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![3
Elementary row operations
The following operations applied to the
augmented matrix [A|b], yield an equivalent
linear system
Interchanges: The order of two rows can be
changed
Scaling: Multiplying a row by a nonzero constant
Replacement: The row can be replaced by the sum
of that row and a nonzero multiple of any other row.](https://image.slidesharecdn.com/solutionoflinearsystemofequations-131010034228-phpapp02/75/Solution-of-linear-system-of-equations-3-2048.jpg)
![16
Back substitution algorithm
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![39
Gauss-Jordan Elimination
Almost 50% more arithmetic operations than
Gaussian elimination
Gauss-Jordan (GJ) Elimination is prefered
when the inverse of a matrix is required.
Apply GJ elimination to convert A into an
identity matrix.
[ ]IA
[ ]1−
AI](https://image.slidesharecdn.com/solutionoflinearsystemofequations-131010034228-phpapp02/75/Solution-of-linear-system-of-equations-39-2048.jpg)
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Similar to Solution of linear system of equations
Solution of linear system of equations
- 1. 1 Solution of linearsystem of equations Circuit analysis (Mesh and node equations) Numerical solution of differential equations (Finite Difference Method) Numerical solution of integral equations (Finite Element Method, Method of Moments) nnnnnn nn nn bxaxaxa bxaxaxa bxaxaxa =+++ =+++ =+++ 2211 22222121 11212111 = nnnnnn n n b b b x x x aaa aaa aaa 2 1 2 1 21 22221 11211 ⇒
- 2. 2 Consistency (Solvability) Thelinear system of equations Ax=b has a solution, or said to be consistent IFF Rank{A}=Rank{A|b} A system is inconsistent when Rank{A}<Rank{A|b} Rank{A} is the maximum number of linearly independent columns or rows of A. Rank can be found by using ERO (Elementary Row Oparations) or ECO (Elementary column operations). ERO⇒# of rows with at least one nonzero entry ECO⇒# of columns with at least one nonzero entry
- 3. 3 Elementary row operations The following operations applied to the augmented matrix [A|b], yield an equivalent linear system Interchanges: The order of two rows can be changed Scaling: Multiplying a row by a nonzero constant Replacement: The row can be replaced by the sum of that row and a nonzero multiple of any other row.
- 4. 4 An inconsistent example = 5 4 42 21 2 1 x x 00 21 Rank{A}=1 Rank{A|b}=2 ERO:Multiplythe first row with -2 and add to the second row − 3 4 0 2 0 1 Then this system of equations is not solvable
- 5. 5 Uniqueness of solutions The system has a unique solution IFF Rank{A}=Rank{A|b}=n n is the order of the system Such systems are called full-rank systems
- 6. 6 Full-rank systems IfRank{A}=n Det{A} ≠ 0 ⇒ A is nonsingular so invertible Unique solution = − 2 4 11 21 2 1 x x
- 7. 7 Rank deficient matrices If Rank{A}=m<n Det{A} = 0 ⇒ A is singular so not invertible infinite number of solutions (n-m free variables) under-determined system = 8 4 42 21 2 1 x x Consistent so solvable Rank{A}=Rank{A|b}=1
- 8. 8 Ill-conditioned system ofequations A small deviation in the entries of A matrix, causes a large deviation in the solution. = 47.1 3 99.048.0 21 2 1 x x = 47.1 3 99.049.0 21 2 1 x x = 1 1 2 1 x x ⇒ = 0 3 2 1 x x ⇒
- 9. 9 Ill-conditioned continued..... Alinear system of equations is said to be “ill-conditioned” if the coefficient matrix tends to be singular
- 10. 10 Types of linearsystem of equations to be studied in this course Coefficient matrix A is square and real The RHS vector b is nonzero and real Consistent system, solvable Full-rank system, unique solution Well-conditioned system
- 11. 11 Solution Techniques Directsolution methods Finds a solution in a finite number of operations by transforming the system into an equivalent system that is ‘easier’ to solve. Diagonal, upper or lower triangular systems are easier to solve Number of operations is a function of system size n. Iterative solution methods Computes succesive approximations of the solution vector for a given A and b, starting from an initial point x0. Total number of operations is uncertain, may not converge.
