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3
Non-Profit Joint Stock Company
The department of
Higher mathematics
MATHEMATICS 1
Methodological Guidelines and Tasks
for carrying out the calculation-graphical work for students of specialties
5В071700 «Heat power engineering»,
5В071800 «Electrical power engineering»,
5В071900 «Radio engineering, electronics and telecommunications»
Part 1
Almaty 2014
ALMATY UNIVERSITY OF
POWER ENGINEERING
AND
TELECOMMUNICATIONS
4
Compiler: Kim R.E., Mathematics 1. Methodological Guidelines and Tasks
for carrying out the calculation-graphical work for students of specialties
5В071700 «Heat power engineering», 5В071800 «Electrical power
engineering», 5В071900 «Radio engineering, electronics and
telecommunications». Part 1. – Almaty, 2014 – 33 p.
Methodological Guidelines and Tasks for carrying out the calculation-
graphical work contain sections of the first semester program of course of
mathematics for students of AUPET: elements of linear algebra, analytic geometry
and complex numbers.
The basic theoretical questions of the program and the solution of an
exemplary embodiment are given.
Tables 11, figures 6, bibl. 3
Reviewer: candidate of sciences in philology, V.S. Kozlov
Printed according to the Publishing plan of Non-Profit Joint Stock Company
"Almaty University of Power Engineering and Telecommunications " for 2014
© Non-Profit Joint Stock Company
"Almaty University of Power Engineering and Telecommunications", 2014
5
Introduction
Mathematics plays an important role in engineering studies. It is not only a
quantitative calculation device, but also mathematics is the means of accurate
research and extremely precise formulation of concepts and problems.
Mathematical methods have become an integral part of any technical discipline.
These facts lead to the need to strengthen the applied orientation of course of
mathematics and improve basic mathematical training. CGW is performed in a
separate thin notebook. In the number of each task the second digit indicates the
variant.
Calculation-graphical work.
Elements of linear algebra, analytic geometry and complex numbers
Purpose: to master the fundamental concepts and methods of the theory of
algebra and geometry. Get the skills in calculation of determinants and operations
on matrices and vectors, which are used for solving systems of equations and used
in analytic geometry. By the analytical curve equations determine their geometric
properties and relative location on the plane or in space.
Perform all arithmetic operations with complex numbers.
Theoretical questions
1. Determinants and their properties, the calculation.
2. Matrices, operations on them, the inverse matrix.
3. Vectors, their length, linear operations on vectors. Collinearity.
4. Scalar, vector, mixed products of vectors and their applications.
5. Different types of equation of the straight line on the plane and in space.
6. Equations of a plane.
7. Angle between the lines, planes, straight line and plane.
8. The distance from a point to the straight line and to the plane.
9. Ellipse, hyperbola, parabola. Their canonical equations.
10. Second-order surfaces.
11. Bringing of the general equations of the second-order curves and surfaces
to the canonical form.
12. Various methods for solving systems of linear equations:
а) Cramer's rule;
b) by using the inverse matrix.
13. Complex numbers, the module and the main part of the complex
numbers, operations over the complex numbers, De Moivre formula.
Calculated tasks
Task1. The third order determinant is given.
It is required to calculate:
a) the minor 23M and the algebraic complement 23A of the element 23a ;
b) the determinant by the triangle rule (rule of Sarrus);
6
c) the determinant by the method of expansion along the i-th row and the j-th
column.
1.1 1,2,
121
201
213
ji 1.2 3,2,
314
131
232
ji 1.3 2,1,
122
013
376
ji
1.4 1,3,
234
084
371
ji 1.5 2,1,
110
231
462
ji 1.6 2,3,
7110
111
496
ji
1.7 1,2,
812
713
301
ji 1.8 2,3,
148
131
215
ji 1.9 1,3,
134
633
522
ji
1.10 2,3,
465
103
553
ji 1.11 3,2,
121
423
211
ji 1.12 1,3,
323
504
234
ji
1.13 3,2,
243
678
312
ji 1.14 2,1,
121
453
212
ji 1.15 1,2,
113
342
653
ji
1.16 1,3,
354
013
582
ji 1.17 3,2,
321
642
063
ji 1.18 3,3,
231
720
103
ji
1.19 2,3,
730
529
1116
ji 1.20 3,1,
434
603
521
ji 1.21 2,3,
121
332
111
ji
1.22 2,2,
503
421
245
ji 1.23 1,3,
221
173
545
ji 1.24 1,2,
722
234
013
ji
1.25 3,2,
161
135
072
ji 1.26 3,1,
2310
155
118
ji 1.27 1,1,
201
123
523
ji
1.28 3,2,
210
123
674
ji 1.29 3,1,
321
135
157
ji 1.30 2,3,
753
421
121
ji
Task 2. Given matrices А, В, С.
Find:
а) the product of matrices АВ or ВС, or СВ (if possible). Explain why if it is
impossible;
b) the matrix 1A , the inverse matrix for A.
7
2.1
А= ,
324
381
273
В= ,
1
7
2
С=
310
411
2.2
А= ,
512
142
350
В= ,042
113
С=
4
2
5
2.3
,
172
394
412
А ,124В
33
12
17
С
2.4
,
634
142
111
А ,
34
71
42
В 234 С
2.5
,
234
352
401
А ,20
13
В
6517
2413С
2.6
,
353
421
121
А ,
56
11
20
14
В
20
17С
2.7
,
321
135
175
А ,
2
6
3
В
301
332С
2.8
,
332
154
043
А ,153
224
В
3
1
6
С
2.9
А= ,
112
620
171
,135В
22
03
18
С
2.10
173
232
201
А , ,
05
74
13
В 245 С
2.11
,
231
713
103
А ,31
24
В
1651
3524С
8
2.12
,
121
642
414
А ,
47
02
31
25
В
10
28С
2.13
А= ,
221
152
110
В= ,
1
3
4
С=
731
420
2.14
А= ,
112
111
496
В= ,
742
321 С=
1
8
5
2.15
А= ,
213
812
301
В= ,426 С=
22
17
63
2.16
А= ,
124
301
154
В=
57
24
23
, С= 432
2.17
А= ,
212
434
633
В= ,21
54
С=
2291
8132
2.18
А= ,
301
234
352
В= ,
32
21
94
13
С=
12
65
2.19
А= ,
602
734
214
В= ,
9
1
8
С=
342
516
2.20
А= ,
412
271
603
В= ,
114
483 С=
8
6
2
2.21
А= ,
524
301
713
В= ,047 С=
43
91
32
2.22
А= ,
264
532
143
В= ,
35
22
16
С= 560
2.23
А= ,
212
043
351
В= ,
12
93 С=
2931
4132
9
2.24
А= ,
314
712
163
В= ,
53
20
15
27
С=
42
26
2.25
А= ,
172
514
332
В= ,
4
5
3
С=
960
142
2.26
А= ,
322
813
465
В= ,210
186
С=
3
6
7
2.27
А= ,
452
271
532
В= ,415 С=
13
72
19
2.28
А= ,
164
042
315
В= ,
30
14
85
С= 825
2.29
А= ,
325
113
2010
В= ,24
37
С=
6314
1732
2.30
А= ,
323
914
132
В= ,
32
73
61
34
С=
24
15
Task 3. Given points А and В; and vectors b and c .
