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Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems

< 0,

> 0.

= ( ) < 0.

where < 0,

Consequently, the value

(1.109) when 0 we get

(t) ( ) y

(t)]dt = c1

l , | c1

|< .

[

Then

(t)dt < ,

( )

(t)dt <

From

	Hence	it follows that		lim y3 (t) = 0.	Next	applying theorem		7,	we get
				t
lim	y1 (t) =	0, lim y2	(t) = 0. So when < 0,		< 0,	> 0, 0 trivial solution of
t		t
the	system (1.105)		is absolutely stable and Iserman’s problem has a solution. As
shown above value			0 = .	So, in the sector [ , 0		= ) Iserman’s problem has a
solution.
	8. Absolute stability. As follows from (1.106)–(1.108) values N						0	= ,	N = 0,
							0		1

N	2	=		0,
	2
	(		2		2
	(

N	3	= (

2

2	2
	2

M	0	=
	0
)			1	,
)			0	,
	1		0

	1	,
1	0

M	3	=
	3

		2					2	2	1
M		=					2		1	,
M	1	=	1						0	,
	1		1					1	0
(				2	)		(1 2
(					)	1	(1 2
						1

M		= (			2	)
M	2	= (				)	1
	2						1
	2		2	2 )					1	.
	2								1
								0
						1		0

Now conditions of theorem 20 will be written as:

2 N2 M 2

						2 N0		M0 0 0 :													1				2		0;
						2 N0		M0 0 0 :						2 1 0								0					0;								(1.110)
			N M								0 :					2							2		2				2		0;
																					1		2						2						(1.111)
		2			1			1		1			1						1		0							1							(1.111)
0 :							(				2	)		1	(		2		2	2 2								2			)			2	2		0;
0 :							(					)			(					2 2											)				2		0;

	2					1						1		0																				2	1	3
																																			(1.112)
	M				> 0 :							( )							(									2		)
3	M			3	> 0 :						2	( )							(											)
3				3			3				2					2		1																	(1.113)
								1	(1		2			2		2	2							2		2							> 0,		(1.113)
									(1		2						2									2							> 0,

						1		0																3					0			2

where < 0, < 0, > 0, , , > 0 is a sufficiently small number. Hence, when 0, we get (v. (1.110)–(1.113)):

	N		M		0 :			1	2	0;
		0		0							(1.114)
2					0	2	1	0	0
			2 N1 M1 1			0 : 1 = 0;					(1.115)

Lectures on the stability of the solution of an equation with differential inclusions


						2	N	2	M			2	0 : 2			2	3	0;
						2		2				2	2	1	2	1	3
	N		M			> 0 :					( )			( )				1	2	2		> 0.
2	N	3	M	3		> 0 :				2	( )			( )					3	2	2	> 0.
2		3		3	3					2			1			1	0		3	0	2

(1.116)

(1.117)

It can be shown that when

< 0,

> 0,

ratios are performed

(1.114)–(1.117) when

	2

< 0,

	1


		2
	2	0

= 0,

	2

2 ( ) 32 > 0.

	1 0,	1			0	= ,
Wherein	1 0,	0	0.	Consequently,	0	= ,	i.e. for the system (1.105)

Iserman’s problem has a solution.

9. The effectiveness of the proposed method. The effectiveness of the proposed method will be shown for the case when = 1, = 2, = 0,1, = 1. In this case ratios (1.110)–(1.113) will be written as:

								− − Γ					≥ 0: 0.1			+			−1 − 2							≥ 0;				(1.118)
						2	0		0			0			2			1			0				0
								N M						0 :						1				2	0;					(1.119)
								N M			1			0 :	1					0			1		0;
							2		1		1		1		1		1			0			1
						−			− Γ	≥ 0: 3				+ 2.2			−1 − 2							+ 2				≥ 0;		(1.120)
				2	2			2	2				1			1	0					2				1	3
			−		− Γ > 0: −					2		− 2.09 − 3.19								−1 − 2						+ 2			> 0. (1.121)
	2	3		3		3				2				1				1			0				3			0	2
From			ratios		(1.118)–(1.121)								we	find		the				limit					value		0 = 3.96			when
2	= 0.01 2			, 2		= 3.0386 2,							= 0.1 ,			= 0.7982									2, −1			= 0.2018 2.
0			2		2				1			3		1	1									1		1	0			1

