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# 12Non-Homogeneous Linear System

By Victoria Robertson,2014-05-27 15:05
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12Non-Homogeneous Linear System

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II. System of Non-Homogeneous Linear Equations

a11 x1 + a12 x2 + L + a1n xn = b1 a x + a x +L+ a x = b 21 1 22 2 2n n 2 LLLL a m1 x1 + a m 2 x2 + L + a mn xn = bm

a11 a12 a 21 a 22 A= M M a m1 a m 2

Coefficient Matrix

??1??

L a1n x1 b1 L a2n x2 b2 L M X = M B= M x b L a mn n m

Matrix form Of equations

AX = B

AX = O

Guiding system

Use a11 a12 Vectors a 21 a 22 ?Á1 = ?Á2 = M M

a m1 a m2

L

a1n a2n ?Án = M a mn

b1 b2 ?Â = M b m

x1?Á1 + x2?Á 2 + L + xn?Á n = ?Â

Vector equation of the system

Determine solutions of nonhomogeneous system

System (1) has ?Â can be linearly represented by ?Á1 , ?Á 2 , L , ?Á n ; solutions A = (?Á , ?Á , L , ?Á , ?Â ), r ( A) = r ( A )

1 2 n

The matrix of system (1) can be expanded as A = (?Á1 , ?Á 2 , L , ?Á n , ?Â ).

Methods to solve nonhomogeneous system

1. Properties of nonhomogeneouse system??s solutions

Property 1: The margin of two solutions of system (1) is a solution of guiding system.

A?Ç1 = B, A?Ç 2 = B A(?Ç1 ?Ç 2 ) = O

Property 2?ºThe sum of a solution of nonhomogeneous system(1) and a solution of guiding system is a solution of nonhomogeneous system(1).

A?Ç = B, A?Î = O A(?Ç + ?Î ) = B 2. General solution of nomhomogeneous system

Theory : Let ?Ç is a special solution of nonhomogen eous system,

?Î1, ?Î 2 ,L, ?Î n r is fundamental solution system of its guiding system .

Then general solution of system (1) is

k1 , k 2 , L, k n r

Deduction

?Ç + k1?Î1 + k 2?Î 2 + L + k n r?Î n r are constants, and r = r ( A).

(i ) When r ( A) = r ( A ) = n, system ??1?? a unique solution; has

(ii ) When r ( A ) = r ( A ) < n , system????has infinite solutions, and its 1

General solution is

?Ç + k1?Î1 + k 2?Î 2 + L + k n r?Î n r

(iii ) When r ( A) ?Ù r ( A ), system 1 has no solution. ????

e.g.1. Solve the system

1 1 0 ?ú 0 2 0 1 0 5 M 3 1 ?ú 0 1 3 M 2 ?ú 0 0 0 0 M 0 0

Equivalent system

1 2 1 M A = 2 3 1 M 4 7 1 M

x1 + 2 x2 x3 = 1 2 x1 + 3x2 + x3 = 0 4 x + 7 x x = 2 2 3 1

r ( A ) = r ( A) Has solutions

2 1 M 1 1 2 1 M 1 1 3 M 2 ?ú 0 1 3 M 2 0 0 0 M 0 1 3 M 2 0 5 M 3 1 3M 2 0 0 M 0

x1 = 5 x3 3 x 2 = 3 x3 + 2

x1 = 3 x3 = 0, x2 = 2 x1 = 5 x3 = 1, x2 = 3

General Solution :

?à?Ç = (3,2,0)T is a special solution.

So, the fundamental ?Î = (5,3,1) T system is

?Ç + k?Î

x1 = 5 x3 x 2 = 3 x3

e.g.2. Find the general solution.

x1 x2 x3 + x4 = 0 x1 x2 + x3 3x4 = 1 x x 2 x + 3 x = 1 / 2 2 3 4 1

0 0 1 1 1 1 1 1 1 1 A = 1 1 1 3 1 ?ú 0 0 2 4 1 1 1 2 3 1 / 2 0 0 1 2 1/ 2

1 1 1 1 0 1 1 0 1 1/ 2 1 2 1/ 2 ?ú 0 0 1 2 1/ 2 ?ú 0 0 0 0 0 0 0 0 0 0 0 0

Equivalent system

x1 = x2 + x4 + 1 / 2 x3 = 2 x 4 + 1 / 2

r ( A ) = r ( A) Has solutions

x2 = x4 = 0 x1 = x3 = 1 / 2 ,?Ç = (1 / 2, 0, 1 / 2, 0)T is a special solution.

