50 CHAPTER 2. LINEAR TRANSFORMATIONS

If it is possible to do row operations and end up with Arow operations→ I, then the above

argument shows that A has an inverse. Conversely, if A has an inverse, can it be found bythe above procedure? In this case there exists a unique solution x to the equation Ax = y.In fact it is just x = Ix = A−1y. Thus in terms of augmented matrices, you would expectto obtain

(A|y) →(I|A−1y

)That is, you would expect to be able to do row operations to A and end up with I.

The details will be explained fully when a more careful discussion is given which is basedon more fundamental considerations. For now, it suffices to observe that whenever the aboveprocedure works, it finds the inverse.

Example 2.1.26 Let A =

 1 0 1

1 −1 1

1 1 −1

. Find A−1.

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

1 −1 1 0 1 0

1 1 −1 0 0 1

 .

Now do row operations until the n×n matrix on the left becomes the identity matrix. Thisyields after some computations, 1 0 0 0 1

212

0 1 0 1 −1 0

0 0 1 1 − 12 − 1

2

and so the inverse of A is the matrix on the right, 0 1

212

1 −1 0

1 − 12 − 1

2

 .

Checking the answer is easy. Just multiply the matrices and see if it works. 1 0 1

1 −1 1

1 1 −1

 0 1

212

1 −1 0

1 − 12 − 1

2

 =

 1 0 0

0 1 0

0 0 1

 .

Always check your answer because if you are like some of us, you will usually have made amistake.

Example 2.1.27 Let A =

 1 2 2

1 0 2

3 1 −1

. Find A−1.

Set up the augmented matrix (A|I) 1 2 2 1 0 0

1 0 2 0 1 0

3 1 −1 0 0 1

