## Evaluating a Third Order Determinant

A determinant of order 3 is associated with a 3 x 3 matrix. Given

"31

a 12

a22 2

FIGURE 5,1

FIGURE 5,1

its determinant has the value

 an d\2 «1 3 ¿Ï23 a?2 «23 «21 + ^13 «21 «22 Gi\ an ail = an - aX2 Q%2 «33 «31 «31 «32 «31 «33

= «I1«22"33 ~ «1l«23«32 + «12«23«M -«12^21 «33

Looking first at the lower line of (5.6). wc sec the value of \A | expressed as a sum of six product terms, three of which are prefixed by minus signs and three by plus signs. Complicated as this sum may appear, there is nonetheless a very easy way of "catching" all these six terms from a given third-order determinant. This is best explained diagram-matically (Fig, 5.1). In the determinant shown in Fig. 5.1, each element in the top row has been linked with two other elements via two solid arrows as follows: a a an 1 a 12 «23 «31 > and tf 13 «21■ Each triplet of elements so linked can be multi plied out, and their product taken as one of the six product terms in (5.6). The solid-arrow product terms are to be prefixed with plus signs.

On the other hand, each top-row element has also been connected with two other elements via two broken arrows as follows: an a32 —► «23, a 12 azi «33 * and «22 «31- kach triplet of elements so connectcd can also be multiplied out, and their product taken as one of the six terms in (5.6). Such products are prefixed by minus signs. The sum of all the six products will then be the value of the determinant.

Example 2

= (2)(5)(9) + (1 )(6)(7) + (3)(8)(4) - (2)(8)(fi) - (1 )(4)(9) - (3)<5)(7) = -9

Example 3

= (-7)(1 )<S) + (0)(4)(0) + (3)(6)(9) - (—7)(6)(4) - (0)(9)(5) - (3)(1 )(0)

C h a pte r 5 Linear Models and Matrix AIge bra (Continued) 91

This method of cross-diagonal multiplication provides a handy way of evaluating a third-order determinant, but unfortunately it is not applicable to determinants of orders higher than 3. For the latter, we must resort to the so-called Laplace expansion of the determinant.