J. Comp. & Math. Sci. Vol.2 (2), 266-273 (2011)
Generalization of a New Technique for Finding the Determinant of Matrices JAYANTA DUTTA1 and S. C. PAL2 1
Dept. of Mathematics, Siliguri Institute of Technology, Siliguri-734009, India. 2 Dept. of Computer Science and Application,University of North Bengal, Siliguri-7340013,India. email: mathjd651@yahoo.com ABSTRACT In this paper determinant of a matrix is evaluated, which generalizes the recursion procedure developed by Rezaifar and Rezaee1 and reduces the arithmetic operations if there are zero elements presents in the matrix. Keywords: Determinant; Matrix; Recursion; Numerical Linear Algebra; Arithmetic Operation.
1. INTRODUCTION Matrices play an important role in all branches of science, engineering, social science and management. Matrices are used for data classification by which many problems are solved using computers. Algebraic analysis of matrices is necessary for solving these problems by matrices in computer. A system of linear equations can be written in matrix form and for solving the system inversion of matrices are necessary. For this purpose, it is necessary to find out the determinant of matrices. A square matrix is singular if and only if its determinant is zero. Determinants are used in many areas of pure and applied mathematics. There are so many existing methods by which the value of the determinant can be evaluated. Recently, Rezaifar and Rezaee1 developed a
recursion technique to evaluate the determinant of a matrix. For some cases this procedure will fail to evaluate the value of the determinant. For A[aij] a (3Ă—3) matrix, if a22 element is zero then M11,33 = 0 and according to Rezaifar and Rezaee procedure, it fails to evaluate the value of the determinant unless rows are changed and as a result the determinant altered. In this paper we have generalized the procedure developed in1 by selecting ith row and ith column and jth row and jth column, ( i, j can be any valid value for the given determinant) instead of selection of first row first column and last row and last column as done in1 to form five other determinants which will give up the determinant value of the given matrix. Impact is to segregate most of the non-zero elements which makes easier necessary
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Jayanta Dutta, et al., J. Comp. & Math. Sci. Vol.2 (2), 266-273 (2011)
M =
x x x x x x
M ij =
x x x x x
arithmetic operations and possibility of M11,33 = 0 can be avoided. 2. EXISTING METHODS There are several direct and non-direct methods are available by which the determinant value of the matrices are evaluated. Some of the direct methods are Basket weave method, Pivotel condensation method (Chio method) and Expanding method. The non-direct methods are Gauss elimination method, LU decomposition method, QR decomposition method and Cholesky decomposition method etc. All these methods are well established and details can be found in several literatures and books. 3. DESCRIPTION OF OUR TECHNIQUE This method is based on a recursive algorithm, which gives value of the determinant of an n×n matrix. To evaluate the determinant of matrices, we describe the following steps:
M ii , jj
x x x = x
x x x
x x x
x x x
x x x
x x x
x x x
x x x
x x x
x x
x x
x x
x
x
x
x
x
x
x
x
x
x x
x x
x
x
x
x
x x x x x x x x x x x x x x x
a. M is an n×n matrix.
b. M ij is a matrix, which is given by the eliminations of ith row and the jth column of the matrix M i.e. ith row and jth column must be ommited from M to obtain M ij . And M ii , jj is a matrix, which is given by the eliminations of ith row ith column and jth row jth column of the matrix M i.e. ith row and ith column and jth row jth column must be ommited from M to obtain M ii , jj
c. The determinant can be expressed as the product of two terms. One is the determinant of size 2×2 where each individual determinant of size (n-1)×(n-1) and the other is the reciprocal of the determinant of size (n-2)×(n-2) as follows:
M =
M ii
M ij
M ji
M jj
i ≠ j and M ii , jj
×
1 , M ii , jj ≠0
Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)
(1)
268
Jayanta Dutta, et al., J. Comp. & Math. Sci. Vol.2 (2), 266-273 (2011) n
M jj = ∑ a is (− 1) M ji ,is
Here M ij presented in (1) is the value of the determinant of the matrix M ij . This method is a recursive method on n×n matrices. By this method a n×n matrix like M is reduced by four (n-1)×(n-1) matrices and one (n-2)×(n-2) matrix. Then the value of the determinant of M is given by expanding the 2×2 determinant made by representing M ij of matrix M , dividing by the determinant of M ii , jj .
