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Файл:Mathematics (Математика). Учебное пособие
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91
Definition. The rank of a matrix is the order of the largest nonzero
-
-
minor. Denote: r(A), R(A), Rang A.
Example 6.
000
To find the rank of the matrix
æ
ç
=
A
ç
è
ö
÷
. Because all minors are
÷
000
ø
zero, then r(A)=0.
Example 7.
001
æ
ç
=
To find the rank of the matrix
B
ç
ç
è
The matrix contains a single nonzero element –
ö
÷
000
.
÷
÷
000
ø
which is a
,1
=b
11
minor 1-st order. All determinants of higher order, composed of the elements of this matrix will contain the 0-th line, and therefore equal to 0.
Therefore, r(B)=1.
Example 8.
001
ö
÷
-=
-
.
542
÷
÷
543
ø
To find the rank of the matrix
æ
ç
С
ç
ç
è
The only minor 3rd is the determinant of the matrix With, but it is
equal to 0, because it contains proportional columns. therefore, r(C)<3.
In order to prove that r(C)=2, it is enough to specify at least one mi-
nor 2nd order, not 0, for example,
01
42
-
so, r(C)=2.
.04
¹-=
Example 9.
001
ö
÷
010
.
÷
÷
100
ø
To find the rank of the matrix
,01. ¹=D
E
so, r(E)=3.
æ
ç
=
E
ç
ç
è
Example 10.
21211
ö
÷
11321
÷
.
÷
-
10510
÷
÷
---
32132
ø
To find the rank of the matrix
æ
ç
ç
=
A
ç
ç
ç
è
--
92
Theoretically, the rank of this matrix can take values from 1 to 4, be-
~
~
~
~
cause the elements of the matrix you can create the minors for 4-th order
inclusive. But instead of calculating all possible minors of the 4th, 3rd,
etc. order, applicable to the matrix And the equivalent transformation.
First, make sure the first column all elements except the first, would be
equal to 0. To do this, we write instead of the second line the amount
from the first, and instead of the third difference between the third and
first double:
--
21211
æ
ç
~
ç
=
A
ç
ç
ç
è
--
ö
÷
-
10510
÷
÷
-
10510
÷
÷
10510
ø
.
Then from the third row subtract the second and the fourth will add a
second:
--
21211
æ
ç
~
~
ç
=
A
ç
ç
ç
è
ö
÷
-
10510
÷
÷
00000
÷
÷
00000
ø
.
After crossing out the zero rows will get matrix dimensions 2 × 5 for
which the maximum order of minors, and, consequently, the maximum
value of the rank is 2:
--
21211
æ
~
ç
=
A
ç
è
ö
÷
÷
-
10510
ø
.
The minor of this matrix is
11
-
10
consequently,
,01
¹=
~
( == ArAr
.2)()
Solve and check yourself using the answers:
--
34532
ö
÷
51530
. Answer: 2
÷
÷
60000
ø
1. To find the rank of the matrix А=
æ
ç
ç
ç
è
-
93
-
-
18213
ö
÷
12530
÷
. Answer: 2
÷
70000
÷
÷
20000
ø
205723
ö
÷
. Answer: 3
244501
÷
÷
020000
ø
2. To find the rank of the matrix А=
æ
3. To find the rank of the matrix А=
ç
ç
ç
è
æ
ç
ç
ç
ç
ç
è
-
--
Topic 4. Matrix Inverses
and Systems of Linear Equations
Introduction. This workshop is devoted to the discussion of the following concepts: nonsingular matrix, a formula for the inverse, a minor.
For a full understanding of the topic you need to have an idea about matrices and determinants of the matrix.
2 -2 1
æ
ç
Example 1. For matrix А =
2 1 -2
ç
ç
1 2 2
è
Solution. Find first the determinant of the matrix
2 -2 1
А D = det А =
2 1 -2
1 2 2
= 27 ¹ 0, so, the inverse matrix exists and
ö
÷
find inverse matrix.
