Page 466 - Matrix Analysis & Applied Linear Algebra

462 Chapter 6 Determinants

Table 6.1.1

a
a
p =(p 1 ,p 2 ,p 3 ) σ(p) a 1p 1 2p 2 3p 3
(1, 2, 3) + 1 × 5 × 9=45
(1, 3, 2) − 1 × 6 × 8=48
(2, 1, 3) − 2 × 4 × 9=72

(2, 3, 1) + 2 × 6 × 7=84

(3, 1, 2) + 3 × 4 × 8=96
(3, 2, 1) − 3 × 5 × 7 = 105

Therefore,

det (A)= σ(p)a 1p 1 2p 2 3p 3 =45 − 48 − 72+84+96 − 105=0.
a
a
p
Perhaps you have seen rules for computing 3 × 3 determinants that involve
running up, down, and around various diagonal lines. These rules do not easily
generalize to matrices of order greater than three, and in case you have forgotten
(or never knew) them, do not worry about it. Remember the 2 × 2 rule given
in (6.1.2) as well as the following statement concerning triangular matrices and
let it go at that.

Triangular Determinants

The determinant of a triangular matrix is the product of its diagonal
entries. In other words,

t 11 t 12 ··· t 1n
0 t 22 ··· t 2n

. . . = t 11 t 22 ··· t nn . (6.1.3)
. . . .

. . . .
.
0 0 ··· t nn

Proof. Recall from the definition (6.1.1) that each term t 1p 1 2p 2 ··· t np n con-
t
tains exactly one entry from each row and each column. This means that there
is only one term in the expansion of the determinant that does not contain an
entry below the diagonal, and this term is t 11 t 22 ··· t nn .