7.1. THE DETERMINANT 137
Denote by Ai j the (n−1)× (n−1) matrix obtained by deleting the ith row and the jth
column of A. Thus cof(A)i j ≡ (−1)i+ j det(Ai j). At this point, recall that from Proposition
7.1.3, when two rows or two columns in a matrix M, are switched, this results in multiplyingthe determinant of the old matrix by−1 to get the determinant of the new matrix. Therefore,by Lemma 7.1.11,
det(B j) = (−1)n− j (−1)n−i det
((Ai j ∗0 ai j
))
= (−1)i+ j det
((Ai j ∗0 ai j
))= ai j cof(A)i j .
Therefore,
det(A) =n
∑j=1
ai j cof(A)i j
which is the formula for expanding det(A) along the ith row. Also,
det(A) = det(AT )= n
∑j=1
aTi j cof
(AT )
i j
=n
∑j=1
a ji cof(A) ji
which is the formula for expanding det(A) along the ith column. ■
7.1.8 Formula For The InverseNote that this gives an easy way to write a formula for the inverse of an n×n matrix.
Theorem 7.1.14 A−1 exists if and only if det(A) ̸= 0. If det(A) ̸= 0, then A−1 =(
a−1i j
)where
a−1i j = det(A)−1 cof(A) ji
for cof(A)i j the i jth cofactor of A.
Proof: By Theorem 7.1.13 and letting (air) = A, if det(A) ̸= 0,
n
∑i=1
air cof(A)ir det(A)−1 = det(A)det(A)−1 = 1.
Now considern
∑i=1
air cof(A)ik det(A)−1
when k ̸= r. Replace the kth column with the rth column to obtain a matrix Bk whosedeterminant equals zero by Corollary 7.1.6. However, expanding this matrix along the kth
column yields
0 = det(Bk)det(A)−1 =n
∑i=1
air cof(A)ik det(A)−1