Notice that when	= 1, = 2,

equations of regulated system have a form

= 0,1

values

= 1.1,

= 1,

= 1

where

̇ = − + ( ), ̇ =
1	1	2
( ) = ( ) = ( ),

	, ̇ = ( ), = 1.1		−	2	− ,
3	3	1			3
= 1.		Then equation (1.86) can be written as:

̇ = 0.1 −

−

+ ( ), ̇ =

, ̇ = 1.1

− − + ( ). (1.122)

For comparison we define the value

0 according to the frequency method of

( )

V.M. Popov [15]. The transfer function from input

to the output is equal to

−0.1 2 + 2 + 1

−

= ( ) =

3 + 0.9 2 + 2 + 1

, ÷ , ̃ ÷ .

Frequency characteristic W (i ) = ReW (i ) iImW (i ),

where

( ) =

−2.09 2 + 3.2 + 1

( ) =

(− 2 + 0.1 4)

, = 2.

= 2.2 + 1

+ 2.2 + 1

− 3.19

Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems

Modified frequency response W* (i ) = X iY , where

X = ReW (i ), Y = ImW (i ).

Frequency condition of absolute stability of V.M. Popov has a form	1	> 0,
	X qY 0
, 0. Condition of V.M. Popov is performed for values 0	= 3.84, = 2.58.

So, the value 0 = 3.84 according to the method of V.M. Popov, and the value0 = 3.96 > 3.84 by the proposed method. It means that proposed method highlights in space of constructive parameters of the system (1.105) a wider area, than that known from the method of V.M. Popov.

Comments

A new method of study of absolute stability of regulated systems with limited resources based on an assessment of improper integrals along the solution of the system is proposed. Nonsingular transformation allowing us to use information about nonlinearity properties is found. The class of regulated systems for which Iserman’s problem has a solution is highlighted. For this class of regulated systems necessary and sufficient conditions of absolute stability is obtained. A difference of the proposed method of study of absolute stability from the known method is that the conditions of absolute stability are obtained without involving Lyapunov functions and frequency theorem. For systems with limited resources, phase variables are limited and are uniformly continuous functions. These properties were used for obtaining stability conditions, and also for evaluating improper integrals. This evaluation allows us to significantly expand the area of absolute stability in space of constructive parameters of the system rather than known results and in some cases we can get necessary and sufficient conditions of absolute stability.

A new method of study of absolute stability of the equilibrium state of regulated systems with limited nonlinearity based on the assessment of improper integrals along the solution of the system was developed.

The proposed method allows us to significantly expand the area of absolute stability in space of constructive parameters of the system rather than known results and in some cases we can get necessary and sufficient conditions of absolute stability.

Highlighted the nonlinear class of regulated systems with limited resources for which Iserman’s problem has a solution. For this class of regulated systems obtained necessary and sufficient conditions of absolute stability.

Iserman’s problem has a solution for regulated systems with limited nonlinearities and it can’t be resolved by the frequency method of V.M. Popov.

It is shown that inefficiency of the frequency method of V.M. Popov for the study of absolute stability of regulated systems with limited resources is revealed: firstly, when deriving the frequency condition, the boundedness of the improper integral is

not taken into account I1 ; secondly, boundedness of solutions of systems with

bounded nonlinearities is not used; thirdly, properties of an improper integral are not taken into account 3.

Literature

1.Lurie A.I. Nekotorye nelinejnye zadachi teorii avtomaticheskogo regulirovanija. – M.: Gostehizdat, 1951. – 216 s.

Lectures on the stability of the solution of an equation with differential inclusions

2.Ajzerman M.A., Gantmaher F.R. Absolute stability reguliruemyh sistem. – М.: Izdatel'stvo AN SSSR, 1963. – 240 s.

3.Gelig A.H., Leonov G.A., Jakubovich V.A. Ustojchivost' nelinejnyh sistem s needinstvennym sostojaniem ravnovesija. – M.: Nauka, 1978. – 400 s.

4.Ajsagaliev S.A. Ob opredelenii oblasti absoljutnoj ustojchivosti vynuzhdennyh dvizhenij v nelinejnyh sistemah. – М.: Izv. AN SSSR. Tehnicheskaja kibernetika. – № 5, 1969. – S. 38–48.

