a combinatorial approach to determinants
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Eva Apriyani 0902304
Laily Herni K 0902166
Noviawati 0902076
Rully Febrayanty 0902200
THE DETERMINANT
FUNCTION
DEFINITION
A permutation of the set of integers(1,2,. . .,n) is an arrangement of these integers in some
order without omissions or repetitions
There are six different permutations of the set of integers {1, 2, 3}. These are
EXAMPLE 1
)3,1,2( )2,1,3()3,2,1(
)1,3,2( )1,2,3()2,3,1(
EXAMPLE 2
Permutations of Four
Integers
List all permutations of the set of integers {1, 2, 3, 4}!
1
33
3
3
34
4
4
4
4
2
33
3
3
34
4
4
4
4
3
22
2
2
24
4
4
4
4
4
22
2
2
23
3
3
3
3
So , the solution of the problem :
1,2,3,4 2,1,3,4 3,1,2,4 4,1,2,3
1,2,4,3 2,1,4,3 3,1,4,2 4,1,3,2
1,3,2,4 2,3,1,4 3,2,1,4 4,2,1,3
1,3,4,2 2,3,4,1 3,2,4,1 4,2,3,1
1,4,2,3 2,4,1,3 3,4,1,2 4,3,1,2
1,4,3,2 2,4,3,1 3,4,2,1 4,3,2,1
We will denote a general permutation of the set (1,2,...,n) by (j1,j2,...,jn). Here, j1 is the first integer in
the permutation, j2 is the second, and so on. Aninversion is said to occur in a permutation (j1, j2, ...,
jn) whenever a larger integer precedes a smallerone. The total number of inversions occurring in a permutation can be obtained as follows: (1) find
the number of integersthat are less than j1 and that follow j1 in the
permutation; (2) find the number of integers thatare less than j2 and that follow j2 in the
permutation. Continue this counting process for j3, . . . ,jn-1 . The sum of these numbers will be thetotal number of inversions in the permutation.
(j1, j2, j3...,jn).
as the first integer in the permutation
as the second integer in the permutation
as the third integer in the permutation
inversion is said to occur in a permutation(j1, j2, ..., jn) whenever a larger integer precedes
a smaller one
Example 3Determine the number of inversion in the following permutation:
(a) (6, 1, 3, 4, 5, 2)
The number of inversion is
(6, 1, 3, 4, 5, 2)
85 +0 + 1 + 1 + 1 =
(b) (2, 4, 1, 3)
The Number of inversion is
(2, 4, 1, 3)
3
(c) (1, 2, 3, 4)
1 + 2 + 0 =
There are zero inversion in this permutation
A permutation is called even if the total number of inversions is an even integer and is called odd if the total number of
inversions is an odd integer.
DEFINITION
Example 4
The following table classifies the various permutations of {1, 2, 3} as even or odd
Permutation Number of Inversion Classification
(1, 2, 3) 0 Even
(1, 3, 2) 1 Odd
(2, 1, 3) 1 Odd
(2, 3, 1) 2 Even
(3, 1, 2) 2 Even
(3, 2, 1) 3 Odd
Combinatorial Definition of The Determinant
By an elementary product from an n x m matrix A we shall mean any product of n entries from A. No two of which come from same row column
Example 5
List all elementary products from the matrices
(a)
2221
1211
aa
aa
Since each elementary product has two factors, and since each factor comes from a different row, an elementary product can be written in the form
Solution (a)
21aa
where the blanks designate column numbers. Since no two factors in the product come from the same column, the column numbers must be 12 or 21. Thus the only elementary products are a11 a22 and a12 a21 .
(b)
333231
232221
131211
aaa
aaa
aaa
Since each elementary product has three factors, each of which comes from a different row, an elementary product can be written in the form
Solution (b)
321aaa
Since no two factors in the product come from the same column, the column numbers have no repetitions; consequently, they must form a permutation of the set {1, 2, 3}. These 3! = 6 permutations yield the following list of elementary products
332211aaa
332112aaa
322113aaa
322311aaa
312312aaa
312213aaa
As this example points out, an n x n matrix A has n! elementary products. They are the products of the formwhere is a permutation of the set (1,2,…,n). By a signed elementary product from A we shall mean an elementary product multiplied by +1 or -1. We use the (+) if is an even permutation and the (-) ifis an odd permutation
jnjj naaa ...
21 21
njjj ,...,,
21
jnjj naaa ...
