Добавил:

Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.

Вуз:

Национальный авиационный университет

Предмет:

[НЕСОРТИРОВАННОЕ]

Файл:

Higher_Mathematics_Part_1

.pdf

Скачиваний:

1052

Добавлен:

19.02.2016

Размер:

6.63 Mб

Скачать

☆

<<< < Предыдущая 1 2 3 4 56 / 286 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 > Следующая >>>

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

	3	2	−1	0

	2	−1	3	0
B =					.
	1	3	−4	0

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

3 2 −1		0	1 3 −4		0	1 3 −4		0		1 3			−4		0
3 2 −1		0	1 3 −4		0	1 3 −4		0		1 3			−4		0
	2 −1 3	0	~	2 −1 3	0	~	0 −7 11	0	~		0	−7 11			0	.
											0	−7 11			0
	1 3 −4	0		1 3 −4	0		0 0 0	0


As the angular minor is not equal to zero, i.e. Μ(2) =													1	3			= −7 ≠ 0 , the
As the angular minor is not equal to zero, i.e. Μ(2) =													1	3			= −7 ≠ 0 , the
													0		−7
rank equals 2 (2<3). Thus the system has infinitely many solutions.
By the last matrix we write down the equivalent system
				x1 +3x2 − 4x3 = 0,
					-7x2		+ 11x3 = 0.
					-7x2		+ 11x3 = 0.

The values x3 = 7t , x2 = 11t , where t R , satisfy	the second equation.
And now from the first equation we obtain x1 = 4x3 − 3x2	= 28t − 33t = −5t .

Answer: x1 = −5t , x2 = 11t , x3 = 7t , where t R .

Example 5. Find egenvalues and eigenvectors of the matrix

3		−1	1
A =	−1 5 −1 .
	1	−1	3

Solution. Let’s solve the characteristic equation for the given matrix

3 − λ	−1	1
3 − λ	−1	1
−1	5 − λ	−1	= 0 ,
1	−1	3 − λ

(3 − λ)(5 − λ)(3 − λ) + 1	+ 1− (5 − λ) − (3 − λ) − (3 − λ) = 0,
(λ − 3)(λ − 2)(λ − 6)	= 0, λ1 = 2, λ2 = 3, λ3 = 6.

Thus, the matrix A has three egenvalues. Now let’s find eigenvectors, substituting in turn the values λ1, λ2 , λ3 into the system

(3 − λ)x1 − x2 + x3 = 0,

− x1 + (5 − λ)x2 − x3 = 0,x1 − x2 + (3 − λ)x3 = 0.

The eigenvector

corresponds to the egenvalue λ1 = 2 , which coordinates

satisfy the system

(3 − 2)x1 − x2 + x3 = 0,

x1 − x2 + x3 = 0,

+ (5 − 2)x2 − x3 = 0,

+ 3x2 − x3 = 0,

− x1

x − x

+ (3 − 2)x = 0,

− x

+ x = 0.

2x2 = 0,

Hence

and the eigenvector

X1 = C1

x1 = − x3

−1

By the analogy the eigenvectors X

= C

and

= C

−2

are found,

where C1 , C2 , C3 are non-zero real numbers.

Micromodule 3

CLASS AND HOME ASSIGNMENT

Solve the system of equations:

а) by the matrixм method; b) by the Cramer’s formulas.

x1 + 3x2 − x3 = 4,	2x1 + x2 + x3 = 2,
1. − x1 − x2 + 3x3 = 6,	2. 3x1 + 3x2 + 2x3 = 5,

2x1 + 6x2 − x3 = 11.	5x1 + 3x2 + 4x3 = 3.

Solve the system of equations using Gauss` method.

x1 + 2x2 + x3 = 5,		2x1 − x2 + x3 − x4 = 1,
3. 2x1 + x2 + 3x3 = 9,
3. 2x1 + x2 + 3x3 = 9,	4.	3x1 + 2x2 − x3 − x4 = 0,

		− x1 − 2x2 + 2x4 = 1,
−3x1 + x2 + 4x3 = 6.		− x1 − 2x2 + 2x4 = 1,
		x	− 3x	+ x + x	= 2.
x1 + 2x2 + 2x3 + x4 = 0,		1	2	3	4
x1 + 2x2 + 2x3 + x4 = 0,
5. 2x1 + 5x2 + 5x3 + 2x4 = 1,
	= 2.
3x1 + 7x2 + 7x3 + 3x4	= 2.

