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Burden - Numerical Analysis 5e (PWS, 1993) Episode 2 Part 6

Tham khảo tài liệu 'burden - numerical analysis 5e (pws, 1993) episode 2 part 6', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 3Ố8 CHAPTER 6 Direct Methods for Solving Linear Systems is a 3 X 3 permutation matrix. For any 3X3 matrix A multiplying on the left by p has the effect of interchanging the second and third rows of A 1 0 0 au 12 13 11 12 13 PA -- 0 0 1 a2ỉ 22 23 31 32 33 _0 1 0_ 31 32 33 _ 21 22 23 _ Pij Similarly multiplying on the right by p interchanges the second and third columns of A. H There are two useful properties of permutation matrices that relate to Gaussian elimination. The first of these is illustrated in the previous example and states that if kn is a permutation of the integers 1 . . . n and the permutation matrix p Pij is defined by 1 if j kị 0 otherwise then PA permutes the rows of A that is Ề 1 fci 2 aki H DA _ ữfc2 1 ứ 2 2 ứlt2 jTzl . r Ề ak l ak - 2 ak i The second result is that if p is a permutation matrix then p1 exists andP-1 Ph At the end of Section 6.4 we saw that for any nonsingular matrix A the linear system Ax b can be solved by Gaussian elimination with the possibility of row interchanges. Hence there is a rearrangement of the equations in the system that permits Gaussian elimination to proceed without row interchanges. This implies that for any nonsingular matrix A a permutation matrix p exists for which the system PAx Pb can be solved without row interchanges. But then the matrix PA can be factored into PA - LU where L is lower triangular and u is upper triangular. Since P 1 pr we have the factorization A PrL ĩ . However unless p Ỉ the matrix pr L is not lower triangular. EXAMPLE 3 Since an 0 the matrix 0 0-1 1 2 3 3 _1 2 1 does not have an LU factorization. However using the row interchange Ef - Ef f followed by 3 - Ej E3 and E4 - Ex Ef produces . ji. 6.5 Matrix Factorization 369 1 1-12 0 0-1 1 0 0 11 0 10 1 Then the row interchange 2 4 followed by 4 3 7 4 gives the matrix 1 1 -1 2 u 0 1 0 1 0 0 1 1 _0 0 0 2_ The permutation matrix associated with the row interchanges 1 - 2 and 2 4 is . - 0 10 0 0 0 0 1 F 0 0 1 0 _1 0 0 0_ Gaussian elimination can