ch.1_ linear system of equations

8/2/2019 Ch.1_ Linear System of Equations

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LINEAR SYSTEM OF EQUATIONS

Learning outcomes : by the end of this chapter

1. You should know,a) What is a linear system of equations b) What is a homogeneous system

c) How to represent a linear system in matrix formd) What is a coefficient matrix

e) What is an augmented matrix2. You should be able to solve a linear system of equations using:

a) Inverse matrix method

b) Cramers rule

c) Row operations:

i. Gauss-elimination method (REF)ii.

Gauss-Jordan method (RREF)

Definition of a Linear Equation in n Variables:A linear equation in n variable n x x x ,,, 21 has the form

b xa xa xa nn =+++ 2211 ,where the coefficients baaa n ,,,, 21 are real numbers (usually known). Thenumber of 1a is the leading coefficient and 1 x is the leading variable . The collection of several linear equations is referred to as the system of linear

equations . Definition of System of m Linear Equation in n Variables:1


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A system of m linear equations in n variables is a set of m equations,each of which is linear in the same n variables:

mnmnmm

nn

nn

b xa xa xa

b xa xa xa

b xa xa xa

=+++

=+++=+++

2211

22222121

11212111

where ,,1,,,2,1 ,, n jmiba iij == are constants.

Example:1Consider the following system of linear equations:

67

832

3,4 2523

43

31

321

321

321

==+

===+=+

x x

x x x

nm x x x

x x x

Example: 2Which of the following are linear equations?2

1 2 3

1 2 3

( ) 3 2 7 ( ) (sin ) 4 (log5) ( ) 231( ) 2 4 ( ) sin 2 3 0 ( ) 4 x

a x y b x x x e c x y

d e y e x x x f x y

+ = + = + =

= + = + =

( ) and ( ) are linear equations.a b ( ),( ), ( ), and ( ) are not linear.c d e f Number of Solutions of a System of Linear EquationsConsider the following systems of linear equations

(a)13

x y

x y

=+ = (b)

42

x y

x y

+ =+ = (c)

6 2 83 4

x y

x y

+ = =

For a system of linear equations, precisely one of the following is tru(a) The system hasexactly one solution.(b) The system hasno solution.(c)

The system hasinfinitely many solutions .Consistent and Inconsistent2

) sol

1 x y =

3 x y+ = 2 x y+ =

4 x y+ =


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A system of linear equations is calledconsistent if it has at least one solutioandinconsistent if it has no solution.

EquivalentTwo systems of linear equations are said to beequivalent if they have the sameset of solutions.Back SubstitutionWhich of the following systems is easier to solve?

2 3 9( ) 3 7 6 22

2 5 5 17

x y z

a x y z

x y z

+ = + =

+ =

2 3 9( ) 3 5

2

x y z

b y z

z

+ =+ =

=System (b) is said to be inrow-echelon form . To solve such a system, use procedure calledback substitution.Augmented Matrices and Coefficient Matrices

Consider them n linear system

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

...

n n

n n

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

+ + + =+ + + =

+ + + =

LL

M

Let

[ ]

11 12 1 1 11 12 1 1

21 22 2 2 21 22 2 2

1 2 1 2

, , |

n n

n n

m m mn m m m mn m

a a a b a a a b

a a a b a a a b A b B Ab

a a a b a a a b

= = = =

L K L K

M M M M M M O M ML K

A is called the coefficient matrix of the system.

B is called the augmented matrix of the system.b is called theconstant matrix of the system.It is possible to write the system in the following matrix form

3


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B X A

b

b

b

m

2

1

2

1

21

22221

11211

=

nmnmm

n

n

x

x

x

aaa

aaa

aaa

Example:

65y-213

==

x

y x

B XA61 y

x 5213 =

Row-Equivalent

Twom n matrices are said to be row-equivalent if one can be obtaineother by a series of elementary row operations.

Now we are in the stage to tacklethe question.

How to solve a linear system AX =B?

First:Row operationsThe key to solve a system of linear equations is to transform theaugmented matrix to some matrix with some properties via a fewelementaryrow operations. As a matter of fact, we can solveany system of linear equation by transforming the associate augmented matrix to a matrix in someform. The form is referred to as thereduced row echelon form .

Definition of elementary row operation:There are 3 elementary row operations:1. Interchange two rows

2. Multiply a row by some nonzero constant

4


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Add a multiple of a row to another rowDefinition of a matrix in reduced row echelon form:

A matrix in reduced row echelon form has the following properties:

1. All rows consisting entirely of 0 are at the bottom of the matrix.2. For each nonzero row, the first entry is 1. The first entry is called aleading 1.

