Page 334 - 35Linear Algebra
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334                                                                    Sample Second Midterm


                            Solutions

                               1.  (a) Whenever detM = ad − bc 6= 0.
                                   (b) Unit determinant bit matrices:


                                               1 0      1 1      1 0      0 1      1 1      0 1
                                                     ,        ,        ,        ,        ,        .
                                               0 1      0 1      1 1      1 0      1 0      1 1
                                   (c) Bit matrices with vanishing determinant:


                                                    0 0     1 0      0 1      0 0      0 0
                                                          ,        ,        ,        ,        ,
                                                    0 0     0 0      0 0      1 0      0 1

                                                    1 1     0 0      1 0      0 1      1 1
                                                          ,        ,       ,         ,        .
                                                    0 0     1 1      1 0      0 1      1 1
                                       As a check, count that the total number of 2×2 bit matrices is 2 (number of entries)  =
                                        4
                                       2 = 16.
                                   (d) To disprove this statement, we just need to find a single counterexam-
                                       ple. All the unit determinant examples above are actually row equiva-
                                       lent to the identity matrix, so focus on the bit matrices with vanishing
                                       determinant. Then notice (for example), that


                                                                 1 1      0 0
                                                                       ∼/        .
                                                                 0 0      0 0
                                       So we have found a pair of matrices that are not row equivalent but
                                       do have the same determinant. It follows that the statement is false.

                               2.
                                          detA = 1.(2.6 − 3.5) − 1.(2.6 − 3.4) + 1.(2.5 − 2.4) = −1 .
                                  (i) Since detA 6= 0, the homogeneous system AX = 0 only has the solution
                                  X = 0. (ii) It is efficient to compute the adjoint

                                                                      T                
                                                         −3     0    2        −3 −1      1
                                               adj A =   −1    2 −1    =    0    2 −1  
                                                           1 −1      0         2 −1      0

                                  Hence
                                                                             
                                                                    3    1 −1
                                                         A −1  =    0 −2    1   .
                                                                  −2     1   0


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