2. NUMERICS METHODS IN ENGINEERING The method of Gauss, also well-known as method of simple elimination of Gauss, it is a technique employee for the resolution of systems of equations. The method of Gauss is divided in two phases: Elimination of the incognito Substitution back 0 0 0 0 0 0 0 0 0 0 0 0
3. NUMERICS METHODS IN ENGINEERING To multiply or to divide a line for an number real different from zero. To add or to subtract to a line another multiplied for an number real not null. Elementary transformation is called in a matrix to: To exchange the place of two lines among if.
4. NUMERICS METHODS IN ENGINEERING GAUSS FOR SYSTEMS OF EQUATIONS Example: Applying the method of elimination of Gauss and using six significant digits, solve the following system of lineal equations: 3x1 – 0.1x2 – 0.2x3 = 7.85 (1) 0.1x1 + 7x2 – 0.3x3 = -19.3 (2) 0.3x1 – 0.2x2 + 10x3 = 71.4 (3)
5. NUMERICS METHODS IN ENGINEERING 1. It multiply the equation (1) for 0.1/3 and it is subtracted of the equation (2) being obtained: 2. Then it is carried out the product of the equation (1) for 0.3/3 and it is subtracted of the equation (3) to eliminate x1. As a result of these operations, one has the following result: 3x1 – 0.1x2 – 0.2x3 = 7.85 7.00333x2 – 0.293333x3 = -19.5617 – 0.1900002x2 + 10.0200x3 = 70.6150
6. NUMERICS METHODS IN ENGINEERING 3. Once made the above-mentioned, you proceeds to eliminate x2 of the equation (3). For it, it is carried out the product of the second equation for -0.190000/7.00333 and the result is subtracted of the equation (3). This process eliminates to x2 of the equation (3), completing the elimination phase. 4.Obteniendose this way the x3 =7.00003 3x1 – 0.1x2 – 0.2x3 = 7.85 (4) 7.00333x2 – 0.293333x3 = -19.5617 (5) 10.0200x3 = 70.0843 (6)
7. NUMERICS METHODS IN ENGINEERING 5. For I finish the values of x2 of the equation they are calculated (5) x2 = -2.50000, and the x1 value with the equation (4) x1 =3.00000 X 1 =7.00003 X2 =-2.50000 X3 =3.00000
8. NUMERICS METHODS IN ENGINEERING The method of elimination of Gauss, it can face the following difficulties: Rounding error. Taking into account that the true solutions of the system are x1 = 3, x2 = -2.5 and x3 = 7, it is observed that there is a small difference with the results obtained by the method of elimination of Gauss. Division among zero. It has been called to the method like method of simple Gauss, because with him it is possible to incur in the division among zero, for example, to solve the following system:
9. NUMERICS METHODS IN ENGINEERING Not well conditioned systems. They are those where small changes in the coefficients generate big variations in the solution. Example of a not well conditioned system: Solving, one has thatx1 = 8 y x2 = 1. Modifying the second equation of the system lightly: One has that x1 = 4 y x2 = 3.
