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−23

2.4.13. 1

1−2

2.4.15. 3

3−1

2.4.17. 0

2.4.19. −2

−1

−32

2.4.21. −1

−2

2.4.23. 3

2.4.25. −3

3−2

2.4.27. 1

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

1	3	1	−1 .
−2	−1	−2	1
2	1	1	3

2	−1	2	3
0	1	3	2
					.
−1	2	−4	−6
1	2	1
			−1
1	2	1	3
1	3	−1	2
				.
−3	−1	1	0
−1	4	1	5

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

0	1	1	0
1	2	2	0
2	2	2	8
−1	2	−1	2
					.
1	−2	2	−2
2	2	3	8

3	0	−1	5
1	1	2	2

−2	0	−3	−4
2	1	−2	3

−32

2.4.14. 1

1−2

2.4.16. 2

−21

2.4.18. 3

2−1

2.4.20. 3

4−2

2.4.22. 0

3−2

2.4.24. 3

−13

2.4.26. −1

1−2

2.4.28. 2

1		2	−1		3
3		−1	1		2
3		−1	1		2	.
−2		3	2		1
2		4	2		6
2		4	2		6
0	1	−1		2

1	0	1	−2 .
1	1	2		3
2	2	2		3
2	2	2		3
3		2	1		3
−2		1	−1		0
−2		1	−1		0		.
1		−1	0		−1
2		2	0		2
2		2	0		2
3		2	−1		2
1		0	2		3
1		0	2		3		.
−2	−1		0		−2
2		1	1		3
2		1	1		3
−2		3	1		3
−1		−2	1
−1		−2	1		−3
1		1	−1		1
−2		2	1		1
0		2	2	5

1		−1	0	−2 .
−2		1	1	4
−1		2	3	7
−1		2	3	7
1		2	−2		−4
1	−1		3		4
1	−1		3		4		.
−1		1	0		−1
1		2	1		−1
1		2	1		−1
1		1	1	4
−1		2	1	0
−1		2	1	0	.
0	−1		1	2
0		2	3	6
0		2	3	6

	−4 2		0	−1 −3				1		−1 2		1		3
	3	−1	1	2	5				0	3	1		2	6
2.4.29.						.	2.4.30.								.
	2	2	−2 1		3				−2 1		−1		−2	−4
	1	3	−1 2		5				−1 0		1		−1
														−1

Micromodule 3

BASIC THEORETICAL INFORMATION.

SYSTEMS OF LINEAR ALGEBRAIC EQUATIONS

System of linear algebraic equations, its consistency, investigation of the system consistency using matrix rank. Solving the system by the Cramer’s formulas, matrix method, Gauss’ method. Undetermined systems and their solutions. Homogeneous systems of linear algebraic equations. Eigenvalues numbers and eigenvectors of matrix.

Literature: [1,chapter 1], [4, part 2, §2.3], [6, chapter 1, §3], [7, chapter 2, §7], [10, chapter 1, §3], [11, chapter 1, §1].

3.1. Basic definitions

Definition 1.18. System of m equations with n unknowns of the form

a11 x1 + a12 x2 + ... + a1n xn = b1 ,
	+ a22 x2	+ + a2n xn = b2	,
a21 x1				(1.1)
..................................................

a x + a x + ... + a x = b
m1 1	m2 2	mn n	m
is called the system of linear algebraic equations (SLAE), where				x1 , x2 ,..., xn

are unknowns; aij ( i = 1, 2,..., m , j = 1, 2,..., n ) are given coefficients, b1 , b2 , ..., bn

are absolute terms of the system.

To solve the system means to find such values of unknowns x j , which after substituting them into given system transform all of its equations into identities.

Definition 1.19. SLAE is called homogeneous, if all absolute terms are equal to zero and is called non-homogeneous, if at least one of them isn’t equal to zero.

The homogeneous system always has a zero solution.

Definition 1.20. The system of equations is called consistent, if it has at least one solution, and it is called inconsistent, if it has no solutions at all.

Definition 1.21. A consistent system is called determined, if it has only one solution ( x10 , x20 ,…xn0 ), and it is called undetermined, if it has more than one

solution.

Definition 1.20. Matrices

a11

a12

...

a1n

a11

a12

...

a1n

...

and

...

A = 21

B = 21

...

... ...

…

...

are called basic and extended matrices of the system (1.1) correspondingly.

3.2. Methods of solution of systems of linear algebraic equations

3.2.1. Cramer’s method. The given system be

a11 x1 + a12 x2 + ... + a1n xn

= b1 ,

= b2 ,

a21 x1 + a22 x2 + ... + a2n xn

................................................

