MA-305 Homework 2 Solutions


  1. A homogeneous system of linear equations is a system in which the last column of the augmented matrix contains all zero entries, i.e. the right-hand side of each equation is zero. Explain why such a system always has a solution, and describe the geometric properties of the solution set.
    Solution:
    For any homogeneous system of n equations in (n unknowns, the point (0,0,...,0) (n elements) will always be a solution. When a homogeneous system has infinite solutions, the solutions will be a subspace of the larger space (Rn when the equations are over the real numbers). Geometrically speaking, the solution will be a point (the origin itself), a line passing through the origin, a plane passing through the origin, a hyperplane passing through the origin, or some other "flat" space containing the origin.

  2. Let

    A = 
    [ 2 -3 ] 
[ 4  1 ]
    Find
    1. 2A - 3A2
      Solution: 
      [  28 21 ]
[ -28 35 ]
    2. A4 - 2A2 + A - 4I2
      Solution: 
      [  -30 186 ]
[ -248  32 ]

  3. Let

    A = 
    [ 2 4 6 ]
[ 1 1 1 ]
    		andB = 
    [  3 -3  1 ]
[ -2  1 -1 ]
[  0  0  1 ]
    If possible, determine the following. If the calculation is not possible, explain why not.

    1. (AB)T
      Solution: 
      [ -2  1 ]
[ -2 -2 ]
[  4  1 ]
    2. ATBT
      Solution:
      This product is not defined, for # of columns of AT = 2 <> # rows of BT = 3

    3. BBT
      Solution: 
      [  19 -10  1 ]
[ -10   6 -1 ]
[   1  -1  1 ]
    4. AAT
      Solution: 
      [ 56 12 ]
[ 12  3 ]
    5. ATA
      Solution: 
      [  5  9 13 ]
[  9 17 25 ]
[ 13 25 37 ]
    6. BTAT
      Solution: 
      [ -2 1 ]
[ -2 2 ]
[  4 1 ]

  4. Consider Fibonacci's famous problem under different conditions. Suppose you start with one pair of immature (newborn) rabbits in a closed pen. If it takes three months for the rabbits to mature, and one additional month to produce one pair of offspring, how many rabbits are present in the pen after 15 months?
    (Hint: Use matrix multiplication to relate fi, the number of rabbits present after i months.)
    Solution:
    Let fi represent the number of rabbits present ofter i months. Then fn+1 = fn + fn-3. If we let
    F(n) = 
    [ fn-3 ] 
[ fn-2 ] 
[ fn-1 ] 
[ fn   ]
    be the vector containing the number of rabbits at times n-3, n-2, n-1, and n, then the number of rabbits present at time n+1 can be found by multiplying the matrix
    K = 
    [ 0 1 0 0 ] 
[ 0 0 1 0 ] 
[ 0 0 0 1 ] 
[ 1 0 0 1 ]
    times F(n). We see that
    F(3) = 
    [ 1 ] 
[ 1 ] 
[ 1 ] 
[ 1 ]
    Thus,
    F(15) = K12F(3) = 
    [ f12 ] 
[ f13 ] 
[ f14 ] 
[ f15 ]
    		=
    [ 26 ] 
[ 36 ] 
[ 50 ] 
[ 69 ]
    i.e. There are 69 rabbits present after 15 months.