Find:
a) the module (length) of the vector ABa and the midpoint of the
segment AB;
b) the projection of the vector a on c ;
c) the area of the parallelogram, obtained from the vectors b and c ;
d) volume of the pyramid constructed from the vectors cba ,, .
3.1 А(5, -4, 3), В(1, 2, -8),
в =(0, 1, 4), с =(5, 2, -3)
3.2 А(-3, 1, 0), В(7, 1, -5),
в =(7, 1, 4), с =(5, 8, -3)
3.3 А(0, 4, 5), В(3, -2, 1),
в =(6, 4, 4), с =(0, 2, -2)
3.4 А(3, -2, 5), В(4, 5, 7),
в =(5, 1, 4), с =(5, -3, -3)
3.5 А(2, -3, 7), В(3, 2, 8),
в =(6, 4, 6), с =(0, 6, -2)
3.6 А(2, -1, 7), В(6, 3, 4) ,
в =(4, 5, 4), с =(7, 8, 5)
10
3.7 А(3, 1, 7), В(2, -3, 9),
в =(9, 1, 4), с =(8, 2, -3)
3.8 А(2, 1, -6), В(1, 4, 9) ,
в =(1, 1, 8), с =(5, -3, 9)
3.9 А(2, -4, 8), В(5, 4, 7),
в =(6, 3, 4), с =(7, 2, -2)
3.10 А(3, 2, 5), В(4, 0, -3) ,
в =(7, 2, 4), с =(5, 8, 4)
3.11 А(2, 3, -1),В(-6, 4, 2),
в =(2, 9, 6), с =(0, 6, 4)
3.12 А(-4, 2, 3), В(8, 7, -2),
в =(5, 3, 4), с =(5, 3, 4)
3.13 А(5, 3, 6), В(-2, 3, 5),
в =(6, 2, 4), с =(7, 2, -7)
3.14 А(0, 6, 0), В(5, 3, -4),
в =(8, 5, 3), с =(7, 8, 7)
3.15 А(4, 2, 0), В(1, -7, 8),
в =(2, 8, 6), с =(0, 6, 3)
3.16 А(4, 2, 5), В(-1, 0, 6),
в =(2, 1, 8), с =(5, 9, 9)
3.17 А(3,-5, 8), В(6, 3, 9),
в =(7, 3, 4), с =(3, 2, -2)
3.18 А(7, 2, 2), В(-5, 7, -7) ,
в =(6, 1, 3), с =(1, 8, 7)
3.19 А(5, -3, 1), В(2, 3, 7),
в =(2, 7, 6), с =(7, 6, 4)
3.20 А(8, -6, 4), В(10, 5, 1),
в =(9, 1, 8), с =(4, 9, 1)
3.21 А(5, 6,-8), В(8, 10,7),
в =(1, 3, 4), с =(3, 6, -2)
3.22 А(1, -1, 3), В(6, 5, 8),
в =(7, 8, 3), с =(1, 8, 9)
3.23 А(3, 5,-7), В(8, 4, 1),
в =(2, 5, 6), с =(7, 6, 8)
3.24 А(6, -6, 5), В(4, 9, 5),
в =( 6, 1, 4), с =(0, 9, 1)
3.25 А(4,6,11), В(9,3,-4),
в =(3, 7, 6), с =(8, 6, 4)
3.26 А(5, 7, 4),В(4,-10, 9),
в =(8, 1, 7), с =(4, 5, 1)
3.27 А(-9, 8, 9), В(7, 1,-2),
в =(7, 2, 4), с =(5, 2, -1)
3.28 А(5, 2, 6), В(1, 8, -2),
в =(4, 2, 3), с =(2, 3, 7)
3.29 А(2, 8, -9), В(7, 5,-5),
в =(8, 1, 6), с =(6, 1, 4)
3.30 А(-2, 7, 0), В(6, 3, 5),
в =(2, -3, 8), с =(3, 9, 2)
Task 4. Given points A1, A2 on a plane and the equation of the line L1.
Write the equation of a straight line:
a) )( 212 AAL – passing through these points;
b) 2L – as a general equation of straight line;
c) 2L – in the form of equation of straight line with slope;
d) 2L – in the form of equation of straight line in segments;
e) 3L , passing through the point 2A and perpendicular to 1L .
