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Emre Instructor: Mengi Math 170A (Winter 2009) Study Guide for Week 4 Before attempting homework 4 please make sure that you have understood the topics listed below. We mainly focused on the Gaussian elimination and LU factorization (section 1.7 in the textbook) during the fourth week. Augmented matrix (section 1.7, page 71) Three row operations and the fact that they preserve the solution set of a linear system when applied to the associated augmented matrix (section 1.7, pages 70-71) The row-reduction of the coe cient matrix into an upper triangular form (this process is called Gaussian elimination, section 1.7, pages 71-76) Computation of an LU factorization assuming no row-interchange is needed during the rowreduction process (section 1.7, pages 76-83) How to use the LU factorization to solve a linear system (see question 3 below, section 1.7, page 79) Homework 4 (due on February 11th, Wednesday by 3pm) In Matlab question (question 3) attach the Matlab outputs. Also please include a print-out of your Matlab code for the LU factorization. Question 1 is possibly a challenging question. It may be a good idea to attempt it in the end. 1.(*) Recall the row reduction process for LU factorization. Suppose A is an n n matrix and reducible to an upper triangular matrix U by only applying row-replacement operations (which the textbook calls type 1 row operations). To row-reduce A into U we proceed column by column from left to right. At the kth step we process column k by making the entries below the diagonal entry on the kth column zero by means of the row-replacement operations. Let A(k) be the matrix obtained from A after processing columns 1, 2, . . . , k 1 and just before processing the kth column. The row-reduction from A into U is depicted below. (Below an x corresponds to an entry possibly non-zero.) 2 6 6 6 6 A=6 6 6 4 | x x x . . . x x x x x . . . x x x x x . . . x x ... ... ... x x x . . . x x x x x . . . x x 3 2 x 0 0 . . . 0 0 x x x . . . x x x x x . . . x x ... ... ... x x x . . . x x x x x . . . x x 3 2 x 0 0 . . . 0 0 x x 0 . . . 0 0 x x x . . . x x ... ... ... x x x . . . x x x x x . . . x x 3 7 7 7 7 7 7 7 5 } 2 6 6 6 6 6 6 6 4 | x 0 0 . . . 0 0 x x 0 . . . 0 0 x x x ... ... .. 0 0 . x 0 x x x x x x . . . x x 3 7 7 7 7 7=U 7 7 5 } A(1) ... ... {z 6 7 6 7 6 7 6 7 6 7 6 7 6 7 4 5 } | A(2) ... ... {z 6 7 6 7 6 7 6 7 6 7 6 7 6 7 4 5 } | A(3) ... ... {z A(n) ... {z De ne the multiplier mjk := ajk /akk for j > k. Equivalently mjk is the scalar such that aj (k+1) (k) (k) = aj mjk ak . (k+1) (k) (k) (1) (That is mjk is chosen to set the entry ajk = 0 when processing the kth column. This is achieved by multiplying the kth row by mjk and subtracting it from the jth row.) Note that aj denotes the jth row of A(k) . Denote also the jth rows of A and U by aj and uj , (1) respectively. By de nition A = A , so for all j = 1, . . . , n it follows that aj = aj . (1) (k) (2) A more subtle observation is that the jth row is updated only at steps k = 1, . . . , j 1, but not modi ed at the jth and later steps meaning for j = 1, . . . , n we have aj = u j . By using (1) equation and properties (2) and (3) prove by induction (on k) that k 1 (k) aj (j) (3) = aj mj1 u1 mj2 u2 mj(k 1) uk 1 = aj i=1 mji ui (4) for any j k. Remark: We used equation (4) in class (with 1 0 a1 a2 m21 1 a3 m31 m32 . = . . . . . an 1 m(n 1)1 m(n 1)2 mn1 mn2 an A k = j) to deduce the LU factorization of A, 0 ... 0 u1 0 ... 0 u2 1 0 u3 . . .. . . . . . 1 0 un 1 un mn(n 1) 1 L U 2. Consider the linear systems 2 2 2 0 2 3 0 5 2 4 4 x = 4 and 4 4 3 x = 7 . 2 4 6 6 0 4 4 8 A1 b1 A2 b2 (a) Compute the LU factorizations of A1 and A2 by hand. (b) Exploiting the LU factorizations from part (a) solve the linear systems A1 x = b1 and A2 x = b2 . (Note : you should only solve triangular system; more precisely rst a lower triangular system by forward substitution followed by an upper triangular system by back substitution. See question 3 below for details.) (c) Compute the determinants of A1 and A2 by exploiting their LU factorizations. 3. Given a non-singular n n dense matrix A (dense in this context means most of the entries, for instance more than half of the entries, of A are nonzero). The standard approach to solve the linear system Ax = b is as follows. 1. Compute the LU factorization A = LU . 2. Let x := U x. Solve the lower triangular system L = b by forward substitution. x 3. Solve the upper triangular system U x = x by back substitution. (a) Write a Matlab function to compute the LU factorization of A. Assume that A can be reduced into an upper triangular matrix U by only applying row-replacement operations. Your function must take a matrix A as input and return a lower triangular matrix L and an upper triangular matrix U . Your Matlab code should look like function [L,U] = lu_factor(A) [n,n1] = size(A); if n ~= n1 error( A must be square ); end L = eye(n); U = zeros(n,n); .... return; In the above code you need to ll in the part in between the statements U = zeros(n,n); and return; The statement U = zeros(n,n); initializes U to the n n zero matrix. You only need to modify the entries of U above and on the main diagonal, that is U (i, j), i j. The statement L = eye(n); sets L to the n n identity matrix initially. You need to modify only the portion of this matrix below the main diagonal, that is L(i, j), i > j. (b) Write another Matlab function to solve the system Ax = b following the steps described above. Here is how your function should look like function x = linear_sys_solver(A,b) [L,U] = lu_factor(A); xhat = forwardsubstitute(L,b); x = backsubstitute(U,xhat); return; (c) Solve the linear systems A1 x = b1 and A2 x = b2 given in question 2 by using the Matlab function linear sys solver that you implemented in part (b). Compare the solutions returned by your Matlab function with the hand-computed solutions from 2.(b).

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