s =1 s≠ j
Now we have to prove that the R.H.S of formula (1) will be the value of the det (M ) Now, n
M ii = ∑ a jt (− 1) n
det(M ) =
∑ a (− 1)
i+s
Again if we expand det (M ) by jth row then
det(M ) =
n
∑ a (− 1)
j +t
= a jj M ii , jj +
M ii , jt + a jj M ii , jj
n
∑ a (− 1)
j +t
jt
t =1 t ≠i, j
M ii , jt
= a jj .C + α where C = M ii , jj
j +t
jt
t =1
M ii , jt
jt
M is
is
s =1
∑ a (− 1)
t =1 t ≠i , j
1
n
j +t
t =1 t ≠i
=
Proof: Following if we expand det (M ) by ith row then
i+s
M jt
n
∑ a (− 1)
where α =
t =1 t ≠i , j
Explicit expression of M ij can be written as
j +t
jt
n
M ii , jt
M jj = ∑ a is (− 1) M jj ,is
follows:
i+s
s =1 n
j +t
i −1
M ii = ∑ a jt (− 1) M ii , jt +
∑ a (− 1)
=
s =1 s ≠i , j
t =1
n
∑ a (− 1)
+
t = i +1
n
=
∑ a (− 1) t =1 t ≠i n
j +t
jt
M ij = ∑ a jt (− 1)
M jj ,is + a ii M jj ,ii
n
∑ a (− 1)
s =1 s ≠i, j
i+s
is
= aii .C + δ where
j +t
M ij , jt
δ =
n
∑ a (− 1)
s =1 s ≠i , j
M ji = ∑ a is (− 1) M ji ,is s =1 s ≠i
= a ii M ii , jj +
M ii , jt
M ii , jt
t =1 t≠ j n
j +t
jt
i+s
is
i+s
n
i+s
is
M ij = ∑ a jt (− 1) t =1 t≠ j
j +t
M jj ,is
M ij , jt
Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)
M jj ,is
269
Jayanta Dutta, et al., J. Comp. & Math. Sci. Vol.2 (2), 266-273 (2011) n
=
∑ a (− 1)
M ij , jt + a ji (− 1)
j +t
i+ j
jt
t =1 t ≠i, j
= a ji (− 1)i + j M ii , jj + = a ji (− 1)
i+ j
n
∑ a jt (− 1)
t =1 t ≠i , j
j +t
n
M ii , jj + β
n
∑ a (− 1)
t =1 t ≠i, j
jt
= a ji (− 1)
i+ j
j +t
i+s
s =1 s ≠i
n
M ij , jt
=
∑ a (− 1)
s =1 s ≠i, j
i+s
is
M ji ,is + aij (− 1)
i+ j
= a ij (− 1)i + j M ii , jj +
where
β=
M ji = ∑ a is (− 1) M ji ,is
M ij , ji
= a ij (− 1)
i+ j
M ij , jt
n
∑ a (− 1)
s =1 s ≠i , j
i+s
is
M ji ,ij
M ji ,is
C +γ
where
C+β
γ =
n
∑ a (− 1)
s =1 s ≠i , j
i+s
is
M ji ,is
Now,
(
1 M ii M jj − M ij M ji M ii , jj
[ [
)
{ {
}{ }{
1 (a jj C + α )(aii C + δ ) − a ji (− 1)i + j C + β aij (− 1)i+ j C + γ C 1 = (aii C + δ )(a jj C + α ) − a ji (− 1)i + j C + β aij (− 1)i + j C + γ C
=
=
[
{
}] }]
]
}
1 i+ j i+ j i+ j a ii (a jj C + α )C + a jj Cδ + αδ − a ij (− 1) a ji (− 1) C + β C − γa ji (− 1) C − βγ C aδ βγ i+ j i+ j i+ j = a ii (a jj C + α ) + a jj δ + − aij (− 1) a ji (− 1) C + β − γa ji (− 1) − C C αδ − βγ i + j +1 i + j +1 = a ii M ii + aij (− 1) M ij + + a jj δ + a ji (− 1) γ C αδ − βγ i + j +1 i + j +1 = a ii M ii + a ij (− 1) (2) M ij + + a jj δ + γ (− 1) a ji C