5.Ajsagaliev S.A. Ob opredelenii oblasti absoljutnoj ustojchivosti sistemy upravlenija s neskol'kimi nelinejnymi jelementami // AN SSSR. Avtomatika and telemehanika. – № 12, 1970.

–S. 83–94.

6.Ajsagaliev S.A. K teorii absoljutnoj ustojchivosti reguliruemyh sistem // Differencial'nye uravnenija. – Minsk-Moskva. – T. 30. – № 5, 1994. – S. 748–757.

7.Ajsagaliev S.A., Ajpanov Sh.A. K teorii global'noj asimptoticheskoj ustojchivosti fazovyh sistem // Differencial'nye uravnenija. – Minsk-Moskva: MGU. – T. 30. – № 8, 1999. – S. 3–11.

8.Ajsagaliev S.A. Problemy kachestvennoj teorii differencial'nyh uravnenij. – Almaty: Qazaq universitetі, 2016. – 420 s.

9.Aisagaliev S.A., Kalimoldayev M.N. Certain problems of Synchronization theory // Journal Inverse Ill Posed Problems. – № 21, 2013. – P. 159–175.

10.Ajzerman M.A. Ob odnoj probleme, kasajushhejsja ustochivosti v "bol'shom" dinamicheskih sistem // UMN. – T. 4. – № 4, 1949. – S. 186–188.

11.Pliss V.A. O probleme Ajzermana dlja sluchaja sistemy treh differencial'nyh uravnenij // Dokl. AN SSSR. – T. 3. – № 121, 1958.

12.Zubov V.I. Lekcii po teorii upravlenija. – M.: Nauka, 1975. – S. 494.

13.Demidovich B.P. Lekcii po matematicheskoj teorii ustojchiosti. – M.: Nauka, 1967. – S. 471.

14.Popov V.M. Giperustojchivost' avtomaticheskih sistem. – M.: Nauka, 1970. – 433 s.

15.Popov V.M. Ob absoljutnoj ustojchivosti nelinejnyh sistem avtomaticheskogo upravlenija // Avtomatika and telemehanika. – T. 8, 1961. – S. 51–63.

16.Ajsagaliev S.A. Problema Ajzermana v teorii absoljutnoj ustojchivosti reguliruemyh sistem // Matematicheskij sbornik. Institut matematiki im. V.A. Steklova, Rossijskoj akademij nauk.

–T. 209. – № 6, 2018. – S. 3–24.

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

Chapter II

ABSOLUTE STABILITY AND AIZERMAN’S PROBLEM OF MULTIDIMENSIONAL REGULATED SYSTEMS

This chapter presents the results of the study of absolute stability of the equilibrium state of regulated systems with many non-linearities in the main, simple critical and critical cases. A class of multidimensional controlled systems is identified for which the Iserman’s problem has a solution. For this class of regulated systems necessary and sufficient conditions for absolute stability are obtained.

Lecture 14.

Absolute stability and Iserman’s problem in the general case.

Problem statement. Nonsingular transformation

Problem statement. Consider an equation of motion of regulated systems of the following type:

x = Ax B ( ),

= Sx,

x(0) = x ,

t I = [0, ),

(2.1)

where

A, B, S

are constant matrices of orders

n n,

n m, m n , respectively, the

matrix

is a Hurwitz matrix, i.e.

Re j

( A) < 0, j = 1, n, j ( A) are eigenvalues of the

matrix

| x0

|< ,

( ) = ( 1( 1), , m ( m )),

= ( 1

, , m ).

Function

( ) = { ( ) = ( ( ), , ( )) C(Rm , Rm ) | 0 ( )

(2.2)

i i i

R1, (0) = 0,

| ( ) | , , Rm , 0 < < },

where 0 = diag( 01, , 0m ) > 0 is a diagonal matrix of order m m,

| |

is a Euclidean

norm,

* = const > 0,

(*) is a transpose sign. Practical systems of automatic control

relate to the systems with limited resources and for such systems vector function ( ) satisfies the condition (2.2).

Regulated systems with an equation of the form x = A0 x B ( ), = Sx, with unlimited nonlinearity ( ) = h ( ) 0 , h = diag(h1, , hm ) are reduced to the system (2.1), (2.2) when the matrix A = A0 BhS is a Hurwitz matrix.