Answers

	1. x1 = 1, x2 = 2, x3 = 3. 2. x1 = 1 ,							x2 = 2 ,			x3 = −2 . 3. x1 = 1 , x2 = 1,
x	= 2 . 4. x	= 3+ 2t , x		=	−6 + 8t	, x =	−3+ 13t		, x		= t , where t R .

3	1	9	2	9		3		9		4

Micromodule 3

SELF-TEST ASSIGNMENTS

3.1. Solve the system of: а) by the matrix method; b) by the Cramer’s formulas.

x1 + x2 − x3 = 0,

3.1.1.2x1 + x2 + 2x3 = 10,

x1 − + x3 = −2.3x2

x1 + 2x2 + 3x3 = −1,

3.1.32x1 + x2 − 4x3 = −3,3x1 + 2x2 + x3 = 1.

2x1 − x2 + 3x3 = −4,

3.1.5.x1 + 2x2 − 4x3 = 19,−3x1 + 4x2 + 2x3 = 3.

x1 + x2 − x3 = 4,

3.1.7.2x1 + x2 + 2x3 = 2,3x1 − x2 + x3 = 0.

x1 + 3x2 − x3 = 0,

3.1.9.2x1 − 4x2 + 4x3 = 6,3x1 + 2x2 + x3 = 4.

3x1 + 4x2 + 3x3 = 3,

3.1.11. 4x1 + 5x2 − 3x3 = 4,

2x1 + 3x2 − 4x3 = 2.

3x1 + x2 + 2x3 = −2,

3.1.13. x1 + 3x2 + 2x3 = 2,

2x1 + x2 − x3 = −1.

2x1 + x2 + x3 = 2,

3.1.2.x1 + 2x2 + x3 = 3,

x1 + x2 + 2x3 = −1.

−= −2,

3.1.4.5x1 + 3x2 − 4x3 = −2,4x1 + 2x2 + 3x3 = 5.4x2 x33x1 +

3x1 − 2x2 + 4x3 = −17,

3.1.6. 4x1 + 3x2 − 2x3 = 18,

3x1 + x2 + 3x3 = −7.

x1 + 2x2 − 2x3 = −3,

3.1.8. 2x1 + x2 + 3x3 = 8,

3x1 − 4x2 + x3 = 5.

2x1 + 3x2 − 3x3 = 2,
3.1.10.	x1 − 4x2 + 5x3 = 2,
		+ 4x2 − 3x3 = 4.
3x1
x1		+	2x3 = 7,
3.1.12. x1		+	3x2	= 7,
2x		− 2x	+ x	= 1.
	1	2	3

3x1 − 2x2 + x3 = 0,
3.1.14. 7x1 + x2 + 4x3 = 0,
	x2 + 5x3 = 4.

x1 − x2 + x3 = 1,
3.1.15. 2x1		+	2x3 = 4,
			= 5.
3x1 + 2x2
x1 + 2x2 + x3 = 2,
3.1.17. 2x1		+	3x3 = −13,
			− 4x3 = 0.
3x1 − 3x2
x1 + 3x2 = 6,
3.1.19.	x1 −	x2 − x3 = 0,
2x +		2x + 3x = −2.
	1	2	3

x1 + 2x2 + 3x3 = 4,

3.1.21.2x1 − x2 + 2x3 = 1,

x1 + 3x2 + x3 = 4.

2x1 − 3x2 + x3 = 8,3x1 + − = 7,

2x1 + − = −12.5x2x2 2x33x33.1.23.

−= −2,

3.1.25.x1 + 2x2 + 3x3 = 7,2x1 + 3x2 + x3 = 1.

x1 + 2x2 + x3 = 4,

3.1.27.3x1 − x2 + 4x3 = 3,2x1 + 5x2 + 6x3 = 6.

x1 + x2 + x3 = 2,

3.1.29.2x1 + x2 − 2x3 = −3,3x1 − 2x2 + x3 = 7.x2 2x33x1 +

2x1 + x2 + x3 = 2,
3.1.16. 3x1 + 3x2 + 2x3 = 5,
	x −	x +	x = −3.
	1	2	3

=4,

3.1.18.x1 + 3x2 + x3 = 6,x1 + x2 + 3x3 = 0.

