180 CHAPTER 8. DETERMINANTS

Proof: By Theorem 8.5.3 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 in the matrix A, replace the kth column with the rth column and then expand alongthe kth column. This yields for k ̸= r,∑n

i=1 air cof(A)ik det(A)−1 = 0 by Corollary 8.3.2 be-cause there are two equal columns. Summarizing, ∑

ni=1 air cof(A)ik det(A)−1 = δ rk.Using

the other formula in Theorem 8.5.3, and similar reasoning, ∑nj=1 ar j cof(A)k j det(A)−1 =

δ rk. This proves that if det(A) ̸= 0, then A−1 exists with A−1 =(

a−1i j

), where a−1

i j =

cof(A) ji det(A)−1 .

Now suppose A−1 exists. Then by Theorem 8.4.4,

1 = det(I) = det(AA−1)= det(A)det

(A−1)

so det(A) ̸= 0. ■The next corollary points out that if an n×n matrix A has a right or a left inverse, then

it has an inverse.

Corollary 8.6.2 Let A be an n×n matrix and suppose there exists an n×n matrix B suchthat BA = I. Then A−1 exists and A−1 = B. Also, if there exists C an n×n matrix such thatAC = I, then A−1 exists and A−1 =C.

Proof: Since BA = I, Theorem 8.4.4 implies detBdetA = 1 and so detA ̸= 0. Thereforefrom Theorem 8.6.1, A−1 exists. Therefore,

A−1 = (BA)A−1 = B(AA−1)= BI = B.

The case where CA = I is handled similarly. ■The conclusion of this corollary is that left inverses, right inverses and inverses are all

the same in the context of n×n matrices.Theorem 8.6.1 says that to find the inverse, take the transpose of the cofactor matrix

and divide by the determinant. The transpose of the cofactor matrix is called the adjugateor sometimes the classical adjoint of the matrix A. It is an abomination to call it the adjointalthough you do sometimes see it referred to in this way. In words, A−1 is equal to one overthe determinant of A times the adjugate matrix of A.

8.6.1 Cramer’s Rule

In case you are solving a system of equations, Ax= y for x, it follows that if A−1 exists,

x=(A−1A

)x= A−1 (Ax) = A−1y

thus solving the system. Now in the case that A−1 exists, there is a formula for A−1 givenabove. Using this formula,

xi =n

∑j=1

a−1i j y j =

n

∑j=1

1det(A)

cof(A) ji y j.