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Theorem 1. 2. The sum of the products of the elements of a row or of a column of a determinant by the cofactors of the respective elements of another row or column is zero.

For example, for the determinant of the third order such equalities are executed:

a11 A21 + a12 A22 + a13 A23 = 0 , a12 A13 + a22 A23 + a23 A33 = 0 .

Remark. The theorems 1.1. and 1.2. hold true for the determinants of any order.

1.4. Concept of determinants of random order

According to theorem 1.1., determinant of the n-th order equals the sum of products of the elements of a row (a column) by their cofactors.

For example, there is the expansion by the elements of the first row for a determinant of order n:

	a11	a12	...	a1n
=	a21	a22	...	a2n	= a	A	+ a	A	+ ... + a	A	.
	... ... ... ...				11	11	12	12	1n	1n
	... ... ... ...
	an1	an2	...	ann

As a result of application of this formula instead of calculation of one determinant of the n-th order it is necessary to calculate n determinants of order (n – 1). But such method of calculation is too bulky.

In practice it is convenient when the part of elements of a row (or a column) is equal to zero (and the more zero elements, the less calculations). For this reason it is better to transform determinant so that all elements of some row (column), except one, were zero. It is necessary to apply properties 1—8 to the determinant (property 8 plays here main role). Then the expansion of a determinant by elements of this row (column) contains only one factor.

We will give formal determination of determinant of the n-th order now.

Lets there is a permutation		j = ( j1 , j2 ,	..., jn ) , where j1 , j2 ,	...,	jn are
numbers 1, 2	,..., п formed in a definite order.		forms the inversion, if
The pair	of elements ( ji ,	jk ) , i < k	forms the inversion, if	ji	> jk .

Number of all such inversions pairs is called as number of inversions in a permutation j = ( j1 , j2 , ..., jn ) and designate as inv( j1 , j2 , ..., jn ) .

The permutation is called even, if the number of all inversions is even and otherwise it is odd. For example inv(3, 2, 1) = 3.

Definition 1.5. The determinant of the n-th order

a11 a12 ... a1n a21 a22 ... a2n

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

an1 an2 ... ann

is the algebraic sum of all possible products which contain one element from every row and every column. The sign of every addend is equal to (−1)t , where t is a number of inversions in other indexes, if the elements (factors aij ) of the addend are in ascending order of first indexes. So,

a11	a12	...	a1n
a21	a22	...	a2n	= ∑ (−1)inv( j1, j2 , ..., jn ) a1 j1 a2 j2 ...anjn .
...	...	...	...	j
an1	an2	...	ann

The number of such products is n! Half of them is taken with a sign plus, and other half is taken with a sign minus.

Remark. The determinant of order n in the triangular form is equal to the product of all elements of the principal diagonal

	a11	a12	...	a1n
	a11	a12	...	a1n
=	0	a22	...	a2n	= a11a22 …ann .
	... ... ... ...
	0	0	...	ann

Micromodule 1

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Calculate determinants:

−2

;

sin α

cos α

sin β

cosβ

Solution.

а)

−2

= 0 7 − 8 (−2) = 16 ;

b)	sin α	cos α			= sin α cosβ − sin β cos α = sin(α − β).
b)	sin α	cos α			= sin α cosβ − sin β cos α = sin(α − β).
	sin β	cosβ
Example 2.			3	2		−4
			3	2		−4
			3		−1	1	.
			1	1		3

Solution. The first method. By the rule of triangles we have

3	2	−4	= 3 (−1) 3 + 2 1 1+ 3 1 (−4) −
3	2	−4
3	−1 1
1	1	3

−1 (−1) (−4) − 2 3 3 − 1 1 3 = −9 + 2 − 12 − 4 − 18 − 3 = −44.

