3.2.1 Example of Gaussian elimination

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1 3.2.1 Example of Gaussian elimination We wish to solve the following matrix equation by Gaussian elimination: (3.31) Towards this goal, we proceed with the different steps as follows: Step M1 On equation (3.31), the operation R1/A(1, 1) (where, A(1, 1) 11) is performed to yield, (3.32) Step M2 On equation (3.32), the operations (R2 R1 A(2, 1)), (R3 R1 A(3, 1)) and (R4 R1 A(4, 1)) are carried out (where A(2, 1) 23, A(3, 1) 22 and A(4, 1) 12) and the resulting matrix equation is given by; (3.33) Step M3 On equation (3.33), the operation R2/A(2, 2) (where, A(2, 2) ) is carried out to get; (3.34) Step M4 On equation (3.34), the operations (R3 R2 A(3, 2)) and (R4 R2 A(4, 2)) are carried out to obtain (where A(3, 2) 2.0 and A(4, 2) ); (3.35) 90

2 Step M5 On equation (3.35), the operation R3/A(3, 3) (where A(3, 3) ) is carried out to obtain the matrix equation shown below: (3.36) Step M6 Lastly, on equation (3.36), the operation (R4 R3 A(4, 3)) (where A(4, 3) ) is carried out to get; (3.37) In equation (3.37), the co-efficient matrix has been converted to an upper-triangular matrix. From the last row of this equation, can be calculated as / Back substituting this value of in the third row of equation (3.37) one can obtain Similarly, substitution of the values of and in the second row of equation (3.37) yields the of as Lastly, substituion of, and in the first row of equation (3.37) gives Optimal order of elimination We have seen that the Gaussian Elimination method is quite effective for solving a large set of spare linear equations without having to invert the co-efficient matrix. Moreover, at every stage, if the calculations pertaining to Gaussian elimination is carried out only using the non-zero terms, great saving in the computational burden can be achieved. However, if the elimination process is carried out in the normal sequence, at any stage of elimination, the original zero-elements may the concerted into a non-zero element. This is normally termed as fill-in phenomenon. On the other hand, instead of following the normal sequence, if the elimination process is carried out in an appropriate order, then the occurrence of fill-in can be avoided to a great extent. A simple example given below illustrates this point. In equation (3.38), an initial co-efficient matrix is shown at the left hand side (part a ) and the structure of the co-efficient matrix after step-1 is shown at the right hand side (part b ). It is to be noted that in this equation, only the positions of non-zero terms (denoted by ) and zero terms (denoted by o ) are shown. As can be seen in equation (3.38), after step-1, all the original zero elements have been converted to non-zero terms (denoted by ), or, in other words, significant 91

3 level of fill-in has occurred o o 2 o 3 o o 3 o 4 o o 4 o a) Initial A matrix b) A matrix after step 1 (3.38) Now, if the original co-efficient matrix shown in part (a) of equation (3.38) is re-arranged as shown in part (a) of equation (3.39), then after step 1, there would be no fill-in as can be observed in part (b) of equation (3.39) o o 4 1 o o 2 o o 2 o o 3 o o 3 o o 1 1 o a) Rearranged A matrix b) Rearranged A matrix after step 1 (3.39) From the above example, it is apparent that if the rows are eliminated in an optimal order, then the number of fill-in would be minimum. However, an ideal optimal order is very difficult to develop and perhaps is impossible. As an alternative, various near optimal ordering schemes have been developed. Some of them are discussed below: Scheme 1 In this scheme, before elimination, number the rows of the co-efficient matrix A according to the number of non-zero, off-diagonal terms. Thus, the rows with only one off-diagonal, non-zero term are numbered first, those with two non-zero, off- diagonal terms are numbered second and so on. However, this scheme does not take into account the changes occurring in the co-efficient matrix during the elimination process. Therefore, this scheme in quite easy and straight forward to implement. Scheme 2 In this scheme the rows of the co-efficient matrix A are numbered such that at each step of the elimination procedure, the row with the fewest number of non-zero off-diagonal terms would be operated next. If more than one row meets this criterion, then any one row is chosen. Therefore, this scheme requires the simulation of the elimination procedure to estimate the changes occurring in the co-efficient matrix in advance. Thus, this method takes longer time as compared to scheme 1 to compute the solution, but is definitely better than scheme 1. 92

4 Scheme 3 In this scheme, the rows are numbered in such a way so that the row which will introduce fewest nonzero off-diagonal terms would be operated upon next. If more than one row satisfies this criterion, choose any one row. Again, this scheme also requires the simulation of the elimination process to study its effects on the co-efficient matrix in advance. Hence, this method also takes longer time than scheme 1. Let us now look at another technique for solving a set of linear equations without the need of inverting the co-efficient matrix, namely, the triangular factorization or LU decomposition. 3.4 Triangular factorization: In triangular factorization or decomposition method, a square matrix A is expressed as a product of two triangular matrices as A LU, where L is a lower triangular matrix and U is an upper triangular matrix. As an example, let the matrix A be a 4 4 (N 4) matrix. Upon triangular factorization (or LU decomposition), the matrix A is represented as A LU. Or, a 11 a 12 a 13 a 14 α β 11 β 12 β 13 β 14 a 21 a 22 a 23 a 24 α 21 α β 22 β 23 β 24 a 31 a 32 a 33 a 34 α 31 α 32 α β 33 β 34 a 41 a 42 a 43 a 44 α 41 α 42 α 43 α β 34 (3.40) With this decomposition, the equation Ax b can be written as, Ax b or, (LU)x b or, L(Ux) b (3.41) Or, Ly b (3.42) In equation (3.42), y Ux. Expanding equation (3.42) we get, α α 21 α α 31 α 32 α 33 0 α 41 α 42 α 43 α 44 y 1 y 2 y 3 y 4 b 1 b 2 b 3 b 4 (3.43) From equation (3.43), the intermediate vector y can be calculated as, y 1 y i b 1 α 11 i 1 1 [b i α ii j1 α ij y j ] ; i 2, 3, N (3.44) 93

5 Again, expanding the expression y Ux we get, β 11 β 12 β 13 β 14 0 β 22 β 23 β β 33 β β 34 y 1 y 2 y 3 y 4 (3.45) Now, With the knowledge of the intermediate vector y, from equation (3.45), the solution vector x can be calculated as, x N x i y n β NN N 1 [y i β ij x j ] ; i (N 1), (N 2), 1 (3.46) β ii j1+1 We will now look into the basic procedure of obtaining the LU decomposition in the next lecturel. 94

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