3. For two successive nonzero rows, the leading 1 in the higher rowappears farther to the left than the leading 1 in the lower row.

4. If a column contains a leading 1, then all other entries in that columnare 0 .

Note: a matrix is in row echelon form as the matrix has the first 3properties.

Example 3

And

00

10

0

00

00

3

00

01

0

00

00

0

00

00

1

are the matrices in reduce

row echelon form.The matrix

0000

2100

5210

4321

is not in reduced row echelon form butin row echelon form since the matrixhas the first 3 properties and all the other entries above the leading 1 column are not 0. The matrix

0000

2210

5210

4301

arenot in row echelon form (alsonot in reduced row echelon form) since5

00012

00100

00010

00002

00001


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leading 1 in the second row is not in the left of the leading 1 in theand all the other entries above the leading 1 in the third column are n

Definition of elementary row operation:There are 3 elementary row operations:3. Interchange two rows4. Multiply a row by some nonzero constant5. Add a multiple of a row to another row.

1.1 Gauss-elimination method

(REF)Step 1: Form augmented matrix[ ]b A : Step 2: Transform[ ]b A : to row echelon form matrix[ ] DC : using rowoperations

Step 3: Solve the system corresponding to[ ] DC

: , using back substitution

Example: Solve the following system of equations3-z-y2-x3

1yx

5z3y-2

==+=+ x

using

Gauss elimination.Sol.

3:1231:0115:312

11 R 21 R

3:1231:01125:23211

R33R R R

13212

R

R

2/21:2/112/102

3:23

230

25:23211

33

22R 2R 32

R

R

21:11101:11025:23211

323 R R R

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22:12001:1102

5:23

21133 R 121 R

611:1001:1102

5:23

211

So, z = 11/6 , y = 5/6, x = 1/6

1.2 Gauss-Jordan Reduction

Method (RREF)Step 1: Form augmented matrix[ ]b A : Step 2: Transform[ ]b A : to reduced row echelon form matrix[ ] F H : usingrow operationsStep 3: for each nonzero row in[ ] F H : , solve the corresponding equations

Example: Solve the following linear system of equations3 z -x3

8 z y2x

9z32y

==+=++ x

using

Gauss-Jordan reduction methodSol.

3:103

8:1129:321

313

212

R 3R -

R 2R

R

R

24:1060

10:550

9:321

22 R 51 R

24:10602:110

9:321

121

323

R 2R

R 6R

+

R

R

12:4002:110

5:101

33 R 41 R

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3:100

2:110

5:101

131

232

R R

R R

R

R 3:100

1:010

2:001

So, x = 2, y = -1, z = 3

1.3Using inverse matrix methodThe Inverse of matrix A:

Definition: A square matrix A is said to be nonsingular (invertible) if therea matrix B such that:

Anxn Bnxn = B nxn Anxn = I nxn

B is called the inverse of A and denoted by A-1 so A A-1= A-1A = I If B does not exist, then we say that A is singular (noninvertible) matrix

Example: = 4321

A

Nonsingular = 4221

A

SingularProperties:

Let A be a nonsingular matrix, then:1. ( A-1 )-1 = A2. ( A-1 )T = ( AT )-1

3. ( AB)-1 = B-1 A-1

4. ( A1 A2 A3 An )-1 = An-1 An-1-1 A2-1 A1

How to find A-1?Basically there are two methods

1. Row operation

2. Using determinant and adjoint matrix8


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Procedure of computing 1 A using row operations

Let A be annxn matrix

Step1:Form the)2( nn

matrix[ ] I A

|Step2:Use elementary row operations to transform[ ] I A | to the form[ ] B I |Step3:If the form[ ] B I | is possible the B A =1 otherwise A is singular.

Example: Use row operations to find the inverse of 101-1452321

= A

If A is a nonsingular matrix, then A-1 exist hence we can get the solution as

1

1

11

B A X

B A X I

B A A X A

-

-

-

===

Example: Using inverse matrix method, find the solution of system

52y3x

1y- 2

=+= x

51 y

x 231-2 = 23

1-2 A= ,

Finding the Inverse of a 2 X 2 matrix

Let a b A c d = and suppose det(A) =ad cb 0. Then 1 A exists and is

given by 1 1 d b A c a D =

.

det (A) = 7 0. So, A-1exist

=23-

12

7

1 A 1-

yx

1

1

7

7

7

1

5

1

23-

12

7

1 BA 1-

===== X

9


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So the solution is x = y = 1Calculation of Inverse Matrix:Example

To find the inverse of

=

531

532

211

A , we can employ the procedu

introduced above.