3x1 + 3x2 + 4x3 = 4,

3.1.20.5x1 − 7x2 + 8x3 = 20,4x1 + 5x2 − 7x3 = −8.

x1 − 2x2 + x3 = 4,

3.1.22.3x1 + x2 + 2x3 = 3,3x1 + 8x2 − 3x3 = 8.

x1 − 2x2 + 3x3 = 3,

3.1.24.2x1 + 3x2 − x3 = 13,3x1 − x2 − 2x3 = 8.x2 + x33x1 +

3x1 + 2x2 + x3 = 7,

3.1.26. 5x1 + x2 − x3 = 4,

2x1 − 7x2 − 3x3 = −11.

+= 1,

3.1.28.5x1 − 2x2 − 3x3 = 5,3x1 + x2 + x3 = 7.

x1 − 2x2 + 2x3 = −2,

3.1.30.2x1 + x2 + 3x3 = 10,3x1 − 4x2 + 6x3 = 0.3x2 x32x1 +

3.2. Investigate SLAE given in the extended matrix on consistency and in the case of consistency find its general solution.

	1	1	−3	0	2		1	−2 0 1			−3
	1	1	−3	0	2		1	−2 0 1			−3
							3	−1	−2	0	1
3.2.1.	1	−1	1	2	−1 .	3.2.2.	3	−1	−2	0	1	.
	3	3	−9	0	6		2	1	−2	−2	4

	1	2	−5	−1	3		1	3	−2	−2	7
	1	2	−5	−1	3		1	3	−2	−2	7

	1		1	1	1		1						1
	1		1	1	1		1						1
	1		1	4	3		5						2
3.2.3.	1		1	4	3		5						2			.
3.2.3.	2		1	9	2		7						3			.
	2		1	9	2		7						3

	4		4	7	6		8						5
	4		4	7	6		8						5

	1		2	3	4								0
	1		2	3	4								0
	7		14	20	27								0
3.2.5.	7		14	20	27								0					.
3.2.5.	5		10	16	19							−2						.
	5		10	16	19							−2

	3		5	6	13								5
	3		5	6	13								5

	2		−1	2	−1					0
	2		−1	2	−1					0
3.2.7.	1		5	4		3				1 .
	5		3	8	1					1
	5		3	8	1					1

	1		−1	1	−1				−2
	1		−1	1	−1				−2
3.2.9.	1		2	−2	−1					−5
3.2.9.	1		2	−2	−1					−5					.
	2		−1	−3	2					−1

	1		2	3	−6				−10
	1		2	3	−6				−10

		3	2	4	4		5								2
		3	2	4	4		5								2
		7	5	9	8		9								3
3.2.11.		7	5	9	8		9								3			.
		5	3	7	9		4								3

		6	5	7	5				−5					−3
		6	5	7	5				−5					−3

		7	8	−5	6						−4					5
		7	8	−5	6						−4					5
3.2.13.		6	7	−4	4								3			5
3.2.13.		6	7	−4	4								3			5
		5	9	7	5								7			3
		5	9	7	5								7			3

		2	3	4	3			0
		2	3	4	3			0
3.2.15.		4	6	9	8	−3 .
		6	9	9	4		8
		6	9	9	4		8

		5	3	4	−2								3					1
		5	3	4	−2								3					1
		8	5	5	−4								4					2
3.2.17.		8	5	5	−4								4					2
		7	4	7	−3								7				−1

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

	2	3	2			7
	2	3	2			7
	5	4	3			12
3.2.4.	5	4	3			12	.
	1	1	1			3

	7	1	1			9
	7	1	1			9

	1	2	3		14
	1	2	3		14
	3	2	1		10
3.2.6.	3	2	1		10			.
3.2.6.	1	1	1		6			.
	1	1	1		6

	1	1	0		3
	1	1	0		3

	1	2	1	0			8
	1	2	1	0			8
3.2.8.	0	1	3	1			15
3.2.8.	0	1	3	1			15		.
	4	0	1	1			11

	1	1	0	5			23
	1	1	0	5			23

	2	7		3		1						5
	2	7		3		1						5
	1	3		5			−2					3
3.2.10.	1	3		5			−2					3	.
	1	5	−9			8						1

