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Файл:Mathematics (Математика). Учебное пособие
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181
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;
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.
182
Topic 3. Determinant of a Matrix.
81021
5212
1038
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5021
5212
19232
æö
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èø
7101
522
10238
æö
ç÷
ç÷
ç÷
-
èø
101021
51212
1032
æö
ç÷
ç÷
ç÷
-
èø
7101
522
028
-
æö
ç÷
--
ç÷
ç÷
èø
7112
582
028
-
æö
ç÷
--
ç÷
ç÷
èø
321
Example. To calculate the determinant
Solution. To calculate the determinant of the third order, we will use
the well-known formula of Sarrus (usually triangles), which can be written by the following formula:
aaa
131211
aaa
232221
aaa
333231
321
654
987
Answer: 0.
The exercises: To calculate the determinant of the matrix
-
-
-
-
-
-
--
1)
2)
3)
21100
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12125
÷
; 8)
÷
282310
ø
21110
ö
÷
121215
÷
; 9)
÷
282310
ø
2110130
ö
÷
12125
÷
; 10)
÷
282310
ø
654
.
987
---++=
06*8*12*4*93*5*74*8*32*6*79*5*1
=---++=
;
;
;
aaaaaaaaaaaaaaaaaa
233211122133132231213213122331332211
4)
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ç
5)
ç
è
; 11)
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ö
÷
12125
÷
; 12)
÷
282310
ø
;
-
;
-
183
6)
711
522
020
-
æö
ç÷
ç÷
ç÷
èø
2101
522
023
--
æö
ç÷
ç÷
ç÷
-
èø
7101
522
028
-
æö
ç÷
--
ç÷
ç÷
èø
7)
211067
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æ
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è
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÷
12125
÷
; 14)
÷
282310
ø
211090
ö
÷
12125
÷
; 15)
÷
282310
ø
;
-
;
16)
;
-
Topic 4. Matrix Inverses
and Systems of Linear Equations
Example. Solving the system via matrix method.
=+-
123
ì
ï
í
ï
î
Solution:
Solve the system matrix method to calculate this inverse matrix
where Aij – This is the algebraic additions to the elements of the matrix.
-
æ
ç
=
A
ç
ç
è
xxx
321
=++
32325
xxx
321
=++
xxx
321
113
ö
÷
215
÷
÷
211
ø
;
AAA
æ
ç
1
-
1
=
A
ç
A
D
ç
è
ö
131211
÷
AAA
÷
232221
÷
AAA
333231
ø
,
184
113
-
=DA
215
0126101526
¹=-+-+-=
11
12
13
21
22
23
31
32
211
21
11
+
1
)(A
=-=
0
;
;
21
25
21
+
1
;
8210
-=--=-=
)()(A
21
15
31
+
1
)(A
;
415
=-=-=
11
+
1
-=
;
312
=---=
)()(A
11
-
12
21
+
1
)(A
;
516
=-=-=
13
22
21
+
1
-=
)()(A
;
413
-=+-=
13
-
32
11
+
1
-=
)(A
;
312
-=--=
11
-
13
21
+
1
)()(A
;
156
-=--=-=
13
23
25
+
1
-=
12
=
1
æ
ç
ç
ç
è
12
)(A
15
-
330
æ
ç
1
ç
ç
è
-
-
330
ö
÷
--
158
÷
-
÷
844
ø
ö
÷
--
158
÷
;
÷
844
ø
12
ö
æ
÷
ç
×
ç
ç
è
1
=
3
÷
12
÷
3
ø
33
-
1
A
=
X
;
853
=+=
×-××
3333120
)(
æ
ç
ç
ç
è
)(*
ö
÷
×-××-
3135128
)(
÷
÷
××-
3834124
ø
12
0
æ
ç
1
-=
84
ç
ç
60
è
0
ö
æ
ö
÷
÷
÷
ø
÷
ç
-=
7
÷
ç
÷
ç
5
ø
è
.
13
-
33
185
The exercises. To solve the system using the inverse matrix.
=+-
42
1)
2)
3)
4)
5)
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
zyx
=-+
73
zyx
zyx
zyx
zyx
=+
zy
; 6)
=+-
1243
=-+
3432
zyx
-=+-
5243
zyx
zyx
zyx
zyx
zyx
=-+
=+-
=++
zyx
zyx
zyx
; 7)
=-+
13572
=+-
9572
-=-+
255
; 8)
=+-
24724
032
2942
; 9)
1743
=+-
16532
; 10)
=-+
743
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
=-+
6422
zyx
=-+
653
zyx
zyx
zyx
zyx
;
=+-
6623
zyx
=++
22543
zyx
-=--
963
;
=-+
10442
zyx
=+-
322
xxx
321
-=++
xxx
=--
zyx
zyx
=+-
zyx
zyx
42
321
xxx
321
;
-=++
344
142
=-+
11243
;
=+-
11423
123
=++
642
.
=++
325
Topic 5. Systems of Equations, Algebraic
Procedures (Gauss-Jordan procedure)
Example. Solve the system by Gauss, for this will make the augmented matrix of the system and simplify its reduction to triangular
form.
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æ
æ
-
ç
ç
ç
è
Thus, this system is equivalent to the system
ö
12
113
ç
÷
3
215
÷
÷
3
211
ø
-
ç
~
ç
è
ö
3
211
ç
÷
12
113
ç
÷
~
ç
÷
3
215
ø
è
ö
3
211
æ
÷
ç
--
--
ç
3
540
3
840
10
÷
~
ç
÷
ç
ø
è
ö
3
211
÷
3
5
÷
.