1

0

0

0

1

0

0

0

1

5

5

2

3

3

1

1

2

1

. =

+=

)1(*2)2()2()1()3()3(r r r

r r r

1

0

0

0

1

0

1

2

1

3

1

2

2

1

1

0

0

1

= )2(*1)2( r r 1

00

0

10

1

21

3

12

2

11

0

01

=

+=

)2(*2)3()3()2()1()1(r r r

r r r 1

00

2

11

3

23

1

11

0

10

0

01

=

+=

)3()2()2()3()1()1(

r r r r r r

1

1

1

2

3

1

3

5

0

1

0

0

0

1

0

0

0

1

The inverse of A is

123135

110

.

2. Using the adjoint)( Aadj

of a matrix to calculate A-1

As 0)det( A , then )det()(1

A Aadj

A = .

Note: If 0)det( A A is nonsingular

1.4 Using Cramers ruleThis Method is to solve a linear system of equations AX = B, if A isnxn

matrix nonsingular matrixStep 1: Find det (A). If det A = 0, then the method cant be applied

Step 2: If det (A) 0, then(A)detAdet i=i x

10


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where Ai is the matrix obtained from A by replacing the ith column of A by B If n 4 Cramers rule is computationally inefficient.

Example: Use Cramers rule to solve the system:

3-zy-2x-

4 z-2yx

1 z-3y2x-

=+=+=+

,1121211-32-

= A =3-41

B

2- -1(3)(-1)3-(1)2-

4)(-1)(-12)-3(1-1)-(22-

= +=

++=

2 AA x4- (2)(-1)3(1)-)1(1

11-3-1-241-31

A xx ===+==

3 A

y6- (5)(-1) (-1)1-(1)2- 13-2-1-411-12-

A yy ===+== A

4AA z 8- 1(3)3(5)-2(-2)-

3-1-2-421132-

A zz ===+==

Hence the solution is x = 2, y = 3, z = 4

DeterminantsCofactor Expansion

This is an other method for evaluating the determinant of nxn matrix, where det11

11-2-

1-21

1-32-

= A


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( A) is defined as:

AaAa )det( i2i2i1i1 inni Aa A +++= where 1 i n expansion over therow i or

AaAa )det( 2j2j1j1j njnj Aa A +++=

where 1 j n expansion over thecolumn j

What is Aij? Aijis thecofactorof a ijgiven by Aij = (-1)i+jdet ( M ij), where M ij is the (n-1)x(n-1) submatrix of A obtained bydeleting the ith row and the jth column.

Whenever you want to use cofactor method choose the row or the colthe highest number of zeroes

Example: Use cofactor method to evaluate the determinant of =732501-321

A

A21= (-1)2+1det M21 = 7332 21 M 55x(-1)A 1221 == +

A23= (-1)2+3det M23 =32

21 23 M 11-x(-1)A 3223 == +

Since 022 =a no need to evaluate 022 = A and 101551-AaAdet 23232121 =+=+= AaProperties of Determinant

1. det ( AB ) = det A det B2. det ( AT) = det ( A)

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3. ( )11 A 1 == A

A

4. If r i = 0 orci = 0 , then det(A) = 0

152

000

5-31

= A r2 =0 So, A = 0

5. if r i = r j orci =c j, then det(A) = 01-51-222131

= A c1 = c3 So, A =

0

6. if we interchange : ri r j or ci c j then )det()det( A A New =

71521-1213

= A r3 r1 21321-1715

= B if det A = c then

det B = - det (A) = - c7. if kr i r i or kci ci , then det(A New) = kxdet(A ), ( k 0)

1051-129-3 421 = A c3 c3 =

551-69-3 221 B det (B) = (1/2)x det (A)

8. if kr j + r i r i or kc j +c j c j , then det(A New) = det(A )Inverse of Matrix a Second Method

How can we use determinant to find the inverse of a matrix?