	5	18		4		5						12
	5	18		4		5						12

	5	1		−3	−6
	5	1		−3	−6
	2	−5		7		9
3.2.12.	2	−5		7		9		.
	4	2		−4	−7

	5	−2		2		1
	5	−2		2		1

	1	3	3	5			−1
	1	3	3	5			−1
3.2.14.	2	6	5	6			1
3.2.14.	2	6	5	6			1		.
	3	7	4	8			2

	3	5	1	9			1
	3	5	1	9			1

	2	−1		1			−1			1
	2	−1		1			−1			1
3.2.16.	2	−1		0			−3			2
3.2.16.	2	−1		0			−3			2			.
	3	0	−1			1						−3

	2	2	−2			5						−6
	2	2	−2			5						−6

	1	−2		3		4					4
	1	−2		3		4					4
	0	1		−1		1						−3
3.2.18.	0	1		−1		1						−3		.
	1	3		0			−3				1

	0	−7		3		1						−3
	0	−7		3		1						−3

		1	2	3	−2					4
		1	2	3	−2					4
		2	−1	2	−3					1
3.2.19.		2	−1	2	−3					1			.
3.2.19.		1	3	1	2					4			.
		1	3	1	2					4

		4	4	6	−3					9
		4	4	6	−3					9

		1	2	−3	4						−1			2
		1	2	−3	4						−1			2
3.2.21.		5	1	3	4							2		12
3.2.21.		5	1	3	4							2		12	.
		7	5	−3	12							0		16

		2	−5	12	8							5		6
		2	−5	12	8							5		6

		1	1	1	1				0
		1	1	1	1				0
3.2.23.		2	1	−2	−2				1
3.2.23.		2	1	−2	−2				1
		3	−2	1	−3				13

		1	−2	−2	1						−3
		1	−2	−2	1						−3

	1 −2 1				1			1
	1 −2 1				1			1
3.2.25. 1			−2	1	−1		−1 .
			−2	1	5			5
	1		−2	1	5			5
	1
		1	2	3	−1	1
		1	2	3	−1	1
		3	2	1	−1
3.2.27.		3	2	1	−1	1					.
3.2.27.		2	3	1	1	1					.
		2	3	1	1	1

		2	2	2	−1
		2	2	2		1
						1
		3	4	−5	7						1
		3	4	−5	7						1
3.2.29.		2	−3	3	−2
3.2.29.		2	−3	3	−2						−1 .
		4	11	−13	16						2

		7	−2	1	3						3
		7	−2	1	3						3

	1		−3	2	−1						11
	1		−3	2	−1						11
3.2.20.	2		1	−3	4						−5
3.2.20.	2		1	−3	4						−5			.
	2		−13	11	−8						49

	4		9	−13 14							−37
	4		9	−13 14							−37

		3	2	1	1						7
		3	2	1	1						7
		5	1	−1	2						4
3.2.22.		5	1	−1	2						4		.
		1	−3	−2	−5						2

		2	−7
		2	−7	−3 −4					−11
				−3 −4					−11
	1		−2	3	2	−2
	1		−2	3	2	−2
	2		1	−2	3	10
3.2.24.	2		1	−2	3	10					.
	3		−4	1	6		0

	1		2	2	1		9
	1		2	2	1		9

		1	−2	3	−4			4
		1	−2	3	−4			4
3.2.26.		0	1	−1	1
3.2.26.		0	1	−1	1			−3 .
		1	3	0	−3			1

		0	−7	3	1
		0	−7	3	1			−3
								−3
	2		1	−1	−1						1		1
	2		1	−1	−1						1		1
	1		−1	1	1					−2			0
3.2.28.	1		−1	1	1					−2			0		.
	3		3	−3	−3						4		2

	4		5	−5	−5						7		3
	4		5	−5	−5						7		3

	3		1	1	−2					3		4
	3		1	1	−2					3		4
3.2.30.	2		3	−2	1				−4			5
3.2.30.	2		3	−2	1				−4			5		.
	1		2	3	4					1		3

	1		−2	3	−3					7		1
	1		−2	3	−3					7		1

3.3. Solve the homogeneous system of linear algebraic equations.

		x2 − x3 + 2x4 = 0,			x1 + x2 − 2x3 + x4 = 0,
−2x + x + 3x − x = 0,					− x		+ 2x	+ x	− x = 0,
3.3.1.	1	2	3	4	3.3.2.	1	2	3	4
3x1 − 2x2 + x3 + x4 = 0,					3x1		+	x3 + x4 = 0,
x	+		3x	+ 2x = 0.			+ 3x2	+ x3 + x4 = 0.
1			3	4	3x1		+ 3x2	+ x3 + x4 = 0.