Lectures on the stability of the solution of an equation with differential inclusions

As known from the scientific and technical literature on automatic control all nodes of regulated systems (sensors, amplifiers, actuators) have limited resources by capacities, by generalized forces and moments caused by their limit values and satu-

ration effects. Consequently, inclusion of (2.2), where the value * > 0

is an arbitrarily

large number, * < covers all kinds of nonlinearities in regulated systems. Equilibrium state of the system (2.1), (2.2) is determined from the solution of

algebraic equations Ax B (			) = 0,	*	= Sx .	If the matrix			A	is a Hurwitz matrix,
	*	*		*	*
the function,	( ) 0	turns into zero only when					= 0,	then the system (2.1), (2.2)

has the only equilibrium state x* = 0.
Notice, that equilibrium state corresponds to a trivial solution x(t) 0,	t I	of the

system (2.1), (2.2). The asymptotic stability of an unperturbed motion as a whole is

studied

x(t) 0,

t I

for all

( )	.
0

We assume with a sufficiently small neighborhood of the point

= * = 0,

func-

tion ( ) 0 can be approximated by a linear function			( ) = ,	= diag(	, ,	m	),
tion ( ) 0 can be approximated by a linear function				1		m
0 < i 0i , i = 1, m. Consequently, when	\| \|< ,	> 0 is a sufficiently small number,
0 < i 0i , i = 1, m. Consequently, when		> 0 is a sufficiently small number,
perturbed motion equation has the form

x = Ax B Sx = A ( )x,

x(0) = x

, | x

, t I,

where A ( ) = A B S,

0 <

, i = 1, m is a limit value

i = 1, m

determined from Hurwitz of the matrix

( ).

If the matrix

( ),

0 < i

< 0i

i = 1, m

is a Hurwitz matrix, then there is a

number

> 0

such as

| x(t) |< 1

when

x0 |< 1,

moreover lim x(t) = 0.

Thus, when

matrix

A ( ),

where

0 <

i =

1, m is a Hurwitz matrix, then the trivial

solution

x(t) 0,

t I

is asymptotically stable by Lyapunov when

t .

Definition 1. The trivial solution

x(t) 0,

t I

of the system (2.1), (2.2) is called

absolutely

stable,

if:

1) matrices

A1( )

are

Hurwitz

matrices,

where

						, ,
0 <	0i	<	0i	,	= diag(	, ,	m	),
i	0i		0i		1		m
solution	of	the			differential	equation

\| x	\|< .
0

0 = diag( 01, , 0m ); 2)	for all ( ) 0	the
(2.1) has the property	lim x(t;0, x0 , ) = 0,
	t

In other words, the trivial solution

x(t) 0,

t I of the

absolutely stable, if it is asymptotically stable generally for any

system (2.1), (2.2) is

( )	.
0

Definition 2. Conditions of absolute stability of the system (2.1), (2.2) are called relations, binding constructive parameters of the system (A, B, S, 0 ), under which

equilibrium state x* = 0 is absolutely stable.

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

Problem 1. Find the condition of absolute stability of equilibrium state of the

system (2.1), (2.2).

Function

(t) = Sx(t),

t I is a control formed by the principle of feedback, and

the matrix S

of order m n is called a feedback matrix. Iserman’s problem consists

in how to choose a feedback matrix

S ,

so that from asymptotic stability of the trivial

solution

x(t) 0,

t I

a linear system

x = Ax B Sx = A ( )x

for

any

0 ,

= diag( 1, , m )

there follows absolute stability of the trivial solution

x(t) 0,

t I

of the system (2.1), (2.2), where

0 is a limit value of the Hurwitz

matrix A1( ),

are arbitrarily small numbers.

Definition

We assume that

the sector

[0,

]

Iserman’s problem

has a

solution, if: 1) there exists a feedback matrix

such as

0 = 0 ,

where

is a

limit value of the Hurwitz matrix

( ),

> 0

is an arbitrarily small number; 2) for

any ( ) = ,

solution of the system (2.1) is asymptotically stable;

3) for any 0 ( )

Problem 2.

0 trivial solution of the system (2.1), (2.2) is absolutely stable. Find a sector [0, 0 ], where Iserman’s problem has a solution.

