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200    CHAPTER 7  Matrices and Linear Systems

                                    First, interchange rows two and three of A to form
                                                                ⎛              ⎞
                                                                  −2  1  6 −3
                                                             B = ⎝ 0  9 3 −7 ⎠   .
                                                                  1   1 2    5
                                 Perform this row operation on I 3 to obtain
                                                                   ⎛        ⎞
                                                                     1  0  0
                                                               E 1 = 0  0  1 ⎠ .
                                                                   ⎝
                                                                     0  1  0
                                 Then
                                                 ⎛        ⎞⎛              ⎞   ⎛              ⎞
                                                   100       −2   1  6 −3       −2  1  6 −3
                                           E 1 A = 001 ⎠⎝ 1       1 2   5 ⎠  =  ⎝ 0  9 3 −7 ⎠  = B.
                                                 ⎝
                                                   010        0   9 3 −7        1   1 2    5
                                 Next multiply row three of A by 7 to form
                                                               ⎛                ⎞
                                                                −2   1   6   −3
                                                           C =  ⎝ 1  1   2    5 ⎠  .
                                                                 0   63  21 −49
                                 Perform this row operation on I 3 to obtain
                                                                   ⎛        ⎞
                                                                     1  0  0
                                                               E 2 = 0  1  0 ⎠ .
                                                                   ⎝
                                                                     0  0  7
                                 Then
                                               ⎛        ⎞⎛               ⎞  ⎛                 ⎞
                                                 100        −2  1  6 −3       −2   1   6   −3
                                          E 2 A = 010 ⎠⎝ 1      1 2    5 ⎠  =  ⎝ 1  1  2    5 ⎠  = C.
                                               ⎝
                                                 007        0   9 3 −7         0   63  21 −49
                                 Finally, add 2 times row one to row two to form
                                                                ⎛              ⎞
                                                                 −21     6   −3
                                                            D = −3    3  14  −1 ⎠  .
                                                                ⎝
                                                                  0   9  3   −7
                                 This operation can be achieved by the elementary matrix
                                                                   ⎛        ⎞
                                                                     1  0  0
                                                               E 3 = 2  1  0 ⎠ .
                                                                   ⎝
                                                                     0  0  1
                                 As a check,
                                               ⎛        ⎞⎛              ⎞   ⎛               ⎞
                                                 100       −2   1  6 −3       −2  1   6  −3
                                          E 3 A = 210 ⎠⎝ 1      1 2    5 ⎠  = −3  3  14  −1 ⎠ = D.
                                                                            ⎝
                                               ⎝
                                                 001        0   9 3 −7         0  9   3  −7
                                    This result has an important consequence. Suppose we form B from A by performing a
                                 sequence of elementary row operations in succession. That is, we perform operation O 1 on A to
                                 obtain A 1 , then O 2 on A 1 to form A 2 , and so on until we perform O r on A r−1 to form A r = B.We
                                 may envision this process
                                                               O 1  O 2   O 3
                                                             A −→ A 1 −→ A 2 −→ A 3 →
                                                                O r−1   O r
                                                             ··· −−→ A r−1 −→ A r = B.
                                 We can perform each elementary operation O j by multiplying on the left by the elementary
                                 matrix E j formed by performing that operation on I n . Then




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                                   October 14, 2010  14:23  THM/NEIL   Page-200        27410_07_ch07_p187-246
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