New Method to Compute the Determinant of a 3x3 Matrix

International Journal of Algebra, Vol. 3, 2009, no. 5, 211 - 219

New Method to Compute the Determinant of a 3x3 Matrix

Dardan Hajrizaj

Department of Telecommunication, Faculty of Electrical and Computer Engineering, University of Prishtina, Bregu i Diellit p.n., 10000 Prishtina, Kosovo dardanhajrizi@

Abstract

In this paper we will present a new method to compute the determinants of a 3x3 matrix. The advantages of this method comparing to other known methods are:

- quick computation, so it creates an easy scheme to compute the determinants of a 3x3 matrix,

- very quick appropriation, so we describe only four elements of the determinants, where it is created a new easy scheme to compute.

This new method creates opportunities to find other new methods to compute determinants of higher orders that will be our paper in the future.

Mathematics Subject Classification: 15A15, 11C20, 65F40

Keywords: methods to compute the determinant of a 3x3 matrix

1 Introduction

a11 a12 ... a1n

a21

a22

...

a

2n

Let A be an nxn matrix. A = . . ... .

.

.

...

.

a1n a2n ... ann

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D. Hajrizaj

Definition 1. - Determinant of n order will be called the sum, which has n! different terms j1, j2 ,... jn a1 j1 a2 j2 ...anjn which will be formed of the matrix A elements, see [4],[7],[8],[9].

a11 a12 ... a1n

a21 a22 ... a2n

D = det A = A = . .

.

...

.=

a a ...a j1, j2 ,... jn 1 j1 2 j2

njn

. ... . Sn

a1n a2n ... ann

Where

j1 , j2 ,... jn

=

+ -

1 1

, ,

if if

j1, j2 ,... jn j1, j2 ,... jn

is an even permutation is an odd permutation

2 Methods to compute the determinants of third order

In base of definition 1, determinant of the third order (for n=3) can be computed in this way, see [4], [7], [8], [9]:

a11 a12 a13 det A = A = a21 a22 a23 = 123a11a22 a33 + 132 a11a23a32 + 312 a13a a 21 32 +

a31 a32 a33 + a 321 13a22 a31 + a 231 12 a23a31 + 213a12 a21a33 = = a11a22 a33 - a11a23a32 + a13a a 21 32 - a13a22 a31 + a12 a23a31 - a12 a a 21 33 While the determinant of the third order expansion by the elements of whatever row we have, see [4], [5], [6], [8]:

a11 a21 a31

a12 a22 a32

a13 a23 a33

=

a11

a22 a32

a23 a33

- a12

a21 a31

a23 a33

+

a13

a21 a31

a22 = a32

=

-a21

a12 a32

a13 a33

+ a22

a11 a31

a13 a33

-

a23

a11 a31

a12 = a32

=

a31

a12 a22

a13 a23

- a32

a11 a21

a13 a23

+

a33

a11 a21

a12 a22

2.1 Sarrus's rule

Method to compute a determinant

213

Using Sarrus's rule, we have those schemes, see [4], [8], [10], [11]:

++ +

---

a11 a12 a13 a11 a12

a21 a22 a23 a21 a22

a31 a32 a33 a31 a32

Scheme 1.

+ +

a11

a12

a13

-

+

a21 a31

a22 a32

a23 a33

-

a11 a12 a13

a21 a22 a23

Scheme 2.

From the description of two first columns of the determinants (first and second columns) will be formed Scheme 1. respectively two rows ( first and second rows) will be formed Scheme 2.The terms, which will be formed by the products of diagonal elements in the left side in both of scheme 1 and 2 become the "-"sign. In this way we get the Sarrus' rule, which are valuable just to compute the determinants of the third order. In base of the Sarrus's rule we have:

++ +

---

a11 A = a21

a12 a22

a13 a23 =

a11 a21

a12 a22

a13 a11 a23 a21

a12 a22

=

a31 a32 a33

a31 a32 a33 a31 a32

= a11a22 a33 + a12 a23a31 + a13a21a32 - a13a22 a31 - a11a23a32 - a12 a a 21 33

2.2 Triangle's rule

The triangle's rule will be formed with this scheme:

a11 a12 a13 a21 a22 a23 a31 a32 a33

a11 a12 a13 a21 a22 a23 a31 a32 a33

Scheme 3.

