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.