{
}
Again,
αδ − βγ C Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)
Jayanta Dutta, et al., J. Comp. & Math. Sci. Vol.2 (2), 266-273 (2011)
270
n n n 1 n j +t i+s j +t i+s = ∑ a jt ( −1) M ii , jt ∑ a is (−1) M jj ,is − ∑ a jt (−1) M ij , jt ∑ a is ( −1) M ji ,is C t =1 s =1 t =1 s =1 s ≠ i , j t ≠i , j s ≠i , j t ≠ i , j
1 = C
n n n n i + j +t + s i + j + s +t M ii , jt M jj ,is − ∑ ∑ a jt a is ( −1) M ij , jt M ji ,is ∑ ∑ a jt a is ( −1) t =1 s =1 tt =≠1i , j ss =≠1i , j t ≠ i , j s ≠i , j n
n
∑ ∑a
=
t =1 s =1 t ≠i, j s ≠i, j
jt
ais (−1) i + j + t + s (M ii , jt M jj ,is − M ij , jt M ji ,is ) n
∑a
t =1 t ≠i , j n
n
∑ ∑a
=
t =1 s =1 t ≠i, j s ≠i, j
jt
j −1t
ais (−1) i + j + t + s (M ii , jt M jj ,is − M jj ,it M ii , js ) n
∑a
t =1 t ≠i , j n
=
(−1) j −1+ t M ii , jj , j −1t
∑a
s =1 s ≠i, j
is
j −1t
(−1) j −1+ t M ii , jj , j −1t
(−1) i + s a j , j − s + i M js ,i ( j − s +i )
Therefore from (2) we have
(
1 M ii M jj − M ij M ji M ii , jj = a ii M ii + a ij (− 1)
i + j +1
) n
M ij + a jj
∑ a (− 1)
i+ s
is
M jj ,is +
s =1 s ≠i, j
∑ {a (− 1) n
+
i+ s
is
s =1 s ≠i, j
}
M ji ,is (− 1)
= a ii M ii + a ij (− 1)
i + j +1
∑ a (− 1) {a n
+
i+s
is
s =1 s ≠i, j
i + j +1
αδ − βγ a ji + c
M ij +
M jj ,is − (− 1)
i+ j
jj
}
αδ − βγ a ji M ji ,is + c
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Jayanta Dutta, et al., J. Comp. & Math. Sci. Vol.2 (2), 266-273 (2011)
= a ii M ii + a ij (− 1)
i + j +1
∑ a (− 1) {a n
M ij +
i+s
is
s =1 s ≠i, j n
+
∑a
s =1 s ≠i, j
is
M jj ,is − (− 1)
i+ j
jj
}
a ji M ji ,is +
(−1) i + s a jj − s +i M js ,i ( j − s + i )
= a ii M ii + a ij (− 1)
i + j +1
n
+
∑a
is
s =1 s ≠i, j
M ij +
{
(−1) i + s a jj M jj ,is − (−1) i + j a ji M ji ,is + a j , j − s + i M js ,i ( j − s + i )
= a ii M ii + a ij (− 1)
i + j +1
n
M ij +
∑a
is
}
(−1) i + s M is
s =1 s ≠i, j
= M . 4. NUMERICAL EXAMPLES
A =
−72 −54 123 −108
2 5 0 6 Example-1.Let A =
0 4 3 0 5 2 6 4
, where
×123) ×
×
1 = (72 ×108 + 54 − 18
1 14418 = − = − 801 . − 18 18
4 7 0 3 A is a matrix.
