2. Linear Equation System Pertemuan 4 Matakuliah: S0262-Analisis Numerik Tahun: 2010

3. Material Outline Linear equation system –Gauss Elimination –Gauss-Jordan Elimination

4. 14-Dec-15 DR. Paston Sidauruk4 Linear Equations The following equation is the general form of linear equation: where: a 1, a 2,…, a n, and b are constants and x 1, x 2,…, x n are variables which also called as the unknowns. Linear Systems A linear system is a finite set of linear equations in the variables x 1, x 2,…, x n. A sequence s 1, s 2,… s n is a solution to linear system if x 1 = s 1, x 2 = s 2 … x n =s n is a solution to every equation in the system.

5. 5 System of Linear Equations (Linear system) Note: Every linear system has either no solution, exactly one solution, or infinitely many solutions Examples: A linear system with infinite many solutions, i.e, x 1 =1, x 2 =2, and x 3 =-1 A linear system with exactly one solution that is: x 1 =2, and x 2 =1 A linear system with no solution

6. 6 A general form of a linear system m equations in n unknowns A linear system in Augmented Matrices

7. 7 Example 3 equations in 3 unknowns The basic method of solving an LS is to replace the given system with a new system which is easier to solve. The following steps are commonly used  Multiply an equation by a non-zero constant  Interchange two equations  Add a multiple of one equation to another Note: the rows in Augmented Matrix correspond to the equations in the system, the above steps can be applied to AM in which equation replaced by row Augmented matrix

8. 8 How to use row operations to find a solution of Linear System: 1. Subtract 2 times 1 st row to 2 nd row and 3 times 1 st row to the third row 2. Multiply 2 nd row by ½ etc., finally we obtain the following

9. 9 Gauss Elimination(GE) GE is a systematic procedure to solve LS by reducing the augmented matrix of given LS to a simple augmented matrix that allow the solution of LS sought by inspection. Procedure: Reduced the Augmented matrix  row-echelon form. Row-echelon form:  The 1 st non-zero number in the row is a 1 called a leading 1  All rows that consist entirely zero are put at the bottom of the matrix  The leading 1 in the lower row occurs farther to the right of the leading 1 in the higher row. A matrix in reduced row-echelon form is a matrix in row- echelon form that each column that contains a leading 1 has zero every where else.

10. 10 Matrices in row-echelon form Examples: Matrices in reduced row-echelon form Examples:

11. 11 Steps in Gauss Elimination  Interchange the top row with another row, if necessary, such that the top of the most left column is non-zero entry  Suppose that the top of left most column entry is a non-zero entry then multiply the first row to introduce a leading 1  Add suitable multiples of the top row to the rows below such that all entries in the column below leading 1 are zero  Apply the steps 1, 2, and 3 to the remaining sub matrix to produce a matrix in row-echelon form Note: Gauss Elimination will produce matrix in row-echelon form Gauss Jordan elimination will produce matrix in reduced row- echelon form

12. 12 Steps in Gauss-Jordan Elimination  Steps 1, 2, 3, and 4 the same as steps 1, 2, 3, and 4 in Gauss Elimination, respectively.  Beginning with the last nonzero row, add suitable multiples of each row to the rows above to introduce all zeros above the leading 1 Note: Gauss Elimination will produce matrix in row- echelon form Gauss Jordan elimination will produce matrix in reduced row-echelon form

13. 13 Note: 1. Gauss Elimination will produce matrix in row-echelon form 2. Gauss Jordan elimination will produce matrix in reduced row- echelon form. 3. The next step is after finding the row-echelon or reduced row- echelon form matrix begin with the last row used the back substitution to find the solution of the given LS. Example: Find the solution of the following linear system using Gauss and Gauss-Jordan eliminations: