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1.4.4.

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1.4.10.

 

 

 

 

 

 

 

 

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1.4.13.

 

 

 

 

 

 

 

 

 

 

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1.4.16.

 

 

 

 

 

 

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1.4.19.

 

 

 

 

 

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1.4.22.

 

 

 

 

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1.4.25.

 

 

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1.4.5.

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1.4.8.

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1

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1.4.11.

 

 

 

 

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5

3

0

 

 

 

 

 

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1.4.14.

 

 

 

 

1

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1.4.17.

 

 

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1.4.20.

 

 

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1.4.23.

 

 

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1.4.26.2 1 1 1 2 1

0 0 2

0 11 . 1

1 21 . 4

0 21 . 3

1 02 . 0

0 11 . 3

1

21 . 0

0 11 . 1

1

21 . 1

 

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1.4.6.

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1.4.9.

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1

 

 

 

 

 

 

 

 

 

 

 

 

1.4.12.

 

 

 

 

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1.4.15.

 

 

 

 

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1.4.18.

 

 

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1.4.21.

 

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1.4.24.

 

 

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1.4.27.

 

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21

 

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1 1

 

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2 4

 

 

 

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1.4.28.

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1.4.29.

1

2 2

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.

1.4.30.

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2 1

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Micromodule 2

BASIC THEORETICAL INFORMATION. MATRICES

The definition of a matrix, operations with matrices. Inverse matrix. Matrix equations. Rank of a matrix and its properties.

Literature: [1, chapter 2—3], [4, part 2, p.p.2.2—2.4], [6, chapter 1, §2], [7, chapter 2, §7], [10, chapter 1, §2], [11, chapter 1, §1].

2.1. The basic concepts

Definition 1.6. A rectangular table of numbers written in the form

a11

a12

...

a1n

 

 

a

a

...

a

 

,

A = 21

22

...

2n

...

...

...

 

 

 

a

...

a

 

 

a

 

 

m1

m2

 

mn

 

contains m × n numbers aij

(i = 1, 2,..., m, j = 1, 2,...n)) is called a matrix. In a

brief form, the matrix can be written as

A = (aij )

(i = 1, 2, ..., m, j = 1, 2, ..., n) ,

where aij are elements of this matrix.

Elements of a matrix form rows and columns. The first index i shows the number of the row and the second index j shows the number of the column

whose intersection is occupied by the element aij . For example, we have a matrix as

a11

a12

a13

 

 

a22

a23

 

B = a21

.

a

a

a

 

31

32

33

 

This matrix has three rows and three columns.

The product m × n is called the matrix dimension.

22

Definition 1.7. If the number of rows in a matrix is equal to the number of columns (m = n), then it is a square matrix, otherwise it is called a rectangular

matrix. Thus matrix B is square, of order, or dimension, three.

The characteristic of a square matrix is a determinant. Only a square matrix has a determinant.

The elements a11 , a22 , ..., ann form a principal (or a leading) diagonal of a square matrix, and the elements a1n , a2 n1 , ..., an1 form a secondary diagonal.

Definition 1.8. Any matrix in which all elements are equal to zero is called a matrix zero.

Definition 1.9. If m = 1 and n > 1, then we get a single-row matrix

A = (a1, a2 , ..., an )

which is known as a row vector, or row matrix, and if m > 1 and n = 1, we get a single-column matrix

a1a A = 2

...

am

which is known as a column vector, or column matrix.

Definition 1.10. A square matrix is called a triangle matrix if all elements placed upon (under) the principal diagonal are equal to zero and there are non zero elements among the other ones.

Definition 1.11. A square matrix is called a diagonal matrix if all its nondiagonal elements are equal to zero.

Definition 1.12. A square matrix is called a unit matrix if all diagonal elements are equal to 1, and non-diagonal elements are equal to 0. It is denoted as:

1

0

0 ...

0

 

 

0

1

0 ...

0

 

 

0

0

1 ...

0

 

I =

.

... ... ...

...

 

0

0

0 ...

1

 

 

 

Definition 1.13. If the rows of a matrix AT are the columns of matrix A and

the columns of a matrix called a transpose matrix

AT are the rows of matrix A then this matrix AT is for matrix A.

23

For any square matrix

a11

a12

...

a1n

 

 

a

a

...

a

 

 

 

A = 21

22

 

2n

 

 

...

...

...

...

