Handout CIV 208 Analisis Numerik CIV 208 P4 5 6

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Respect, Professionalism, & Entrepreneurship

Metode Eliminasi

Pertemuan – 4, 5, 6

Mata Kuliah : Analisis Numerik Kode : CIV - 208

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• _{Sub Pokok Bahasan :}

 _{Eliminasi Gauss}

 _{Eliminasi Gauss Jordan}  _{Dekomposisi LU}

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• _{This topic deals with simultaneous linear algebraic} equations that can be represented generally as

• _{where the a’s are constant coefficients and the b’s are} constants.

• _{The technique described in this chapter is called Gauss} elimination because it involves combining equations to eliminate unknowns.

a₁₁.x₁ + a₁₂.x₂ + … + a_1n.x_n = b₁ a₂₁.x₁ + a₂₂.x₂ + … + a_2n.x_n = b₂

……

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• _{Before proceeding to the computer methods,}

we will describe several methods that are appropriate for solving small (n≤3) sets of simultaneous equations and that do not require a computer.

• _{These are the graphical method, Cramer’s}

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• _{For example : solve}

3x₁ + 2x₂ = 18 - x₁ + 2x₂ = 2

• _{The solution is the}

intersection of the two lines

at x₁=4 and x₂=3.

• _{This result can be checked by}

substituting these values into the original equations to yield

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• _{Another method is use elimination}

• _{The elimination of unknowns by combining}

equations is an algebraic approach that can be illustrated for a set of two equations:

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• Eq. (1) might be multiplied by a₂₁and Eq. (2)

by a₁₁to give

a₁₁a₂₁x₁+a₁₂a₂₁x₂=b₁a₂₁ (3) a₂₁a₁₁x₁+a₂₂a₁₁x₂=b₂a₁₁ (4)

• _{Subtracting Eq. (3) from Eq. (4) will, therefore,}

eliminate the x₁ term from the equations to yield

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• _{which can be solved for}

• _x₂_{can then be substituted into Eq. (1), which can be}

solved for

• _{example : Use the elimination to solve}

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Gauss Elimination

• _{In the previous section, the elimination of unknowns was used to solve}

a pair of simultaneous equations. The procedure consisted of two steps: 1. The equations were manipulated to eliminate one of the unknowns from the equations. The result of this elimination step was that we had one equation with one unknown.

2. Consequently, this equation could be solved directly and the result back-substituted into one of the original equations to solve for the remaining unknown.

• _{This basic approach can be extended to large sets of equations by}

developing a systematic scheme or algorithm to eliminate unknowns and to back-substitute.

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• _{The Gauss Elimination is designed to solve a}

general set of n equations:

• _{As was the case with the solution of two}

equations, the technique for n equations

consists of two phases: elimination of unknowns and solution through back substitution.

a₁₁.x₁ + a₁₂.x₂ + … + a_1n.x_n = b₁(5.a)

a₂₁.x₁ + a₂₂.x₂ + … + a_2n.x_n = b₂ (5.b) ……

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• _{The initial step will be to eliminate the first unknown, x}₁_{, from} the second through the nth equations.

• _{To do this, multiply Eq. (5.a) by a}₂₁_/a₁₁_{to give}

• _{Subtracted from Eq. (5.b) to give}

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• _{The procedure is then repeated for the}

remaining equations.

• _{For instance, Eq. (5.a) can be multiplied by}

a₃₁/a₁₁and the result subtracted from the third equation.

• _{Repeating the procedure for the remaining}

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• Eq. (5.a) is called the pivot equation and a₁₁is

called the pivot coefficient or element

a₁₁.x₁ + a₁₂.x₂ + a₁₃.x₃ + …. + a_1n.x_n = b₁(6.a) a/

22.x2 + a/23.x3 + …. + a/2n.xn = b/2 (6.b)

a/

32.x2 + a/33.x3 + …. + a/3n.xn = b/3 (6.c)

……… a/

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• _{Now repeat the above to eliminate the}

second unknown from Eq. (6.c) through (6.d).

• _{To do this multiply Eq. (6.b) by a’}₃₂_/a’₂₂_{, and}

subtract the result from Eq. (6.c).

• _{Perform a similar elimination for the}

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• _{The procedure can be continued using the}

remaining pivot equations.

