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a
=
1
; if the object
Y
appearances slightly exceeds those of the object
Y
,
()
()
ik
k
i
. If the characteristic
X
appearances of the object
then
a
=
3
Y
exceed those of
()
ik
i
the object
Y
to the extent “more”, “noticeably more” or “much more”, then
()
k
a
=
5
7
9
. Evaluations
a
=
2
4
6
8
are intermediate.
ik
ik
Let us assume
a
=
1
/
a
, compose a matrix of paired comparisons
ki
ik
⎛
1
a
a
...
a
⎞
12
13
1
j
⎜
⎟
1
⎜
⎟
1
a
...
a
23
2
j
⎜
a
⎟
12
⎜
⎟
1
1
A
=
⎜
1
...
a
⎟
m
3
j
a
a
⎜
⎟
13
23
⎜
.
.
.
...
.
⎟
1
1
1
⎜
⎟
...
1
⎜
⎟
a
a
a
⎝
⎠
1
j
2
j
3
j
and determine its eigenvalues. For this purpose let us equate a determinant to zero.
1
−
λ
a
a
...
a
12
13
1
j
1
1
−
λ
a
...
a
23
2
j
a
12
1
1
=
0
.
1
−
λ
...
a
3
j
a
a
13
23
.
.
.
...
.
1
1
1
...
1
−
λ
a
a
a
1
j
2
j
3
j
λ
Let us choose maximum eigenvalue
and determine a corresponding
max
(
)
eigenvector
. For this purpose we'll solve a set of equations
ω
=
ω
,
,...,
ω
m
m
m
,
j
written in the matrix form
1
−
λ
a
a
...
a
⎛
⎞
max
12
13
1
j
⎜
⎟
1
ω
⎛
⎞
⎜
⎟
1
−
λ
a
...
a
m
,
⎜
⎟
max
23
2
j
⎜
⎟
a
ω
⎜
⎟
12
m
,
2
⎜
⎟
1
1
⎜
⎟
ω
1
−
λ
...
a
=
0
⎜
⎟
⎜
⎟
m
,
3
max
3
j
a
a
⎜
⎟
13
23
.
⎜
⎟
⎜
.
.
.
...
.
⎟
⎜
⎟
⎜
1
1
1
⎟
ω
⎝
⎠
...
1
−
λ
m
,
j
⎜
⎟
max
a
a
a
⎝
⎠
1
j
2
j
3
j
(2.3)
It is known [15], that the system (2.3) has a solution and