 

 

 

 

a

...

a

 

 

 

a

 

 

 

n1

n2

 

nn

 

 

we can correspondently set a determinant det(A) (or

(A) ):

 

 

a11

a12

...

a1n

 

 

 

 

 

det(A) =

a21

a22

...

a2n

.

 

 

...

... ... ...

 

 

 

an1

an2

...

ann

 

 

 

 

 

Definition 1.14. A square matrix for which

det(A) 0

is called a non-

singular matrix.

 

 

 

 

 

 

 

 

If the det(A) = 0, then matrix is called a singular matrix.

 

Definition 1.15. Two matrices

A = (aij )

and

B = (bij )

are equal if the

elements occupying the same places are equal, i.e. if aij = bij for all i and j (in

this case the number of rows (and, similarly, columns) of the matrices A and B must be equal).

2.2. Operations with matrices

Let matrix A and matrix B be the same order.

 

B = (bij ) is a matrix

1. The Sum of two matrix A = (aij )

and

С = (сij ) whose elements are defined by the relation

 

aij + bij = cij (i = 1, 2, ..., m,

j = 1, 2, ..., n).

The notation is A + B = C.

2. The product of the matrix A = (aij ) by any real number λ is a matrix

whose every element is equal to the product of the respective element of the matrix A by the number λ, i.e.

λA = λ(aij ) = (λaij ) (i = 1, 2, ..., m, j = 1, 2, ..., n).

3. The product of the matrix A = (aij ) , which has m rows and k columns by the matrix B = (bij ) which has k rows and n columns is a matrix С = (сij )

24

which has m rows and n columns and whose element сij is equal to the sum of

the products of the elements of the i-th row of the matrix A by the j-th column of the matrix B, i.e.

Am×k

Bk×n = Cm×n

 

 

 

must be the same

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

size of product

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where

сij = ai1b1 j + ai2b2 j + ... + aik bkj

 

 

 

 

(i = 1, 2, ..., m,

j = 1, 2, ..., n) .

In this the number k of columns of the matrix A must be equal to the number of rows of the matrix B, otherwise the product is not defined.

Remark. In general case AB BA . The multiplication of matrices is not commutative.

Properties of operations with matrices

1. A + B = B + A.

6.

(αA)B = A(αB).

 

 

 

2. A + (B + C) = (A + B) + C .

7.

(A + B)C = AC + BC .

 

 

 

3. (α + β)A = αA + βA .

8.

(AB)C = A(BC) .

 

 

 

4. α(A + B) = αA + αB.

9.

(AT )T = A .

5. α(βA) = (αβ)A .

10. (A + B)T = AT + BT .

2.3. Inverse matrix

An inverse matrix A1 exists if and only if matrix A is non-singular А ( det(A) 0 ).

Definition 1.16. The inverse matrix A1 is a matrix which satisfies the conditions

A1 A = A A1 = I,

were I is an identity matrix

25

The inverse matrix can be calculated by the formula

 

 

A11

A21

...

An1

 

 

 

1

 

A

A

...

A

 

 

A1 =

 

12

22

 

n2

 

,

det(A)

 

 

...

...

 

... ...

 

 

 

 

 

A1n

A2n

...

 

 

 

 

 

 

Ann

 

where Aij is the cofactor of the element aij .

Remark. Pay the special attention to the order of indexes.

2.4. Matrices equations

Let’s consider the system of n linear equations with n unknowns

a11x1 + a12 x2 + ... + a1n xn

= b1,

 

 

 

 

+ a22 x2

+ ... + a2n xn

= b2 ,

a21x1

 

− − − − − − − − − − − − − − −

 

a

n1

x

+ a

n2

x

+ ... + a

nn

x

= b .

 

 

1

 

2

 

n

n

We write the above equations in the matrix form. We introduce the following designations

a11

a12

...

a1n

 

x1

 

 

b1

 

 

 

a22

...

 

 

 

 

 

 

 

 

 

a21

a2n

 

x2

 

 

b2

 

A =

 

 

 

....

...

 

,

X =

 

and

B =

.

... ....

 

 

...

 

 

...

 

a

n1

a

n2

...

a

 

 

 

x

 

 

b

 

 

 

 

 

nn

 

n

 

 

n

 

The matrix A is called the matrix of system. X is the matrix of unknowns. B is the matrix of constant terms.

Then applying the rule of multiplication of matrices the given system can be written by one matrix equation

A X = B.

Solution.

Let the matrix A is non-singular matrix (det A ≠ 0) . Therefore the inverse matrix A1 exists. Multiply both parts of the matrix equation by A1 we get

A1 A X = A1 B, (property A1 A = I );

26

I X = A1 B, (property I X = X ).