Definition : Let A benxn nonsingular matrix. The inverse of A is given by13


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(A)adj (A)det

1 1 = A

How to find adj (A)? Theadjointof A , is the matrix whose (i, j )th element is the cofactor A jiof

a ji =nnnn

n

n

A A A

A A A

A

Aadj

21

22212

12111 AA

)(

A matrix is nonsingular if and only if det (A) 0. If det A = 0, A-1does not exist and we say that A is singular

HW: Find the inverse of the matrix

=217

654

21-3

A

Cofactor Expansion, Inverse Matrix And Determinants:

Definition of cofactor:

Let ija A= be nn matrix. The cofactor of ija is defined as

( ) )det(1 ij ji

ij M A+= ,

where ij M is the )1()1( nn submatrix of A by deleting the i th row of

14


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j th column o .f A

Definition of adjoint:

The nn matrix )( Aadj , called the adjoint of A, isT

nnnn

n

n

nnnn

n

n

A A A

A A A

A A A

A A A

A A A

A A A

Aadj ==

21

22221

11211

21

22212

12111

)( .

Important result:

n I A A Aadj Aadj A )det()()( ==and

15

)det()(1

A Aadj

A =


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Example:Let

=

531

241

302

A

Then,

=

=

=

31

41 ,

51

21 ,

53

24131211 M M M ,

==

=

31

02 ,

51

32 ,

53

30232221 M M M ,

=

=

=

41

02 ,

21

32 ,

24

30333231 M M M

Thus,

( ) [ ]( ) ( ) [ ]( ) ( ) [ ]( ) ( ) [ ]( ) ( ) [ ]( )

1 111

1 2 1 212

1 3 1 313

2 1 2 1

11

12

13

221

2 2 2 222

2 323

1

22

23

1 1 4 5 ( 2) ( 3) 14,

1 1 ( 1) 5 ( 2) 1 3

1 1 ( 1) ( 3) 4 1 1

1 1 0 5 ( 3) 3 9

1 1 2

det( )

det( )

det( )

det( )

det( )

de

5 3 1 7

1 t(

M

M

M

A

A

A

A

A

A

M

M

M

+

+ +

+ +

+ +

+ +

+

= = == = == = == = = = = == ( ) [ ]

( ) ( ) [ ]( ) ( ) [ ]( ) ( ) [ ]

2 3

3 1 3 1

31

3 2 3 232

3 3 3 333

31

32

33

)

det( )det( )

det

1 2 ( 3) 0 1 6

1 1 0 ( 2) 3 4 121 1 2 ( 2) 3 ( 1) 1

1 1 2 4 0 ( 1) 8( )

M M

A A

M A

+

+ +

+ +

+ +

= =

= = = = = == = =

==

861

173

12914

)(

332313

322212

312111

A A A

A A A

A A A

Aadj

and

16


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==

861

173

12914

251

)det()(1

A Aadj

A .

Homework:

=

32-023-0033124-43-21

A Let

1. Evaluate det A2. Evaluate A-1 using adj A

Use REF & RRFE to solve the following system:

5-w7 z2 y3-x-11w3 z -yx23-w9- z -y5x2

3 w5- z2 y

=++=++=+=++ x

Ex.6 Solve the following system by Gaussian-Jordan elimination.2 8 102 2

7 17 7 1

y z

x y z

x y z

+ = + =+ + =

.

17

) sol


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0 2 8 10 1 2 1 2 1 2 1 21 2 1 2 0 2 8 10 0 2 8 107 17 7 1 7 17 7 1 0 3 14 15

1 2 1 2 1 2 1 2 1 2 0 2 1 0 0 120 2 8 10 0 1 4 5 0 1 0 5 0 1 0 50 0 2 0 0 0 1 0 0 0 1 0 0 0 1 0

The

solution is 12, 5, 0. x y z = = =Ex.7 Find a condition on numbers a, b, and c such that the followingsystem is consistent. When that condition is satisfied, find all solutions .

323 7

x y z a x y z b

x y z c

+ + = + =+ =

1 3 1 1 3 1 1 3 11 2 1 0 1 2 0 1 2

3 7 1 0 2 4 3 0 0 0 2

a a a

b b a a b

c c a c a b

+ + +

The

system is consistent 2 0c a b + = .When 2 0c a b + = , the last matrix becomes

1 0 5 2 3

0 1 2

0 0 0 0

a b

a b

+ Let z t = . The general solution is

5 (2 3 )2 ( ) : arbitrary

x t a b

y t a b t

z t

= += + +=

Number of solutions of a non-homogeneous system of linear

equations

For a non-homogeneous system , precisely one of the following is

true.

(1) the system has a single (unique) solution;

(2) the system has more than one solution;18

) sol


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(3) the system has no solution at all.

