Volume 04 Issue 06-2024
8
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
8-14
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
A
BSTRACT
Certain methods used in this paper to calculate numerical determinants are computationally intensive. For
certain forms of literal and numerical determinants, some methods of their calculation have been
developed.
K
EYWORDS
Determinant, triangle method, recurrent relations, major minor, diagonal view, determinant order.
I
NTRODUCTION
The main idea of the method of bringing the
determinant to the form of a triangle is that all
elements on one side of the diagonal are reduced
to zero by performing elementary substitutions. If
the elements lying on one side of the main
diagonal are equal to zero, then such a
determinant is equal to the product of all
elements on the main diagonal. If all the elements
lying on one side of the auxiliary diagonal of the
determinant are equal to zero, then such a
determinant is
2
)
1
(
)
1
(
−
−
n
n
equal to the product of
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Copyright:
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Research Article
SOME METHODS OF CALCULATING N-ORDER
DETERMINANTS AND WAYS TO SOLVE EXAMPLES RELATED
TO THEM
Submission Date:
May 31,
2024,
Accepted Date:
June 05, 2024,
Published Date:
June 10, 2024
Crossref doi:
https://doi.org/10.37547/ijasr-04-06-02
Saidova Nilufar Rozimorotovna
Ph.D., Associate Professor of Navoi University of Innovations, Uzbekistan
Abdurakhmanov Ghulam Erkinovich
Teacher of Navoi University of Innovations, Uzbekistan
Volume 04 Issue 06-2024
9
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
8-14
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
all the elements of the diagonal taken with the
sign.
Example 1
. Calculate the nth-order determinant.
a
a
a
x
a
a
x
a
a
a
x
a
a
a
a
d
...
...
...
...
...
...
...
+
+
+
=
Solving. We add all previous columns to the last
column:
x
na
a
a
x
a
x
na
x
a
a
a
x
na
a
a
a
d
+
+
+
+
+
=
...
...
...
...
...
...
...
The common multiplier in the last column under
the determinant sign is -
we subtract na + x. We subtract the last column
multiplied by a from each of the previous
columns. As a result, we arrive at the determinant
in the form of a triangle, with all elements above
the auxiliary diagonal consisting of zeros:
1
0
...
0
1
0
...
0
...
...
...
...
...
1
...
0
0
1
0
...
0
0
)
(
x
x
x
x
na
d
+
=
So,
1
2
)
1
(
)
(
)
1
(
−
−
+
−
=
n
n
n
x
na
x
d
The main idea of the method of separation of
linear multipliers is to treat the n-order
determinant as an m-order polynomial of one or
more variables. One can find m mutually radical
linear multipliers that divide the determinant
directly or by performing certain substitutions. In
that case, the determinant constant multiplier S is
equal to the product of these linear multipliers.
The constant number S is found as a result of
comparing the term of the determinant and the
term in the product of linear multipliers,
respectively.
Volume 04 Issue 06-2024
10
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
8-14
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
Example 2
. Calculate the nth-order determinant.
a
x
n
a
x
n
a
x
n
d
+
+
+
=
...
3
2
1
...
...
...
...
...
...
2
1
...
3
1
...
3
2
1
Solving. The product of elements on the diagonal
of the determinant keeps x at the largest - (n
–
1)
level. So, this determinant is a polynomial of (n
–
1) degree x. At x = 2 - a, x = 3 - a, ..., x = n - a, the 1st
and 2nd, 1st and 3rd,. , 1st and nth lines of this
determinant are the same will be the same, and as
a result, the determinant will be zero. Thus, d
determinant is divided by x + a - 2, x + a - 3, ..., x +
a - n, and therefore,
d = c ( x + a -2)( x + a -3)...( x + a
–
n ) (*)
To find the number c, we compare the term xn-1
formed by multiplying the elements of the main
diagonal with the term c xn-1 on the right side of
(*). Given that these terms are equal, s = 1 and as
a result
We form d = ( x + a
–
2 ) ( x + a
–
3 )...( x + a
–
n ).
In the method of recurrent relations, the given
determinant is reduced to one or more
determinants of the same order of small order.
For this, the determinant is spread over a row or
column. In some cases, the determinant is made
convenient by making certain substitutions and
then spreading it over rows or columns. An
equality that expresses a determinant through
one or more lower-order determinants in the
same form is called recurrent or return equality.
Using the method of mathematical induction, the
general expression of the given determinant is
derived from the recurrent equation.
This method can also be used in the following
modified form: in the recurrent equation
expressed by n-order determinants, the
expression when replacing n in this recurrent
equation with (n
–
1) is given; similarly ( n
–
2 )-
order expression, etc. will be posted. As a result,
the general view of the n-order determinant is
formed. The correctness of this expression is
checked using the method of mathematical
induction.
Example 3
. Calculate the nth-order determinant.
Volume 04 Issue 06-2024
11
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
8-14
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
7
3
...
0
0
0
0
...
...
...
...
...
...
...
0
0
...
4
7
3
0
0
0
...
0
4
7
3
0
0
...
0
0
4
7
=
n
d
Solving.
Spread
along
the
first
line,
2
1
12
7
−
−
−
=
n
n
n
d
d
d
we generate . This is
consistent with the recurrence relation
0
12
7
2
=
+
−
x
x
quadratic
equation
)
(
4
,
3
=
=
has
roots.
