Page 273 - Advanced engineering mathematics

8.2 Evaluation of Determinants I 253

But this is really just a permutation of the n − 1 numbers 2,3,··· ,n, since 1 is fixed and only
2,3,··· ,n are acted upon. In the definition of equation (8.1), we may therefore sum over only
the permutations q of 2,3,··· ,n, and factor a 11 from all of the terms of the sum, to obtain

|A|= a 11 a 2q(2) a 3q(3) ···a nq(n) =|A 11 |.
q q
This is a 11 times the determinant of the n − 1 × n − 1 matrix formed by deleting row one and
column one of A.
In the general case that a kr is an element of a row or column whose other elements are all
zero, we can interchange k −1 rows and then r −1 columns to obtain a new matrix with a kr in the
1,1 position of a row or column having its other elements equal to zero. Since each interchange
incurs a factor of −1 in the determinant, then by the preceding result,
|A|= (−1) k−1+r−1 a kr |A kr |= (−1) k+r a kr |A kr .
We are rarely lucky enough to encounter a matrix A having a row or column with all but
possibly one element equal to zero. However, we can use elementary row and column operations
to obtain such a matrix B from A. Furthermore from properties (3), (5), and (8) of determinants,
we can track the effect of each row and column operation on the value of the determinant. This
and the lemma enable us to reduce the evaluation of an n × n determinant to a constant times
an n − 1 × n − 1 determinant. We can then repeat this strategy, eventually obtaining a constant
times a determinant small enough to evaluate conveniently.

EXAMPLE 8.2
Let
⎛ ⎞
4 2 −3
A = 3 4 6 ⎠ .
⎝
2 −6 8
We want |A|. This is a simple example, but illustrates the point. We can get two zeros in column
two by adding −2 times row one to row two, then 3 times row one to row three. Since this
elementary row operation does not change the value of the determinant, then |A|=|B|, where
⎛ ⎞
4 2 −2
B = −5 0 10 ⎠ .
⎝
14 0 2
Exploiting the zeros in all but the 1,2 place in column two, then
|A|=|B|= (−1) 1+2 (2)|B 12 |

−5 10
=−2
14 2
=−2(−10 − 140) = 300.

EXAMPLE 8.3
Let
⎛ ⎞
−6 0 1 3 2
⎜ −1 5 0 1 7 ⎟
⎜ ⎟
A = ⎜ 8 3 2 1 7 .
⎟
⎜ ⎟
⎝ 0 1 5 −3 2 ⎠
1 15 −3 9 4

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October 14, 2010 14:26 THM/NEIL Page-253 27410_08_ch08_p247-266