The second method. By the theorem 1.1 (expansion by the elements of the first row) we get

	3	2	−4			1+1		−1 1	1+2	3	1



	3	−1 1		= 3 (−1)				1 3	+ 2 (−1)	1	3	+
	1	1	3		3	−1
	1	1	3
			1+3				= 3 (−4)		− 2 8 − 4 4 = −44.
			1+3
+(−4) (−1)					1	1

The third method. Using properties of a determinant we obtain

	3		2	−4		+ III (−3)				0		−1	−13
	3		2	−4		+ III (−3)				0		−1	−13
	3		−1	1		+ III (−3) =				0		−4	−8	= a31 A31 =
	1		1	3						1	1		3
			= 1 (−1)3+1					−1	−13			= 8 − 52 = −44.
			= 1 (−1)3+1					−1	−13			= 8 − 52 = −44.
								−4	−8
		1	0	−1	2
		1	0	−1	2
Example 3.		1	2	−2	0		.
		−1	3	0	2
		2	1	−1	3

Solution. The determinant has some zero elements, but it is better, when the zero elements are in one row or a column. For example, let’s make zero all elements of the first column, except for the first element. For this purpose to a column 3 we shall add a column 1, and to a column 4 we shall add a column 1, multiplied by(-2), we get

	1	0	−1	2
	1	0	−1	2
=	1	2	− 2	0	=
	−1	3	0	2
	2	1	−1	3

		+ I + I (−2)
				1 0 −1+ 1 2 − 2											1 0 0 0
				1 0 −1+ 1 2 − 2											1 0 0 0
		=		1 2 −2 + 1 0 − 2								=			1 2 −1 −2								.
			−1 3 0 − 1 2 + 2												−1 3 −1 4
				2 1 −1+ 2 3 − 4											2 1 1 −1
Using the theorem 1.1. we obtain
		= a11 A11 = 1 (−1)1+1 M11 =															2		−1		−2
																	2		−1		−2
																	3 −1 4					.
																	1	1			−1
Let’s make two zeros in the second column. We have
						2	−1	−2	+ III					3	0				−3	.
						2	−1	−2	+ III					3	0				−3
		=				3	−1	4		+ III =				4	0			3
						1	1	−1						1	1				−1
We expand a determinant by the elements of the second column
	= a32 A32 = 1 (−1)3+2 M32									= −			3		−3			= −(9 + 12) = −21.
	= a32 A32 = 1 (−1)3+2 M32									= −			3		−3			= −(9 + 12) = −21.
													4		3
Answer.		= –21.
Example 4. Solve equation
					−2 − k			2				0
					−2 − k			2				0
					2			4 − k				6				= 0 .
					1			2			3 − k
Solution. We expand a determinant by the elements of the first row
−2 − k	2	0						1+1			4 − k				6						1+2			2 6



2	4 − k	6		= (−2 − k)(−1)							2				3 − k				+ 2(−1)					1 3 − k	+ 0 =
1	2	3 − k
1	2	3 − k

=(−2 − k)((4 − k)(3 − k) − 12) − 2(2(3 − k) − 6) = −(2 + k)(k2 − 7k) + 4k =

=−(k3 + 2k 2 − 7k 2 − 14k) + 4k = −k3 + 5k 2 + 18k .

So, the given equation is equivalent to the equation

−k3 + 5k 2 + 18k = 0 .

We have

											−k(k 2 − 5k − 18) = 0 ;
k = 0 or k2 − 5k − 18 = 0 , from here k =																	5 ±		97	.
k = 0 or k2 − 5k − 18 = 0 , from here k =																2				.
																2
Answer: k = 0;							5 ±			97	.
Answer: k = 0;										2	.
										2
												Micromodule 1
								CLASS AND HOME ASSIGMENTS
Calculate determinants:
1.	6		−4							2.		5		0	.							3.		121		110			.
1.	6		−4	.						2.		5		0	.							3.		121		110			.
	9	5										11		−22										132		121
	sin α			− cosα								log2		5 − log9					16				2			4
																							2			4

4.									.	5.										.		6.	3			5		.
4.	cosα				sinα				.	5.		log8		3		log5 2				.		6.	1			3		.
	cosα				sinα							log8		3		log5 2							1			3
																								2			2
		1			4							1		−1			−3							5		1			−16


7.	1 2			3						8.												9.
7.	1 2			3		.				8.		−2		−1 7				.				9.		−4		−2 13				.
		2			2							4		−5		−10								8		−4			−23
		7		3								4		−5		−10								8		−4			−23
		7		3									−1 1 −2 2												1 −3				−1 3
		1		−2			−2


													−4	3			−1 −1								−1	4			−1 −1
10.		−6		3			22		.	11.			−4	3			−1 −1				.	12.			−1	4			−1 −1		.
		4		−11			−3						11	−8		34			16						1	0		27 20
		4		−11			−3						10	−8 8 −1											2	−11 10 −3
													10	−8 8 −1											2	−11 10 −3

Answers

1. 66. 2. –110. 3. 121. 4. 1. 5. 5/3. 6. 0,6. 7. 13/21. 8. –5. 9. –10. 10. –15. 11.

