146 CHAPTER 8. MATRICES

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

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

)(8.16)

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

)(8.17)

for the second. Lets 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, 8.17 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? Note that exactly the same row operationswere used in both systems. In each case, the end result was something of the form (I|v)

where I is the identity and v gave a column of the inverse. In the above,

(xy

), the first

column of the inverse was obtained first and then the second column

(zw

).

To simplify this procedure, you could have written(1 1 | 1 01 2 | 0 1

)