	−2x1 + 2x2 + x3 − x4 = 0,
		3x1 + x2											+			x4 = 0,
3.3.3.		3x1 + x2											+			x4 = 0,
3.3.3.		− x				+ x						+ 2x				+ 2x		= 0,
			1						2						3		4
								4x2				+ 3x3 + 2x4 = 0.
								4x2				+ 3x3 + 2x4 = 0.
	3x1 + x2 + 3x3 + 2x4 = 0,
				+ 2x2								+ 3x3 + x4
3.3.5. − x1				+ 2x2								+ 3x3 + x4						= 0,
								x2 − x3 + 2x4 = 0,
	2x + 4x + 5x + 5x = 0.
			1					2						3			4
	2x1 + x2 − 2x3 + x4 = 0,
		x1 − x2										+				x4 = 0,
3.3.7.		x1 − x2										+				x4 = 0,
3.3.7.	− x			+ 2x							+ x				−	x	= 0,
			1					2					3			4
	2x			+ 2x							− x +					2x = 0.
			1					2					3			4
	− x1							+					x3 − x4 = 0,
3.3.9.	−2x					+ x				−			2x		+ x		= 0,
3.3.9.			1					2					3			4
	2x +						3x		−				x		+ x = 0,
			1					2					3			4
	− x			+			4x			−			2x		+ x		= 0.
			1					2					3			4
	4x1 + x2 + 2x3 − 3x4 = 0,

3.3.11.	x1 − 2x2 + x3 = 0,
3.3.11.			− 3x					− x				2	− 2x + x					= 0,
							1					2				3		4
	2x					− 2x						+ x			− 2x			= 0.
			1						2				3			4
	−2x1 + 2 x2 + x3 − x4 = 0,

3.3.13.	3x1 − x3 + x4 = 0,
3.3.13.	x					− x			+ 3x						+ 2x			= 0,
			1					2					3			4
			2x					+ x				+ 3x				+ 2x		= 0.
					1				2						3		4
	x1 + 3x2 − x3 + 2x4 = 0,
	−2x							+ x				+ 3x				+ x		= 0,
3.3.15.				1					2						3	4
3.3.15.	3x				− 2x				2			− x			− 2x			= 0,
			1						2				3			4
	2x + 2x + x + x = 0.
			1						2				3			4
	3x1 + 2x2 − x3 + 2x4 = 0,
3.3.17.	− x				+ x				+			3x		= 0,
3.3.17.			1					3					4
	− x					+ 2x						− 4x		4		= 0,
			2						3					4
	2x			+ x					+			2x		+ x = 0.
			1					2					3			4

	x1 − x2 + 2x3 + x4 = 0,
						+ 5x3 + 4x4 = 0,
3.3.4.	5x2 − 2x2					+ 5x3 + 4x4 = 0,
3.3.4.	−2x			+ x		− x	− 2x		= 0,
		1		2		3	4
	−2x + x − x − 2x = 0.
		1		2		3	4
	x1 + x2 + x3 − x4 = 0,
3.3.6.	− x + x − x − x = 0,
3.3.6.		x	− x		+ 2x		+ 3x		= 0,
		1		2		3	4
		1		2		3	4
		x	+ x		+ 2x		+ x	= 0.
		1		2		3	4
	3x1 − x2 + 2x3 + 3x4 = 0,
						− x3	− 2x4 = 0,
3.3.8.	−2x1 + x2					− x3	− 2x4 = 0,
3.3.8.		x + 2x			+ x − x				= 0,
		1		2		3	4
	2x + 2x + 2x								= 0.
		1		2		3

x1 − 2x2 + 2x3 − x4 = 0,

3.3.10.3x1 + x2 − x3 + x4 = 0,

2x1 + 3x3 + x4 = 0,6 = 0.+ x4x2 + 4x3x1 −

x1 − x2 + 2x3 − x4 = 0,

3.3.12.3x1 + 2x2 − x3 + x4 = 0,

−2x1 + x2 + x3 + 2x4 = 0,

x1 + x2 + x3 + x4 = 0.