+ a

... + a

= b .

nn n

This system can be reduced to the form

x1 =

1 ,

2 ,

................

n ,

where

a11

a12

...

a1n

a12

...

a1n

a21

a22

...

a2n

a22

...

a2n

...

... ... ...

... ... ... ...

an1

an2

...

ann

an2

...

ann

a11

...

a1n

a11

a12

...

2 =

a21

...

a2n

…,

n =

a21

a22

...

... ... ... ...

...

... ... ...

an1

...

ann

an1

an2

...

(1.2)

Definition 1.23. If ≠ 0 , then SLAE (1.2) has a unique solution, which can be found by the Cramer’ formulas:

	x1 =	1	, x2 =	2	, …, xn =		n	.
	x1 =		, x2 =		, …, xn =			.


Definition 1.24.
If	= 0 , and at least one of the determinants					i	≠ 0 , then the system is
(1.2) inconsistent.
If	= 0 and all the determinants			1 , 2 , ....,		n	are equal to zero, then
the system (1.2) has infinitely many solutions.

3.2.2. Matrix method. Using the concept of matrix product, the system (1.1) can be written in the form

AX = B,

where

a11	a12	...	a1n	x1			b1
a	a	...	a	x		,	b
A = 21	22	...	2n	, X =	2	,	B =	2	.
...	...	...	...		...			...
	a	...	a
a	a	...	a	x			b
m1	m2		mn		n			n

If in SLAE п = т and the determinant of the system		(A) ≠ 0 , then unique
solution of the system can be determined by the formula

	X = A−1 B.	(1.3)

3.2.3. Gauss method is used for solving SLAE of an arbitrary form. This method is based on the elementary transformation of rows of the system, i.e. the system remains equivalent to the given system, if:

1)interchange two equations;

2)multiply both sides of the equation by the non-zero multiplier;

3)add elements of one equation to the corresponding elements of another equation multiplied by the same number.

With help of such transformations the system (1.1) can be reduced to the trapezium form

There can be only two cases when r < n:

1.if at least one of the numbers br +1, ...,bm isn’t equal to zero. Then the system (3.1) has no solutions;

2.if all the numbers br +1, ...,bm are equal to zero. Then the system (3.1) has infinitely many solutions.

+ a

...

+ a

a22 x2

...

+ a2n xn =

b2 ,

a33 x3 + ...

+ a3n xn =

b3 ,

− − − − − − − − − − − − − − − − − − −

(1.4)

ars xs

+ ...

+ arn xn = br ,

0 = br +1,

− − − − −

0 =

If r = n , then the system (1.4) acquires triangular form

+ c

x + ...

11 1

a22 x2 + ...

+a2n xn

b2 ,

..................................

amn xn = bm .

The obtained system and then the given system (1.1), has the unique solution, which is determined in such way. At first let’s find xn from the last

equation, after that from the next to last equation let’s find xn−1 ; climbing onto the system from the last equation to the first one, we’ll find all the unknowns.

Remark. Since extended matrix one-to-one corresponds to SLAE, then elementary transformations of the equations are equivalent to the transformations of rows of the extended matrix. Therefore, further on when solving SLAE by the Gauss method we’ll operate only with extended matrix.

3.3. Criteria of SLAE consistency

Let we have a system in the form

a11 x1 + a12 x2 + ... + a1n xn = b1 ,
a x + a x ...+ + a x = b ,
	21 1	22 2	2n n	2
................................................

a x + a x ...+ + a x = b
	m1 1	m2 2	mn n	m

Let’s construct the basic and the extended matrices of the given system

a11

a12

...

a1n

a11

a12

...

a1n

...

A = 21

...

B = 21

...

... ...

…

...

Theorem 1.3. (Kronecker-Capelli) In order to make SLAE consistent, it
is necessary and sufficient, the rank			r(A) of the basic matrix А be equal to
the rank r(B) of the extended matrix В.
If rank of the basic matrix equals the rank of the extended matrix and is
equal to the number of unknowns, i.e. r(A) = r(B) then the system has a
unique solution.
If rank of the basic matrix equals the rank of the extended matrix, but it is
less than the number of unknowns, i.e. r(A) = r(B) < n then the system
has infinitely many solutions.
If r(B) > r(A) , then the system is inconsistent

3.4. Homogeneous system of linear algebraic equations
Let’s consider homogeneous system of the n–th order
a11 x1 + a12 x2 + ... + a1n xn						= 0;
		+ a22 x2		+ ... + a2n xn		= 0;
a21 x1		+ a22 x2		+ ... + a2n xn		= 0;	(1.5)
..........................................							(1.5)

a	x	+ a	x	+ ... + a	x	= 0.
n1	1	n2	2	nn	n

Since the extended matrix В of homogeneous system has no additional nonzero minors (in comparison with matrix Α ), then r(A) = r(B) , i.e. homogeneous system is always consistent.