We have

X = A1 B.

Consequence. To decide a system it is sufficient to find the inverse matrix of system and multiply it by the matrix of constant terms B on the right.

2.5. Rank of a matrix

Let’s consider a matrix A of m × n

a11

a12

...

a1n

a

a

...

a

 

 

A = 21

22

....

 

2n .

...

....

...

 

a

a

...

a

 

 

m1

m2

 

mn

We choose in the given matrix A k rows and k columns, where number k m and k n. The determinant of order k, which consists of the cross elements

of chosen rows and columns, is called a minor of order k of the matrix A.

Definition 1.17. Rank of matrix A, designate as r(A) = rank(A), is called the greatest order of the minor if this minor isn’t equal to zero

1. The rank of a matrix exists for any matrix and

0r(A) min(m, n).

2.r(A) = 0 if and only if A is zero matrix.

3.r(A) = n for square matrix A of order n if and only if the matrix A is non-

singular (det A 0).

For example, the matrix

1

3

1

 

4

2

1

 

A =

 

 

2

4

0

 

 

 

has nine minors of order 2, among which there are also such:

 

1

3

 

,

 

1

1

 

,

 

 

 

 

 

 

 

 

 

4

2

 

 

 

4

1

 

 

 

3 1

 

. The minor of the third order of the given matrix is its determinant.

 

 

 

 

 

 

 

 

 

4 0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

27

We can calculate a rank of a matrix in the following way. If there is a minor of order k is not equal to zero and all minors of order (k+1) are equal to zero, then r(A) = k.

On the practice to find a rank of the highest orders it is useful to apply another method:

A rank of a matrix will not change if we use elementary transformations:

1)Interchanging of two rows (columns);

2)Multiplication of each element of a row (column) by a number k;

3)Addition elements of a row (column) and corresponding elements of another row (column) multiplying by a number k.

4)Elimination of a zero row from the matrix

Using elementary transformations a matrix can be transformed to the form

when all elements except

only a11,a22 ,...,arr ,

 

where r ≤ min(m, n) , are

equal to zero. Then a rank of matrix is equal to r.

 

 

 

 

Micromodule 2

 

 

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Find the product of matrices AB , if

 

 

1

3

2

2

0

1

 

5 1

 

A =

4

1

,

B =

1 .

 

3

 

4

3

 

 

 

 

 

 

2

Solution. We have A2×3

and B3×3 . As the number of columns of a matrix A

is equal to the number of rows of a matrix B. The operation of multiplication A B has the sense and we calculate the product of matrices in the following way:

AB =

 

(

2

)

+ 3

 

5

+ 2

 

4 1

 

0 + 3

 

1+ 2

 

3 1

 

1+ 3

(

)

+ 2

 

 

=

 

21 9 2

.

1

 

 

 

 

 

 

 

 

 

1

 

2

 

 

 

4 (2) + 1 5

+ 3 4 4 0 + 1 1+ 3 3 4 1+ 1 (1) + 3

2

 

9 10 9

 

 

 

 

Example 2. Find

 

f (A) , if A = 2

 

3

,

f (x) = (x2 3x)(3x + 2) .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution. It is necessary to find the value

 

 

 

 

 

 

 

 

 

 

 

 

 

We obtain

 

 

 

 

 

 

f (A) = (A2 3A) (3A + 2E) .

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

2

 

 

2 3

 

 

2 3

 

2 3

 

=

 

4 12 6 15

 

 

 

8 21

 

;

 

 

=

4 5

 

 

=

4 5

 

 

4 5

 

 

8

+ 20

12

+ 25

 

=

28 13

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

28

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

2

8 21

2 3

8 21

6 9 14 12

;

 

3A =

28

 

3

 

 

=

 

 

 

 

=

2

 

 

 

 

13

4

 

5

28

13

12

15

16

 

 

 

 

3A + 2I =

 

2 3

 

 

1 0

6 9

2 0 8 9

 

 

 

 

 

3

 

+ 2

 

=

 

 

+

 

=

;

 

 

 

 

 

 

 

4 5

 

 

0 1 12 15

0 2 12 17

 

 

 

 

 

14

12 8

9

14 8 + (12) 12

14 (9) + (12) 17

=

f (A) =

 

2

 

17

 

=

 

+ (2) 12

 

16 (9) + (2) 17

 

 

 

16

 

12

 

16 8

 

 

 

 

 

 

 

 

 

 

=

256

78

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

104

 

178

 

 

 

 

 

 

 

 

 

 

 

 

256

78

 

 

 

 

 

 

 

 

 

 

Answer: f (A) =

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

104

 

178

 

 

 

 

 

 

 

 

 

 

Remark. If

f (x) = a0 xn + a1 xn1 + an1 x + an

and

A is any square matrix

then

f (A) = a0 An + a1 An1 + an1 A + an E .