4.1 045:),3,2(),1,3( 121 yxLAA 4.2 0132:),2,6(),1,4( 121 yxLAA
4.3 0253:),13,2(),7,4( 121 yxLAA 4.4 013:),2,4(),1,5( 121 yxLAA
4.5 0154:),3,6(),1,2( 121 yxLAA 4.6 0922:),2,1(),3,8( 121 yxLAA
4.7 0243:),1,3(),7,1( 121 yxLAA 4.8 0138:),3,7(),1,2( 121 yxLAA
4.9 047:),3,5(),1,6( 121 yxLAA 4.10 0452:),1,6(),2,4( 121 yxLAA
4.11 023:),2,2(),3,4( 121 yxLAA 4.12 04:),2,4(),1,7( 121 yxLAA
11
4.13 015:),3,1(),1,9( 121 yxLAA 4.14 0852:),3,1(),1,8( 121 yxLAA
4.15 0253:),1,3(),7,5( 121 yxLAA 4.16 0167:),3,7(),7,2( 121 yxLAA
4.17 057:),3,7(),1,5( 121 yxLAA 4.18 0642:),1,2(),8,4( 121 yxLAA
4.19 02:),2,9(),4,4( 121 yxLAA 4.20 012:),1,4(),6,7( 121 yxLAA
4.21 0152:),7,1(),1,8( 121 yxLAA 4.22 0942:),3,1(),1,3( 121 yxLAA
4.23 025:),1,6(),1,5( 121 yxLAA 4.24 047:),4,1(),8,2( 121 yxLAA
4.25 029:),2,9(),2,4( 121 yxLAA 4.26 0455:),1,2(),6,2( 121 yxLAA
4.27 0452:),9,1(),1,6( 121 yxLAA 4.28 094:),5,1(),1,10( 121 yxLAA
4.29 085:),1,1(),1,3( 121 yxLAA 4.30 0127:),5,1(),8,1( 121 yxLAA
Task 5. Given points 321 ,, AAA . It is required to:
a) write the equation of the plane );( 3211 AAAP
b) write 1P as a general equation of a plane;
c) write 1P as equation of a plane in segments;
d) write 1P in a general equation of a plane, passing through the point 1A ;
e) make a canonical equation of the straight line )( 321 AAL ;
f) write a parametric equation of the straight line 1L ;
g) write the equation of the straight line )( 12 NAL perpendicular to the
plane 1P .
5.1 )2,6,4(),1,3,2(),5,1,3( 321 AAA 5.2 )7,2,1(),2,4,1(),1,4,7( 321 AAA
5.3 )7,5,4(),1,9,3(),5,4,3( 321 AAA 5.4 )1,3,4(),1,5,9(),2,4,3( 321 AAA
5.5 )6,5,2(),4,2,3(),5,8,0( 321 AAA 5.6 )2,4,5(),8,7,3(),6,5,2( 321 AAA
5.7 )7,4,4(),1,9,1(),2,4,1( 321 AAA 5.8 )6,8,2(),1,6,2(),1,9,1( 321 AAA
5.9 )2,5,4(),1,5,3(),5,4,1( 321 AAA 5.10 )2,2,5(),8,7,4(),1,5,9( 321 AAA
5.11 )7,5,1(),2,9,3(),5,7,3( 321 AAA 5.12 )6,3,2(),1,7,2(),1,6,1( 321 AAA
5.13 )6,7,2(),4,2,4(),5,9,0( 321 AAA 5.14 )7,2,3(),1,4,2(),1,5,7( 321 AAA
5.15 )7,3,4(),1,7,1(),1,4,3( 321 AAA 5.16 )1,3,4(),1,5,8(),2,4,4( 321 AAA
5.17 )6,3,4(),1,8,1(),1,4,1( 321 AAA 5.18 )2,4,5(),7,2,3(),6,5,2( 321 AAA
5.19 )7,3,5(),1,7,3(),1,4,2( 321 AAA 5.20 )5,8,2(),1,6,2(),2.9,3( 321 AAA
5.21 )6,3,6(),1,8,2(),3,8,1( 321 AAA 5.22 )6,3,1(),1,7,4(),5,6,1( 321 AAA
5.23 )7,2,4(),1,9,1(),1,4,8( 321 AAA 5.24 )7,1,3(),2,3,2(),1,5,8( 321 AAA
5.25 )6,3,4(),1,8,9(),2,4,1( 321 AAA 5.26 )5,3,4(),9,5,8(),2,3,4( 321 AAA
5.27 )2,3,5(),8,7,3(),1,5,2( 321 AAA 5.28 )2,1,5(),7,2,3(),6,5,9( 321 AAA
5.29 )6,3,2(),1,6,2(),1,8,1( 321 AAA 5.30 )5,7,2(),1,6,4(),1,9,3( 321 AAA
12
Task 6. Solve the system
a) by Cramer’s method;
b) by matrix method (using the inverse matrix).
6.1
13
72
yx
yx
6.2
12
34
yx
yx
6.3
32
25
yx
yx
6.4
132
3
yx
yx
6.5
12
38
yx
yx
6.6
42
26
yx
yx
6.7
132
13
yx
yx
6.8
1126
34
yx
yx
6.9
33
29
yx
yx
6.10
13
752
yx
yx
6.11
17
34
yx
yx
6.12
43
27
yx
yx
6.13
13
72
yx
yx
6.14
126
72
yx
yx
6.15
33
265
yx
yx
6.16
132
38
yx
yx
6.17
12
78
yx
yx
6.18
42
66
yx
yx
6.19
13
734
yx
yx
6.20
1136
73
yx
yx
6.21
32
27
yx
yx
6.21
13
132
yx
yx
6.23
62
123
yx
yx
6.24
42
27
yx
yx
6.23
13
72
yx
yx
6.26
62
27
yx
yx
6.27
6
74
yx
yx
6.28
126
32
yx
yx
6.29
62
12
yx
yx
6.30
52
2
yx
yx
Task 7. Solve the system of equations.
a) by Cramer’s method;
b) by matrix method (using the inverse matrix).
