LU Decomposition 0 solutions submitted (max: Unlimited) This problem has four pa

Question

LU Decomposition 0 solutions submitted (max: Unlimited) This problem has four parts. Develop one MATLAB function which outputs the solutions of all parts. (1) Solve the following system of equations using LU factorization with partial pivoting 2x 1-6x 2- x 3 40 -3x 1- x 2 6x 3 - 20 -8x1 + x2-2x3 = 50 Output the lower triangular matrix L, the upper triangular matrix U, the permutation matrix P, the intermediate solution vector d (see Chapra Sec. 10.2.1: Step 2.), and the final solution vectorx. Hint You may use he MATLAB function LuP = u A , which returns lower nangular matrix L upper nangular mat xuan mutation matrix s a p A 2) Evaluate the inverse invA of the matrix of the coefficients A using LU decomposition. Output also the intermediate matrix D-LIP. 3) Evaluate the determinant of A using the LU factors. Output also the determinant or the permutation matrix detP 4) Evaluate the condition number using the Frobenius norm of both the matrix A and its inverse invA you have evaluated above. - - Your Function 1 function [L,U,P,d,x,D,invA, detP,detA, condAHUFactorization() 21%% Output 3 % L: lower triangular coefficients (3x3 matrix) 4 % U: upper triangular coefficients (3x3 matrix) 51% d: intermediate solution (3x1 column vector) 61% x: final solution for unknowns (3x1 column vector) 8 % invA: inverse of matrix A (3x3 matrix) 91% detp: determinant of matrix 1e% detA: determinant of matrix A % condA: condition number of matrix A using Frobenius norm % krite your code here.

Explanation / Answer

1) SAVE THE FOLLOWING CODES IN MATLAB AND GET THE RESULTS ACCORDINGLY-

SAVE THE FUNCTION FILE IN MATLAB-

Part (A)

function[L,U,X]=LU_Parker(A,B)

[m,n]=size(A);

if (m ~= n )

disp ( 'LR2 error: Matrix must be square' );

return;

end;

% Part 2 : Decomposition of matrix into L and U

L=zeros(m,m);

U=zeros(m,m);

for i=1:m

% Finding L

for k=1:i-1

L(i,k)=A(i,k);

for j=1:k-1

L(i,k)= L(i,k)-L(i,j)*U(j,k);

end

L(i,k) = L(i,k)/U(k,k);

end

% Finding U

for k=i:m

U(i,k) = A(i,k);

for j=1:i-1

U(i,k)= U(i,k)-L(i,j)*U(j,k);

end

end

end

for i=1:m

L(i,i)=1;

end

% Program shows U and L

U

L

% Now use a vector y to solve 'Ly=b'

y=zeros(m,1);

y(1)=B(1)/L(1,1);

for i=2:m

y(i)=-L(i,1)*y(1);

for k=2:i-1

y(i)=y(i)-L(i,k)*y(k);

y(i)=(B(i)+y(i))/L(i,i);

end;

end;

% Now we use this y to solve Ux = y

x=zeros(m,1);

x(1)=y(1)/U(1,1);

for i=2:m

x(i)=-U(i,1)*x(1);

for k=i:m

x(i)=x(i)-U(i,k)*x(k);

x(i)=(y(i)+x(i))/U(i,i);

end;

end

2)SAVE THE FOLLOWING SCRIPT FILE IN MATLAB AND HIT ENTER-

Part (B)

>> A=[2 -6 -3;-3 -1 6;-8 1 -2];

>> B=[40;20;50];

>> LU_Parker(A,B);

Part (C)

A_inverse=inv(A);

Part (D)

A_determinant=det(A);

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