21 21
njjj ,...,,
21 njjj ,...,,
21
Example 6
List all signed elementary products from the
matrices
(i) (ii)
aa
aa
2221
1211
aaa
aaa
aaa
333231
232221
131211
Solution
(i)
Elementary
Product
Associated
PermutationEven or Odd
Signed
Elementary
Product
-
a11
a12
a 22
a 21
a11 a 22
a 21a12
(1,2)
(2,1)
even
odd
(ii) Elementary
Product
Associated
Permutation
Even or
Odd
Signed
Elementary
Product
(1, 2, 3) even
(1, 3, 2) odd -
(2, 1, 3) odd -
(2, 3, 1) even
(3, 1, 2) Even
(3, 2, 1) odd -
a11 a 22 a 33
a11 a 23a 32
a12
a12 a 23
a 21 a 33
a 31
a 13 a 21 a 32
a 22a 13 a 31
a11 a 22 a 33
a 23a 32
a 12a 21 a 33
a12 a 31a 23
a 21 a 32a 13
a 22a 13 a 31
a 11
Let A be a square matrix. We define det (A)
to be the sum of all signed elementary
products from .
DEFINITION
Example 7
a11 a12
a 21 a 22
= a11 a 22 - a12a 21
By multiplying the entries on the rightward arrow and
subtracting the product of the entries on the leftward
arrow
(i) det
(ii) deta11 a12 a 13
a 21 a 22
a 32 a 33a 31
a11 a12
a 32a 31
a 22a 21 =
a11a 22a 33 +
a 23
a12 a 31a 23 a 21 a 32a 13+ --
a 22a 13 a 31
a 23a 32a11 -
The determinant is then computed by summing the products
on the rightward arrows and subtracting the products on the
leftward arrows.
a12 a 21 a 33
EXAMPLE 8
Evaluating DeterminantsEvaluate the determinants of
Solution987
654
321
B
8
5
2
7
4
1
987
654
321
Using the earlier method, we get
det(B)=(45)+(84)+(96)-(105)-(-48)-(-72)=240
EVALUATING
DETERMINANTS BY
ROW REDUCTION
• Let A be a square matrix. If has a row of zeros or a
column of zeros, then det (A) = 0
a
a
a
aa
aa
aa
32
22
12
3332
2322
1312
0
0
0
0
0
0
000000
00
000
331232232213
231323123322
aaaaaa
aaaaaa
THEOREM 1
Proof:
Example 9
a
aa
aaa
aaaa
44
3433
242322
14131211
000
00
0
aaaa
aaa
aa
a
44434241
333231
2221
11
0
00
000
The general form of a lower triangle matrices 4×4
The general form of an upper triangle matrices 4×4
Calculate det (A), where
a
aa
aaa
aaaa
44
3433
242322
14131211
000
00
0
Example 10
det (A) = a11 a22…amn
Theorem 2
If A is triangle matrix nxn, then det (A) is the multiplied result of entries in leading diagonal , so det (A) = a11 a22…amn
Example 11
40000
89000
67600
15730
38372
1296)4)(9)(6)(3)(2(
Theorem 3
a. If B is the matrix that results when a single row or single column of A is multiplied by a scalar k , then det (B) = k det (A)
Proof:
let A = let B =
aaa
aaa
aaa
333231
232221
131211
333231
232221
131211
aaa
kakaka
aaa
= k
)det()det( BkA
333231
232221
131211
aaa
kakaka
aaa
aaa
aaa
aaa
333231
232221
131211
b. If B is the matrix that results when two rows or two columns of are interchanged, then det (B) = - det (A)
Proof: The first and second rows of A are interchangedaaa
aaa
aaa
333231
131211
232221
aaa
aaa
aaa
333231
232221
131211
)det()det( AB
c. If B is the matrix that results when a multiple of one row of A is added to another row or when a multiple of one column is added to another column, then det (B) = det (A)
A multiple of the second row of A is added to the first row.
proof :
aaa
aaa
kaakaakaa
333231
232221
231322212111
aaa
aaa
aaa
333231
232221
131211
)det()det( AB
121
410
321
A
121
410
1284
1A
121
321
410
2A121
232
321
3A
If we calculate this matrices using method on example 8 then we have
det (A) = -2, det (A1)= -8, det(A2) = 2, det (A3) = -2
Look at this :
det (A1)= 4 det(A), det(A2) = - det (A),
and det (A3) = det (A)
EXAMPLE 12
EXAMPLE 13
162
963
510
det A
162
510
963
162
510
321
3
5100
510
321
3
5500
510
321
3
100
510
321
)55)(3(
165)1)(55)(3(
the first and second row of A were interchanged
the common factor of 3 from the first row was taken through the determinant sign
-2 times the first row was added to the third row
-10 times the second row was added to the third row
a common factor of -55 from the last row was taken through the determinant sign
8411
5193
8462
4231
A
8411
5193
0000
4231
det A
Example 14
-2R1+R2
So, if square matrix have 2 rows which
comparable then determinant is zero
Next, we don’t reduced because from
theorem 1, we get that det(A)= 0
EXAMPLE 15
Every matrix have two rows which comparable so from examination, every matrix have determinant is zero.
82
41
872
423
872
151239
4185
2526
5413
THANK YOU
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