÷
÷
ø
we can find her by the formula: А-1 = 1/D
(i,j=1,2,3) – co-factors of the elements а
-2
1
+1
A
A
=- = + =( ) ,1
11
=- =- + =-( ) ( ) ,1
12
1
1
+2
2 2
-2
2
1 2
2 4 6
4 2 6
of the original matrix. So then:
i j
A A A
æ
11 21 31
ç
A A A
12 22 32
ç
ç
A A A
è
13 23 33
ö
÷
, where А
÷
÷
ø
i j
94
=- = - =( ) ,1
=-
=- = -=( ) ,1
=- =- + =-( ) ( ) ,1
=-
=- =--- =( ) ( ) ,1
=- = + =( ) ,1
1
2
2
2
3
3
3
+1
+2
+3
+1
+2
+3
1 2
1
-
2
2 2
1
2
1 2
-2
2
1 2
1
-
2
1 -2
1
2
2 -2
-2
2
2 1
27
4 1 3
=--- =( ) ( ) ,1
4 1 3
= - =( ) ,1
4 1 3
2 4 6
6
æ
ç
1
-6 3 6
ç
ç
3 -6 6
è
A
13
A
21
A
22
A
23
A
31
A
32
A
33
as a result А-1 =
Example 2. For a given matrix А =
4 2 6
4 2 6
4 2 6
ö
÷
÷
÷
ø
=
æ
ç
1
ç
9
ç
è
2 6 3
2 1
-2 1 2
1 -2 2
1 2
æ
ç
3 4
è
1
2
+3
det A = 4 – 6 = –2.
M11=4; M12= 3; M21= 2; M22=1
x11= –2; x12= 1; x21= 3/2; x22= –1/2
So, А–1=
æ
ç
è
32 12/ /
ö
.
÷
-
ø
-
2 1
Example 3. To solve the system of equations:
=--
zyx
zyx
zyx
0
ö
æ
÷
ç
14
÷
ç
, A =
÷
ç
16
ø
è
Х =
ì
ï
í
ï
î
x
ö
æ
÷
ç
y
÷
ç
, B =
÷
ç
z
ø
è
ö
÷
.
÷
÷
ø
ö
÷
, find inverse А–1.
ø
05
=++
1432
=++
16234
æ
ç
ç
ç
è
--
115
ö
÷
321
÷
÷
234
ø
95
Find the inverse matrix А–1.
30
30
30
--
115
D = det A =
321
234
32
= –5; M21 =
23
31
24
21
34
5
;
30
10
;
30
5
;
1
æ
ç
6
ç
1
ç
ç
3
ç
1
ç
6
è
;10
-=
M23 =
;5
-=
1
12
1
22
1
32
1
30
7
--
15
19
30
1
30
19
-
M22 =
;
14
30
;
30
15
M11 =
M12 =
M13 =
1
11
1
21
1
31
A–1 =
Let's we check:
5
æ
ç
30
--
115
ö
ç
÷
10
ç
321
÷
ç
30
÷
234
ç
5
ø
ç
30
è
A×A
æ
ç
–1
ç
=
ç
è
Find the matrix X.
=
5(4–9) + 1(2 – 12) – 1(3 – 8) = –25 – 10 +5 = –30.
1
8
11
30
11 --
= 1; M31 =
23
15=-
M32 =
;14
24
15
-
34
1
---
===
aaa
13
30
1
---
;
aaa
23
1
---
-===
aaa
33
ö
÷
=
1
;
16
=-=-=
30
11
M33 =
;19
;
;
÷
÷
;
÷
÷
÷
ø
1
1
30
14
--
30
19
30
ö
÷
30
16
30
11
-
30
æ
÷
ç
1
÷
=
ç
÷
30
ç
÷
è
÷
ø
31
-
32
15=-
15
21
11 --
= –1;
=
;16
;11
+--+-+
111651914551025
ö
÷
-++-+-
333215728115205
÷
÷
-++-+-
2248438424103020
ø
=E.
96
1
1
Х =
x
ö
æ
÷
ç
y
÷
ç
÷
ç
z
ø
è
= А–1В =
1
æ
ç
ç
ç
ç
ç
ç
è
30
6
7
1
--
15
3
19
1
30
6
ö
÷
30
÷
0
æ
8
ç
÷
14
ç
÷
15
×
ç
÷
11
16
è
-
÷
30
ø
1
æ
0
ç
6
ç
ö
1
÷
ç
=
0
ç
3
ç
1
0
ç
6
è
÷
÷
ø
14
30
98
15
266
30
16
ö
++
÷
30
128
+--
15
176
-+
30
1
ö
æ
÷
÷
ç
÷
=
2
÷
ç
÷
.