7.1
2323
932
732
321
321
321
xxx
xxx
xxx
7.2
1643
62
1022
321
321
321
xxx
xxx
xxx
7.3
1725
1342
123
321
321
321
xxx
xxx
xxx
13
7.4
222
93
332
321
321
321
xxx
xxx
xxx
7.5
152
3243
3423
321
321
321
xxx
xxx
xxx
7.6
934
93
13632
321
321
321
xxx
xxx
xxx
7.7
4342
1657
9432
321
321
321
xxx
xxx
xxx
7.8
7632
4
1134
321
321
321
xxx
xxx
xxx
7.9
114
457
2432
321
321
321
xxx
xxx
xxx
7.10
8523
6452
84
321
321
321
xxx
xxx
xxx
7.11
52
1243
3423
321
321
321
xxx
xxx
xxx
7.12
332
7432
752
321
321
321
xxx
xxx
xxx
7.13
1023
922
1544
321
321
321
xxx
xхx
xxx
7.14
72
6434
332
321
321
321
xxx
xxx
xxx
7.15
234
523
622
321
321
321
xxx
xxx
xxx
7.16
423
95
432
321
321
321
xxx
xxx
xxx
7.17
035
6243
232
321
321
321
xxx
xxx
xxx
7.18
524
523
7453
321
321
321
xxx
xxx
xxx
7.19
224
853
53
321
321
321
xxx
xxx
xxx
7.20
842
125
63
321
321
321
xxx
xxx
xxx
7.21
142
13225
93
321
321
321
xxx
xxx
xxx
7.22
1123
7332
332
321
321
321
xxx
xxx
xxx
7.23
1123
743
854
321
321
321
xxx
xxx
xxx
7.24
2523
14432
852
321
321
321
xxx
xxx
xxx
7.25
63
1432
532
321
321
321
xxx
xxx
xxx
7.26
832
543
162
321
321
321
xxx
xxx
xxx
7.27
13
1652
5782
321
321
321
xxx
xxx
xxx
14
7.28
332
432
4425
321
321
321
xxx
xxx
xxx
7.29
955
6743
9232
321
321
321
xxx
xxx
xxx
7.30
132
3323
3347
321
321
321
xxx
xxx
xxx
Task 8. Given point А; radius of the circle R; a, b – semiaxes of curves;
equation of the straight line D. It is required to:
а) write the equation of a circle with center A and radius R;
б) write the canonical equation of an ellipse with semiaxes a and b. Find the
coordinates of it’s foci and the eccentricity;
c) write the canonical equation of the hyperbola with the real semiaxis a and
imaginary semiaxis b. Find the coordinates of its foci, eccentricity, equations of the
asymptotes;
d) write the canonical equation of the parabola with vertex at the origin and if
D is its directrix. Find the coordinates of its focus, eccentricity;
e) make the drawing of ellipse, hyperbola, parabola.
8.1 A(2,-4), R=4, a=1, b=3, D: x=-5 8.2 A(-8,2), R=1, a=6, b=5, D: x=-5
8.3 A(1,-4), R=5, a=8, b=3, D: y=-6 8.4 A(5,-4), R=2, a=6, b=4, D: y=-2
8.5 A(2,-5), R=7, a=3, b=2, D: x=4 8.6 A(1,8), R=5 , a=3, b=2, D: x=-3
8.7 A(3,-4), R=9, a=7, b=6, D: y=-2 8.8 A(10,1), R=8, a=1, b=6, D: y=-4
8.9 A(5,-4), R=1, a=6, b=4, D: x=-5 8.10 A(6,3), R=8 , a=2, b=3, D: x=-5
8.11 A(1,-3), R=5 , a=8, b=2, D: y=6 8.12 A(5,5), R=2, a=1, b=3, D: y=-7
8.13 A(2,-6), R=7, a=3, b=4, D: x=5 8.14 A(12,6), R=7, a=6, b=2, D: x=-5
8.15 A(3,4), R=9 , a=2, b=6, D: y=-8 8.16 A(0,5), R=4, a=6, b=4, D: y=8
8.17 A(2,-9), R=7, a=5, b=2, D: x=6 8.18 A(-5,0), R=7, a=4, b=5, D: x=1
8.19 A(8,4), R=6 , a=8, b=5, D: y=2 8.20 A(5,1), R=2, a=9, b=1, D: x=-1
8.21 A(5,-4), R=4, a=6, b=4, D: x=1 8.22 A(-3,2), R=4, a=8, b=4, D: y=1
8.23 A(1,8), R=5, a=9, b=4, D: y=-6 8.24 A(9,1), R=6, a=4, b=7, D: x=-3
8.25 A(2,-5), R=7, a=7, b=4, D: x=9 8.26 A(-9,2), R=7, a=1, b=8, D: y=7
8.27 A(7,4), R=5, a=1, b=7, D: y=8 8.28 A(11,-4), R=2, a=2, b=4, D: x=8
8.29 A(-2,5), R=5, a=7, b=1, D: x=8 8.30 A(12,4), R=7, a=3, b=5, D: y=-9
Task 9. Determine the type (name) of the second order surface and make a
schematic drawing:
9.1 y
zxb
zyxa 3
63);1
443)
22222
9.2
133
);141616
)22222
zx
bzyx
a
9.3 y
zxb
zyxa
43);1
1163)
22222
9.4
163
);1101010
)22222
zx
bzyx
a
9.5 z
yxb
zyxa 5
91);1
443)
22222
9.6
zyx
bzyx
a 53
);1111
)22222
15
9.7 1
43);1
4925)
22222
zy
bzyx
a 9.8
zzy
bzyx
a 6916
);1444
)22222
9.9 1
63);1
1912)
22222
zx
bzyx
a 9.10
yzx
bzyx
a 249
);11169
)22222
9.11 y
zxb
zyxa 4
49);1
16169)
22222
9.12
zyx
bzyx
a 463
);14912
)22222
9.13 x
zyb
zyxa 2
610);1
483)
22222
9.14
044
);11162
)22222
zx
bzyx
a
9.15 y
zxb
zyxa 2
43);1
1164)
22222
9.16
063
);1493
)22222
yx
bzyx
a
9.17 y
xb
zyxa 3
3);1
543)
2222
9.18
048
);11164
)22222
zx
bzyx
a
9.19 y
zxb
zyxa
49);1
1614)
22222
9.20
163
);1449
)22222
zx
bzyx
a
9.21 z
yxb
zyxa 2
53);1
9163)
22222
9.22
yx
bzyx
a 83
);164169
)2222
9.23 y
zxb
zyxa 2
412);1
4169)
22222
9.24
yz
bzyx
a 46
);14412
)2222
9.25 1
93);1
4259)
22222
zx
bzyx
a 9.26
yzx
bzyx
a 21616
);1161616
)22222
9.27 1
44);1
11625)
22222
zx
bzyx
a 9.28
yzx
bzyx
a 244
);1444
)22222
9.29 z
xyb
zyxa 4
63);1
2519)
22222
9.30
144
);111616
)22222
zx
bzyx
a
Task 10. Lead to the canonical form the equation of the second order,
construct this curve.
10.1 020282 yxx 10.2 0109325449 22 yxyx
10.3 016282 yxx 10.4 0222 22 xyx
10.5 07822 22 yxyx 10.6 06205 2 yxy
10.7 028684 2 yxx 10.8 05262 yxx
10.9 010036409 22 yxyx 10.10 01991864916 22 yxyx
10.11 028824 2 yxy 10.12 05040109 22 yyx
10.13 0201843 22 xyx 10.14 030262 yxx
10.15 016842 yxy 10.16 0101284 2 yxy
16
10.17 030284 2 yxx 10.18 0642424 22 yxyx
10.19 032882 yxy 10.20 098222 yxyx
10.21 09183095 22 yxyx 10.22 0301684 2 yxy
10.23 09282 yxy 10.24 0106189 22 yxyx
10.25 01276436169 22 yxyx 10.26 0301223 2 yxy
10.27 0252122 yxy 10.28 04422 yyx
10.29 02012822 yxyx 10.30 06464416 2 yxy
Task 11. Given complex numbers z1 and z2.
