3.3. THE MATHEMATICAL THEORY OF DETERMINANTS 97

For example if

A =

 a b c

d e f

h i j

and i = 2, then

B1 =

 a b c

d 0 0

h i j

 , B2 =

 a b c

0 e 0

h i j

 , B3 =

 a b c

0 0 f

h i j

Denote by Aij the (n− 1)× (n− 1) matrix obtained by deleting the ith row and the jth

column of A. Thus cof (A)ij ≡ (−1)i+j

det(Aij). At this point, recall that from Proposition

3.3.6, when two rows or two columns in a matrixM, are switched, this results in multiplyingthe determinant of the old matrix by−1 to get the determinant of the new matrix. Therefore,by Lemma 3.3.15,

det (Bj) = (−1)n−j

(−1)n−i

det

((Aij ∗0 aij

))

= (−1)i+j

det

((Aij ∗0 aij

))= aij cof (A)ij .

Therefore,

det (A) =

n∑j=1

aij cof (A)ij

which is the formula for expanding det (A) along the ith row. Also,

det (A) = det(AT)=

n∑j=1

aTij cof(AT)ij=

n∑j=1

aji cof (A)ji

which is the formula for expanding det (A) along the ith column. ■

3.3.6 A Formula for the Inverse

Note that this gives an easy way to write a formula for the inverse of an n×n matrix. Recallthe definition of the inverse of a matrix in Definition 2.1.22 on Page 48.

Theorem 3.3.18 A−1 exists if and only if det(A) ̸= 0. If det(A) ̸= 0, then A−1 =(a−1ij

)where

a−1ij = det(A)−1 cof (A)ji

for cof (A)ij the ijth cofactor of A.

Proof: By Theorem 3.3.17 and letting (air) = A, if det (A) ̸= 0,

n∑i=1

air cof (A)ir det(A)−1 = det(A) det(A)−1 = 1.