÷
ç
3
÷
ø
è
÷
ø
Total solution system: x =1; y = 2; z = 3.
Solve and check yourself using the answers:
1- 1 2
ö
÷
4 2 5
.
÷
÷
2 3 7
ø
1. To find the inverse matrix for the matrix: А=
Answer: А-1 =
-
æ
ç
ç
ç
è
6 5- 8
ö
÷
13- 11 18
.
÷
÷
1- 1 1
ø
æ
ç
ç
ç
è
2. To solve the system matrix method
=+-+
4
ì
ï
ï
í
ï
ï
î
xxxx
4321
=-+-
1232
xxxx
4321
=+-
62
xxx
431
=-+-
03
xxxx
4321
Answer: {1, 2, 3, 4}.
3. To solve the system matrix method
x1 – x2 + x3 = 6,
2x1 + x2 + x3 = 3,
x1 + x2 +2x3 = 5.
Answer: {1, –2, 3}
Topic 5. Systems of Equations, Algebraic
Procedures. (Gauss-Jordan procedure)
Introduction. This workshop is devoted to the discussion of the following concepts: an augmented matrix, augmented matrix as a resulting
from a row operation, Gauss elimination, Gauss-Jordan procedure. For a
97
full understanding of the topic you need to have an idea about methods
2314
51040
--=-
12331233
~05711~05711
01522310012
æöæö
ç÷ç÷
ç÷ç÷
ç÷ç÷
èøèø
----
æöæö
ç÷ç÷
ç÷ç÷
ç÷ç÷
---
èøèø
of solution to the system of equations.
Example 1. Solve the system by Gauss.
=--
05
ì
ï
í
ï
î
zyx
=++
1432
zyx
=++
16234
zyx
Prepare the augmented matrix of the system.
--
0115
æ
æ
ç
ç
ç
è
ö
ç
÷
~
14321
ç
÷
ç
÷
16234
è
ø
14321
æ
ö
ç
÷
~
16234
ç
÷
ç
÷
--
0115
è
ø
14321
æ
ö
ç
÷
---
~
401050
ç
÷
ç
÷
---
7016110
è
ø
14321
---
401050
18600
So, the original system can be represented in the form:
xyz
++=
ì
ï
yz
í
ï
618
z
î
=
, where get: z = 3; y = 2; x = 1.
Example 2. To solve the system of linear equations by Gauss.
=-+
52
ì
ï
í
ï
î
xxx
321
-=+-
332
xxx
321
=-+
107
xxx
321
Prepare the augmented matrix of the system.
А* =
21151233
1233~2115
7111071110
---
---
--
ö
÷
÷
÷
ø
--
So, the original system can be represented in the form:
-=+-
332
ì
ï
í
ï
x
3
î
xxx
321
=-
1175
xx
32
-=-
2
, where get: x3 = 2; x2 = 5; x1 = 1.
Solve and check yourself using the answer. To solve the system composed of equations x + y – 3z=2, 3x – 2y + z = – 1, 2x + y – 2z = 0 by
Gauss:
Answer: {– 0,7; –1,2; –1,3}
98
Topic 6. Systems of Equations
Introduction. This workshop is devoted to the discussion of the following concepts: forms of systems of equations, using the inverse of a
matrix to find the solution of systems of equations.
ì
ï
ï
í
...
...
ï
ï
î
...
2211
Consider the linear system (1): let us introduce the following notation:
aaa
æ
ç
ç
=
A
ç
ç
ç
è
x
ö
æ
1
÷
ç
x
÷
ç
2
=
X
B
÷
ç
...
÷
ç
÷
ç
x
n
ø
è
b
ö
æ
1
÷
ç
b
÷
ç
2
=
– column free members.
÷
ç
...
÷
ç
÷
ç
b
m
ø
è
...
...
...
21
– column unknown,
ö
n
11211
÷
aaa
÷
n
22221
÷
– the main matrix system,
............