It is required to find:
а) the module of the complex numbers z1;
b) the argument of the complex number z1;
c) the representation of the complex number z1 in the trigonometric and
exponential forms;
d) the sum of complex numbers z1 and z2 analytically and graphically;
e) (z2)5;
f) multiplication of z1z2 in trigonometric form;
g) all complex roots 32z (for versions with even numbers) and 4
2z (for
versions with odd numbers). Show the solution in the drawing.
11.1
)2
sin2
(cos28
;88
2
1
iz
iz
11.2
)3
2sin
3
2(cos5
;355
2
1
iz
iz
11.3
)2
3sin
2
3(cos27
;77
2
1
iz
iz
11.4
)4
sin4
(cos16
;10
2
1
iz
z
11.5
)4
5sin
4
5(cos18
;399
2
1
iz
iz
11.6
)4
6sin
4
6(cos23
;33
2
1
iz
iz
11.7
)sin(cos4
;322
2
1
iz
iz
11.8
)6
5sin
6
5(cos27
;2727
2
1
iz
iz
11.9
))4
3sin()
4
3(cos(24
;44
2
1
iz
iz
11.10
)6
2sin
6
2(cos12
;366
2
1
iz
iz
17
11.11
)4
7sin
4
7(cos18
;939
2
1
iz
iz
11.12
)3
5sin
3
5(cos8
;344
2
1
iz
iz
11.13
))3
sin()3
(cos(8
;434
2
1
iz
iz
11.14
)4
7sin
4
7(cos25
;55
2
1
iz
iz
11.15
)4
sin4
(cos9
;9
2
1
iz
iz
11.16
)6
5sin
6
5(cos3
;3
2
1
iz
iz
11.17
)4
7sin
4
7(cos32
;33
2
1
iz
iz
11.18
)3
6sin
3
6(cos2
;22
2
1
iz
iz
11.19
)6
9sin
6
9(cos32
;33
2
1
iz
iz
11.20
))3
sin()3
(cos(28
;88
2
1
iz
iz
11.21
)6
5sin
6
5(cos3
;3
2
1
iz
iz
11.22
)sin(cos64
;64
2
1
iz
z
11.23
)3
5sin
3
5(cos64
;64
2
1
iz
iz
11.24
)8
6sin
8
6(cos6
;333
2
1
iz
iz
11.25
)6
7sin
6
7(cos26
;66
2
1
iz
iz
11.26
)3
2sin
3
2(cos5
;5
2
1
iz
iz
11.27
)6
8sin
6
8(cos32
;33
2
1
iz
iz
11.28
))4
sin()4
(cos(28
;88
2
1
iz
iz
11.29
)4
5sin
4
5(cos2
;3
2
1
iz
iz
11.30
)6
10sin
6
10(cos3
;13
2
1
iz
iz
18
Solution of an exemplary embodiment
Task1. The third order determinant is given 3,2,
251
820
143
ji .
It is required to calculate:
a) the minor 23M and the algebraic complement 23A of the element 23a ;
b) the determinant by the triangle rule (rule of Sarrus);
c) the determinant by the method of expansion along the 2-nd row and the 3-
rd column.
Solution:
a) the minor M ij of the element aij equals to the determinant, obtained from
given by deleting the i-th row and j-th column. Thus cross out second row and the
third column in our determinant, we obtain 11415351
4323 M . The
algebraic complement of the element is calculated by formula Aij=(–1) ji M
ij.
So, 23A =(–1) 32 11= –11;
b) the triangle rule: the third order determinant is the sum of six terms; terms
with a plus sign are obtained by multiplication of three elements of the determinant
taken by the scheme , terms with a negative sign – by the scheme .
So,
3 4 1
0 2 8 3 2 ( 2) 4 8 1 0 5 ( 1) 1 2 ( 1)
1 5 2
;9812023212)2(04583
c) expansion formula along the third column has the form:
333323231313
333231
232221
131211
AaAaAa
aaa
aaa
aaa
,
20
43)1()2(
51
43)1(8
51
20)1()1(
251
820
143333231
= 2 – 88 – 12= – 98;
similarly calculate the determinant expanding along the second row:
19
51
43)1(8
21
13)1(2
25
14)1(0
251
820
143322212
.9888100
Task 2. Given matrices А =
123
432
321
, В =
3333
1111, С =
4
3
2
1
.
Find:
а) the product of matrices АВ or ВС, or СВ (if possible). Explain why if it is
impossible;
b) the matrix 1A , the inverse matrix for A.
Solution:
а) the product of matrices AB is possible if the number of columns of matrix
A equals the number of rows of the matrix В. Matrix sizes: А 33 , В 42 , С 14 . So,
А 33 В 42 = 23 – multiplication is impossible. В 42 С 14 = 44 –
multiplication is possible. С 14 В 42 = 21 – multiplication is impossible. Matrix
multiplication BC is a matrix E, the number of rows is equal to the number of rows
of the matrix B, the number of columns equals the number of columns of the matrix
C: nmnkkm EСВ . Element ije of the matrix E is equal to the sum of products
of i-th row of the matrix B for the j-th column of the matrix C.
So, ВС=
3333
1111
4
3
2
1
=
30
10
43332313
41312111,
where 121442 ECB .
And ECB
30
10;
b) an inverse matrix exists for a square matrix A, if the determinant of the
matrix is not equal to zero; and an inverse matrix does not exist if A =0.
20
The inverse matrix А 1 for matrix А =
333231
232221
131211
aaa
aaa
aaa
is obtained by the
formula
332313
322212
312111
1 1
AAA
AAA
AAA
AA , where A – the determinant of the matrix А;
ijA – the algebraic complement of the elements ija .
Let us find the determinant of the matrix А:
A =
123
432
321
= –30 0, so, А 1 exists.
