# System of Linear Equation

**Topics:**System of linear equations, Numerical linear algebra, Pivot element

**Pages:**18 (3158 words)

**Published:**February 28, 2013

Elankovan Sundararajan School of Information Technology Faculty of Information Science and Technology TR 3923 Elankovan Sundararajan 1

Lecture 3

System of Linear Equations

TR 3923

Elankovan Sundararajan

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Introduction

• Solving sets of linear equations is the most frequently used numerical procedure when real-world situations are modeled. modeled Linear equations are the basis for mathematical models of 1. 2. 2 3. 4. 5. Economics, Computational Biology Comp tational Biolog and Bioinformatics Bioinformatics, Weather prediction, Heat and mass transfer, Statistical analysis, and a myriad of other application.

•

•

TR 3923

The methods for solving ODEs and PDEs also depend on them.

Elankovan Sundararajan

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System of Linear Equations

• Consider the following general set of n equations in n unknowns: a11 x1 a12 x2 a1n xn c1 : R1

a21 x1 a22 x2 a2 n xn c2 , an 1,1 x1 an 1, 2 x2 an 1,n xn cn 1 , an ,1 x1 an , 2 x2 an ,n xn cn . Which can be written in matrix form as:

: R2

: R n-1 : Rn

A x b.

~ ~

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where, A is the nxn matrix, ,

a11 a12 a1n a21 a22 a2 n A . a an 2 ann n1

x iis th ( x 1) column vector, the (n l t

~

x x1 , x2 ,, xn

~

T

x1 x2 . x n 5

TR 3923

Elankovan Sundararajan

• Containing the n unknowns which we seek, and g , the known (n x 1) column,

c

~

is

c c1 , c2 ,, cn

~

T

c1 c2 . c n

• We shall be employing the Gauss elimination scheme to solve for the vector x . Here the forward elimination ~ process leads to an upper triangular system of equations and the solutions are recovered by means of backward substitution substitution.

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Elankovan Sundararajan

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a)

i.

Forward Elimination

First Gauss elimination step Retain R1: a1,1x1+ a1,2x2+ … + a1,nxn=c1. Eliminate x1 from the second to the nth equations

a2,1 R1 , R2 : R2 a1,1

a a a a2,1 2,1 a1,1 x1 a2,2 2,1 a1,2 x2 a2,n 2,1 a1,n xn R2 ~ a1,1 a1,1 a1,1 a c2 2,1 c1. a1,1

a a a a2,2 2,1 a1,2 x2 a2,n 2,1 a1,n xn c2 2,1 c1. i.e. R2 ~ a1,1 a1,1 a1,1

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a2 , 2

Elankovan Sundararajan

a 2 ,n

c2

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i.e. R2 ~ a2,2x2 a2,3x3 a2,n xn c2 . Do the same row operations for R3, R4, … , Rn, i.e.

i.e.

a3,1 R3 : R3 R1 , a1,1 a4,1 R1 , R4 : R4 a1,1 Rn : Rn an ,1 R1. a1,1

i.e.

ai ,1 Ri : Ri R1 , a1,1 i 2, 3, , n. ai ,1 ci : ci c1. a1,1

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Elankovan Sundararajan

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• Therefore at the end of the 1st Gauss elimination Therefore, step, we obtain the following reduced system: R 1 : a1,1 x1 a1, 2 x2 a1,3 x3 a1,n xn c1 , R : 2 R : 3 a2, 2 x2 a2,3 x3 a2,n xn c2 , a3, 2 x2 a3,3 x3 a3,n xn c3 , R -1 : n R : n TR 3923

an 1, 2 x2 an 1,3 x3 an 1,n xn cn 1 , an , 2 x2 an ,3 x3 an ,n xn cn . Elankovan Sundararajan 9

•

The 1st equation is called the pivot equation and a1,1 which appears in the multiplier ai ,1 , i=2,3, … , n is a1,1 called the pivot pivot. ii. Second Gauss Elimination step i.e. Ri: Ri a, 2 i a2 , 2 R2 ,

i 3, 4, , n. ai, 2 ci : ci c2 . a2 , 2

Retain R1, Retain R2, p , and at the end of the 2nd elimination process, we get the equivalent system: TR 3923 Elankovan Sundararajan 10

R 1 : a1,1 x1 a1, 2 x2 a1,3 x3 a1,n xn c1 , R : 2 R : 3 a2, 2 x2 a2,3 x3 a2,n xn c2 , a3,3 x3 a3,n xn c3 , R -1 : n R : n • •

an1,3 x3 an1,n xn cn1 , an,3 x3 an,n xn cn ....

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