• _{The final manipulation in the sequence is to}

use the (n−1)th equation to eliminate the xn−1term from the nth equation.

a₁₁.x₁ + a₁₂.x₂ + a₁₃.x₃ + …. + a_1n.x_n = b₁

a/

22.x2 + a/23.x3 + …. + a/2n.xn = b/2

a//

33.x3 + …. + a//3n.xn = b//3

………

a//

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• _{now solve for x}_n_:

• _{This result can be back-substituted into the}

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• _{The procedure, which is repeated to evaluate}

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Example 1

Use Gauss elimination to solve x1 − 2x2 + 2x3= 1

2x1 + x2 − 3x3= −3

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Gauss-Jordan Elimination

• _{The Gauss-Jordan method is a variation of}

Gauss elimination.

• _{The major difference is that when an unknown}

is eliminated in the Gauss-Jordan method, it is eliminated from all other equations rather

than just the subsequent ones.

• _{In addition, all rows are normalized by dividing}

them by their pivot elements.

• _{Thus, the elimination step results in an}

identity matrix rather than a triangular matrix

• _{Consequently, it is not necessary to employ}

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Example 2

Use Gauss-Jordan elimination to solve

x1 − 2x2 + 2x3= 1 2x1 + x2 − 3x3= −3

−3x1 + x2 − x3 = 4

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LU Decomposition

• _{As described in previous lesson, Gauss elimination} is designed to solve systems of linear algebraic

equations,

[A]{X}={B} (1)

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• _{LU decomposition methods separate the}

time-consuming elimination of the matrix [A] from the manipulations of the right-hand side {B}.

• _{Thus, once [A] has been “decomposed,”}

multiple right-hand-side vectors can be evaluated in an efficient manner.

• _{Interestingly, Gauss elimination itself can be}

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• _{Equation (1) can be rearranged to give}

[A]{X}−{B}=0 (2)

• _{Suppose that Eq. (2) could be expressed as an upper} triangular system:

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• _{Equation (3) can also be expressed in matrix notation and} rearranged to give

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• _{Assume that there is a lower diagonal matrix}

with 1’s on the diagonal, (5)

• _{That has the property that when Eq. (4) is}

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• _{If this equation holds, it follows from the rules}

for matrix multiplication that

[L][U]=[A] (7)

• _and

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• _{Gauss elimination can be used to decompose [A] into [L] and} [U]. For example :

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• _{The first step in Gauss elimination is to multiply row 1 by the} factor

• _{and subtract the result from the second row to eliminate a}₂₁_.

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• _{Similarly, row 1 is multiplied by}

• _{And the result subtracted from the third row to eliminate a}₃₁ • _{The final step is to multiply the modified second row by}

• _{and subtract the result from the third row to eliminate a’}₃₂

11 31 31

a a

f 

22 32 32

a a f



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• The value of f₂₁, f₃₁, f₃₂ actually are the

element of [L].

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Example 1

• _{Solve the equation below, using LU}

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Crout Decomposition

• _{An alternative approach involves a [U]}

matrix with 1’s on the diagonal.

• _{This is called Crout decomposition.}

• _{The Crout decomposition approach}

generates [U] and [L] by sweeping

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• _{Crout Decomposition can be implemented by the following} concise series of formulas

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Example 2

• _{Repeat example 1 using Crout Decomposition}

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Gauss-Seidel Iteration

• _{Iterative or approximate methods provide an}

alternative to the elimination methods

• _{The Gauss-Seidel method is the most}

commonly used iterative method.

• _{Assume that we are given a set of n}

equations:

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• _{Suppose that for conciseness we limit ourselves to a 3×3 set} of equations.

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• _{Now, we can start the solution process by choosing} guesses for the x’s.

• _{A simple way to obtain initial guesses is to assume that} they are all zero.

• _{These zeros can be substituted into Eq. (2.a), which can} be used to calculate a new value for x1 = b1/a11.

• _{Then, we substitute this new value of x}₁_{along with the} previous guess of zero for x3 into Eq. (2.b) to compute a

new value for x2.

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• _{Then we return to the first equation and}

repeat the entire procedure until our solution converges closely enough to the true values.

• _{Convergence can be checked using the}

criterion

s j

i j i j i i,

a . %

x x x



   



100

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Example 1

Use the Gauss-Seidel method to obtain the solution of the system

Recall that the true solution is x1=3, x2=−2.5, and x3=7.

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Jacobi Iteration

• _{An alternative approach, called Jacobi}

iteration, utilizes a somewhat different tactic.

• _{Rather than using the latest available x’s, this}

technique uses Eq. (2) to compute a set of new x’s on the basis of a set of old x’s.

• _{Thus, as new values are generated, they are}

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Example 2

Use the Jacobi Iteration method to obtain the solution of the system

Recall that the true solution is x1=3, x2=−2.5, and x3=7.

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Homework