ExamplesI. Exactly one solution:Solve for the following system:

3 3

82

932

31

321

321

=

=+=++

x x

x x x

x x x

[Solution:]The Gauss-Jordan reduction is as follows:

Step 1: The augmented matrix is

3

8

9

1

1

3

0

1

2

3

2

1

.

Step 2:The matrix in reduced row echelon form is 3

1

2

1

0

0

0

1

0

0

0

1

Step 3: The solution is 3 ,1 ,2 321 === x x x

II. Infinite number of solutions:

Solve for the following system:1 530242

21

321

=+

=+

x x

x x x



1

0

0

2

5

4

3

2

Step 2: The matrix in reduced row echelon form is

19


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12

3

5

1

0

0

1

Step 3: The linear system corresponding to the matrix in reduced row

echelon form is13

25

32

31

=

=+

x x

x x

The solutions are 1 3 22 5 , 1 3 x x x= = +

3 x is free variable or parameter and let 3 , x t t R= therefore

Rt t xt xt x =+== , ,31 ,52 321III.No solution:Solve for the following system:

62 17535422

431

4321

4321

=

=+++

=+++

x x x

x x x x

x x x x


Step 1:The augmented matrix is 611

5

2

7

4

1

5

3

0

3

2

1

1

1

Step 2: The matrix in reduced row echelon form is

1

0

0

0

3

2

0

2

1

0

1

0

0

0

1

Step 3: The linear system corresponding to the matrix in reduced row

echelon form is1 0 032 02

432

431

==++=

x x x

x x x

Since ,10 there is no solutionExample:Solve for the following linear system:

20


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5723 -113 2

-39 52 352

4321

4321

4321

4321

=++

=++

=+

=++

x x x x x x x x x x x x x x x x

[Solution:] The Gauss-Jordan reduction is as follows:


511

33

73

95

21

12

31

51

12

21

Step 2:After elementary row operations, the matrix in reduced row

echelon form is

0

32

5

0

23

2

0

10

0

0

01

0

0

00

1

.

Step 3:The linear system corresponding to the matrix in reduced rowechelon form is

3423 243 2

542 1

==

=+

x x x x

x x

The solutions are 1 4 2 4 35 2 , 2 3 , 3 x x x x x= = +

Rt t xt xt xt x =+=+== , ,23 ,32 ,25 4321Example:Find conditions ona such that the following system has no soluone solution, or infinitely many solutions.

1

( 2) 1

2 2 ( 2) 1

x ay z

x a y z

x y a z

+ = + + =

+ + =

21

) sol


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1 1 1 1 1 1 1 1 11 2 1 1 0 2 2 0 0 0 2 2 0 0

2 2 2 1 0 2 2 1 0 0 11 1 1 1

Case1: 1 0 0 1 10 0 0 01 1 1

Case2: 1 0 1 0 00 0 1

1 0 1 1(a) 0 0 1 0 0

0 0 0 1

a a a

a a a

a a a a

a

a

a

a

a

=

=

1 1 1(b) 0 0 1 0 0

10 0 1

a

a

a

1 : has infinitely many solutions.0 : has no solutions.1 and 0 : has exactly one solution.

a

a

a a

==

Rank of Matrix ADefinition : The dimension of row space of A is called

therow rank of A, the dimension of columnspace of A is called thecolumn rank of A row rank of A= column rank of A

How to find it?Step1: C A RREF Step2:rank A = the number of non zero rowsin CTheorem: Let A be anm x n matrix. Then,

22


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{dimdim dimrowspace nullspace

rank A nullity A n+ =1 2 3 14 2 43

Rank and Singularity

First : Nonhomogeneous system

The rank of square matrix can be used to determine whether the matrix

is singular or nonsingular, as the following theorem.

Theorem: Ann n matrix A isnonsingulariff rank A= n

Corollaries (1) If A is an n n matrix, then rank A = n iff det( ) 0 A .

(2) Let be an n n matrix. The linear system Ax = b hasa unique solution

for every 1n matrixb Iff rank ( A) = n.(3) rank ( A) = n , A is nonsingular so A-1exist Ax = b

AA-1x = A-1 b; x = A-1

Ex.8 Find the rank of the matrix1 1 1 42 1 3 0

3 4 8 20

A =

.

1 1 1 4 1 1 1 4 1 1 1 42 1 3 0 0 1 5 8 0 1 5 83 4 8 20 0 1 5 8 0 0 0 0

A

= Hence 2rankA = .