x1 = x2 + x4 x3 = 2x4

x1 1 1 x2 1 0 = , = , x 0 1 x 0 2 4 3

the fundamental system is

?Î1 = (1,1,0,0)

T

?Î 2 = (1,0,2,1)

T

General solution is ?Ç + k1?Î1 + k2?Î 2

Process of solving homogeneous system 1. Find the A, and change A into echelon form, then find r ( A)

and r ( A ) to determineif it has solutions. How to determine?? 2.If

it has solutions,change A into row simplestform,and computethe real unknownand free unknown,and then find the equivalentsystem;

3.Let free unknown variables are zero, we can figure out the real unkown, and the special solution ?Ç is computable; use some values instead of free unkown, and we get fundamental solutions system and also the general solution.

What to note??

Completement System with parameters

We should determine the parameters before solving the system.?ªthis is obligatory.To determine the parameters, the following form should be satisfied.

r ( A ) = r ( A)

2 x1 + ax2 x3 = 1 e.g.1. What is a, can the system ax1 x2 + x3 = 2 have no solution? 4 x + 5 x 5 x = 1 2 3 1 or unique solution? or infinite solutions? and if it has, figure out the solutions.

Generally, we have two ways. First we can use determinant Second we can use elementary operations.

2 a 1 = 10 + 4a 5a 2 a 1 A = a 1 1 A = a 1 1 4 10 + 5a 2 4 5 5 4 5 5 = 5a 2 a 4 4 A = 0 a = 1, a = ?à a ?Ù 1 and a ?Ù 4 , then the system has unique solution. 5 5

2 x1 + x2 x3 = 1 When a = 1, the system is x1 x2 + x3 = 2 4 x + 5 x 5 x = 1 2 3 1

The system has no parameters any more. The system has no parameters any more.

4 2 x1 5 x2 x3 = 1 4 4 When a = , the system is 5 x1 x2 + x3 = 2 5 4 x1 + 5 x2 5 x3 = 1

x1 x 2 x3 + x 4 = 0 e.g.2. What is ?Ë , can the system x1 x 2 + x3 3 x 4 = 1 have solutions? x x 2 x + 3x = ?Ë 2 3 4 1

1 1 1 1 0 1 1 1 1 0 A = 1 1 1 3 1 ?ú 0 0 2 4 1 1 1 2 3 ?Ë 0 0 1 2

0 1 1 1 1 1 2 1/ 2 ?ú 0 0 0 0 0 0 ?Ë + 1/ 2

1 When ?Ë = ??r ( A ) = r ( A)?? system has solutions. the 2

Question : Can we solve it by determinant?

No ??

Two questions about the system. 1.let nonhomogen eous system AX = B, the rank of its coefficien t

matrix A is 3,?Ç1 ,?Ç2 ,?Ç3 are its threespecialsolutions,and?Ç1 = 2??4?? T?? ?? 3??5??

?Ç2 +?Ç3 = 1??3?? T?? the generalsolutionof AX = B. ?? 2??4?? find

For the hypothesis, there is only one solution vector in the fundamental solution system which is ?Î = 2?Ç1 (?Ç 2 + ?Ç ) = ??2??4?? T 1?? 3?? T = (3,4,5,6)T ?? 2 3?? 5?? ?? 2??4??

?à The general solution is ?Ç1 + k?Î . ??please look at it in reference page 82??

1 ?? ??What are a, b, can ?Â be linearly represented by ?Á1 , ?Á 2 , ?Á 3 ? What is the formula ?

2 . Let ?Á 1 = (1,3,0 ,5 )T , ?Á 2 = (1, 2 ,1, 4 )T , ?Á 3 = (1,1, 2 ,3)T , ?Â = (1, a ,3, b )T .

??2??What are a, b, can ?Â not be linearly represented by ?Á1 , ?Á 2 , ?Á 3 ?

x1 ?Â = (?Á1 , ?Á 2 , ?Á 3 ) x2 = AX x 3

1 3 A = (?Á1 , ?Á 2 , ?Á 3 ) = 0 5

Let ?Â = x1?Á 1 + x 2?Á 2 + x3?Á 3

1 1 x1 2 1 , X = x2 1 2 x 3 4 3

The question what are a, b, can ?Â be linearly represented by ?Á1 , ?Á 2 , ?Á 3 , can be transformed into whether the system AX = ?Â has solutions.

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