- 12. 12 Direct solution Methods Gaussian Elimination By using ERO, matrix A is transformed into an upper triangular matrix (all elements below diagonal 0) Back substitution is used to solve the upper- triangular system = n i n i nnnin iniii ni b b b x x x aaa aaa aaa 11 1 1 1111 ⇒ ERO = n i n i nn inii ni b b b x x x a aa aaa ~ ~ ~00 ~~0 111111 Backsubstitution
- 13. 13 First step ofelimination = = = = )2( )2( 3 )2( 2 )1( 1 3 2 1 )2()2( 3 )2( 2 )2( 3 )2( 33 )2( 32 )2( 2 )2( 23 )2( 22 )1( 1 )1( 13 )1( 12 )1( 11 )1( 11 )1( 11, )1( 11 )1( 311,3 )1( 11 )1( 211,2 0 0 0 / / / nnnnnn n n n nn b b b b x x x x aaa aaa aaa aaaa aam aam aam = )1( )1( 3 )1( 2 )1( 1 3 2 1 )1()1( 3 )1( 2 )1( 1 )1( 3 )1( 33 )1( 32 )1( 31 )1( 2 )1( 23 )1( 22 )1( 21 )1( 1 )1( 13 )1( 12 )1( 11 nnnnnnn n n n b b b b x x x x aaaa aaaa aaaa aaaa Pivotal element
- 14. 14 Second step ofelimination = = = )3( )3( 3 )2( 2 )1( 1 3 2 1 )3()3( 3 )3( 3 )3( 33 )2( 2 )2( 23 )2( 22 )1( 1 )1( 13 )1( 12 )1( 11 )2( 22 )2( 22, )2( 22 )2( 322,3 00 00 0 / / nnnnn n n n nn b b b b x x x x aa aa aaa aaaa aam aam = )2( )2( 3 )2( 2 )1( 1 3 2 1 )2()2( 3 )2( 2 )2( 3 )2( 33 )2( 32 )2( 2 )2( 23 )2( 22 )1( 1 )1( 13 )1( 12 )1( 11 0 0 0 nnnnnn n n n b b b b x x x x aaa aaa aaa aaaa Pivotal element
- 15. 15 Gaussion elimination algorithm Definenumber of steps as p (pivotal row) For p=1,n-1 For r=p+1 to n For c=p+1 to n 0 / )( )()( , = = p rp p pp p rppr a aam )( , )()1( p pcpr p rc p rc amaa ×−=+ )( , )()1( p ppr p r p r bmbb ×−=+
- 16. 16 Back substitution algorithm = − −−−−− )( )1( 1 )3( 3 )2( 2 )1( 1 1 3 2 1 )( )( 1 )( 11 )3( 3 )3( 33 )2( 2 )2( 23 )2( 22 )1( 1 )1( 13 )1( 12 )1( 11 0000 000 00 0 n n n n n n n nn n nn n nn n n n b b b b b x x x x x a aa aa aaa aaaa [] 1,,2,1 1 1 1 )()( )( 1 1 )1( 1)1( 11 1)( )( −−= −= −== ∑+= − − − −− −− − nnixab a x xab a x a b x n ik k i ik i ii ii i n n nn n nn nn nn nn n n n
- 17. 17 Operation count Numberof arithmetic operations required by the algorithm to complete its task. Generally only multiplications and divisions are counted Elimination process Back substitution Total 6 5 23 23 nnn −+ 2 2 nn + 33 2 3 n n n −+ Dominates Not efficient for different RHS vectors
- 18. 18 LU Decomposition A=LU Ax=b ⇒LUx=b DefineUx=y Ly=b Solve y by forward substitution ERO’s must be performed on b as well as A The information about the ERO’s are stored in L Indeed y is obtained by applying ERO’s to b vector Ux=y Solve x by backward substitution
- 19. 19 LU Decomposition byGaussian elimination = −−−−−− )( )( 1 )( 11 )3( 3 )3( 33 )2( 2 )2( 23 )2( 22 )1( 1 )1( 13 )1( 12 )1( 11 4,3,2,1, 3,12,11,1 2,31,3 1,2 0000 000 00 0 1 1 0 001 0001 00001 n nn n nn n nn n n n nnnn nnn a aa aa aaa aaaa mmmm mmm mm m A Compact storage: The diagonal entries of L matrix are all 1’s, they don’t need to be stored. LU is stored in a single matrix. There are infinitely many different ways to decompose A. Most popular one: U=Gaussian eliminated matrix L=Multipliers used for elimination