Many works are devoted to the study of absolute stability of regulated systems in general and critical cases. Among them, we should note monographs [1–3]. There are two approaches to the study of absolute stability of regulated systems: the method of A.I. Lurie [1] and the method of V.M. Popov [2]. The relationship between these methods was established in the works of V.A. Yakubovich and his students [3]. Solution of equations of A.I. Lurie was obtained based on the second Lyapunov’s method by choosing the Lyapunov function in the form of “a quadratic form plus an integral of nonlinearities”. In the end the method of A.I. Lurie leads to checking solvability of matrix inequalities. Naturally, it is rather difficult to apply this approach to solving applied problems due to uncertainty of the choice of arbitrary constants under conditions of absolute stability.

The difficulty of checking frequency conditions, the need to select an area of absolute stability in the space of constructive parameters of the system led to creation of algebraic conditions of absolute stability by reducing the frequency conditions to checking the positivity of polynomials on the positive semiaxis [4, 5].

In 1949, M.A. Iserman formulated the following problem [10]: let the solutions of

					0		,
all linear systems	of the	form x = Ax B ,				0		= Sx		be asymptotically
stable. Will the	system	solutions				= Sx			with	any nonlinearity
stable. Will the	system	solutions	x = Ax B ( ),			= Sx			with	any nonlinearity
( ) = { ( ) C(R , R ) \| 0 ( )					, , R }
	1	1		2				1	possess the property of
			0						possess the property of

asymptotic stability in general? Iserman’s problem was solved for systems of secondorder by I.G. Malkin, N.P. Yerugin, N.N. Krasovsky.

In 1957, R.E. Kalman formulated the following problem: Let the solutions of all

linear systems of the form x =					Ax B x,			0 0 ,		= Sx be asymptotically stable.
Will the system			solutions		x = Ax B ( ),				= Sx with any nonlinearity
( )		= { ( ) C1	(R1, R1 ) \| 0	d ( )			,	, R1}		possess the property of asymp-
	1
					d	0

totic stability in general? Kalman problem has a positive solution when n = 2.

Lectures on the stability of the solution of an equation with differential inclusions

Solutions of the Iserman problem and the Kalman problem remain open for the

case when n > 2.

For

multidimensional regulated

systems, where ( ) = { ( ) = ( 1( 1),

(

) C(Rm , Rm ) | 0

(

)

, Rm }

Iserman’s problem is

0i i

formulated as follows: let the solutions of all linear systems of the form

x = Ax B Sx,

= diag( , ,

= Sx,

= diag(

, ,

where

are

constant matrices of orders

n n, n m,

m n

, respectively, be asymptotically stable.

Will the solutions of nonlinear systems

x = Ax B ( ), = Sx

with any nonlinearity

( ) possess the property of asymptotic stability in general?

Note that as it follows from the work [12] Iserman’s problem not always has a solution. Consequently, to solve Iserman’s problem it is necessary to highlight the class of multidimensional regulated systems, by imposing additional requirements on

the properties of nonlinearities. Function	(t) = Sx(t),	t [0, ) is a control formed by
the principle of feedback, and the matrix	S of order	m n is called a feedback mat-

rix.

An attempt to solve Iserman’s problem for multidimensional regulated systems with limited nonlinearities for the case when n > 2 by selecting a feedback matrix S was proposed.

A new method of study of absolute stability of multidimensional regulated systems based on a priori estimation of improper integrals along the solution of the system as proposed. The condition of effectiveness of absolute stability is achieved by using additional information on non-linearities in the form of their boundedness, and constructiveness is determined by the formulation of the stability condition in the form of positive definiteness of the matrix from the constructive parameters of the system

( A, B, S,	).
0

This work is a continuation of research from [6–9, 11]. Efficiency and constructiveness of the proposed method for one-dimensional systems, when m = 1 are shown in [11].

Nonsingular transformation. To simplify verification of the proposed condition of absolute stability, it is advisable to transform the original equation of motion (2.1).

Let the matrix	B =\|\| B1, , Bm \|\|, where Bi ,
	*	R	n
Lemma 1. Let the row vector i

i*Bi = 1, i*B

i ,

= 1, m are column vectors

i = 1, m such that:

= 0, i = 1, m, i j,

n 1.

