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1 Linear algebra














                  This chapter discusses the solution of sets of linear algebraic equations and defines basic
                  vector/matrix operations. The focus is upon elimination methods such as Gaussian elim-
                  ination, and the related LU and Cholesky factorizations. Following a discussion of these
                  methods, the existence and uniqueness of solutions are considered. Example applications
                  include the modeling of a separation system and the solution of a fluid mechanics boundary
                  value problem. The latter example introduces the need for sparse-matrix methods and the
                  computational advantages of banded matrices. Because linear algebraic systems have, under
                  well-defined conditions, a unique solution, they serve as fundamental building blocks in
                  more-complex algorithms. Thus, linear systems are treated here at a high level of detail, as
                  they will be used often throughout the remainder of the text.



                  Linear systems of algebraic equations

                  We wish to solve a system of N simultaneous linear algebraic equations for the N unknowns
                  x 1 , x 2 ,..., x N , that are expressed in the general form

                                        a 11 x 1 + a 12 x 2 +· · · + a 1N x N = b 1
                                        a 21 x 1 + a 22 x 2 +· · · + a 2N x N = b 2   (1.1)
                                                       .
                                                       .
                                                       .
                                        a N1 x 1 + a N2 x 2 +· · · + a NN x N = b N
                  a ij is the constant coefficient (assumed real) that multiplies the unknown x j in equation
                  i. b i is the constant “right-hand-side” coefficient for equation i, also assumed real. As a
                  particular example, consider the system

                                               x 1 + x 2 + x 3 = 4
                                              2x 1 + x 2 + 3x 3 = 7                   (1.2)
                                              3x 1 + x 2 + 6x 3 = 2
                  for which

                                   a 11 = 1  a 12 = 1   a 13 = 1  b 1 = 4
                                   a 21 = 2  a 22 = 1   a 23 = 3  b 2 = 7             (1.3)
                                   a 31 = 3  a 32 = 1   a 33 = 6  b 3 = 2



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