- 20. 20 Operation count A=LUDecomposition Ly=b forward substitution Ux=y backward substitution Total For different RHS vectors, the system can be efficiently solved. 33 3 nn − 2 2 nn + 2 2 nn − 33 2 3 n n n −+ Done only once
- 21. 21 Pivoting Computer usesfinite-precision arithmetic A small error is introduced in each arithmetic operation, error propagates When the pivotal element is very small, the multipliers will be large. Adding numbers of widely differening magnitude can lead to loss of significance. To reduce error, row interchanges are made to maximise the magnitude of the pivotal element
- 22. 22 Example: Without Pivoting = −93.22 414.6 210.114.24 281.5133.1 2 1 x x − = − 8.113 414.6 7.113000.0 281.5133.1 2 1 x x = 001.1 9956.0 2 1 x x 31.21 133.1 14.24 21 ==m 4-digit arithmetic Loss of significance
- 23. 23 Example: With Pivoting = − 414.6 93.22 281.5133.1 210.114.24 2 1 x x = − 338.5 93.22 338.5000.0 210.114.24 2 1 x x = 000.1 000.1 2 1 x x 04693.0 14.24 133.1 21 ==m
- 24.
- 25. 25 Row pivoting Mostcommonly used partial pivoting procedure Search the pivotal column Find the largest element in magnitude Then switch this row with the pivotal row
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- 29. 29 Row Pivoting inLU Decomposition When two rows of A are interchanged, those rows of b should also be interchanged. Use a pivot vector. Initial pivot vector is integers from 1 to n. When two rows (i and j) of A are interchanged, apply that to pivot vector. = n i jp 3 2 1 = n j ip 3 2 1
- 30. 30 Modifying the bvector When LU decomposition of A is done, the pivot vector tells the order of rows after interchanges Before applying forward substitution to solve Ly=b, modify the order of b vector according to the entries of pivot vector = 9 5 7 6 8 4 2 3 1 p − − − = 9.6 5.3 7.2 2.5 6.9 8.4 2.1 6.8 3.7 b − − − =′ 9.6 6.9 7.2 2.5 5.3 8.4 6.8 2.1 3.7 b
- 31. 31 LU decomposition algorithmwith row pivoting For k=1,n-1 column to be eliminated p=k For r=k+1 to n if if p>k then For c=1 to n For r=k+1 to n For c=k+1 to n kr k rk k kk k rkkr ma aam , )1( )()( , / = = + )( , )()1( k kckr k rc k rc amaa ×−=+ rpthen => pkrk aa taaaat pcpckckc === ,, Column search for maximum entry Interchaning the rows Updating L matrix Updating U matrix
- 32.
- 33. 33 Example continued... = − −− =′ 3 1 2 241 230 124 pA Eliminate a21and a31 by using a11 as pivotal element A=LU in compact form (in a single matrix) = −− −− =′ 3 1 2 75.15.325.0 230 124 pA Multipliers (L matrix)
- 34. 34 Example continued... = −− −− =′ 1 3 2 230 75.15.325.0 124 pA = −− −− =′ 3 1 2 75.15.325.0 230 124 pA Column search:Maximum magnitude at the third row Interchange 2nd and 3rd rows
- 35. 35 Example continued... = −− −− =′ 1 3 2 230 75.15.325.0 124 pA Eliminate a32by using a22 as pivotal element = −− −− =′ 1 3 2 5.35.3/30 75.15.325.0 124 pA Multipliers (L matrix)
- 36.