The product of diagonal elements and product of elements in the both vertex of two triangles of the first determinant get the "+" sign and the product of diagonal elements and product of elements in the both vertex of two triangles of the second determinant get the "-" sign. In base of triangle's rule, we have:

a11 a12 a13 a11 a12 a13 a21 a22 a23 - a21 a22 a23 =

a31 a32 a33 a31 a32 a33

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D. Hajrizaj

= a11a22 a33 + a12 a23a31 + a13a21a32 - a13a22 a31 - a11a23a32 - a12 a21a33 2.3 Another scheme to compute of the determinants of the third order Another scheme to compute the determinants of the third order is Scheme 4, see [5].

+ + a11 a12 a13 a21 a22 a23

-

+ -

a31

a32

a33

-

Scheme 4.

In base of this scheme, we have:

+ + a11 a12 a13 a21 a22 a23

-

+ -

a31

a32

a33

-

= a11a22 a33 + a12 a23a31 + a13a21a32 - a13a22 a31 - a11a23a32 - a12 a21a33

2.4 Chio's condensation method

Chio's condensation is a method for evaluating an nxn determinant in terms of (n -1) x (n -1) determinants, see [1], [3].

a11 a12

A

=

1 a n-2

11

a21 a22 a11 a12

a31 a32 M

a11 a12

an1 an2

a11 a21

a13 a23

L

a11 a21

a1n a2n

a11 a31

a13 a33

L

a11 a31

a1n a3n

M

O

M

a11 an1

a13 an3

K

a11 an1

a1n ann

For n = 3 , we obtain:

Method to compute a determinant

215

a11 A = a21

a31

a12 a22 a32

a13 a23 a33

=1 a11

a11 a21 a11 a31

a12 a22 a12 a32

a11 a13 a21 a23 = 1 a11a22 - a12a21 a11a23 - a13a21 = a11 a13 a11 a11a32 - a12a31 a11a33 - a13a31 a31 a33

=

1 a11

[(a11a22

-

a12a21) (a11a33

-

a13a31 )

-

(a11a32

-

a12a31) (a11a23

-

a13a21)]

=

=

1 a11

[a121a22a33

-

a11a22a13a31

-

a12a21a11a33

+

a12a21a13a31

-

a121a32a23

+

a11a32a13a21

+

+ a12a31a11a23 - a12a31a13a21] = a11a22a33 - a22a13a31 - a12a21a33 - a11a32a23 + a32a13a21 +

+ a12a31a23

2.5 Dodgson's condensation method

The Dodgson's condensation method is a method, which determinants of the order

nxn expansion in determinant of the (n -1) x (n -1) order, than (n - 2) x (n - 2) order

and so one, see [2]. Using the Dodgson's condensation method for the determinants of the third order, we obtain:

a11 A = a21

a31

a12 a22 a32

a13 a23 a33

=

a11 a21 a 21 a31

a12 a22 a 22 a32

a12 a13 a22 a23 = a11a22 - a12 a21 a12 a23 - a13a22 = a22 a23 a21a32 - a22 a31 a22 a33 - a23a32 a32 a33

= (a11a22 - a12 a21 ) (a22 a33 - a23a32 ) - (a21a32 - a22 a31 ) (a12 a23 - a13a22 ) =

= a11a222 a33 - a11a22 a23a32 - a12 a21a22 a33 + a12 a21a23a32 - a a 21 32 a12 a23 + a21a32 a13a22 +

+ a22 a31a12 a23

-

a

2 22

a31a13

=

a11a

2 22

a33

- a11a22 a23a32

- a12 a21a22 a33

+ a21a32 a13a22

+

+

a22 a31a12 a23

-

a

2 22

a31

a13

In base of Dodgson's condensation method the final result will be divided with a22

term, so we have:

A = a11a22 a33 - a11a23a32 - a12 a21a33 + a21a32 a13 + a31a12 a23 - a22 a31a13

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