Let us find the algebraic complements of all elements of the matrix А:
;1013
42)1(;11
12
43)1( 21
12
11
11
АА
;423
21)1(;10
13
31)1(
;812
32)1(;13
23
32)1(
32
23
22
22
12
21
31
13
АА
АА
.732
21)1(
;1042
31)1(;1
43
32)1(
33
33
23
32
13
31
А
АА
Let us form А 1 by the above formula
А 1 = ;
7413
101010
1811
30
1
and then it is need to check:
.
100
010
00111
EAAAA
Task 3. Given points А(7, –9, 3) and В(1, 0, –5); and vectors
)2,1,6(),5,0,1( cb .
Find:
21
a) the module (length) of the vector ABa and the midpoint of the
segment AB;
b) the projection of the vector a on c ;
c) the area of the parallelogram, obtained from the vectors b and c ;
d) volume of the pyramid constructed from the vectors cba ,, .
Solution:
а) for points A ( 111 ,, zyx ) and B( 222 ,, zyx ) coordinates of the vector a AB
are obtained by the formula a AB = ( 121212 zzyyxx ,, ).
So, a (1–7, 0– (–9), –5–3) = (–6, 9, –8);
The module (length) of the vector а : 2
1
2
1
2
1 zyха .
So, AB =222 )8(9)6( = 181 ;
the midpoint of the segment AB has coordinates
С
222
212121 zzyyxx,, = C
2
35,
2
90,
2
17= С(4, –9/2, –1);
b) the projection of the vector a on c is с
саас
Pr ;
where 313131 zzyyxxсa – scalar product of vectors ),,(),,,( 333111 zyxczyxa .
If vectors a and c are orthogonal, then 0сa .
In our case 43281966 )()(сa , as 0сa , so the
vectors а and с are not orthogonal; calculate more
049,141
43
)2(16
43Pr
222
с
саас
;
c) the area of the parallelogram, obtained from the vectors
),,(),,,( 333222 zyxczyxb , is the module of the vector, received from their
vector product: cbd
; at first find the coordinates:
2 2 2 2 2 2 2 2 2 2 2 2
2 2 2
3 3 3 3 3 3 3 3 3 3 3 3
3 3 3
, ,
i j ky z x z x y y z x z x y
b c x y z i j ky z x z x y y z x z x y
x y z
;
0 5 1 5 1 0
1 0 5 5,32,11 2 6 2 6 1
6 1 2
i j k
b c i j k
;
22
further find module of the vector:
1050132)5()1,32,5( 222 cbd
;
d) the volume of pyramid constructed from vectors а , b , c , can be found
from a module of their mixed product
333
222
111
6
1
6
1
zyx
zyx
zyx
cbaV
.3
154326
6
1)1830082700(
6
1
216
501
896
6
1
As 0cba , these vectors are not coplanar.
Task 4. Given points А 1 (4, –2), А 2 (8, 1) on a plane and the equation of the
line L 1 : –x + 4y + 5 = 0.
Write the equation of a straight line:
a) )( 212 AAL – passing through these points;
b) 2L – as a general equation of straight line;
c) 2L – in the form of equation of straight line with slope;
d) 2L – in the form of equation of straight line in segments;
e) 3L , passing through the point 2A and perpendicular to 1L .
Solution:
a) the equation of the straight line passing through two points ),( 111 yxМ ,
),( 222 yxМ , is obtained by the formula 12
1
12
1
yy
yy
xx
xx
.
So, equation of the straight line L 2 is )2(1
)2(
48
4
yx or
3
2
4
4
yx;
b) let us write the equation of the straight line L in the general form
Ax + By + C = 0:
3
2
4
4
yx 0204384123 yxyx , (A = 3, B = –4, C = –20).
The geometric meaning of the coefficients: А, В – are coordinates of the
normal (perpendicular) vector of the straight line L , ie 243 L),()В,А(N ;
c) 2L in the form of equation of straight line with slope y = kx + m:
23
3
2
4
4
yx5
4
32)4(
4
3 xyyx , (k =
4
3, m = –5);
d) 2L in the form of equation of straight line in segments 1b
y
a
x:
1
53/202043
yxyx , ( a = 20/3, b = –5 );
e) the equation of the straight line passing through the point ( 11 , yx ) and
parallel to the vector ),( nmb , has the form: n
yy
m
xx 11
.
If vector 1N = (A;B) = (–1;4) is perpendicular to the straight line 1L and the
straight line 3L is also perpendicular to 1L , then we find in our case that the
directional vector for the straight line 3L is )4;1(),( 1 Nnmb and the
equation has the form 4
1
1
813
yx:LL .
Task 5. Given points А 1 (1, 2, –1), А 2 (3, 3, 2), А 3 (2, –3, 7). It is required to:
a) write the equation of the plane );( 3211 AAAP
b) write 1P as a general equation of a plane;
c) write 1P as equation of a plane in segments;
d) write 1P in a general equation of a plane, passing through the point 1A ;
e) make a canonical equation of the straight line )( 321 AAL ;
f) write a parametric equation of the straight line 1L ;
g) write the equation of the straight line )( 12 NAL perpendicular to the
plane 1P .
Solution:
а) the equation of the plane passing through the points ),,( 1111 zyxМ ,
),,( 2222 zyxМ , ),,( 3333 zyxМ has the form: 0
131313
121212
111
zzyyxx
zzyyxx
zzyyxx
.
Here Р1: 0
851
312
121
0
172312
122313
121
zyxzyx
(*);
b) expanding the determinant in the right-hand side of (*), we obtain the
equation of the plane P1 in the general form Ax + By + Cz + D = 0:
23x – 13y – 11z – 8 = 0, (A = 23, B = –13, C = –11, D = –8).
