In each part use the formula in Theorem 145 to compute the i

In each part, use the formula in Theorem 1.4.5 to compute the inverse of the matrix, and check 35. y0ur result by showing that AA^-1 = A^-1 A = 1 A = [I -2 1 i] A = [2 I 1 0]

Solution

a)

A =   i      -2

        1       i

det(A) = i*i - (-2)*1

           = -1 +2

           = 1

                                   //  inverse of      a      b     = 1/(ad-bc)   [d      -b]  

                                                             c      d                            -c       a

therefore inverse of the matrix = 1/1 [ i        2 ]     = [ i          2 ]

                                                            -1       i            -1           i   

now A* A^-1 = [ i      -2 ] [i        2]

                       1       i    -1         i   

= -1 +(-2)(-1)          2i -2i

    i -i                     2+(-1)

=     1        0

        0        1

A^-1 *A =   [ i       2 ]   [ i      -2 ]

                  -1      i     1          i

       = -1+2            -2i +2i

           -i+i              2 - 1

          =       1     0

                     0      1

therefore A*A^-1 = A^-1 * A = I

b)

A = [ 2       i ]

        1       0

A^-1 = 1/(-i) [0       -i]

                  -1       2

= i [0      -i]                //            1/(-i)      = i

       -1       2

   = 0     1

      -i     2i

A*A^-1 = 2    i        0      1

            1    0       -i      2i

         = 0-(i*i)        2+i(2i)

            0                 1

          = 1              0                       i*i    = -1

               0             1

A^-1 * A = 0         1               2         i

              -i         2i              1         0

= 0 +1           0+0

   -2i+2i          -i*i +0

=      1         0

         0         1