ENGR 1405 Engineering Mathematics 1

Faculty of Engineering and Applied Science
2000 Fall

Problem Set 3
Questions

  1. Find the products AB and BA, if possible, or explain why the product does not exist when

          [ 2  6  1  3 ]        [ 1  0 -1  2  6 ]
  A = [ 0  4  2  5 ] ,  B = [ 4  2  2  4  4 ]
      [ 1  2  2  8 ]        [ 3  1  6  8 -2 ]
                            [ 5  3  1  3  1 ]

  1.  

    1. Determine, if possible, the inverse for each of the following matrices by using both the row reduction technique and the determinant / adjoint method.

    2. Where appropriate also find the product of the inverse and the matrix B.

    1.     U = [  3 -2 ] ,   B = [ 1 ]
        [ -7  5 ]         [ 3 ]

    2.         [ -1  2  4 ]         [  .7  ]
    V = [  2 -4  1 ] ,   B = [  .85 ]
        [  5  6 -3 ]         [ -.53 ]

    3.         [  2  8  6 ]         [ 1 ]
    H = [  3  4  0 ] ,   B = [ 1 ]
        [  0 16 18 ]         [ 1 ]

  1. Another matrix inverse technique.
    By using only the properties of matrix multiplication and the inverse

    1. Find A-1 when A2 - 2A = 5I, and 

          A = [ 1 2 ]
        [ 3 1 ]

    2. Find A-1 when A3 - 2A2 - A - 6I = 0, and 

              [  1  3  2 ]
    A = [ -1 -2  1 ]
        [  0 -1  3 ]

  1. The components in a particular electronic circuit can be described as being in one of three states: “low”, “medium” or “high”.

    After each cycle of operation, the following changes occur:
    20% of all components that were in a “low” state are now in a “medium” state.
    10% of all components that were in a “low” state are now in a “high” state.
    20% of all components that were in a “medium” state are now in a “low” state.
    10% of all components that were in a “medium” state are now in a “high” state.
    10% of all components that were in a “high” state are now in a “low” state.
    10% of all components that were in a “high” state are now in a “medium” state.
    All other components do not change state.

    If, before the first cycle,
          80% of all components were in a “low” state,
          20% of all components were in a “medium” state and
          no components were in a “high” state,
    then find the proportions of components that are in each of the three states

    1. after the first cycle;
    2. after the second cycle;
    3. in the “steady state” (after infinitely many cycles).

    Notes on this question:

    A transition matrix   "A"   can be used to represent the probability that a component in each state will, after one cycle, be in each of the three possible states:

                          From:
                 low  medium  high
           low  [ .7    .2     .1 ]
   To:  medium  [ .2    .7     .1 ]  =  A
          high  [ .1    .1     .8 ]

    Note how the entries in each column must add up to 1.

    The initial proportions of components in each state can also be represented by a matrix (a column vector) "x0": 

               low  [ .8 ]
        medium  [ .2 ]  =  x
          high  [ 0  ]      0

    Again the entries in each column must add up to 1.

    The proportions of components in each state after one cycle can be represented by the matrix equation

    x1 = A x0

    In the same way,

    x2 = A x1 = A (A x0) = A2 x0

    A vector   "x"   in the steady state is unchanged after any cycle, so that

    x = A x       Þ     (A - I)x = 0

    From the solution of this linear system (together with the requirement that the components of vector   x   must add up to 1), the steady state solution can be found.

    This is an example of “Markov chains”, of some importance in the study of probability.

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