24
The geometrical sense of coefficients: А, В, С – are coordinates of a normal
(perpendicular) vector of the plane, ie vector 1111323 P),,(N ;
c) let us move free term –8 in the general equation of plane on the right side
and divide both sides by 8. Obtain the equation of the plane in segments
1c
z
b
y
a
x:
18/ 23 8/13 8/11
x y z
,
where а = 8/23, b= –8/13, с = –8/11 – are sizes of the segments cut off by the plane
on coordinate axes, considering from the origin of coordinates;
d) let us expand the determinant on the left side of (*) along the first row
.0)1(11)2(13)1(23
051
12)1(
81
32)2(
85
31)1(
zyx
zyx
We obtain the equation of plane in the form A(x-x 0 ) + B(y-y 0 ) + C(z-z 0 ) = 0
)1,2,1( 000 zyx ;
e) the canonical equation of the straight line passing through two points
),,( 1111 zyxМ and ),,( 2222 zyxМ , has the form 12
1
12
1
12
1
zz
zz
yy
yy
xx
xx
. So, the
equation of the straight line L 1 has the form 27
2
33
3
32
3
zyx or
5
2
6
3
1
3
zyx;
f) the equation of the straight line in the form p
zz
n
yy
m
xx 000
is
called canonical, where a = (m, n, p) – is the direction vector. In the previous
section we have already found the canonical equations of straight lines. The
equations of the straight line in the form
0
0
0
zptz
ynty
xmtx
are called parametric. To get
the parametric equations of the straight line, we equate the canonical equations to
the parameter t, and from these equalities we find x, y, z.
L 1 : 5
2
6
3
1
3
zyx=
255
2
366
3
31
3
tztz
tyty
txtx
t.
25
So, the parametric equations of L 1 :
25
326
3
tz
ty
tx
;
g) as 11 )( PNA , so vector 1)11,13,23( PN is the direction vector of
the straight line )( 12 NAL . Then the canonical equation of the straight line
)( 12 NAL has the form:11
1
13
2
23
1
zyx.
Task 6. Solve the system :343
12
yx
yx
a) by Cramer’s method;
b) by matrix method (using the inverse matrix).
Solution:
а) the solution of the system
22221
11211
byaxa
byaxa by Cramer's rule has the form
21 , yx , where
2221
1211
aa
aa – determinant of the system,
222
121
1ab
ab ,
221
111
2ba
ba – auxiliary determinants obtained from the determinant of the system
by replacing the first and second columns by column of free terms.
In this case 1043
21
, 2
43
211
, 633
112
.
So, х = –1/5, у =3/5.
The answer can be written as a vector:
53
51
X ;
b) in matrix form solution of the system
22221
11211
byaxa
byaxa is written as:
BAX 1, where
y
xX – column matrix of unknowns,
2
1
b
bB – column
matrix of free terms, 1A – inverse matrix for the matrix of the system
2221
1211
aa
aaA .
26
Let us find the inverse matrix (look example 2b) for the matrix of the system
43
21A . As the determinant of the system 010 А , then the inverse
matrix exists and is equal to
13
24
10
11A .
Then
5/3
5/1
3
1
13
24
10
11 BAXy
x. Or the answer in the
usual form: х = –1/5, у = 3/5.
Task 7. Solve the system of equations. :
733
2432
35
321
321
321
xxx
xxx
xxx
a) by Cramer’s method;
b) by matrix method (using the inverse matrix).
Solution:
а) the coefficient matrix of the system is
133
432
511
A , column of free
terms –
7
2
3
В .
Then the solution of the system by Cramer's rule has the form:
3
32
21
1 ,, xxx ,
where 162124530123
133
432
511
is the determinant of the
system,
16
733
232
311
,32
173
422
531
,64
137
432
513
321
–
auxiliary determinants obtained from the determinant of the system by replacing the
first, second and third column by column of free terms. Finally, we obtain the
solution by Cramer’s method:
27
116
16,2
16
32,4
16
64 33
22
11
xxx or
1
2
4
3
2
1
х
х
х
X ;
b) in matrix form the solution of the system is written:
7
2
3
133
432
5111
3
2
1
х
х
х
X .
Let us find the inverse matrix (see example 2b) , as 016 А , the
inverse matrix exists and has the form
103
61614
111615
16
1
133
432
5111
1A .
So,
1
2
4
16
32
64
16
1
79
423242
773245
16
1
7
2
3
103
61614
111615
16
11
3
2
1
BA
х
х
х
.
Task 8. Given point А(3, –7); radius of the circle R = 6; a = 2; b = 3 – semi-
axes of curves; D: у = –3 – equation of the straight line. It is required to:
а) write the equation of a circle with center A and radius R;
b) write the canonical equation of an ellipse with semiaxes a = 2 and b = 3.
Find the coordinates of it’s foci and the eccentricity;
c) write the canonical equation of the hyperbola with the real semiaxis a = 2
and imaginary semiaxis b = 3. Find the coordinates of it’s foci, eccentricity,
equations of the asymptotes;
d) write the canonical equation of the parabola with vertex at the origin and if
D: y = –3 is its directrix. Find the coordinates of its focus, eccentricity;
e) make the drawing of ellipse, hyperbola, parabola.
Solution:
а) the equation of a circle centered at 0 0( , )x y with the radius R has the form 22
0
2
0 )()( Ryyxx , So, in our version: 36)7()3( 22 yx ;
b) the canonical equation of an ellipse with semiaxes a and b has the form
12
2
2
2
b
y
a
x; So, in our version: 1
94
22
yx
.
28
If ba , then 22 bac and points )0,(),0,( 21 cFcF are foci of the
ellipse, and the eccentricity of the ellipse is а
c .
If ab , then 22 abc and ),0(),,0( 21 cFcF ,
b
c .
Since ab in our version then 54922 abc and the
eccentricity equals 3
5
b
c , ( 1 ). Foci lie on the y-axis:
)5,0(),5,0( 21 FF .
Let us draw the ellipse on the coordinate plane (see Figure 1), where
2121 ,,, BBAA – vertices of the ellipse.
Figure 1
c) the canonical equation of hyperbola with real semiaxis a , imaginary
semiaxis b has the form 12
2
2
2
b
y
a
x; with the real semiaxis b and the imaginary
semiaxis a: 12
2
2
2
b
y
a
x or 1
2
2
2
2
b
y
a
x. For hyperbola with real semiaxis a,
eccentricity equals a
c , where
22 bac ; asymptotes of the hyperbola
equation have the form xa
by ; foci are points )0,(),0,( 21 cFcF , located on the
real axis.
By hypothesis a =2, b =3, so the canonical equation of the hyperbola with the
real semiaxis a is 194
22
yx
. For it half-foci distance 1394 c ;
29
eccentricity equals )1(,2
13
a
c; foci: )0,13(),0,13( 21 FF ; equations
of the asymptotes: xy2
3 .
