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Unformatted text preview: E3101: A study guide and review, Version 1.5 Marc Spiegelman December 7, 2004 Here is a list of subjects that I think weve covered in class (your mileage may vary). If you understand and can do the basic problems in this guide you should be in very good shape. This guide is probably over-thorough. The test itself will have about 6-7 questions covering the whole course but emphasizing the material after the midterm (which implicitly includes all of the previous material). Ill try to avoid anything overly tricky. This initial version may have some mistakes. . . Keep an eye on the version number. 1 Chapter 2: Solution of A x = b for square A Elimination and the LU decomposition For any square matrix A know how to factor A into PA = LU where P is a permutation matrix, L is a lower triangular matrix of multipli- ers l ij with 1s on the diagonal and U is a an upper triangular matrix with Pivots on the diagonal. An example of a matrix that needs a row exchange is 1 2 3 2 4 3 1 0 0 (1) as the second row will definitely cause a zero in the 2,2 position after elimination. There is no unique permutation required, however and you can always get creative as long as it works (i.e. doesnt generate another zero in a pivot position). For example one possible permutation is to move row 3 to row 1, row 1 to row 2 and row 2 to row 3 i.e. PA = 0 0 1 1 0 0 0 1 0 A = 1 0 0 1 2 3 2 4 3 (2) 1 E3101: Study Guide 2004 2 which is now readily factored to PA = LU = 1 0 0 1 1 0 2 2 1 1 0 0 2 3 0 0- 3 (3) Solution of Square A x = b Also be able to solve the basic square problem A x = b by elimi- nation by forming the augmented matrix [ A b ] , eliminating to [ U c ] then solving by back- substitution. Find A- 1 by Gauss Jordan Elimination You may need this to do eigenvalue problems. First form the augmented matrix [ A I ] where I is the identity matrix, then proceed to eliminate the entire matrix downwards, then upwards, then divide by the pivots to end up with [ I A- 1 ] . A- 1 for a 2 by 2 matrix The formula for the inverse of a general 2 2 matrix is useful to remem- ber, particularly for eigenproblems. A = bracketleftbigg a b c d bracketrightbigg A- 1 = 1 ad- bc bracketleftbigg d- b- c a bracketrightbigg (4) Product rules for inverse (and transpose) If matrices A and B are both invertible, then ( AB )- 1 = B- 1 A- 1 ( A and B must both be square to be invertible). Product rule for transpose looks the same although A and B do not have to be square or invertible, i.e ( AB ) T = B T A T . 2 Chapter 3: Solutions of general m n A x = b and the four subspaces General solution of A x = b For a general m n matrix, know how to find all solutions of the linear system A x = b . The general solution is to form the augmented matrix [ A b ] and use Gauss-Jordan elimination to reduce it to Row Reduced Echelon Form [ R d ] . Then 1. Identify the pivot columns and the free columns 2. Determine the rank of the matrix2....
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- Fall '07