solutions to HW #2

# solutions to HW #2 - ChemE 109 Numerical and Mathematical...

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ChemE 109 - Numerical and Mathematical Methods in Chemical and Biological Engineering Solution to Homework Set 2 1. (10 points) From the deﬁnition of a determinant, one can show the following properties: Property 1: A determinant changes in sign but not in magnitude with an interchange of a pair of rows or a pair of columns. Property 2: A determinant remains unchanged by additions of multiples of one row (column) to another row (column). Using these properties, show that if A 0 is the upper triangular matrix formed during the solution of a system of equations Ax = c via Gauss elimination with partial pivoting, then: det A = ( - 1) I det A 0 = ( - 1) I n Y i =1 a 0 ii where I is the number of row interchanges. Proof: In Gauss elimination, the matrix A is undergoing a series of elementary transformations as follows: A A 1 A 2 A 3 ⇒ ··· ⇒ A 0 The transformations can be either interchange of rows(columns) or additions of multiples of one row (column) to another row (column). The A s are the intermediate matrices during the transformations. From Properties 1 and 2, the absolute values of the determinants of these A s are equal. Since only an interchange changes the sign of the determinant and there are totally I times of interchanges, the relationship between the determinants of A and A 0 is det A = ( - 1) I det A 0 The determinant of A 0 is the multiplication of its diagonal elements due to the upper triangular structure: det A 0 = n Y i =1 a 0 ii Therefore the proof is completed. 2. (10 points) (a) Create a program using MATLAB which performs LU decomposition with partial pivoting . Sample program: function [x] = LUSolver(A, b) n=size(A,1); %find the size of A U=A; %Start the U matrix as A 1

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L=eye(size(A)); %Start the L matrix as identity P=eye(n); %Start the permutation matrix as identity y=zeros(size(b));
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solutions to HW #2 - ChemE 109 Numerical and Mathematical...

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