College Algebra 9th txtbk

490 CHAPTER 7 SYSTEMS OF EQUATIONS AND MATRICES
SOLUTION
Cofactor of 2 (1) 11 6 5 `
2 0` ` 6 5
2 0`
(6)(0) (2)(5) 10
Cofactor of 5 (1) 23 2 0 `
1 2` ` 2 0
1 2`
[(2)(2) (1)(0)] 4
2 is a 11
5 is a 23
MATCHED PROBLEM 2 Find the cofactors of 2 and 3 in the determinant in Example 2.
[Note: The sign in front of the minor, (1) ij , can be determined rather mechanically by
using a checkerboard pattern of and signs over the determinant, starting with in
the upper left-hand corner:
Use either the checkerboard or the exponent method—whichever is easier for you—to determine
the sign in front of the minor.]
Theorem 1 will give us a step-by-step procedure for finding third-order determinants
without having to memorize formula (2).
Z THEOREM 1 Computing a Third-Order Determinant
The value of a determinant of order 3 is the sum of three products obtained by
multiplying each element in any row or any column by its cofactor. This is called
expanding along a row or column.
Proving Theorem 1 requires six different calculations: expanding an arbitrary thirdorder
determinant along each of the rows and columns, and showing that the result matches
formula (2). You will be asked to complete a couple of those cases in the exercises.
EXAMPLE 3 Evaluating a Third-Order Determinant
Evaluate †
2 2 0
3 1 2 †
1 3 1
SOLUTION
We can choose any row or column to expand along. We will choose the first row because
of the zero: we won’t need to find that cofactor because it will be multiplied by zero.
†
2 2 0
3 1 2
1 3 1
† a 11 a Cofactor
of a 11
b a 12 a Cofactor
of a
b a 13 a Cofactor
12
of a
b
13
2 c (1) 11 1 2 `
3 1 `d (2) c 3 2
(1)12 `
1 1 `d 0
(2)(1)[(1)(1) (3)(2)] (2)(1)[(3)(1) (1)(2)]
(2)(5) (2)(1) 12