	−3x1 + x2 + 2x3 − x4 = 0,
						+ x4 = 0,
3.3.14.	2x1 + 3x2 − x3					+ x4 = 0,
3.3.14.		x		− 2x	+ 3x	+ 2x	= 0,
		1		2	3		4
				x2 + 2x3		+ x4 = 0.
				x2 + 2x3		+ x4 = 0.
	x1 + 2x3 − x4 = 0,
						= 0,
3.3.16.	−2x1 + x2 + x4					= 0,
3.3.16.	2x			+ x	+ x + 2x		= 0,
			1	2	3	4
	x		+ 2x		+ 2x	+ 2x	= 0.
		1		2	3	4
	−2x1 + 3x2 + 2x3 + x4 = 0,
			− 2x2 + x3 − x4 = 0,
3.3.18. x1			− 2x2 + x3 − x4 = 0,
	3x1 + x2 − x3 = 0,
	x		+ x + x = 0.
		1		2	3

x1 + x2 + 2x3 + x4 = 0,

3.3.19. 2x2 + 3x3 − x4 = 0,

−2x1 − 3x2 − x3 + x4 = 0,− x1 − x2 + 4x3 + x4 = 0.

−3x1 + 2x2 − 2x3 + 2x4 = 0,

3.3.21. 2x1 − x2 + x3 − x4 = 0,

−x1 − x3 + x4 = 0,

−2x1 + x2 − 2x3 + 2x4 = 0.

−x3 = 0,

3.3.23.x1 − x2 + 2x3 − x4 = 0,

3x1 + x3 + x4 = 0,4x1 + x2 + 2x3 + 2x4 = 0.+ 2x4x22

x1 + x2 + x3 + x4 = 0,

3.3.25. 2x1 − x2 + 2x3 − x4 = 0,

−3x1 + x2 − 2x3 + 2x4 = 0,

x1 + 2x2 + 2x3 + 3x4 = 0.

3x1 + 3x2 − x4 = 0,

3.3.27. −2x1 + x2 + x3 + 2x4 = 0,

x1 − 2x2 − 3x4 = 0,

2x1 + 2x2 + x3 − 2x4 = 0.

=0,

3.3.29.3x1 − x2 + x3 + 2x4 = 0,2x1 + 2x2 − 2x3 + x4 = 0,

x1 + 3x2 − x3 + 2x4 = 0.2x2 − x4−4x1 +

=0,

3.3.20.− x1 + x2 + 2x4 = 0,

3x1 − 2x2 − x3 = 0,

4x1 + 2x2 + x3 + x4 = 0.+ 2x3 − x43x22x1 +

	4x1 − 2x2 + 3x3 + x4 = 0,
	−2x − x				2	−	2x			+ x = 0,
3.3.22.				1	2				3	4
3.3.22.	x		+ x − x				= 0,
	2			3		4
	2x − 2x + 2x + x = 0.
		1		2					3	4
	3x1 + 2x3 + 2x4 = 0,

3.3.24. −2x1 + x2 − x3 = 0,
	3x1 − 2x2 + x3 + x4 = 0,
	4x			− x	+ 2x				+ 3x = 0.
		1		2			3			4
	− x1 + x2 + 2x3 − 2x4 = 0,
				+ x2	− x3 + 3x4 = 0,
3.3.26.	3x1			+ x2	− x3 + 3x4 = 0,
3.3.26.	− x			− x	+ x				= 0,
			1	2			3
	x		+ x + 2x					+ x = 0.
	1			2			3			4
	x1 + x2 + x3 + x4 = 0,
	−2x − x + 2x + x = 0,
3.3.28. 2x				− x	+ x			=		0,
				1	2				3	4
		1		3			4
	x		+ x + 3x					= 0.
	1			3		4
	x1 − x2 + 2x3 + x4 = 0,
				+ x3	+ 2x4 = 0,
3.3.30. 3x2				+ x3	+ 2x4 = 0,
	−2x1 + x 2 − x3 − 2x4 = 0,
	− x			+ x	− x				= 0.
			1	3			4

Micromodule 4

BASIC THEORETICAL INFORMATION VECTORS

Vectors, linear operations with vectors. Projection of a vector on an axis. Linear dependence and independence of vectors. Basis and system of coordinates. Vectors in CCS.