11.12. 12.

Micromodule 1

SELF-TEST ASSIGNMENTS

1.1. Calculate determinants, using:

а) method of transformation to a triangular form;

b)expansion of a determinant by the elements of some row (column).

c)rule of triangles.

1.1.1.	2		3	4		.							1.1.2.	3		4	5				.				1.1.3.		4			5		6						.
1.1.1.	4		6	7		.							1.1.2.	6		7	8				.				1.1.3.		7			8		9						.
	8		9	10										9		10	11									10				11		12
1.1.4.	1		4	1					.				1.1.5.	1		5	1		.						1.1.6.		1			6		1				.
	1		4	1										1		5	1										1			6		1
	2		1	5										3		1	6										4			1		7
	1	6 −1												1 7 −1													1		8		−1
1.1.7.	1		1	1	.								1.1.8.	1		1	1	.							1.1.9.		1			1		1	.
	1		1	1										1		1	1										1			1		1
	3		1	4										4		1	5										5			1		6
	5		2	6										6		3	7										7			4		8
1.1.10.		1	−1 1										1.1.11.		1	−1 1									1.1.12.				1		−1 1
		1	−1 1												1	−1 1													1		−1 1
		1	6			−1						.			1	7		−1						.					1		8		−1								.
		4	3	7											5	4	8												6		5	9
1.1.13.		2	3	1						.			1.1.14.		3	4	1			.					1.1.15.				4		5	1					.
		2	3	1											3	4	1												4		5	1
		5	4				1								6	5	1												7		6			1
		6	7	0											7	8	0												8		9	0
1.1.16.		4	3	7							.		1.1.17.		5	4	8					.			1.1.18.				6		5	9								.
		4	3	7											5	4	8												6		5	9
		2	0	5											3	0	6												4		0	7
		1	−1 1												1	−1 1												1			−1 1
1.1.19.		7	2	4							.		1.1.20.		8	3	5						.		1.1.21.				9		4	6							.
		7	2	4											8	3	5												9		4	6
		3	1	5											4	1	6												5		1	7
		1	−1 1												1	−1 1													1	−1 1
1.1.22.		2	1	4				.					1.1.23.		3	1	5			.					1.1.24.				4	1		6			.
		2	1	4											3	1	5												4	1		6
		1	3	1											1	4	1											1		5		1
		5	1	6											6	1	7											7		1		8

1.1.25.		−1	4 3				1.1.26.	−1 5 4					1.1.27.	−1 6 5
		−1	4 3					−1 5 4						−1 6 5
	1		2 −1			.		1	3	−1		.		1	4 −1			.
		−1 5		2				−1 6		3				−1 7		4
1.1.28.		−1 6		1	.		1.1.29.	−1	7	1	.		1.1.30.	−1	8	1	.
		−1 6		1				−1	7	1				−1	8	1
		3	1	4				4	1	5				5	1	6
		−1	5	1				−1	6 1					−1	7	1

