Gaussian Elimination -(3.1) b 1. b 2., b. b n

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1 Gaussian Elimination -() Consider solving a given system of n linear equations in n unknowns: (*) a x a x a n x n b where a ij and b i are constants and x i are unknowns Let a n x a n x a nn x n a a a n b x A a a a n n a ij i,j, b b n b i i and x x x j n j a n a n a nn Then the system (*) in matrix and vector notations is Ax b We know from linear algebra that the system Ax b may have a unique solution, infinitely many solutions or no solution A system is said to be consistent if it has a solution A system that has no solution is said to be inconsistent The system has a unique solution if and only if the matrix A is nonsingular (invertible, ie, A exists) and the corresponding solution is: x A b In this section, we study how to determine systematically if a given matrix A is invertible and how to find the solution when it is unique Idea: If A is an upper triangular matrix, then A is nonsingular and Ax b has a unique solution if a ii for i,,n On the other hand, if a ii forsomei, then Ax b does not have a unique solution How can we reduce A to an upper triangular matrix without the solution? We can use elementary row operations to reduce A to an upper triangular matrix U and the corresponding system Ux b is equivalent to the system Ax b, that is, both systems have the same solution set Let A A b be the augmented matrix of a given system of linear equations Ax b Elementary Row Operations: Let R, R,, R n be rows of A Row operations: a cr i b R i R j c c R i c R j d R i R j Elementary Row Operations: () R i R j () cr i R i () R i cr j R i An elementary row operation is a row operation but a row operation may not be an elementary row operation mainly because of the difference in () and (c) An elementary row operation for A b does not change the solution set of a linear system Ax b That is if A b elementary row operation B b then Ax b and Bx b have the same solution set Example Let R, R a 4R R R : R, R 6 b 4R R R : R 6, R Computational complexity of elementary row operations: number of multiplications ( division is counted as multiplication) and number of additions required: x n

2 elementary row operation number of multiplications number of additions R i R j cr i R i n R i cr j R i n n Gaussian-elimination: Gaussian-elimination is a process of reducing a matrix to an upper triangular matrix for a square matrix or a generalized upper triangular matrix for a rectangular matrix by elementary row operations For example: a a a a a a a a a b b b b b or b a a a a 4 b b b b 4 a a a a 4 a a a a 4 b b b 4 b b 4 The process should be done with using the minimum number of multiplications Elimination process: Consider two rows: R R a a a a 4 a a a a 4 Suppose that we want to eliminate the element a, that is we want the matrix to become the form of How many elementary row operations are needed and what is the number of multiplications required? In the case when a, only one elementary row operation is needed: a a R R R and the number of multiplications is 4 for c a a, ca, ca, ca 4 Gaussian-elimination: Start at the first column if a and eliminate all elements below a using elementary row operations a i a R R i R i, i,,n Continue the process for the nd column, the rd column,, the n th column If for some i n, a ii, then find a ji forj i and switch R i R j Ifsuchaa ji does not exist, then the process is broken down If A is nonsingular, then the matrix can be triangularized without a break down (without considering the round off error) What are the numbers of multiplications and additions needed? Computational complexity of Gaussian elimination for an n n matrix: ith column elementary row operation multiplications additions n n n n n n n n n n n

3 The total number of multiplications: n i n ii i i i The total number of additions: n i n ii i i i n nn 6 n nn 6 Hence, we have the numbers of multiplications n 5 6 n n the numbers of additions n n n n n n n 5 6 n n n n Solving Systems of Linear Equations By Gaussian-elimination: Backward-substitution: n Let U u ij i,j be an upper triangular matrix, ie, u ij whenever i j Consider solving the system of linear equations: u u u u n u n x b Ux b, where U u u u n u n, x x, b b u n n u n n x n Note that the matrix U is nonsingular if and only if u ii fori,,n Assume that U is nonsingular Note that the nth equation: x n contains only one unknown x n and it can be solved easily: x n Now replace x n by in the first n equations in the system and then the n th equations: u n n x n u n n contains only one unknown x n and it can be solved easily: x n u n n u n n Continue the process to compute x n, x n,,x, x Compute x n, and for i n,n,,,, x i u ii b i u ii x i u ii x i u in x n What are the numbers of multiplications and additions needed? Let us count the following x n

4 i multiplications additions n From the table at the left, we have n nn the numbers of multiplications n n n n the numbers of additions n n n n Gaussian elimination with backward substitution for solving Ax b : Step I Gaussian-elimination: A b U b The system does not have a unique solution if the elimination process breaks down Step II Backward-substitution: u ii fori,,n Compute x n, and for i n,n,,,, x i u ii b i u ii x i u ii x i u in x n Example Let A (i) A x b : 7 8 9, A 7 8, b, and b Determine if A x b, A x b and A x b have a unique solution, infinitely many solutions or no solution Find the solution or the general solution if the system is consistent Step I A b 4R R R R R R 6 6 A x b is consistent and has infinitely many solutions Step II Solve the general solution: x is free, let x t, a real number x 6t t x, the general solution: x t t t number MatLab: A[;456;789];b[;;]; B[A b]; rref(b) -- (ii) A x b : Step I A b R R R 7R R R R R R t t t R R R A x b has no solution because the last equation: x x x can never hold 4 6, t is a real 6

5 MatLab: A[;456;789];b[;;-]; B[A b]; rref(b) - (iii) A x b : Step I A b 7 8 4R R R 7R R R R R R 6 A x b is consistent and has a unique solution Step II Solve the solution: x x 6 x, the solution: x Check: A x b 7 8 Use MatLab: A[;456;78];b[;;]; xinv(a)*b Example Find the inverse of A if it exists using Gaussian Elimination and backward 7 8 substitution Let C A C C C We know that AC I Since AC AC AC AC, B can be obtained by solving systems: AC, AC and AC Step I: Gaussian Elimination for A I U B, B B B B 5

6 A I 7 8 4R R R 7R R R R R R 6 4 U B B B Step II: Backward Substitution: solve UC B, UC B, UC B x 6 4, x 4 6, the solution: B x 6, x x 6 x 4, the solution: B 4 6, x x 6, the solution: B x 4 A B Check: AB