So,
n
n
n
c
c
d
4
3
2
1
+
=
. We find the coefficients c1
and c2 from the formulas
)
(
1
2
1
−
−
=
d
d
c
,
)
(
1
2
2
−
−
−
=
d
d
c
.
,
7
,
37
7
3
4
7
1
2
=
=
=
d
d
, .
since c1=-3, c2=4. So, it will be
1
1
3
4
+
+
−
=
n
n
n
d
.
The method of expanding the determinant into
the sum of determinants is sometimes easily
calculated by expressing the n-order determinant
in the form of the sum of two or more
determinants.
Example 4
. Calculate the nth-order determinant.
a
b
a
b
a
b
a
b
a
d
0
...
0
0
0
0
...
0
0
0
0
...
...
...
...
...
...
...
0
0
...
0
0
0
0
...
0
0
0
0
...
0
0
=
Solving. Spread the determinant on the first
column:
Volume 04 Issue 06-2024
12
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
8-14
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
.
)
1
(
)
1
(
...
0
0
0
...
...
...
...
...
0
...
0
0
...
0
0
)
1
(
...
0
0
0
...
...
...
...
...
0
...
0
0
...
0
1
1
1
1
1
n
n
n
n
n
n
n
b
a
b
b
a
a
b
b
a
b
b
a
b
a
b
a
a
d
+
−
+
−
−
−
+
=
−
+
=
=
−
+
=
Both determinants have a triangular form.
The method of changing the elements of the
determinant - in this method, by changing all the
elements of the determinant to one number, it
becomes convenient to calculate the algebraic
complement of all elements. This method is based
on the following property: if we add exactly one
number x to all elements of the determinant d,
then the number x of the determinant is d.
x
a
x
a
x
a
x
a
d
a
a
a
a
d
nn
n
n
nn
n
n
+
+
+
+
=
=
...
...
...
...
...
,
...
...
...
...
...
1
1
11
'
1
1
11
let it be into two determinants with respect to
line 1, and each of them into two determinants
with respect to line 2, etc. we write Determinants
with more than one row of all elements equal to x
are equal to zero, and determinants with one row
of elements equal to x are spread over this row.
Then we form the equality that needs to be
proved
=
+
=
n
j
i
ij
A
x
d
d
1
,
'
. Thus, the calculation
of the d' determinant is reduced to the calculation
of the d determinant and the sum of its algebraic
complements.
Calculating the nth-order determinant to the
Vandermond determinant is the Vandermond
determinant, is called the determinant in the
form of
1
3
2
1
2
3
2
2
2
2
1
1
3
1
2
1
1
...
1
...
...
...
...
...
...
...
1
...
1
−
−
−
=
n
n
n
n
n
n
n
n
x
x
x
x
x
x
x
x
x
x
x
x
V
Volume 04 Issue 06-2024
13
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
8-14
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
It is calculated using the following formula:
.)
(
)
)...(
)...(
)(
)(
)...(
)(
(
1
1
2
2
4
2
3
1
1
3
1
2
−
=
=
−
−
−
−
−
−
−
=
−
k
i
n
k
i
n
n
n
n
n
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
V
Some determinants can be calculated by bringing
them to the Vandermonde determinant.
Example 5
. Calculate the determinant by
multiplying by the Vandermond determinant.
n
n
n
n
n
n
n
n
n
n
n
n
n
d
1
1
1
1
1
2
1
1
2
2
1
1
1
1
1
...
...
...
...
...
...
...
...
+
+
−
+
+
−
−
=
Solving.
Under
the
determinant
sign
n
n
n
n
1
2
1
,...,
,
+
, we subtract the multipliers from
the first, ..., (n+1) lines, respectively. As a result,
we form the Vandermond determinant:
=
−
=
=
+
+
+
+
+
+
j
i
j
j
i
i
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
d
1
2
1
1
1
1
1
2
2
2
2
1
1
1
1
1
2
1
...
...
1
...
...
...
...
...
1
...
1
...
.
)
(
...
1
2
1
+
−
=
−
=
j
i
j
i
j
i
i
j
j
i
j
i
i
j
n
n
n
n
C
ONCLUSION
Currently, we know that in many areas, as
mathematics enters, the issue of calculating it in
the most optimal way is seen. The methods
presented in this article deal with the issue of
calculating the determinants given in a complex
form, simplifying them. As an example, we can say
that when calculating determinants, it is much
easier to calculate them in a convenient way,
except for the principals. In addition, modern
Volume 04 Issue 06-2024
14
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
8-14
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
professions are also used, by including the
algorithm of these methods in the program, it will
be possible to easily calculate many complex-
looking determinants.
R
EFERENCES
1.
Leng S. Algebra. M. Mir, 1968.
2.
Kostrykin A.I. Introduction to algebra. M.,
1977, 495 pages.
3.
Leng S. Algebra. M. Mir, 1968.
4.
Narzullayev U.Kh., Soleyev A.S. Algebra i
theory chisel. I-II chast, Samarkand,
5.
Faddeyev D.K., Sominsky I.S. Sbornik zadach
po vyshey algebre. M., Nauka, 1977.
6.
B.L. Van der Waerden. Algebra. M., Nauka,
1976
7.
Sbornik zadach po algebre pod redaksiyey. A.I.
Kostrikina, M., Nauka, 1985.
8.
Khojiyev J., Feinleb A.S. Algebra and number
theory course, Tashkent, "Uzbekistan", 2001.
9.
2002 Faddeyev D.K. Lexii po algebra. M.,
Nauka, 1984, 415 st.