1.2. Find the real roots of the given equation:

1.2.1.		1− k		2	3						1.2.2.		2 − k			1	3
		1− k		2	3								2 − k			1	3
		3		4 − k	7		= 0 .						3			3 − k	6		= 0 .
			−1	2	1− k										−2	1	−1− k
1.2.3.	1− k			4	5						1.2.4.	−1− k				2	1
	1− k			4	5							−1− k				2	1
	2			1− k	3			= 0 .					1			4 − k	5		= 0
		−4		3	−1− k								3			2	5 − k
1.2.5.		1− k		1	2						1.2.6.	−2 − k				2	0
		1− k		1	2							−2 − k				2	0
			−2	3 − k	1			= 0 .					2			4 − k	6		= 0 .
		4		2	6 − k								1			2	3 − k
1.2.7.		2 − k		−1	1						1.2.8.	−3 − k				3	0
		2 − k		−1	1							−3 − k				3	0
		5		1− k	6	= 0 .									−1	3 − k	2		= 0 .
			−2	2	−k								2			1	3 − k
1.2.9.		4 − k		−2	2						1.2.10.				−3 − k	4	1				= 0 .
		4 − k		−2	2										−3 − k	4	1
			−1	3 − k	2			= 0 .							−3	1− k	−2
		1		2	3 − k										1	−2 −1− k
1.2.11.			1− k	0		1					1.2.12.				−3 − k	−1	−4
			1− k	0		1									−3 − k	−1	−4
			4	−2 − k		2				= 0 .					1	2 − k	3			= 0 .
			3	−2	1− k										−1	3	2 − k
1.2.13.			2 − k	1		3					1.2.14.			1− k		3	4	= 0 .
			2 − k	1		3								1− k		3	4
			−3	2 − k		−1			= 0 .					0		2 − k	2
			0	2	2 − k									3		−2 1− k

1.2.15.		1− k 2			3							1.2.16.		−1− k			−2		−3
		1− k 2			3									−1− k			−2		−3
			−1	−k	−1	= 0 .									3		2 − k		5				= 0 .
			3	−2 1− k											2		0		2 − k
1.2.17.		1− k		−1	0							1.2.18.		−2 − k			2		0
		1− k		−1	0									−2 − k			2		0
			−1	2 − k	1				= 0 .						2		4 − k		6				= 0 .
			5	−2 3 − k											1		2		3 − k
1.2.19.		3 − k		−1	2							1.2.20.		−2 − k			2		0
		3 − k		−1	2									−2 − k			2		0
			−1	3 − k	2		= 0 .								2		4 − k		6				= 0 .
			2	−2	−k										1		2		3 − k
1.2.21.		2 − k		0	2							1.2.22.		−1− k			2		1
		2 − k		0	2									−1− k			2		1
			−1	2 − k	1					= 0 .					3		−4 − k		−1			= 0 .
			3	1	4 − k										2		−2		−k
1.2.23.		1− k		−3	−2							1.2.24.		−1− k			2		1
		1− k		−3	−2									−1− k			2		1
			−1	3 − k	2						= 0 .				1		4 − k		5					= 0 .
			−3	2	−1− k										1		−2 −1− k
1.2.25.		2 − k		1	3							1.2.26.		−k		2		2
		2 − k		1	3									−k		2		2
			0	3 − k	3				= 0 .					−2		4 − k		2		= 0 .
			3	−2 1− k									1			2	3 − k
1.2.27.		2 − k		3	5							1.2.28.		−1− k 3				2
		2 − k		3	5									−1− k 3				2
			−1	3 − k	2				= 0 .						4		−k	4			= 0 .
			−1	2	1− k										1		2 3 − k
1.2.29.		1− k		3	4							1.2.30.		−2 − k			2		0
		1− k		3	4									−2 − k			2		0
			−1	2 − k	1			= 0 .							2		4 − k		6				= 0 .
			3	−2 1− k											1		2		3 − k
1.3. Calculate:
	1 −2 −1 3													1 −2 −1 3
	1 −2 −1 3													1 −2 −1 3
1.3.1.	−1 3			−1	−1		.					1.3.2.		−1		3	−1		−1		.
	3		−8	1	5									3		−8	4		6
	2		3	−14	21									2		2	−12		19

	1	−2	−1 3					1	−2	−1 3
	1	−2	−1 3					1	−2	−1 3
1.3.3.	−1	3	−1	−1		.	1.3.4.	−1	3	−1 −1		.
	3	−8	7	7				3	−8	10	8
	2	1	−10	17				2	0	−8	15
	1	−2	−1 3					1	−2	−1 3
	1	−2	−1 3					1	−2	−1 3
1.3.5.	−1	3	−1 −1		.		1.3.6.	−1	3	−1 −1			.
	3	−8	13	9				3	−8	16	10
	2	−1 −6 13						2	−2 −4 11
	1	−2	−1 3					1	−2	−1 3
	1	−2	−1 3					1	−2	−1 3
1.3.7.	−1	3	−1 −1		.		1.3.8.	−1	3	−1 −1				.
	3	−8	19	11				3	−8	22	12
	2	−3	−2 9					2	−4	−10 7

