Page 278 - Advanced engineering mathematics

258 CHAPTER 8 Determinants

If we expand by column two, we get
3
i+2
|A|= (−1) a i2 M i2
i=1

12 −9 2+2 −6 7

1+2
=(−1) (3) + (−1) (−5)
2 −6 2 −6

−6 7

3+2
+ (−1) (4)
12 −9

=(−3)(−72 + 18) − 5(36 − 14) − 4(54 − 84) = 172.
Sometimes we use row and column operations to produce a row or column with some zero
elements, then write a cofactor expansion by that row or column. Each zero element eliminates
one term from the cofactor expansion.
SECTION 8.3 PROBLEMS

−5
In each of Problems 1 through 10, evaluate the determi- 4 1 7

nant using a cofactor expansion by a row and again by a 9. −9 3 2 −5
column. Elementary row and/or column operations may be −2 0 −1 1
performed first to simplify the cofactor expansion. 1 14 0 3

−8 5 1 7
2
−4 2 −8
0 1 3 5 −6
1. 1 1 0
10. 2 2 1 5 3
1 −3 0

0 4 3 7 2

1 1 6 1 1 −7 −6 5

2. 2 −2 1 11. Show that

3 −1 4

1 a 2
a
7 −3
1
1 b b = (a − b)(c − a)(b − c).
2

3. 1 −2 4
1 c c 2

−3 1 0

This is called Vandermonde’s determinant.

5 −4 3

4. −1 1 6 12. Show that

−2 −2 4
a b c d

−5 0 1 b c d a
6

2 −1 3 7 c d a b

5. d a b c
4 4 −5 −8

1 −1 6 2 0 1 −1 1

4 3 −5 6 1 c d a
= (a + b + c + d)(b − a + d − c) .
1 −5 15 2 1 d a b

6.
0 −5 1 7 1 a b c

8 9 0 15
13. Prove that the points (x 1 , y 1 ), (x 2 , y 2 ),and (x 3 , y 3 ) in

−3 1 14 the plane are collinear (lie on a line) if and only if

7. 0 1 16
1
2 −3 4 x 1 y 1

1 = 0.
x 2 y 2
14 13 −2 5 1
x 3 y 3
7 1 1 7

8.
0 2 12 3 Hint: This determinant is zero exactly when one row

1 −6 5 23 or column is a linear combination of the others.

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October 14, 2010 14:26 THM/NEIL Page-258 27410_08_ch08_p247-266