30 CHAPTER 1. REVIEW OF SOME LINEAR ALGEBRA

and (2 −1−1 1

)(1 11 2

)=

(1 00 1

),

showing that this matrix is indeed the inverse of A.

In the last example, how would you find A−1? You wish to find a matrix(

x zy w

)such that (

1 11 2

)(x zy w

)=

(1 00 1

).

This requires the solution of the systems of equations,

x+ y = 1,x+2y = 0

andz+w = 0,z+2w = 1.

Writing the augmented matrix for these two systems gives(1 1 | 11 2 | 0

)(1.12)

for the first system and (1 1 | 01 2 | 1

)(1.13)

for the second. Let’s solve the first system. Take (−1) times the first row and add to thesecond to get (

1 1 | 10 1 | −1

)Now take (−1) times the second row and add to the first to get(

1 0 | 20 1 | −1

).

Putting in the variables, this says x = 2 and y =−1.Now solve the second system, (1.13) to find z and w. Take (−1) times the first row and

add to the second to get (1 1 | 00 1 | 1

).

Now take (−1) times the second row and add to the first to get(1 0 | −10 1 | 1

).

Putting in the variables, this says z =−1 and w = 1. Therefore, the inverse is(2 −1−1 1

).

Didn’t the above seem rather repetitive? Exactly the same row operations were used inboth systems. In each case, the end result was something of the form (I|v) where I is the