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SYSTEM OF EQUATIONS María Isabel Cadena Veloza Numerical Methods
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CRAMER: Cramer's rule is a process that helps you solve systems of linear equations having the same number of equations and variables. A method that applies the determinants.An example with all its steps.Solve the following system of equations: 3x - 2y = 4 6x + y = 13 GRAPHICAL METHODS
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Find first determinant of the coefficients of the variables. You call the primary determinant and you name with a D. D = (3) (a) - (6) (-2) = 15 You notice that the determinant of the coefficient matrix gave us 15. Continue with the process. Look at the procedure for finding the determinant for the variable x.Replaces the column of coefficients in the variable x with the values of the constants. Then observed the process: Dx = (4) (a) - (13) (-2) = 4 + 26 = 30
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To find the value of x, divide the value determined by the primary determinant Dx D. That is, calculateNow look at how you find the value of y.Dy is calculated using the determinant D = (3) (13) - (6) (4) = 39-24 = 15 Note that in thisdecisivechange in thesecondcolumnbytheconstant. To find the value and divide the value found for D by the major determinant D. That is, calculate y = 1.
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Conclude that the solution of the system is (2,1). This means that the two lines represented by the original equations intersect at a point with coordinates (2,1). Remember that if the system is in intersecting lines call it consistent.
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ELIMINATION OF MYSTERIES: In The Method of Elimination (Addition and Subtraction), as its name says, we must remove one of the unknowns, are these, for example: x, y, z; this is done by multiplying one of the equations by any number, positive or negative (all components of the equation will be affected by this issue), and already making another addition with the unknown equation chosen excess must be removed (= 0).
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Example: 2x +3 y = 2 - -> equation 1x-2y = 8 - -> equation 2 SOLUTION: to eliminate the unknown coefficients are reversed as shown equations acontinuación: striking out “y“ 2 (2x +3 y = 2)3 (x-2y = 8) 2 is the coefieciente of “y" two and the equation 3 is the ratio of “y" the one. being so: 4x +6 y = 43x-6y = 24 Now if you can remove the "y" and now clears the "x" to get its value 7x = 28x = 28 / 7x = 4
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replaced only after the value "x" to any of the two equations to get the "y“ 2x +3 y = 2 2 (4) +3 y = 2 8 +3 y = 2 3y = 2-8 3y =- 6 y =- 6 / 3 y = 2
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SIMPLE GAUSS It is quite hard to solve non-linear systems of equations, while linear systems are quite easy to study. There are numerical techniques which help to approximate nonlinear systems with linear ones in the hope that the solutions of the linear systems are close enough to the solutions of the nonlinear systems. We will not discuss this here. Instead, we will focus our attention on linear systems. For the sake of simplicity, we will restrict ourselves to three, at most four, unknowns. The reader interested in the case of more unknowns may easily extend the following ideas. Definition. The equation ax + by + cz + dw = h where a , b , c , d , and h are known numbers, while x , y , z , and w are unknown numbers, is called a linear equation . If h =0, the linear equation is said to be homogeneous . A linear system is a set of linear equations and a homogeneous linear system is a set of homogeneous linear equations.
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For example, and are linear systems, while is a nonlinear system (because of y 2 ). The system is an homogeneous linear system.
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Gaussian Elimination. Consider a linear system. 1.Construct the augmented matrix for the system; 2.Use elementary row operations to transform the augmented matrix into a triangular one; 3.Write down the new linear system for which the triangular matrix is the associated augmented matrix; 4.Solve the new system. You may need to assign some parametric values to some unknowns, and then apply the method of back substitution to solve the new system. Let us summarize the procedure:
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Example. Use Gaussian elimination to solve the linear system The associated augmented matrix is We keep the first row and subtract the first row multiplied by 2 from the second row. We get This is a triangular matrix. The associated system is Clearly the second equation implies that this system has no solution. Therefore this linear system has no solution.
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This procedure differs from the Gaussian method in which when you delete an unknown, is removed from all remaining equations, ie, the preceding equation as well as pivot to follow.Solve the following set of equations 3.0 X1 - 0.1 X2 - 0.2 X3 = 7.8500 0.1 X1 + 7.0 X2 - 0.3 X3 = - 19.3 0.3 X1- 0.2 X2 + 10 X3 = 71.4000 GAUSS JORDAN
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First we express the coefficients and the vector of independent terms as an augmented matrix. The first line is normalized by dividing by 3 to obtain:
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The term X1 can be removed from the second row by subtracting 0.1 times the first in the second row. In a similar way, subtracting 0.3 times the first in the third line delete the term with the third row X1. Then, the second line is normalized by dividing by 7.00333:
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Reducing X2 terms in the first and third equation is obtained: The third line is normalized dividing by 10 010:
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Finally, the terms with X3 be reduced in the first and second equation to get: Note that back substitution is required for the solution.
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