	1 −2 −1 3										1 −2 −1
	1 −2 −1 3										1 −2 −1
1.3.9.	−1		3	−1 −1		.				1.3.10.	−1	3	−1
	3		−8	25 13							3	−8	28
	2		−5	2	5						2	−6	4
		1	1	0	−2						1	2	0
		1	1	0	−2						1	2	0
1.3.11.		1	2	1	−4		.			1.3.12.	2	5	1
		3	13	11	−18						3	15	11
		−2	2	5	1						−2	0	6
		1	3	0	−2						1	4	0
		1	3	0	−2						1	4	0
1.3.13.		3	10	1	−8			.		1.3.14.	4	17	1
		3	17	11	−14						3	19	11
		−2 −2 7			1						−2 −4 8
		1	5	0	−2						1	6	0
		1	5	0	−2						1	6	0
1.3.15.		5	26	1	−12			.		1.3.16.	6	37	1
		3	21	11	−10						3	23	11
		−2 −6 9			1						−2 −8 10
		1	7	0		−2					1	8	0
		1	7	0		−2					1	8	0
1.3.17.		7	50	1	−16				.	1.3.18.	8	65	1
		3	25	11		−6					3	27	11
		−2	−10 11		1						−2	−12	12

14−1 .

−2

−6

−16 . 1

−2

−10 −12 .

−2

−14

−8 . 1

−2

−18

−4 .

		1	−2	2	4				1	9		0	−2
		1	−2	2	4				1	9		0	−2
1.3.19.		2	−3	3	5		.	1.3.20.	9	82		1	−20					.
		−1	10	2	−27				3	29 11 −2
		2 0 4 −2							−2 −14 13 1
	1		−2	2	1				1	−2		2	2
	1		−2	2	1				1	−2		2	2
1.3.21.	2		−3	3	−1	.		1.3.22.	2	−3		3	1			.
	−1 4 −1 −6								−1	6	0		−13
	2		0	1	−8				2	0		2	−6
	1		−2	2	3				1	10		0	−2
	1		−2	2	3				1	10		0	−2
1.3.23.	2		−3	3	3	.		1.3.24.	9	91		1	−20	.
	−1		8	1	−20				3	31	11		0
	2		0	3	−4				1	15		25	1
	1		−2	2	5				1	−2		2	6
	1		−2	2	5				1	−2		2	6
1.3.25.	2		−3	3	7	.		1.3.26.	2	−3		3	9		.
	−1		12	3	−34				−1	14		4	−41
	2		0	5	0				2	0		6	2
	1		−2	2	7				1	−2		2	8
	1		−2	2	7				1	−2		2	8
1.3.27.	2		−3	3	11	.		1.3.28.	2	−3		3	13			.
	−1		16	5	−48				−1	18		6	−55
	2		0	7	4				2	0		8	6
	1		−2	2	9				1	−2		2	10
	1		−2	2	9				1	−2		2	10
1.3.29.	2		−3	3	15	.		1.3.30.	2	−3		3	17				.
	−1		20	7	−62				−1	22		8	−69
	2		0	9	8				2	0		10	10

1.4. Calculate determinants, using:

а) method of transformation to a triangular form;

b) expansion of a determinant by the elements of some row (column).

	1	2	−1	−1			2	1	−1	−1			2	1	−3
	1	2	−1	−1			2	1	−1	−1			2	1	−3
1.4.1.	0	−1	1	2	.	1.4.2.	0	−2 2		2	.	1.4.3.	0	−2 2
1.4.1.	−1	1	1	−1	.	1.4.2.	−3	0	1	−1	.	1.4.3.	−3	4	1
	2	−1	0	0			1	−1 2		0			2	−1 0

0 −11 . 2

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