Page 150 - 35Linear Algebra
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150                                                                                      Matrices


                                  (b) We saw in Chapter 1 that the operator B = u× (cross product
                                       with a vector) is a linear operator. It can therefore be written as
                                       a matrix (given an ordered basis such as the standard basis). How
                                       is it that composing such linear operators is non-associative even
                                       though matrix multiplication is associative?


                            7.5     Inverse Matrix


                            Definition A square matrix M is invertible (or nonsingular) if there
                            exists a matrix M −1  such that

                                                       M  −1 M = I = MM    −1 .

                            If M has no inverse, we say M is singular or non-invertible.

                            Inverse of a 2 × 2 Matrix Let M and N be the matrices:


                                                         a  b               d −b
                                                  M =          ,    N =
                                                         c d               −c    a
                            Multiplying these matrices gives:

                                                        ad − bc       0
                                               MN =                       = (ad − bc)I .
                                                             0 ad − bc

                                                 d −b
                                           1
                            Then M  −1  =                , so long as ad − bc 6= 0.
                                         ad−bc  −c    a

                            7.5.1    Three Properties of the Inverse

                               1. If A is a square matrix and B is the inverse of A, then A is the inverse
                                  of B, since AB = I = BA. So we have the identity

                                                                 −1 −1
                                                               (A )    = A.

                                                −1
                                                               −1
                                                    −1
                                                                                 −1
                               2. Notice that B A AB = B IB = I = ABB A              −1  so
                                                                         −1
                                                            (AB) −1  = B A   −1


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