HW5

HW#5 Problem 1. We need to find p 3 given p 0 = [ 0 0 1 ] p 1 = [ 0 0 1 ] [ 0.95 0 0.05 0.25 0.5 0.25 0.1 0 0.9 ] ¿ [ 0 ∗ 0.95 + 0 ∗ 0.25 + 1 ∗ 0.1 0 ∗ 0 + 0 ∗ 0.5 + 1 ∗ 0 0 ∗ 0.05 + 0 ∗ 0.25 + 1 ∗ 0.9 ] ¿ [ 0.1 0 0.9 ] p 2 = [ 0.1 0 0.9 ] [ 0.95 0 0.05 0.25 0.5 0.25 0.1 0 0.9 ] ¿ [ 0.1 ∗ 0.95 + 0 ∗ 0.25 + 0.9 ∗ 0.1 0 0.1 ∗ 0.05 + 0 ∗ 0.25 + 0.9 ∗ 0.9 ] ¿ [ 0.185 0 0.815 ] p 3 = [ 0.185 0 0.815 ] [ 0.95 0 0.05 0.25 0.5 0.25 0.1 0 0.9 ] ¿ [ 0.25725 0 0.74275 ] p 3 ¿ [ 0.25725 0 0.74275 ] Problem 2 . [ 2 0 1 0 2 0 0 1 0 2 1 0 3 1 0 0 ][ 29 5 3 8 ] R 3 – 1 2 R 1 [ 2 0 0 0 2 0 − 1 1 0 2 1 0 3 1 − 1.50 0 ][ 29 5 − 11.5 8 ]

R 4  R 2 [ 2 0 0 0 2 1 − 1 0 0 0 1 2 3 0 − 1.50 1 ][ 29 8 − 11.5 5 ] R 3 + R 2 [ 2 0 0 0 2 1 0 0 0 0 1 2 3 0 − 1.50 1 ][ 29 8 − 3.5 5 ] R 4 − ¿ 2R 3 [ 2 0 0 0 2 1 0 0 0 0 1 0 3 0 − 1.50 4 ][ 29 8 − 3.5 12 ] Back-substitution { 2 x 1 + 2 x 2 + 3 x 4 = 29 x 2 = 8 x 3 − 1.5 x 4 =− 3.5 4 x 4 = 12 { x 1 = 29 − 2 ∗ 8 − 3 ∗ 3 2 = 2 x 3 = 1.5 ∗ 3 − 3.5 = 1 x 4 = 3 x 2 = 8 Amount of each compound contained in the sample. NaMgSi 2 O 6 2 CaFeSi 2 O 6 8 NaMgAl 2 O 6 1 KALSi 3 O 6 3 Problem 3 .

Problem 4. function [ x ] = gauss_elim( A, b ) %%gauss_elim uses Gauss elimination with partial pivoting to solve Ax = b %%Inputs: A - A matrix % b - b vector %%Outputs: x - solution to Ax = b %%Author: Aleksandr Mnatsakanyan n = size(A,1); % the size of the A matrix x = zeros(n,1); % preallocation of the x vector sum = 0; % used for back-substitution, could have used backslash but did it this way anyways for i = 1:n-1 % for loop to go through all the columns up to one before the last one [~,max_val] = max(abs(A(i:n,i))); % getting the index of the biggest absolute value of the column i max_val = max_val + i - 1; % correcting the index % if the max value index matches the current i value that means that % our pivot element is the greatest if not, we need to pivot if max_val ~= i pivot = [i max_val]; % creating the pivot vector else pivot = [i i]; % creating the pivot vector

end if pivot(1) ~= pivot(2) % if pivoting is needed, we create temporary variables for A and b TempA = A; Tempb = b; % then we use them as reference for the new pivoted A and b A(pivot(1),:) = TempA(pivot(2),:); A(pivot(2),:) = TempA(pivot(1),:); b(pivot(1)) = Tempb(pivot(2)); b(pivot(2)) = Tempb(pivot(1)); end % eliminating the variables below the current pivot element for ii = i+1:n b(ii) = b(ii)-(A(ii,i)/A(i,i)).*b(i); A(ii,:) = A(ii,:)-(A(ii,i)/A(i,i)).*A(i,:); end end % backsubstitution for k = n:-1:1 for kk = 1:n sum = sum + A(k,kk)*x(kk); % this is the sum of all elements of a row other than the needed x(i) so the row looks like sum + a(ij) * x(i) = b(i) end x(k) = (b(k) - sum)/A(k,k); % solving the row for x(i) sum = 0; % resetting the value of the sum end Problem 5. function [ L, U, P ] = lu_fact( A ) %lu_fact performs LU factorization with partial pivoting %Input: A - A matrix to be factored %Outputs: L - lower-triangular matrix % U - upper-triangular matrix % P - permutation matrix associated with pivots %Author: Aleksandr Mnatsakanyan L = diag(ones(size(A,1),1)); P = L; n = size(A,1); % this part of the code is the same as gauss_elim with the addition of an % if statement. for i = 1:n-1 [~,max_val] = max(abs(A(i:n,i))); max_val = max_val + i - 1; if max_val ~= i pivot = [i max_val]; else

Problem 7. function [norm_val] = vector_norm(x,normnum) %%calc_norm computes the 1-, 2-, or infinity-norm of a vector %%Inputs: x - vector whose norm is to be calculated % normnum - which norm is to be computed %%Output: norm_val - computed norm value %%Author: Aleksandr Mnatsakanyan % I think this code is pretty self-explanatory if normnum == 1 norm_val = sum(abs(x)); elseif normnum == 2 norm_val = sqrt(sum(x.^2)); else norm_val = max(abs(x)); end end