23

) sol


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Ex.9 Find the rank of the matrix21 1 2

1 1 2 02 2 6 4

a

A aa a

=

2 2 2

2 2

2 2

2

2

1 1 2 1 1 2 1 1 2

1 1 2 0 0 0 0 0

2 2 6 4 0 2 4 2 0 0 2 4

1 1 2 0

Case1: 0 0 0 1 2 20 0 0 0

1 1 2

Case2: 0 0 1 0

0 0 2 4

(a

a a a

A a a a a a

a a a a a a a

a rankA

a

a a

a a

=

= =

2

1 1 2 4

) 2 0 1 0 2 2

0 0 0 0

1 1 2

(b) 2 0 1 0 3

0 0 1 2

2 if 0 or 2Hence

3 if 0 and 2

a rankA

a

a a rankA

a

a arankA

a a

= = = +

= ==

(a) Either or 1 1rankA rankB r rankA rankB r = = = = .(b) { } { }min , 1 , min ,r rankB m n rankA m n= + .

24

) sol


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(c) If m n< then r m n < , hence either the system is inconsistent or infinitely many solutions.

Ex.10:Find conditions ona such that the following system has no solutio

solution, or infinitely many solutions.1

( 2) 1

2 2 ( 2) 1

x ay z

x a y z

x y a z

+ = + + =

+ + =

1 1 1 1 1 1 1 1 11 2 1 1 0 2 2 0 0 0 2 2 0 0

2 2 2 1 0 2 2 1 0 0 11 1 1 1

Case1: 1 0 0 1 10 0 0 01 1 1

Case2: 1 0 1 0 00 0 1

1 0 1 1(a) 0 0 1 0 0

0 0 0 1

a a a

a a a

a a a a

a

a

a

a

a

=

= 1 1 1

(b) 0 0 1 0 010 0 1

a

a

a

1 : has infinitely many solutions.0 : has no solutions.1 and 0 : has exactly one solution.

a

a

a a

==

Theorem 3Let be the coefficient matrix of an linear system.The following statements are equivalent:(a) The system has a unique solution.(b)(c) is row-equivalent to .n

A n n

rankA n

A I

=

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) sol


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SECOND: Homogeneous system

Theorem:

If A ann n

matrix, the homogeneous system0 Ax = has a nontrivial solution iff A issingular.PROOFSuppose that A is nonsingular. Then A-1

1 1

1

( ) 0

( ) 0

0

0

n

A Ax A

A A x

I x

x

==

=

=Then the only solution to HS is x=0

1. The system Ax = 0 has a nontrivial solution:a) If rank A < n b) rank = n =1 det A = 0

c) det A 0 A-1does not exist (nontrivial)

Ax = 0 A-1Ax = 0 x = 0 (trivial)6) The linear system B AX = has a solution iff ]|[ B Arank rankA =

7) The system B AX = is inconsistent (no solution) iff B is not in the column spacTrivial Solution

Clearly, 1 2 ... 0n x x x= = = = is a solution to such a system, it is called the

solution. Any solution in which at least one variable has a nonzercalled a nontrivial solution.26

11 1 12 2 1

21 1 22 2 2

1 1 2 2

0

0

0

n n

n n

m m mn n

a x a x a x

a x a x a x

a x a x a x

+ + + =+ + + =

+ + + =

L

L

M

L


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Ex.1Show that the following homogeneous system has nontrivial 1 2 3 4

1 2 4

1 2 3 4

2 02 2 03 2 0

x x x x

x x x

x x x x

+ + =+ =+ + + =

1 1 2 1 0 1 1 2 1 0 1 1 2 1 0

2 2 0 1 0 0 4 4 3 0 0 4 4 3 0

3 1 2 1 0 0 4 4 2 0 0 0 0 1 0

1 1 2 1 0 1 1 2 0 0 1 0 1 0 03

0 1 1 0 0 1 1 0 0 0 1 1 0 040 0 0 1 0 0 0 0 1 00 0 0 1 0

Let 3 x t = . The general solution is

1 2 3 4, , , 0 : x t x t x t x t arbitrary= = = = .By taking 0t , we get a nontrivial solution.Theorem 1:Anm n homogeneous system is consistent. Moreover if m n< thenit has infinitely many solutions.Theorem 2:Let A be the coefficient matrix of ann n homogeneoussystem.The following statements are equivalent:(a) The system has only the trivial solution.(b) rankA n= .(c) A is row-equivalent ton I .

27

) sol

ch.1_ linear system of equations

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