4
4
÷
÷
--- 15
300
ø
186
=++
4
454
1
=--
=
1
2
=
32
ì
ï
ï
í
ï
ï
î
2
x
Answer:
3
xxx
321
xx
x
28
,
3
, calculate
-=+
32
3
05712
4
-=-
153
7
-=-=×--=x
.
x
7
-=x
,
3
3
5
.
5
4
5
0
=x
Topic 6. Systems of Equations
The exercises. Solve the system by Gauss.
=+-
42
zyx
=-+
73
zyx
zyx
zyx
zyx
=+
zy
; 6)
=+-
1243
=-+
3432
zyx
-=+-
5243
zyx
zyx
zyx
zyx
zyx
=-+
=+-
=++
zyx
zyx
zyx
; 7)
=-+
13572
=+-
9572
-=-+
255
; 8)
=+-
24724
032
2942
; 9)
1743
=+-
16532
; 10)
=-+
743
1)
2)
3)
4)
5)
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
5
=x
,
=-+
6422
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
zyx
=-+
653
zyx
zyx
zyx
zyx
;
=+-
6623
zyx
=++
22543
zyx
-=--
963
;
=-+
10442
zyx
=+-
322
xxx
321
-=++
xxx
=--
zyx
zyx
=+-
zyx
zyx
42
321
xxx
321
;
-=++
344
142
=-+
11243
;
=+-
11423
123
=++
642
.
=++
325
187
Topic 7. The Cramer’s rule
51231(1)(1)1(1)120
-
D==×-×+-×-+×-=¹
1232
1232
D==×-×+-×-+
12(1)1(1)01281284
+×-+×-=-×+=-
Solve the system by the method of Kramer. The main determinant of
the system is:
113 -
=D
Write out the determinant by the first row of elements using the formula:
aaa
131211
aaa
232221
aaa
333231
311
( )
112
121211
125251
11
+
Record and calculate auxiliary determinants
1211
-
11
312121(1)(1)
x
1
( )
+
312
215
211
.
AaAaAa
×+×+×=
131312121111
.
1213
++
12
+
31
13
1(1)0
+×-=
+
31
3121
32
11
53231
x
D==×-×+
2
+
()
32
132
5253
1213
++
1213
188
3112
(1)(1)12(1)311212460
-
+-×-+×-=×++×=
12
12
12
(
)
32
11
51331
x
D==×-×+
3
+
( )
32
113
5351
1213
++
1311
D
x
0
1
=
x
So
1
D
D
x
=
x
2
D
D
x
=
x
3
D
84
2
60
3
Answer:
The exercises. Solve the system by Cramer’s rule.
ì
ï
1)
í
ï
î
ì
ï
2)
í
ï
î
ì
ï
3)
í
ï
î
ì
ï
4)
í
ï
î
zy
ì
ï
5)
í
ï
î
0
==
;
7
-=-=
;
5
==
;
57;-;0
.
=+-
42
zyx
; 6)
=-+
73
zyx
=+-
1243
zyx
=-+
3432
zyx
; 7)
-=+-
5243
zyx
=-+
13572
zyx
=+-
9572
zyx
; 8)
-=-+
255
zyx
=+-
24724
zyx
=-+
032
zyx
; 9)
=+-
2942
zyx
=+
=++
1743
zyx
; 10)
=+-
16532
zyx
=-+
743
zyx
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
ì
ï
í
ï
î
=-+
6422
zyx
;
=-+
653
zyx
=+-
6623
zyx
=++
22543
zyx
;
-=--
963
zyx
=-+
10442
zyx
=+-
322
xxx
321
;
-=++
42
xxx
321
-=++
344
xxx
321
=--
142
zyx
;
=-+
11243
zyx
=+-
11423
zyx
=+-
123
zyx
.
=++
642
zyx
=++
325
zyx
189
Topic 8. Аnalytical geometry
{}{
}
7;1;2,3;2;1
=-=-
r
r
(
)
xxyyzz
abababab
=×+×+×
r
r
{}{
}
;;,;;
xyzxyz
aaaabbbb
r
r
Þ
(
)
,731221210
=-×+×+×-=-¹Þ
r
r
{}{}{
}
1;2;1,0;2;1,2;0;3
r
rr
xyz
xyz
xyz
aaa
abcbbb
ccc
{}{}{
}
===Þ
r
rr
02164040020
==-++----=¹Þ
{}{
}
1;3;1,3;2;3.
r
r
{}{
}
2;1;4,4;1;3.
r
r
Example 1. To prove that the vectors are perpendicular
ab
Solution: Two vectors are perpendicular if their dot product is 0, the
scalar product of the vectors defined by the projections on the coordinate
?
axes is calculated by the formula:
where
ab
dicular.
abc=--==
is 0, the mixed product of vectors is calculated by the formula:
where
mixed product of vectors:
r
rr
abc
coplanar.
three-element vectors.
==
( )
Example 2. To prove that the vectors are coplanar
Solution: Three vectors are coplanar if the mixed product of vectors
r
rr
=
,
aaaabbbbcccc
--
;;,;;,;;
xyzxyzxyz
121
( )
203
The exercises. Сalculate the scalar and vector product of two given
,
calculate the scalar product:
the vectors are not perpen-
?
calculate the
vectors are not
ab=-=-
1)
ab==
2)
190
{}{
}
0;1;2,1;3;2.
ab
==-
r
r
{}{
}
1;2;1,3;1;2.
r
r
{}{
}
2;1;7,2;4;3.
==-
r
r
{}{
}
4;1;5,1;3;1.
r
r
{}{
}
3;1;2,2;3;1.
=-=-
r
r
{}{
}
4;1;5,1;3;1.
r
r
{}{
}
9;1;2,1;1;4.
r
r
{}{
}
8;2;3,2;8;0.
r
r
3)
ab==
4)
ab
5)
ab=-=
6)
ab
7)
ab=--=-
8)
ab==-
9)
ab==-
10)
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