Page 270 - Advanced engineering mathematics

250 CHAPTER 8 Determinants

Proof Conclusion (1) follows from the observation that each term in the sum of equation (8.1)
is a product of matrix elements, one element from each row and one from each column. We
t
therefore obtain the same terms from both A and A .
The reason for conclusion (2) is that a zero row or column puts a zero factor in each term of
the defining sum in equation (8.1).
Conclusion (3) states that interchanging two rows, or two columns, changes the sign of the
determinant. We will illustrate this for the 3 × 3 case. Let A =[a ij ] be 3 × 3 matrix and let
B =[b ij ] be formed by interchanging rows one and three of A. Then

b 11 = a 31 ,b 12 = a 32 ,b 13 = a 33 ,
b 21 = a 21 ,b 22 = a 22 ,b 23 = a 23 ,

and

b 31 = a 11 ,b 32 = a 12 ,b 33 = a 13 .

From Example 8.1,

|B|= b 11 b 22 b 33 − b 11 b 23 b 32 + b 12 b 23 b 31

=−b 12 b 21 b 33 + b 13 b 21 b 32 − b 13 b 22 b 31
= a 31 a 22 a 13 − a 31 a 23 a 12 + a 32 a 23 a 11
=−a 32 a 21 a 13 + a 33 a 21 a 12 − a 33 a 22 a 11
=−|A|.

Conclusion (4) follows immediately from (3). Form B from A by interchanging the two
identical rows or columns. Since A = B, |A|=|B|. But by (3), |A|=−|B|=|A|. Then |A|= 0.
Conclusion (5) is true because multiplying a row or column of A by α puts a factor of α in
every term of the sum (8.1) defining the determinant.
Conclusion (6) follows from (2) if α =0, so suppose that α =0. Now the conclusion follows
from (4) and (5). Suppose that row k of A is α times row i.Form B from A by multiplying
row k by 1/α. Then B has two identical rows, hence zero determinant by (4). But by (5), |B|=
(1/α)|A|= 0, so |A|= 0.
Conclusion (7) follows by replacing each a kj in the defining sum (8.1) with b kj + c kj .Note
here that k is fixed, so only one factor in each term of (8.1) is replaced. In particular, generally
the determinant of a sum is not the sum of the determinants. Conclusion (7) also holds if each
element of a specified column is written as a sum of two terms.
Conclusion (8) follows from (4) and (7). To see this we will deal with rows to be specific.
Suppose α times row i is added to row k of A to form D. On the right side of equation (8.2),
replace each b kj with αa ij , and each c kj with a kj , resulting in the following:

⎛ ⎞
a 11 a 12 ··· a 1n
⎜ ··· ··· ··· ··· ⎟
⎜ ⎟
⎜ ··· ⎟
a i1 a i2 a in
⎜ ⎟
D = ⎜ ··· ··· ··· ··· ⎟
⎜ ⎟
⎜ ⎟
αa i1 + a k1 αa i2 + a k2 ··· αa in + a kn
⎜ ⎟
··· ··· ··· ···
⎝ ⎠
a n1 a n2 ··· a in
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October 14, 2010 14:26 THM/NEIL Page-250 27410_08_ch08_p247-266