Question

please answer by giving Matlab code for this question

Solve the following linear system of equations. 

x₁+3 x₂-x₃+2 x₄+3 x₅=5

x₁+2 x₂-x₃+4 x₄+9 x₅=-3

3 x₁-5 x₂+3 x₃+2 x₄-x₅=7

2 x₂+x₃+2 x₅=-3

x₁-x₂+2 x₅=1

Using 

-LU Factorization. 

-Cramer's rule.

Answer 1

MATLAB code:

% Defining the equation given in matrix form

% Ax = b form

A = [1 3 -1 2 3; 1 2 -1 4 9; 3 -5 3 2 -1; 0 2 1 0 2; 1 -1 0 0 2];

b = [5; -3; 7; -3; 1];

% Finding LU decomposition of A matrix

[L , U] = lu(A);

% To get solution using LU decomposition

% We can modify the solution as LUx = b

% Lets assume Ux = y

% then we will have Ly = b

% we will solve 'y' first and then x

y = inv(L) * b;

x_lu = inv(U) * y;

% Displaying the result

disp("The value of x using LU decomposition are")

disp(x_lu);

% Part b)

% using cramers rule

det_A = det(A);

det_x = [det([b A(: , 2:end)]); det([A(: , 1) b A(: , 3:end)]); det([A(: , 1:2) b A(: , 4:end)]); ...

mdet([A(: , 1:3) b A(: , 5)]); det([A(: , 1:4) b])];

x_c = det_x ./ det_A ;

% Displaying the result

disp("The value of x using cramer method are")

disp(x_c);

1 2 3 % Defining the equation given in matrix form % Ax = b form A = [1 3 -1 2 3; 1 2 -1 4 9; 3-5 3 2 -1; 0 2 1 0 2; 1 -1 0 0 2]; b = [5; -3; 7; -3; 1]; % Finding LU decomposition of A matrix [L, U] = lu(A); 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 - 24 25 26 - 27 28 29 % To get solution using LU decomposition % We can modify the solution as LUX = b % Lets assume Ux = y % then we will have Ly = b % we will solve 'y' first and then x y = inv(L) *b; x_lu = inv(0) * y; % Displaying the result disp("The value of x using LU decomposition are") disp(x_lu); % Part b) % using cramers rule det_A = det(A); det_x = [det([b A: , 2:end)]); det([A(:,1) b A: , 3:end)]); det([A(:, 1:2) b A(: , 4:end)]); ... mdet([AC: 1:3) b A: , 5)]); det([A(:, 1:4) b])]; X_C = det_x ./ det_A ; % Displaying the result disp("The value of x using cramer method are") disp(X_C);

OUTPUT:

The value of x using LU decomposition are 4.4286 0.7013 -1.6753 0.4416 -1.3636 The value of x using cramer method are 4.4286 0.7013 -1.6753 0.4416 -1.3636

Answer 2

nice code but check line 25 of the code image and remove m from mdet......