5.4. THE DETERMINANT 75

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

5.4.3, 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 5.4.11,

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.

5.4.8 Formula For The InverseNote that this gives an easy way to write a formula for the inverse of an n×n matrix.

Theorem 5.4.14 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 5.4.13 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 5.4.6. 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