152 CHAPTER 8. RANK OF A MATRIX

The first nonzero column is the second in the matrix. We switch the third and first rowsto obtain  0 1 1 5

0 2 0 10 0 2 3

Now we multiply the top row by −2 and add to the second. 0 1 1 5

0 0 −2 −90 0 2 3

Next, add the second row to the bottom and then divide the bottom row by −6 0 1 1 5

0 0 −2 −90 0 0 1

Next use the bottom row to obtain zeros in the last column above the 1 and divide thesecond row by −2  0 1 1 0

0 0 1 00 0 0 1

Finally, add −1 times the middle row to the top. 0 1 0 0

0 0 1 00 0 0 1

 .

This is in row reduced echelon form.

Example 8.2.10 Find the row reduced echelon form for the matrix 1 2 0 2−1 3 4 30 5 4 5

▶▶You should verify that the row reduced echelon form is 1 0 − 8

5 00 1 4

5 10 0 0 0

 .

Having developed the row reduced echelon form, it is now easy to verify that the rightinverse found earlier using the Gauss Jordan procedure is the inverse.