Chapter: 3b System of Linear Equations Dr. Asaf Varol asvarol@mail.wvu.edu

Gaussian Elimination In the Gaussian Elimination Method, Elementary Row Operations (E.R.O.'s) are applied in a specific order to transform an augmented matrix into triangular echelon form as efficiently as possible [6]. This is the essence of the method: Given a system of m equations in n variables or unknowns, pick the first equation and subtract suitable multiples of it from the remaining m-1 equations. In each case choose the multiple so that the subtraction cancels or eliminates the same variable, say x1. The result is that the remaining m-1 equations contain only n-1 unknowns (x1 no longer appears) [6]. Now set aside the first equation and repeat the above process with the remaining m-1 equations in n-1 unknowns [6]. Continue repeating the process. Each cycle reduces the number of variables and the number of equations. The process stops when either:

Gaussian Elimination (Cont’d) There remains one equation in one variable. In that case, there is a unique solution and back-substitution is used to find the values of the other variables [6]. There remain variables but no equations. In that case there is no unique solution [6]. There remain equations but no variables (ie. the lowest row(s) of the augmented matrix contain only zeros on the left side of the vertical line). This indicates that either the system of equations is inconsistent or redundant. In the case of inconsistency the information contained in the equations is contradictory. In the case of redundancy, there may still be a unique solution and back-substitution can be used to find the values of the other variables [6].

Algorithm for Gaussian Elimination Transform the columns of the augmented matrix, one at a time, into triangular echelon form. The column presently being transformed is called the pivot column. Proceed from left to right, letting the pivot column be the first column, then the second column, etc. and finally the last column before the vertical line. For each pivot column, do the following two steps before moving on to the next pivot column [6]: Locate the diagonal element in the pivot column. This element is called the pivot. The row containing the pivot is called the pivot row. Divide every element in the pivot row by the pivot (ie. use E.R.O. #1) to get a new pivot row with a 1 in the pivot position [6]. Get a 0 in each position below the pivot position by subtracting a suitable multiple of the pivot row from each of the rows below it (ie. by using E.R.O. #2). Upon completion of this procedure the augmented matrix will be in triangular echelon form and may be solved by back-substitution [6].

Example Use Gaussian elimination to solve the system of equations[6]:

Solution Perform this sequence of E.R.O.'s on the augmented matrix. Set the pivot column to column 1. Get a 1 in the diagonal position (underlined):

Solution (Cont’d)

Solution (Cont’d)

Results It is solved by back-substitution. Substituting z = 3 from the third equation into the second equation gives y = 5, and substituting z = 3 and y = 5 into the first equation gives x = 7. Thus the complete solution is [6]: {x = 7, y = 5, z = 3}.

Example Express the following system in augmented matrix form and find an equivalent upper- triangular system and the solution [4].

Example (Cont’d) The augmented matrix is

Example (Cont’d) The first row is used to eliminate elements in the first column below the diagonal. We refer to the first row as pivotal row and the element a11=1 is called the pivotal element. The values mk1 are the multiples of row 1 then are to be subtracted from row k for k=2,3,4. The result after elimination is [4]

Example (Cont’d) The second row is used to eliminate elements in the second column that lie below the diagonal. The second row is the pivotal row and the values mk2 are the multiples of row 2 that are to be subtracted from row k for k=3,4. The result after elimination is [4]

Example (Cont’d) Finally, the multiple m43=-1.9 of the third row is subtracted from the fourth row, and the result is the upper- triangular system [4].

Example (Cont’d) The back-substitution algorithm can be used to solve the previous matrix, and we get X4=2 X3=4 X2=-1 X1=3

Example a) Use MATLAB to construct the augmented matrix for the linear system of the below given matrix. b) Use the max command to find the element of greatest magnitude in the first column of the coefficient matrix A. C) Break the augmented matrix into the coefficient matrix U and constant matrix Y of the upper-triangular system UX=Y [4].

Answers a) >>A=[1 2 1 4;2 0 4 3;4 2 2 1;-3 1 3 2]; >>B=[13 28 20 6]’; >>Aug=[A B]

Answer b) In the following MATLAB display, a is the element of greatest magnitude in the first column of A and j is the row number >>[a,j]=max(abs(A(1:4,1)))

Answer c) Let Augup=[U|Y] be the upper-triangular matrix.

References Celik, Ismail, B., “Introductory Numerical Methods for Engineering Applications”, Ararat Books & Publishing, LCC., Morgantown, 2001 Fausett, Laurene, V. “Numerical Methods, Algorithms and Applications”, Prentice Hall, 2003 by Pearson Education, Inc., Upper Saddle River, NJ 07458 Rao, Singiresu, S., “Applied Numerical Methods for Engineers and Scientists, 2002 Prentice Hall, Upper Saddle River, NJ 07458 Mathews, John, H.; Fink, Kurtis, D., “Numerical Methods Using MATLAB” Fourth Edition, 2004 Prentice Hall, Upper Saddle River, NJ 07458 Varol, A., “Sayisal Analiz (Numerical Analysis), in Turkish, Course notes, Firat University, 2001 http://mathonweb.com/help/backgd3e.htm