Literature: [1, chapter 4], [4, part 3, item 3.2], [6, chapter 2, §§ 1—3], [7, chapter 1, § 3], [10, chapter 1, § 2], [11, chapter 1, § 2].

4.1. Vectors. Basic concepts

The vector is defined by the numerical value and a direction (speed, acceleration, force, etc.)

Geometrically the vector is a directed line segment (Fig. 1.2, а) and is

designated as a or AB , where the point A is the beginning or initial point of the vector, and В is its end or terminal point.

Definition 1.24. A distance between the beginning of a vector and its end is called the length (or the module) of a vector and is designated as | a | or | AB | .

Definition 1.25. Vectors which lie on one straight line or parallel straight lines are called collinear.

Vectors a and b are equal if they are collinear, have identical modules and identical directions (Fig. 1.2, b).

Two vectors are opposite if they are collinear, have identical modules and opposite directions.

A vector whose beginning and end coincide is called a zero vector. Its direction is not determined.

The vector whose length is equal to a unit is called a unit vector.

Definition 1.26. A unit vector whose direction coincides with a direction of a vector a is called an ort of the vector a and is designated as a0 .

Definition 1.27. Three vectors are called coplanar if they lie on the same or parallel planes

Three vectors are coplanar if two of them or all three are collinear or if one of them is a zero vector.

4.2.Linear operations on vectors

1)Addition (subtraction) of vectors.

2)Multiplication of a vector by a number (scalar).

We define the sum a + b of two vectors a and b as follows: we position vectors a and b so that the terminal point of a coincides with the initial point of b . Then the sum a + b is represented by an arrow directed from the initial

point of a to the terminal point of b (Fig. 1.2, c) (a rule of a triangle).

The sum of two vectors can also be constructed with help of the rule of parallelogram (Fig. 1.2, d).

Subtraction of vectors is defined as an operation opposite to addition. We define a difference a - b of two vectors a and b as follows: we position vectors a and b so that the initial point of a coincides with the initial point of b .Then

the difference a - b is represented by a vector directed from the terminal point of b to the terminal point of a . (Fig. 1.2, e).

	В			b		a – b
a	a			a	a + b	a – b
	a	b	a	a + b		a
A	a b	b		a + b
					b
						b
						b
				c	d	e
				Fig. 1.2	d
Definition 1.28. If			a is a vector and λ is a scalar then the product λ a is a
vector having a magnitude equal to the product \| λa \|=\| λ \|\| a \|						and having the same

direction as a if λ > 0 and the direction opposite to a if λ < 0 .

From definition of multiplication of a vector on number it follows that when vectors are collinear there is a unique number λ such that b = λ a , and on the

contrary, if b = λ a ,		a and b are collinear.
		Properties of operations on vectors
1.	a + b = b + a .		2.	(a + b) + c = a + (b + c) .
3.	λ(a + b) = λa + λb .		4.	a + (−a) = 0 .
5.	a + 0 = a .		6.	a(λ + μ) = λa + μa.

7. λ(μa) = λμa.

4.3. Projection of a vector on axis

Definition 1.29. A directed straight line with a given initial point and a unit of length is called an axis.

A basis of perpendicular AA1 dropped from the point A on an axis l is called

a projection of the point A on the axis l (Fig. 1.3, а).

Suppose A1 is the projection of the point A onto the axis l, B1 is the projection of a point B onto the axis l. We suppose a = AB .

<<< < Предыдущая 1 2 3 4 56 / 286 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 > Следующая >>>

Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]

#
05.09.2019478.78 Кб6gtfnmТ.docx
#
15.08.2019637.53 Кб3Gulenko.docx
#
12.11.2018765.44 Кб6Hagakure_-_Sokrytoe_v_listve.doc
#
19.02.20166.95 Mб201Higher Mathematics. Part 3.pdf
#
19.02.20169 Mб95Higher_Mathematics_Part_1.pdf
#
19.02.20166.63 Mб1052Higher_Mathematics_Part_1.pdf
#
19.02.20167.48 Mб408Higher_Mathematics_Part_2.pdf
#
19.02.20169.73 Mб45Higher_Mathematics_Part_2.pdf
#
19.02.20167.31 Mб123Higher_Mathematics_Part_2.pdf
#
19.02.20169.11 Mб304Higher_Mathematics_Part_3.pdf
#
19.02.2016694.82 Кб11HiraganaKatakanaWorkSheet (1).pdf