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222    CHAPTER 7  Matrices and Linear Systems

                                 Step 1. Find the general solution H of AX = O.
                                 Step 2. Find any one solution U p of AX = B.
                                 Step 3. The general solution AX = B is then H + U p .

                                    We know how to carry out step (1). We will outline a procedure for step (2).
                                    To find a particular solution U p , proceed as follows.
                                                                          .
                                                                          .
                                 Step 1. Define the n × m + 1 augmented matrix [A.B] by adjoining the column matrix B as an
                                       additional column to A. The augmented matrix contains the coefficients of the unknowns
                                       of the system (in the first m columns), as well as the numbers on the right side of the
                                       equations (elements of B).
                                                .
                                                .
                                 Step 2. Reduce [A.B]. Since we reduce a matrix to obtain leading entries of 1 wherever possible
                                       from upper left toward the lower right, this results eventually in a reduced matrix
                                                                     .        .
                                                                              .
                                                                     .
                                                                   [A.B] R =[A R .C],
                                       in which the first m columns are the reduced form of A, and the last column is whatever
                                                                                                       .
                                                                                                       .
                                       results from B after the row operations used to reduce A have been applied to [A.B].
                                           Solutions of the reduced system A R X = C are the same as solutions of the original
                                       system AX = B because the operations performed on the coefficients of the unknowns
                                       are also performed on the b j ’s.
                                               .
                                               .
                                 Step 3. From [A R .C], read a particular solution U p . When added to the general solution H of the
                                       associated homogeneous system, we have the general solution of AX = B.
                                           We will look at some examples. Example 7.24 suggests how this augmented matrix
                                       procedure tells us when the system has no solution.



                         EXAMPLE 7.23
                                 We will solve the system

                                                          ⎛            ⎞     ⎛   ⎞
                                                            −3   2    2        8
                                                          ⎝ 1    4   −6 ⎠ X =  ⎝ 1 ⎠  .
                                                             0  −2    2        −2
                                 The first step is to reduce the augmented matrix

                                                                              .
                                                               ⎛                   ⎞
                                                                −3    2   2   . .  8
                                                          .    ⎜                   ⎟
                                                          .    ⎜              . .  ⎟
                                                                 1    4   −6 .   1  ⎟.
                                                        [A.B]= ⎜
                                                               ⎝                   ⎠
                                                                              .
                                                                 0   −2   2   . .  −2
                                 Carrying out the reduction procedure on this 3 × 4 augmented matrix, we obtain
                                                                     .
                                                          ⎛               ⎞
                                                            1  0  0 . .  0
                                                    .     ⎜               ⎟      .      .
                                                    .     ⎜          . .  ⎟      .      .
                                                  [A.B] R = ⎜ 0  1  0 .  5/2 ⎟ =[A R .C]=[I 3 .C].
                                                                     .
                                                          ⎝               ⎠
                                                            0  0  1 . .  3/2
                                                                                                         .
                                                                                                         .
                                 C is whatever results in the fourth column when we reduce A, the first three columns of [A.B]).

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