Example 3. Find the inverse matrix, if

1

2

1

 

1

3

1

 

A =

.

 

2

1

0

 

 

 

Solution. Firstly we calculate the determinant of matrix А:

det(A) =

 

1

2

1

 

 

 

1

2

1

 

= 10 0 .

 

 

 

 

 

1

3

1

 

=

 

0

5

0

 

 

 

2

1

0

 

 

 

2

1

0

 

 

Then the inverse matrix exists because А is a non-singular matrix. We find cofactors:

A

1+1

 

3 1

 

= 1; A = (1)

2+1

 

 

2 1

 

= 1; A

 

= (1)

3+1

 

2 1

 

= 5;

 

 

 

 

 

 

 

= (1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

11

 

 

1 0

 

 

 

 

21

 

 

 

 

 

1

0

 

31

 

 

 

 

 

 

 

 

 

 

 

3

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A = −

 

1 1

 

= 2; A =

 

1 1

 

= 2; A = −

 

 

1 1

 

= 0;

 

 

 

 

 

 

 

 

 

 

 

 

 

 

12

 

 

 

 

 

2

0

 

 

22

 

2

 

0

 

 

 

 

 

32

 

1

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A =

 

1 3

 

= −5; A = −

 

1 2

 

= 5; A =

 

1

2

 

= 5.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

13

 

 

 

2

 

1

 

 

23

 

 

 

2

 

1

 

 

 

33

 

 

 

1

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

29

So, the inverse matrix can be written in the form

 

 

 

 

 

 

1

1

5

 

0,1

0,1

0, 5

 

A

1

 

1

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

2

2

0

 

=

0, 2

0, 2

0

.

 

10

 

 

 

 

5 5

5

 

 

0, 5 0, 5

0, 5

 

 

 

 

 

 

 

 

 

 

Example 4. Solve the matrix equation

 

 

 

2

2

1

4

,

X A B = C , if A =

,

B =

 

С = 1

2

)

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

1

2

7

 

(

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution. Consequently we have

 

 

 

 

 

 

 

 

 

 

 

X A B = C , X A BB1 = CB1 , X A E = CB1 , X A = CB1 ,

 

 

 

 

 

X AA1 = CB1 A1 , X E = CB1 A1 , X = CB1 A1 .

 

 

 

Find the inverse matrices A1 and B1 :

 

 

 

 

 

 

 

 

(A) =

 

2 2

 

= −4 0 , A = 1 , A = −2 , A = −3 , A = 2 .

 

 

 

 

 

 

 

 

 

 

 

 

 

3

1

 

 

 

 

 

 

11

 

 

21

 

12

22

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

1

= −

1

 

1

2

;

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

1

4

 

 

 

 

 

 

2

 

 

 

 

 

 

 

(B) =

= −1 , B = −7 , B = −4 , B = −2 , B = −1 .

 

 

 

 

 

 

2

7

 

 

 

11

 

 

 

21

 

12

22

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B

1

= −

7 4 7 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

1

2

 

 

 

 

 

Then

 

 

 

 

 

 

 

 

 

 

1

 

1

 

 

7 4

 

 

 

 

1

1 2

 

 

 

 

 

 

 

 

 

X = CB

 

 

A

 

= (1

2)

 

 

 

 

 

 

 

3 2

 

=

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 1

 

 

 

 

 

 

 

 

 

 

 

 

1

 

(1 7

 

 

 

 

2 1 4 2 1)

1 2

 

 

 

 

 

1

 

1 2

= −

 

 

 

2

 

 

 

= −

 

 

 

(3 2)

 

 

 

=

4

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3 2

 

 

 

 

 

 

3 2

= −

1

(3 1

+ 2

(3) 3 (2) + 2 2) = −

1

 

(3 2) =

 

3

 

1

 

 

 

 

 

 

 

 

 

.

4

4

 

4

 

 

 

 

3

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

Answer: X

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Example 5. Find the rank of a matrix

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

2

0

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 3 1 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Α =

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

1

1

 

6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

30

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