÷
÷
aaa
mnmm
ø
Theorem (the theorem of Kronecker-Kapelly). The system (1) has
a solution if and only if the rank of the system matrix is equal to the rank
of the extended matrix. If (1) has a solution (one or more) then it is a
consistent system
Example 1.
To solve the system of equations.
=+
832
ì
í
î
yx
-=-
95
yx
The only solution is a pair of numbers х = 1, у = 2.
=+++
bxaxaxa
11212111
nn
=+++
bxaxaxa
22222121
nn
,
(1)
............................................
=+++
bxaxaxa
mnmnmm
.
99
Example 2.
x1
+ 2x
+ 3x
–
2x4
= 6,
8 4
To solve the system of equations.
í
î
=+
622
yx
.
=+
3
yx
ì
The solution to this system will be any two numbers x and y satisfying the condition у = 3 – х. For example, х=1, у=2; х=0, у=3 и т. д.
Example 3.
To solve the system of equations.
=-
0
ух
ì
í
=-
1
ух
î
.
It is obvious that this system has no solutions, since the difference between two numbers can't take two different values.
Example 4.
Using the theorem of Kronecker clarify the consistency of the system
of equations:
2
3
2x1 + 4x2 – 2x3 – 3x4 = 18,
3x1 + 2x2 – x3 + 2x4 = 4,
2x1 – 3x2 + 2x3 + x4 = – 8,
Using elementary transformations to transform the augmented matrix
of this system
-
6
2321
ö
÷
6
÷
÷
-
14
÷
÷
-
20
ø
6
2321
ö
÷
8
÷
~
÷
6
÷
÷
---
14
ø
ö
÷
÷
~
÷
÷
÷
ø
~
~
æ
ç
ç
ç
ç
ç
è
æ
ç
ç
ç
ç
ç
è
æ
ç
ç
ç
ç
ç
è
-
-
--
-
-
-
-
1800
-
ö
-
6
2321
÷
--
18
3242
÷
~
÷
4
2123
÷
÷
--
8
1232
ø
-
6
2321
ö
÷
6
1800
81040
2321
111610
365400
111610
18
÷
÷
-
14
÷
÷
8
ø
6
ö
÷
8
÷
~
÷
6
÷
÷
ø
~
: 6
æ
ç
ç
ç
ç
ç
è
æ
ç
ç
ç
ç
ç
è
æ
ç
ç
ç
ç
ç
è
-
1800
--
81040
--
5470
-
-
111610
-
1800
81040
-
6
2321
-
8
111610
-
6
1800
-
3
6900
~
100
-
6
2321
ö
÷
-
8
111610
÷
~
÷
-
9
5100
÷
÷
6
1800
ø
-
-
-
ö
6
2321
÷
8
111610
÷
÷
9
5100
÷
÷
-
2
1000
ø
~
~
æ
ç
ç
ç
ç
ç
è
æ
ç
ç
ç
ç
ç
è
-
-
-
ö
-
6
2321
÷
-
8
111610
÷
~
÷
6
1800
÷
÷
-
9
5100
ø
-
6
2321
ö
÷
8
111610
÷
~
÷
78
: –39
9
÷
÷
ø
-
5100
39000
æ
ç
ç
ç
ç
ç
è
æ
ç
ç
ç
ç
ç
è
-
Rank A=Rank A1=4, then the system of equations has a solution.
Moreover, if r=Rank A=Rank A1=n then system has a single solution.
0...
=+++
ì
ï
ï
í
ï
ï
î
2211
xaxaxa
1212111
nn
0...
=+++
xaxaxa
2222121
nn
.
............................................
0...
=+++
xaxaxa
nmnmm
(2)
Note that the system is always consistent, because it has the zero so-
lution
21
xxx
n
called trivial.
,0...
====
When r<n, the system is uncertain. It could be inconsistent system
(has not any solutions) or it has an infinitely many solutions.
Solve and check yourself using the answer:
=+
532
yx
=-
123
yx
Find the sum of the roots of the system of equations
ì
í
î
Answer: 2
Topic 7. The Cramer’s rule
Introduction. This workshop is devoted to the discussion of the following concepts: the Cramer’s rule and Procedure of The Cramer’s rule.
Example 1. To find a solution to the system of equations by the
method of Kramer:
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