2.1. MATRICES 51

Next take (−1) times the first row and add to the second followed by (−3) times the firstrow added to the last. This yields 1 2 2 1 0 0

0 −2 0 −1 1 0

0 −5 −7 −3 0 1

 .

Then take 5 times the second row and add to −2 times the last row. 1 2 2 1 0 0

0 −10 0 −5 5 0

0 0 14 1 5 −2

Next take the last row and add to (−7) times the top row. This yields −7 −14 0 −6 5 −2

0 −10 0 −5 5 0

0 0 14 1 5 −2

 .

Now take (−7/5) times the second row and add to the top. −7 0 0 1 −2 −2

0 −10 0 −5 5 0

0 0 14 1 5 −2

 .

Finally divide the top row by −7, the second row by -10 and the bottom row by 14 whichyields  1 0 0 − 1

727

27

0 1 0 12 − 1

2 0

0 0 1 114

514 − 1

7

 .

Therefore, the inverse is  − 17

27

27

12 − 1

2 0114

514 − 1

7



Example 2.1.28 Let A =

 1 2 2

1 0 2

2 2 4

. Find A−1.

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

1 0 2 0 1 0

2 2 4 0 0 1

and proceed to do row operations attempting to obtain

(I|A−1

). Take (−1) times the top

row and add to the second. Then take (−2) times the top row and add to the bottom. 1 2 2 1 0 0

0 −2 0 −1 1 0

0 −2 0 −2 0 1

