MATH 2050 Linear Algebra

(Section 4)
2009 Winter

Assignment 2   –   Solutions

  1. x1 + 2x3 + 3x5 = 0 x1 + 4x3 + 2x4 = 1 2x1 + 4x3 + 7x5 = 2

    [A|b] = [ 1 0 2 0 3 | 0 ] [ 1 0 4 2 0 | 1 ] [ 2 0 4 0 7 | 2 ]
    row-echelon form = [ 1 0 2 0 3 | 0 ] [ 0 0 1 1 -3/2 | 1/2 ] [ 0 0 0 0 1 | 2 ]
    reduced row-echelon form = [ 1 0 0 -2 0 | -13 ] [ 0 0 1 1 0 | 7/2 ] [ 0 0 0 0 1 | 2 ]
    The output from the linear system reduction program is available here.
    r = rank(A) = rank(A|b) = 3   but   n = 5
    p = n – r = 5 – 3 = 2
    There is a two-parameter family of solutions.
    x1, x3 and x5 are leading variables and x2 and x4 are the free parameters (call them s and t).
    Reading the reduced row-echelon form:
    x1 - 2 x4 = -13 x3 + x4 = 7/2 x5 = 2

    Therefore there are

      infinitely many solutions:
      x = (-13, 0, 7/2, 0, 2) + (0,1,0,0,0)s + (2,0,-1,1,0)t  

  1. x - 2y + z = 3 2x - 4y + 3z = 7 -5x + 10y - 2z = -12 3x - 6y - z = 5

    [A|b] = [ 1 -2 1 | 3 ] [ 2 -4 3 | 7 ] [-5 10 -2 | -12 ] [ 3 -6 -1 | 5 ]
    reduced row-echelon form = [ 1 -2 0 | 2 ] [ 0 0 1 | 1 ] [ 0 0 0 | 0 ] [ 0 0 0 | 0 ]
    The output from the linear system reduction program is available here.
    r = rank(A) = rank(A|b) = 2   but   n = 3
    p = n – r = 3 – 2 = 1
    There is a one-parameter family of solutions.
    x and z are leading variables and y is a free parameter.
    Reading the reduced row-echelon form:
    1 x – 2y + 0 z = 2
    0 x – 0y + 1 z = 1
    0 x – 0y + 0 z = 0   and
    0 x – 0y + 0 z = 0
    Let   y = t   then   x = 2 + 2t.   The third and fourth equations are redundant (they do not add any information not already provided by the first two equations).   Therefore there are

      infinitely many solutions:
      (x,y,z) = (2,0,1) + (2,1,0)t  

  1. x - 2y + z = 3 2x - 3y + 3z = 7 -5x + 10y - 2z = -12 3x - 6y - z = 5

    [row reduction]
    [row reduction]
    reduced row-echelon form = [ 1 0 0 | 2 ] [ 0 1 0 | 0 ] [ 0 0 1 | 1 ] [ 0 0 0 | 0 ]
    The output from the linear system reduction program is available here.
    r = rank(A) = rank(A|b) = n = 3
    There is therefore a

      unique solution,  
      (x, y, z) = (2, 0, 1)  

    [An over-determined system can have infinitely many solutions, a unique solution or no solutions.   An under-determined system cannot have a unique solution (only infinitely many solutions or none at all).]

  1. For the homogeneous linear system
          x + y - z = 0 ky + z = 0 x + y + kz = 0
    find all values of k for which there are non-trivial solutions.

    [reduction to triangular form]
    If and only if every column of the row-echelon form of matrix A has a leading one, then   rank A = n   and the homogeneous system has a unique solution (the trivial solution).   This will happen if and only if k not=0 and k + 1 not= 0
    Therefore non-trivial solutions exist for

      k = 0   or   k = –1   only  

  1. When any of the three types of row operation is applied to a homogeneous linear system, the resulting linear system is also homogeneous.   Prove that this statement is true.

    The augmented matrix for a linear system is   [A|b].
    If and only if the linear system is homogeneous, the vector b is the zero vector.
    The final column of the augmented matrix of any homogenoeus linear system is therefore a column of zeros only.

    Row operation I (swap two rows) will interchange two zeros with each other in the final column, no matter which two rows are chosen.
    Row operation II (multiply a row by a non-zero constant) will leave the zero entries in the final column unchanged.   The product of 0 with any finite number is 0.
    Row operation III (add a multiple of one row to a different row) will add a multiple of one zero to another zero in the final column.

    Thus the application of any row operation to a homogeneous linear system will leave the final column of zeros unchanged (and will leave any other entirely zero column unchanged also). The resulting system is therefore also homogeneous.

    In the reduction of a homogeneous linear system, one may therefore omit the final column of zeros safely.

  1. Find, where possible, conditions on a and b, such that the system has no solution, one solution or infinitely many solutions.
                  x + ay + bz = 1 ax + 4y + z = 4b 2x - 4y + 2bz = b

    Reducing the system to row-echelon form:
    [A|b] = [ 1 a b | 1 ] [ 0 4-a^2 1-ab | 4b-a ] [ 0 -4-2a 0 | b-2 ]

    Special case   a = 2:
    [A|b] = [ 1 2 b | 1 ] [ 0 0 1-2b | 4b-2 ] [ 0 -8 0 | b-2 ]
    [A|b] = [ 1 2 b | 1 ] [ 0 1 0 | (2-b)/8 ] [ 0 0 1-2b | 4b-2 ]
    If   b = 1/2 then the row-echelon form is
    [A|b] = [ 1 2 1/2 | 1 ] [ 0 1 0 | 3/16 ] [ 0 0 0 | 0 ]
    rank A = rank [A | b] = 2 but n = 3, so there is a one-parameter family of solutions.
    x and y are the leading variables and z is a free parameter.
    If b is not 1/2, then the row-echelon form is
    [A|b] = [ 1 2 b | 1 ] [ 0 1 0 | (2-b)/8 ] [ 0 0 1 | -2 ]
    rank A = rank [A | b] = n = 3, so there is a unique solution.

    Special case   a = –2:
    [A|b] = [ 1 -2 b | 1 ] [ 0 0 1+2b | 4b+2 ] [ 0 0 0 | b-2 ]
    Clearly this system will be inconsistent (leading one in the column for right side constants) unless b = 2.
    In the case b = 2, the echelon form is
    [A|b] = [ 1 -2 2 | 1 ] [ 0 0 1 | 2 ] [ 0 0 0 | 0 ]
    This time x and z are the leading variables and y is a free parameter, so that the system has a one-parameter family of solutions.

    When a2 is not 4, the row reduction continues:
    [row reduction]
    [row reduction]
    Two possibilities arise:
    If   ab = 1   then the row-echelon form is
    [row reduction]
    [A|b] = [ 1 a 2 | 1 ] [ 0 1 0 | * ] [ 0 0 0 | 1 ]
    which is an inconsistent system (no solutions).
    ab = 1 ==> divisor not zero
    If   ab is not 1   then the row-echelon form is
    [A|b] = [ 1 a b | 1 ] [ 0 1 * | * ] [ 0 0 1 | * ]
    which clearly has a unique solution.

    Combining all of these results together, the system has

      no solutions if   (ab = 1 and a not equal ±2) or (a = –2 and b not equal 2)
      a one-parameter family of solutions if   (a = –2 and b = 2) or (a = +2 and b = 1/2)  
      and a unique solution otherwise.

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      Created 2008 12 31 and most recently modified 2009 01 02 by Dr. G.H. George