Linear Algebra, Theory And Applications, 2012a

3.3. THE MATHEMATICAL THEORY OF DETERMINANTS 91
Proof: Denote M by (m ij ) . Thus in the first case, m nn = a and m ni =0ifi ≠ n while
in the second case, m nn = a and m in =0ifi ≠ n. From the definition of the determinant,
det (M) ≡
∑
sgn n (k 1 , ··· ,k n ) m 1k1 ···m nkn
(k 1 ,··· ,k n )
Letting θ denote the position of n in the ordered list, (k 1 , ··· ,k n ) then using the earlier
conventions used to prove Lemma 3.3.1, det (M) equals
∑
(k 1,··· ,k n)
)
(−1) n−θ θ
sgn n−1
(k 1 , ··· ,k θ−1 , k θ+1 , ··· , n−1
k n m 1k1 ···m nkn
Now suppose (3.14). Then if k n ≠ n, the term involving m nkn in the above expression
equals zero. Therefore, the only terms which survive are those for which θ = n or in other
words, those for which k n = n. Therefore, the above expression reduces to
∑
a sgn n−1 (k 1 , ···k n−1 ) m 1k1 ···m (n−1)kn−1 = a det (A) .
(k 1 ,··· ,k n−1 )
To get the assertion in the situation of (3.13) use Corollary 3.3.8 and (3.14) to write
det (M) =det ( M T ) (( ))
A
T
0
=det
= a det ( A T ) = a det (A) .
∗ a
In terms of the theory of determinants, arguably the most important idea is that of
Laplace expansion along a row or a column. This will follow from the above definition of a
determinant.
Definition 3.3.16 Let A =(a ij ) be an n×n matrix. Then a new matrix called the cofactor
matrix cof (A) is defined by cof (A) =(c ij ) where to obtain c ij delete the i th row and the
j th column of A, take the determinant of the (n − 1) × (n − 1) matrix which results, (This
is called the ij th minor of A. ) and then multiply this number by (−1) i+j . To make the
formulas easier to remember, cof (A) ij
will denote the ij th entry of the cofactor matrix.
The following is the main result. Earlier this was given as a definition and the outrageous
totally unjustified assertion was made that the same number would be obtained by expanding
the determinant along any row or column. The following theorem proves this assertion.
Theorem 3.3.17 Let A be an n × n matrix where n ≥ 2. Then
det (A) =
n∑
a ij cof (A) ij
=
j=1
n∑
a ij cof (A) ij
. (3.15)
The first formula consists of expanding the determinant along the i th row and the second
expands the determinant along the j th column.
Proof: Let (a i1 , ··· ,a in )bethei th row of A. LetB j be the matrix obtained from A by
leaving every row the same except the i th row which in B j equals (0, ··· , 0,a ij , 0, ··· , 0) .
Then by Corollary 3.3.9,
n∑
det (A) = det (B j )
j=1
i=1