MA-305 Homework 3 Solutions

Due at 2:35 pm, Thursday, February 26, 1998


  1. Find the inverse of the following matrix:

       [ 1 1 0 3 ]
   [ 0 2 0 2 ]
   [ 0 0 2 5 ]
   [ 0 0 0 1 ]
    The inverse of the matrix is

       [ 1 -1/2  0   -2  ]
   [ 0  1/2  0   -1  ]
   [ 0   0  1/2 -5/2 ]
   [ 0   0   0    1  ]

  2. Find the inverse of the following matrix, where a and b are non-zero.

       [ 1 a 0 0 ]
   [ a 1 b 0 ]
   [ 0 b 1 a ]
   [ 0 0 a 1 ]
    The inverse of the matrix is

        [        2    2               2                              2         ]
    [  -1 + a  + b       a (-1 + a )         a b                a  b       ]
    [ --------------- ,- -------------- ,   -----       ,    - ------      ]
    [        d                d               d                   d        ]
 
    [           2                 2                                        ]
    [  a (-1 + a )          -1 + a            b                  a b       ]
    [ -------------- ,   - --------     ,  - ---        ,       -----      ]
    [        d                 d              d                   d        ]

    [                                              2                  2    ]
    [      a b                 b             -1 + a          a (-1 + a )   ]
    [     -----      ,      - ---       , - ---------   ,   -------------  ]
    [       d                  d                d                  d       ]

    [       2                                       2              2    2  ]
    [      a  b               a b          a (-1 + a )       -1 + a  + b   ]
    [  - --------    ,       -----      , ------------- , - -------------- ]
    [        d                 d                d                  d       ]

                 2    2    4
where d = 1 - 2 a  - b  + a

  3. Find matrices T and U (U row-echelon form) such that TA = U, for

    A = 
       [   2  -2    2 ]
   [ -12   6   -1 ]
   [  -6  -2  -10 ]

    For A above
    T = 
       [  1   0  0 ]
   [  6   1  0 ]
   [ -5 -4/3 1 ]
    		   U = 
       [ 2 -2    2  ]
   [ 0 -6   11  ]
   [ 0  0 -56/3 ]

  4. Find T-1 for the matrix T in Problem 3 above.

    For T above
    T-1 = 
       [  1   0  0 ]
   [ -6   1  0 ]
   [ -3 -4/3 1 ]

  5. Suppose A is a 3x3 matrix with LU factors

    [ 1
[ l2,1
[ l3,1
     0
 1
 l3,2
     0 ]
 0 ]
 1 ]
    [ u1,1
[ 0
[ 0
     u1,2
 u2,2
 0
     u1,3 ]
 u2,3 ]
 u3,3 ]
    Find det(A).

    det(A) = det(LU) = det(L)det(U) = 1 * (u1,1u2,2u3,3) = u1,1u2,2u3,3

  6. For
    A = 
       [ a1,1 a1,2 a1,3 ]
   [ a2,1 a2,2 a2,3 ]
   [ a3,1 a3,2 a3,3 ]
    		B = 
       [ 1 0 0 ]
   [ 0 0 1 ]
   [ 0 1 0 ]
    verify the following:
    1. det(A) = det(AT)

           det(A) = a1,1a2,2a3,3 + a1,2a2,3a3,1 + a1,3a2,1a3,2
                     - a3,1a2,2a1,3 - a3,2a2,3a1,1 - a3,3a2,1a1,2
                    =det(AT)

    2. det(AB) = det(A)det(B)

      We have det(A) above.

      det(B) = -1     (as B is a type I elementary matrix)
      AB = 
         [ a1,1 a1,2 a1,3 ]
   [ a3,1 a3,2 a3,3 ]
   [ a2,1 a2,2 a2,3 ]
      and

           det(AB) = - a1,1a2,2a3,3 - a1,2a2,3a3,1 - a1,3a2,1a3,2
                     + a3,1a2,2a1,3 + a3,2a2,3a1,1 + a3,3a2,1a1,2
                    =det(A)det(B)

  7. For square matrices in general, does det(AB) = det(BA)? Explain.

    We have, by a theorem discussed in class, that det(AB) = det(A)det(B), and as determinants of matrices are polynomials, determinants commute under multiplication. So we have
         det(AB) = det(A)det(B) = det(B)det(A) = det(BA)