If r(A) = n	or, that is the same, (A) ≠ 0 , then the system has a unique
trivial solution	x1 = x2 = ... = xn = 0 , of what it is easily to check it, for instance,

applying Cramer’s rule to the system.

If r(A) = k < n , then the homogeneous system (1.5) has infinitely many

solutions, which are determined, for example, by the Gauss` method.

Remark. Universal method of solving systems of linear algebraic equations both determined and undetermined, homogeneous and nonhomogeneous is the Gauss method.

3.5. Eigenvalues and eigenvectors of matrix

Definition 1.25. Any non-zero column vector X , which satisfies the con-

dition

A X = λ X ,

where λ is a real number, is called eigenvector of the matrix A , and number λ

is the egenvalue of the matrix A , which corresponds to the vector X . The vector X is ambiguously determined (accurate to non-zero scalar multiplier).

Egenvalues of matrix A are the roots of its characteristic equation (A − λE) = 0 , or in the unfolded form (for example of the third order matrix)

a11 − λ	a12	a13
a11 − λ	a12	a13
a21	a22 − λ	a23	= 0 .
a31	a32	a33 − λ

Eigenvector X = x2 , which corresponds to the egenvalue λ , is deter-

mined from the system of equations

(a11 − λ)x1 + a12 x2 + a13 x3 = 0,a21 x1 + (a22 − λ)x2 + a23 x3 = 0,a31 x1 + a32 x2 + (a33 − λ)x3 = 0.

When λ is multiple root of the characteristic equation, the system can define several eigenvectors.

Micromodule 3

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Solve the system of equations

2x1 + x2 − 3x3 = 3,

x1 + x2 + 2x3 = 2,

− x1 + x2 + x3 = 0.

by the matrix method and by the Cramer’s formulas.

system. Using properties of determinants, we’ll calculate the determinant of the basic matrix:

(A) =	2	1	−3	=	2	3	−1	=	0	3	−1	=
	2	1	−3		2	3	−1		0	3	−1
	1	1	2		1	2	3		0	2	3
	−1 1		1		−1 0		0		−1 0		0

= 0 − 9 + 0 − (2 + 0 + 0) = −11 ≠ 0

As (A) ≠ 0 , then there exists an inverse matrix. We find it

A =		1 2						= −1 ,						A = −							1 −3					= −4 , A =								1 −3					= 5 ,
A =		1 2						= −1 ,						A = −							1 −3					= −4 , A =								1 −3					= 5 ,
11		1				1							21								1				1					31				1		2
		1				1															1				1									1		2
A = −				1 2								= −3 , A =												2 −3				= −1 ,			A = −							2 −3				= −7 ,
A = −				1 2								= −3 , A =												2 −3				= −1 ,			A = −							2 −3				= −7 ,
12				−1					1								22							−1		1						32						1		2
				−1					1															−1		1												1		2
A =			1			1			= 2 ,					A = −								2 1				= −3 , A =							2 1				= 1 .
A =			1			1			= 2 ,					A = −								2 1				= −3 , A =							2 1				= 1 .
13			−1			1							23									−1			1					33			1			1
			−1			1																−1			1								1			1
The inverse matrix has such form
																									−1	−1				−4		5
																		A	−1				=		−1		−3			−1	−7
																																			.
																									11										.
																									11		2			−3		1
																											2			−3		1
Then using the formula (1.3) we obtain
					x1									1		−1 −4 5 3																1			−1 3 − 4 2 + 5 0
	X =				x						= −			1			−3 −1 −7												2	= −		1				−3 3 − 1 2 − 7 0									=
	X =				x						= −						−3 −1 −7												2	= −						−3 3 − 1 2 − 7 0									=
					2																															2 3 − 3 2 + 1 0
					x								11					2 −3 1											0			11				2 3 − 3 2 + 1 0
					3
																									1		−11				1
																									1
																				= −							−11			=	1	.
																				= −					11		−11			=	1	.
																									11		0				0
																											0				0
Thus, x1 = x2										= 1,				x3 = 0					is a solution of the given system.
Cramer’s method. Now let’s use Cramer’s formulas:
					= −11,								1 =					3	1						−3	= −11,						=				2		3		−3			= −11 ,
																		3	1						−3											2		3		−3
																		2	1						2						2					1		2			2
																		0	1						1										−1			0			1
				=			2			1			3		= 0 ; x								=			1 = 1 , x					=				2	= 1 , x					=		3	= 0.
							2			1			3
	3						1					1 2
							1					1 2
								−1		1			0						1											2								3
								−1		1			0
Example 2. Solve the system of equations
																						x1 + 2x2 + 3x3 = 12,

																						2x1 + x2 + 2x3 = 9,
																						−3x				+ x			+ 4x		= 10,
																									1		2			3

using Gauss` method.