2 0 4 0 1
By the method the value of the determinant can be evaluated as
A=
138 177 168 111 168) ×
×
1 = (138 × 111 - 177 × 18
1 14418 = − = − 801 . 18 18 rd
nd
As in A the 3 row and 2 column is non-zero, so if we choose i = 3 and j = 2 then by our generalized method the determinant of A can be obtained as
Example-2. Let B =
4 6 2 3 3 4 5 0
, where
0 2 0 5 B is a matrix. By the method described in1 the determinant of B can be evaluated as
1 B = 80 − 28 × = (80 × 36 + 28 × 48) 22 48 36 ×
1 4224 = = 192. 22 22
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Jayanta Dutta, et al., J. Comp. & Math. Sci. Vol.2 (2), 266-273 (2011)
As in B the 2nd row is non-zero, so if
× 54) ×
we choose i = 2 and j = 3 then the determinant of B can be obtained using generalized method
B =
− 10 40
×
− 60 48 ×
1 = (60 × 40 - 10 × 48) 10
1 1920 = = 192 10 10
Example-3.
4 2 7 0 0 2 4 5 6 0 Let C = 0
2 6 7 5 , where C is a
0 3 2 5 0 0 6 4 6 3 matrix. By the method1 the determinant of C can be evaluated as
C =
245 112 − 60 − 78
×
1 = (112 × 60 - 245 × 35
1 12390 78) × = − = − 354 . 35 35 As in C 3rd column is non-zero, so if we choose i = 2 and j = 3 then by generalized method the determinant of C can be obtained as
C =
32 −372 −54 −36
×
1 = (- 32 × 36 - 372 60
1 21240 = − = − 354 . 60 60
5. COMPARISON In Rezaifar and Rezaee method1 if the division by zero appears, rows should be changed, as a result the determinant altered. But in our generalized method rows and columns can be selected according to our choice so that division of zero can be avoided and there is no question of alteration of rows. One more advantage of our method is that we can segregate most of the non-zero element which reduces the arithmetic operation than that of method1. If Tn denotes the total number of scalar multiplication required to evaluate an n×n determinant then for the expanding method, Tn = nTn-1 + n and for the method1 and our generalized method, Tn = 4Tn-1 + Tn-2 + 3. Table-I: Total number of scalar multiplication and their comparison
Example
Method1
Example-1 Example-2 Example-3
49 49 185
Generalized Method 43 41 124
From Table-1, it follows that if there are zero elements presents in the matrix then generalized method requires less number of scalar multiplications than that of the method1. Moreover, this differences in scalar multiplication increases with the order of the matrix (n). In the Table-I, the location of one zero element in the matrix is considered as
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Jayanta Dutta, et al., J. Comp. & Math. Sci. Vol.2 (2), 266-273 (2011)
equivalent to one arithmetic operation and subsequently when an nĂ—n determinant is multiplied by zero this is also considered as equivalent to one arithmetic operation. CONCLUSION If all the elements of the matrix are non zero then the number of scalar multiplication required in expanding method is very large compared to the method developed in1 and the generalized method. As in case of a expanding method, n number of (n-1)Ă—(n-1) have to be evaluated where as in the method1 and the generalized method the number of evaluated (n-1)Ă—(n-1) determinant is fixed to 4 and also in this case the method developed in1 and the generalized method requires same number of scalar multiplication. But this will be
reversed if there are sufficient numbers of zero elements presents in the matrix. In this case the number of scalar multiplication in expanding method is rapidly decreased compared to the method1 and the generalized method, and the generalized method requires less number of scalar multiplications than that of the method1 which asserts from the Table-1. Further, in particular if i = 1 and j = n then the formula (1) coincides with the formula developed in1. REFERENCES 1. Omid Rezaifar and Hossein Rezaee, A new approach for finding the determinant of matrices, Applied Mathematics and Computation, 188, 1445-1454, 2007.
Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)