4.2. FINDING THE INVERSE OF A MATRIX 83

Proof: ⇒ If A−1 exists, then A−1A = I and so Ax= 0 if and only if x= 0. Why? Butthis says that the columns of A are linearly independent.⇐ Say the columns are linearly independent. Then they form a basis for Fn. Thus

there exists bi ∈ Fn such thatAbi = ei

where ei is the column vector with 1 in the ith position and zeros elsewhere. Then from theway we multiply matrices,

A(

b1 · · · bn

)=(

e1 · · · en

)= I

Thus A has a right inverse. Now letting B ≡(

b1 · · · bn

), it follows that Bx= 0

if and only if x= 0. However, this is nothing but a statement that the columns of B arelinearly independent. Hence, by what was just shown, B has a right inverse C,BC = I. Thenfrom AB = I, it follows that

A = A(BC) = (AB)C = IC =C

and so AB = BC = BA = I. Thus the inverse exists.Finally, if AB = I, then Bx= 0 if and only if x= 0 and so the columns of B are a

linearly independent set in Fn. Therefore, it has a right inverse C which by a repeat of theabove argument is A. Thus AB = BA = I. ■

Similarly, if A has a left inverse then it has an inverse which is the same as the leftinverse.

The theorem gives a condition for the existence of the inverse and the above proceduregives a method for finding it.

Example 4.2.4 Let A =

 1 0 11 2 11 1 2

. Find A−1 in arithmetic of Z3.

Form the augmented matrix 1 0 1 1 0 01 2 1 0 1 01 1 2 0 0 1

 .

Now do row operations in Z3 until the n×n matrix on the left becomes the identity matrix.This yields after some computations, 1 0 0 0 2 2

0 1 0 1 2 00 0 1 1 1 1

and so the inverse of A is the matrix on the right, 0 2 2

1 2 01 1 1

 .