Solution. We write down the extended matrix of the given system

	1	2	3	12

B =	2	1	2	9

	−3	1	4	10

and, performing elementary transformations of the rows, we reduce it to the trapezium form:

1		2	3	12	1 2			3	12	1 2			3	12

	2	1	2	9		0	−3 −4		−15		0	−3 −4		−15	~
					~					~
	−3 1		4	10		0	7	13	46		0	1	5	16

1		2	3	12	1		2	3	12

~	0	1	5	16	~	0	1	5	16 .
	0	−3	−4			0	0	11
				−15					33

The last row corresponds to the equation 11x3 = 33 , from which x3 = 3 . Then we write down the equations:

x2 + 5x3 = 16 , thus x2	= 16 − 5x3 = 16 − 15 = 1 , x2 = 1.
x1 + 2x2 + 3x3 = 12 , thus	x1 = 12 − 2x2 − 3x3 = 12 − 2 − 9 = 1 .

Hence, the solution of the system is: x1 = 1 , x2 = 1, x3 = 3 .

Example 3. Investigate the system of equations

x1 + 2x2 + 3x3 − x4 = 1,

3x1 + 2x2 + x3 − x4 = 1,2x1 + 3x2 + x3 + x4 = 1,

2x1 + 2x2 + 2x3 − x4 = 1.

on consistency and in the case of consistency find the solution. Solution. We write down the extended matrix of the given system

	1	2	3	−1	1

	3	2	1	−1
B =					1	.
	2	3	1	1	1


	2	2	2	−1
					1

We find the rank of this matrix (and simultaneously the rank of the basic matrix), performing the elementary transformations of the rows:

1	2	3		−1			1			1	2 3 −1				1			1	2 3 −1			1
1	2	3		−1			1			1	2 3 −1				1			1	2 3 −1			1
3	2	1		−1						0	−4 −8 2					−2		0	−2 −4 1
3	2	1		−1			1	~		0	−4 −8 2					−2	~	0	−2 −4 1			−1	~
2	3	1			1		1	~		0	−1 −5 3					−1	~	0	−1 −5 3			−1	~
2	3	1			1		1			0	−1 −5 3					−1		0	−1 −5 3			−1

2	2	2		−1						0	−1 1		−2		0			0	−1 1		−2	0
2	2	2					1			0	−1 1		−2		0			0	−1 1		−2	0
							1
					1	2		3 −1				1		1		2 3 −1				1
					1	2		3 −1				1		1		2 3 −1				1
					0	0		6 −5				1		0		−1 −5 3
			~		0	0		6 −5				1	~	0		−1 −5 3				−1	.
			~		0		−1		−5 3			−1	~	0		0 6 −5				1	.
					0		−1		−5 3			−1		0		0 6 −5				1

					0	0		6 −5				1		0		0 0 0				0
					0	0		6 −5				1		0		0 0 0				0

From the form of the last matrix we make the conclusion that the rank of a basic matrix equals 3. The rank of an extended matrix also equals 3, since the rank of an extended matrix isn’t less than rank of a basic matrix, and the last row, which contains only zero elements, doesn’t increase a rank (all minors of the fourth order are equal to zero).

Conclusion. The given system has infinitely many solutions. We find these solutions the following way. According to the form of the last matrix we write down the system

x1 + 2x2 + 3x3 − x4 = 1,
	− x2 − 5x3 + 3x4 = −1,

	6x3 − 5x4 = 1,

which is equivalent to the given system.

Moving step by step from the last equation to the first one, we find:

x =		1+ 5x4	;	x = −5x + 3x + 1 = −5						1+ 5x4			+ 3x + 1 =					1− 7x4				;
x =			;	x = −5x + 3x + 1 = −5									+ 3x + 1 =									;
3	6			2	3		4		6						4			6
	6								6									6
x = 1− 2x − 3x + x = 1− 2								1− 7x4	− 3			1+ 5x4		+ x =			1+ 5x4				.
x = 1− 2x − 3x + x = 1− 2									− 3					+ x =							.
	1		2	3	4		6		6						4			6
							6		6									6
We can write the equation in such form: x									=		1+ 5t		, x =			1− 7t			, x	3	=		1 + 5t	,
							1		6				2		6					3	6
x4 = t , where t R .									6						6						6
x4 = t , where t R .
Example 4. Find all solutions of the system
					3x1 + 2x2 − x3 = 0,
							− x2 + 3x3 = 0,
					2x1		− x2 + 3x3 = 0,
						x	+ 3x − 4x = 0.
						1	2		3
50

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