It is easy to construct a hyperbola as follows: construct a rectangle with sides
byax , (in our case 3,2 yx ). The diagonals of a rectangle are the
asymptotes of the hyperbola, the point of intersection of the rectangle with the real
axis of the hyperbola – vertices of the hyperbola (see Figure 2):
Figure 2
d) by hypothesis, if the directrix of the parabola y = –p/2, then the axis of
symmetry of the parabola with vertex at the origin is the y-axis. Hence, its canonical
equation is pyx 22 ; if the directrix of the parabola x = –p/2, then the axis of
symmetry of the parabola with vertex at the origin is the x-axis. Hence, its canonical
equation has the form pxy 22 . As the directrix of the parabola has the equation у
= –3, then 632
pp
and the equation of the parabola is:
yxyx 1262 22 . The focus of the parabola is the point )2,0(p
F , lying on
the axis of symmetry. In our case, the focus is )3,0(F . Let us construct the parabola
(see Figure 3).
30
Figure 3
Task 9. Determine the type (name) of the second order surface and make a
schematic drawing:
a) 14212
222
zyx
; b) 0423
222
zyx
.
Solution:
а) 14212
222
zyx
is the canonical equation of one-sheeted hyperboloid
with the symmetry axis Ox. Here its schematic drawing (see Figure 4);
Figure 4
b) 0423
222
zyx
is the canonical equation of the cone of the second
order with the symmetry axis Oz and vertex at the origin. Here its schematic
drawing (see Figure 5);
31
Figure 5
Task 10. Lead to the canonical form the equation of the second order
0196418169 22 yxyx and construct this curve.
Solution: the equation of the form
1)()(
2
2
0
2
2
0
b
yy
а
xx
defines, respectively, the ellipse or the hyperbola.
The equation of the form: 2
0
2
00
2
0 )(2)(),(2)( xxpyyyypxx
defines the parabola. These curves have symmetry center (for the ellipse and
hyperbola) or vertex of the parabola at the point ),( 00 yxC . To determine the type
of the curve, we apply the method of allocating of perfect squares:
.019)4(16)2(9019)6416()189( 2222 yyxxyyxx
Supplement the terms containing x, and the terms containing y, to complete
the squares:
019)444(16)112(9 22 yyxx ,
0)4(161919)44(16)12(9 22 yyxx ,
36)2(16)1(9 22 yx ,
1)16/(36
)2(
9/36
)1( 22
yx, 1
4/9
)2(
4
)1( 22
yx
,
so we have the hyperbola with center С(–1; –2), real semi-axis b = 3/2 (along the
axis Oy) and the imaginary semi-axis a = 2. Let us construct its schematic drawing
(see Figure 6)
32
Figure 6
Task 11. Given complex numbers:
).4
sin4
(cos9;99 21
iziz
It is required to find:
а) the module of the complex numbers z1;
b) the argument of the complex number z1;
c) the representation of the complex number z1 in the trigonometric and
exponential forms;
d) the sum of complex numbers z1 and z2 analytically and graphically;
e) (z2)5;
f) multiplication of z1z2 in trigonometric form;
g) all complex roots 32z using De Moivre formula.
Solution:
а) if the complex number in algebraic form is:
,9;9,)9,9(99),( 1 whereiziz
then the module of the complex number z1:
;29)9(9 22
1
22 zorz
b) the argument of the complex number z1:
);09;09(4
19
9)arg()arg( 11
arctgarctgzarctgz
c) in the trigonometric and exponential form complex number has the form:
;)sin(cos iezziz
;29))
4sin()
4(cos(2999 4
1
i
eiiz
33
d) addition of complex numbers z1 и z2:
;63,236,15)92
29()
2
299()
2
2
2
2(999
),2
2
2
2(9)
4sin
4(cos9
21
2
iiiizz
iiz
e) for calculating (z2)5 we use De Moivre formula:
);2
2
2
2(29))
4
5sin()
4
5(cos(29
;)sin(cos
555
2 iiz
eznnzz ninnn
f) the multiplication of z1z2 in algebraic form is:
;2812
281)1(
2
281
2
281
2
281
2
281
2
281)
2
2
2
2(9)99(
2
2
21
iiiiizz
but multiplication z1z2 in trigonometric form is easily calculated:
;281)0sin0(cos281))
44sin()
44(cos(929
)),sin()(cos(
21
21212121
iizz
izzzz
g) all complex roots of the n-th degree are calculated by De Moivre formula:
.2,1,0,3
24sin
3
24cos29
,1,,3,2,1,0,2
sin2
cos
332
k
k
i
k
z
nkn
ki
n
kzz nn
Write all the roots obtained for all k:
));12
sin12
(cos29))3
4sin3
4(cos29:0 33
1
iizk
));12
9sin
12
9(cos29))
3
24sin
3
24(cos29:1 33
1
iizk
)).
12
17sin
12
17(cos29))
3
44sin
3
44(cos29:2 33
1
iizk
On the graph all the cube roots are located at the vertices of an equilateral
triangle.
34
Bibliography
1. Khasseinov K.A. Canons of Mathematics. – Moscow: Nauka, 2007. –
592 p. – Almaty, Akbar, 2011 – 592 p., 2nd
edition.
2. Индивидуальные задания по высшей математике: Ч.1 Линейная и
векторная алгебра. Аналитическая геометрия. Дифференциальное исчисление
функции одной переменной/под ред. А.П. Рябушко. – Мн.: Выш. шк., 2007. –
304 с.
3. Индивидуальные задания по высшей математике: Ч.2 Комплексные
числа. Неопределенный и определенный интегралы. Функции нескольких
переменных Обыкновенные дифференциальные уравнения /под ред. А.П.
Рябушко – Мн.: Выш. шк., 2007. – 304 с.
Contents
Introduction 3
Calculation-graphical work. Elements of linear algebra,
analytic geometry and complex numbers 3
Theoretical questions 3
Calculated tasks 3
Solution of an exemplary embodiment 16
Bibliography 32
35
Summary plan 2014, pos.226
Kim Regina Evgenievna
MATHEMATICS 1
Methodological Guidelines and Tasks
for carrying out the calculation-graphical work for students of specialties
5В071700 «Heat power engineering»,
5В071800 «Electrical power engineering»,
5В071900 «Radio engineering, electronics and telecommunications»
Part 1
Editor V.S. Kozlov
Specialist on standardization N.K. Moldabekova
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