = A(i,j); A(i,j) = A(j,i); A(j,i) = temp; temp = b(i); b(i) = b(j); b(j) = temp; end end end % Undo column scaling for i = 1:n x(i) = b(i) / A(i,n); end end ``` This function takes the matrix `A` of coefficients and the right-hand side vectors `b` as inputs, and returns the solution vector `x`. The function first checks that the number of RHS vectors matches the number of equations. It then performs Gaussian elimination to eliminate the pivot columns from the matrix `A`, back-substitutes to solve for the remaining variables, and finally undoes the column scaling to return the solution vector `x`. Assistant 5's response: Here is a MATLAB function that solves a system of linear equations using Gaussian Elimination: ```css function [x] = solveLinearEquations(A, b) % Check if the matrix is square and of full rank if size(A, 2) ~= size(A, 1) || rank(A) ~= size(A, 1) error('Matrix must be square and of full rank'); end % Prepare the augmented matrix [c, d] = gemv(A, b); Aug = [0 1 c; 0 0 d]; % Perform Gaussian elimination [x] = reduce(Aug, 1); end ``` You can call this function by passing in the matrix `A` representing the system of linear equations and the vector `b` representing the right-hand side of the equations, like this: ```scss x = solveLinearEquations([1 2 3; 4 5 6], [10; 20; 30]); ``` This will return the vector `x` that solves the system of linear equations. Assistant 6's response: Here's an example MATLAB function that solves a system of linear equations using Gaussian elimination: ``` function [x] = gaussElim(A,b) % This function solves a system of linear equations using Gaussian elimination % Input: % A - coefficient matrix of the system % b - vector of constants % Output: % x - vector of solutions % The function first computes the row echelon form of A and then solves the system of equations using back substitution. n = size(A,1); for i=1:n-1 for j=i+1:n if abs(A(i,j)) > abs(A(j,i)) A(i,j) = -A(j,i); b(j) = -b(j); end end for k=i+1:n A(i,k) = A(i,k) / A(i,i); end end row = true(n,1); col = true(n,1); for i=1:n if ~row(i) for j=i+1:n if ~col(j) A(i,j) = -A(i,j); b(j) = -b(j); row(i) = false; col(j) = false; end end end row(i) = true; col(i) = true; end for i=1:n b(i) = b(i) / A(i,i); end x = A \ b; end ``` To use this function, simply call it with the coefficient matrix A and the vector of constants b as arguments: ``` [x] = gaussElim([1 2; 3 4], [5; 6]) ``` This would solve the system of equations: ``` [1 2; 3 4] * [x1, x2] = [5; 6] ``` and return the solutions: ``` x1 = 2 x2 = -1 ```