- 37. 37 Example continued... = 3 2 1 3 2 1 x x x Backward substitution − = 5.10 75.1 5 3 2 1 y y y Forward substitution − = − −− 5.10 75.1 5 5.300 75.15.30 124 3 2 1 x x x Ux=y − = − 12 3 5 15.3/30 0125.0 001 3 2 1 y y y Ly=b’
- 38. 38 Gauss-Jordan elimination Theelements above the diagonal are made zero at the same time that zeros are created below the diagonal )1()1()1( 2 )1( 1 )1( 2 )1( 2 )1( 22 )1( 21 )1( 1 )1( 1 )1( 12 )1( 11 nnnnn n n baaa baaa baaa )2()2()2( 2 )2( 2 )2( 2 )2( 22 )1( 1 )1( 1 )1( 12 )1( 11 0 0 nnnn n n baa baa baaa − − )()( )1( 2 )2( 22 )1( 1 )1( 11 00 00 00 n n n nn n n ba ba ba )3()3( )2( 2 )2()2( 22 )2( 1 )2()1( 11 00 0 0 nnn nn nn ba baa baa
- 39. 39 Gauss-Jordan Elimination Almost50% more arithmetic operations than Gaussian elimination Gauss-Jordan (GJ) Elimination is prefered when the inverse of a matrix is required. Apply GJ elimination to convert A into an identity matrix. [ ]IA [ ]1− AI
- 40. 40 Different forms ofLU factorization Doolittle form Obtained by Gaussian elimination Crout form Cholesky form = 33 2322 131211 3231 21 333231 232221 131211 00 0 1 01 001 u uu uuu ll l aaa aaa aaa = 100 10 1 0 00 23 1312 333231 2221 11 333231 232221 131211 u uu lll ll l aaa aaa aaa 100 10 1 00 00 00 1 01 001 23 1312 33 22 11 3231 21 u uu d d d ll l
- 41. 41 Crout form Firstcolumn of L is computed Then first row of U is computed The columns of L and rows of U are computed alternately 11 ii al = 11 1 1 l a u j j = njji l ula u niijulal ii i k kjikij ij j k kjikijij ,,3,2, ,,2,1, 1 1 1 1 =≤ − = =≤−= ∑ ∑ − = − =
- 42. 42 Crout reduction sequence = 1000 100 10 1 0 00 000 34 2423 141312 44434241 333231 2221 11 44434241 34333231 24232221 14131211 u uu uuu llll lll ll l aaaa aaaa aaaa aaaa Anentry of A matrix is used only once to compute the corresponding entry of L or U matrix So columns of L and rows of U can be stored in A matrix 1 2 3 4 5 6 7
- 43. 43 Cholesky form A=LDU(Diagonals of L and U are 1) If A is symmetric L=UT ⇒ A= UT DU= UT D1/2 D1/2 U U’=D1/2 U ⇒ A= U’T U’ This factorization is also called square root factorization. Only U’ need to be stored
- 44. 44 Solution of ComplexLinear System of Equations Cz=w C=A+jB Z=x+jy w=u+jv (A+jB)(x+jy)=(u+jv) (Ax-By)+j(Bx+Ay)=u+jv = − v u y x AB BA Real linear system of equations
- 45. 45 Large and SparseSystems When the linear system is large and sparse (a lot of zero entries), direct methods become inefficient due to fill-in terms. Fill-in terms are those which turn out to be nonzero during elimination 5553 444241 353331 2422 141311 000 00 00 000 00 aa aaa aaa aa aaa ′ ′′ ′′′ 55 4544 353433 2422 141311 0000 000 00 000 00 a aa aaa aa aaa Elimination Fill-in terms
- 46. 46 Sparse Matrices Nodeequation matrix is a sparse matrix. Sparse matrices are stored very efficiently by storing only the nonzero entries When the system is very large (n=10,000) the fill-in terms considerable increases storage requirements In such cases iterative solution methods should be prefered instead of direct solution methods