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252 CHAPTER 8 Determinants

4. Evaluate |I n | for n = 2,3,···. Hint:Inthe sumof 6. Show that an upper or lower triangular matrix is non-
equation (8.1), the only term that does not have a singular if and only if it has nonzero main diagonal
zero factor corresponds to the identity permutation elements.
p : 1,2,··· ,n → 1,2,···n.
7. Let B be n × m. We know that we can achieve each
5. Show that the determinant of an upper or lower triangu- elementary row operation by multiplying on the left
lar matrix is the product of its main diagonal elements. by the matrix formed by performing the operation
Hint: Every term but one of the sum (8.1) contains a on I n . Show that each elementary column operation
factor a ij with i > j and a term a ij with i < j, and one can be performed by multiplying on the right by the
of these terms must be zero if the matrix is upper or matrix obtained by performing the column operation
lower triangular. The exceptional term corresponds to on I m .
the permutation p that leaves every number 1,2,··· ,n
unmoved.

8.2 Evaluation of Determinants I

The more zero elements a matrix has, the easier it is to evaluate its determinant. The reason for
this is that every zero element causes some terms in the sum of equation (8.1) to vanish. For
example, in Example 8.1, if a 12 = a 13 = 0,
⎛ ⎞
0 0
a 11
A = a 21 a 22 a 23 ⎠
⎝
a 31 a 32 a 33
and

a 22 a 23

= a 11 (a 22 a 33 − a 23 a 32 )
|A|= a 11
a 32 a 33
with four of the six terms of |A| being 0 cancelling because of the zeroes in the first row of A.
A generalization of this observation will form the basis of a useful method for evaluating
determinants.

LEMMA 8.1

Let A be n × n, and suppose row k or column r has all zero elements, except perhaps for a kr .
Then
|A|= (−1) k+r a kr |A kr |, (8.3)

where A kr is the n − 1 × n − 1 matrix formed by deleting row k and column r of A.
This reduces the problem of evaluating an n × n determinant to one of evaluating a smaller,
n −1×n −1, determinant. To see why the lemma is true, begin with the case that all the elements
of row one, except perhaps a 11 , are zero. Then
0 0 ··· 0
⎛ ⎞
a 11
···
a 12 a 22 a 23 a 2n
⎜ ⎟
⎜
⎟
A = ⎜ . . . . . ⎟.
⎝ . . . . . . . . . . ⎠
a n1 a n2 a n3 ··· a nn
In the sum of equation (8.1), the factor a 1p(1) is zero if p(1) = 1, because all the other elements
of row one are zero. This means we need only consider the sum over permutations p of the form
p : 1,2,3,··· ,n → 1, p(2), p(3),··· , p(n).

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October 14, 2010 14:26 THM/NEIL Page-252 27410_08_ch08_p247-266