(2.3)

where Bi B j , i j, (*) tion (2.1), an identity holds

is a transpose sign. Then along the solution of the equa-

x(t) = Ax(t) (
x(t) = Ax(t) (			(t)),	t [0, ), i = 1, m.	(2.4)
i	i	i i

If besides rank B* = m and Gram determinant

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

			<	,		>	<	,		>	<	,		m		>
			1		1		1		2		1			m
(	, ,	) =														0,
(	, ,	) =														0,
1	m		<		,	>	<		,	>	<		,			>
			<		,	>	<		,	>	<		,		m	>
			m		1		m		2		m				m

(2.5)

then vectors i ,

i = 1, m

exist

and they are linearly independent, where < i , j > is a

scalar product of vectors i ,

j ,

j = 1, m.

Proof. As

i = 1, m,

then by multiplying the identity from the

i = ( i1, , in ),

left x(t) = Ax(t) B ( (t)), t I

we have

x(t) =

Ax(t) B ( (t)),

t I, i = 1, m,

Hence taking into account (2.3), we get (2.4). Notice,

where i B = ( i

B1, , i

Bm ).

that ratio (2.3) can be written as linear algebraic equations

i 2

= 0,

i 2

= 1,

i 2

= 0, i = 1, m.

i1 m1

If rank B* = m, then this system of equations has a solution,

i = 1, m.

From the

condition (2.5) it follows that vectors

i ,

i = 1, m

are linearly independent. The Lem-

ma is proved.

Lemma 2. Let the vector

0 R

such that

Bi B j ,

0 Bi = 0, i = 1, m, where

i j. Then along the solution of equation (2.1) an identity holds

Ax(t),

t I.

(2.6)

0 x(t) = 0

If besides rank B* = m and Gram determinant

< 01, 02 >

< 01, 0n m >

( 01, , 0n m ) =

(2.7)

< 0n m , 01 >

< 0n m , 0n m >

then vectors 01, , 0n m

exist and they are linearly independent, where 0i ,

i = 1, n m

are obtained from 0 by choosing (n m) , are arbitrary vector components 0 .

Lectures on the stability of the solution of an equation with differential inclusions

Proof. Let the vector written as

	B
10	11

0	n	*
0	= ( 10 , , n0 ) R	. Then ratio 0

	B	= 0, ,	B		B
n0	1n	10	m1	n0	mn

Bi =

= 0, 0.

i = 1, m

can be

If rank

= m,

then this system has a solution

(

, ,

where

m 1,0

n,0

, ,

n0 are any numbers. Define the vectors 01 R

, , 0n m

by choosing

m 1,0

arbitrary numbers m 1,0 , , n0 .

In particular

(1,0,

,0),

02 = 0 (0,1,0, ,0),

0n m

(0, ,0,1).

By multiplying the identity on the left (2.1) by

0 we get (2.6). By the hypothesis

of the lemma, the following inequality holds (2.7). Consequently, vectors

Bi = 0,

i = 1, n m,

then identity (2.6) is

i = 1, n m are linearly independent. As 0i

equivalent to

x(t) =

Ax(t), t

I, i =1, n m.

The Lemma is proved.

Lemma 3. Let the conditions of the lemmas 1, 2 be satisfied, and let, moreover, the rank of the matrix

R =|| 1, , m 01, , 0n m ||

(2.8)

of order n n of equations

be equal to

Then equation (2.1) is equivalent to the following system

y1 = c11 y1 c1n yn 1 ( 1 ), , ym = cm1 y1 cmn yn m ( m ),
ym 1 = cm 1,1 y1 cm 1,n yn , , yn = cn1 y1 cnn yn ,	(2.9)
1 = d11 y1 d1n yn , , m = dm1 y1 dmn yn ,

where

i = 1, m,

= i x,

ym i = 0i x,

i = 1, n m.

Proof. As rank R = n, then from (2.8) it follows that vectors

i = 1, n

m form the basis in

Then vectors

* Ax = c

* x c

x c *

11 1

1m m

1m 1 01

1n 0n m

Ax = c

* x c

* x c *

m1 1

m,m m

m,m 1 01

mn 0n m

Ax = c

* x c

m 1,1

m 1,m

m 1,m 1 01

m 1,n

0n m

Ax = c

* x c * x c

x c

0n m

n1 1

nm m

n,m 1 01

n,n 0n m

i = 1, m,

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			1		1		1		2		1			m
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			m		1		m		2		m				m

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			1		1		1		2		1			m
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			m		1		m		2		m				m

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			1		1		1		2		1			m
(	, ,	) =														0,
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			m		1		m		2		m				m