MATH 2050 Linear Algebra (Section 4)

Department of Mathematics and Statistics
2009 Winter

Term Test 1 - Solutions

  1. For matrices   A = [ 2 1 ; 0 -1 ] ; B = [ 6 0 -2 ; 3 1 4 ] find, where possible, the products   AB, BA and BTAT and the matrix   A-1.

    AB = [ 15 1 0 ; -3 -1 -4 ]

    Matrix   B   has dimensions (2×3).
    Matrix   A   has dimensions (2×2).
    The number of columns in   B   does not match the number of rows in   A .
    Therefore   BA   does not exist.

    BT AT = [ 15 -3 ; 1 -1 ; 0 -4 ]
    [Note that there is no need to evaluate the product BTAT directly!]

    inverse of A = [ 1/2 1/2 ; 0 -1 ]
    or one may use Gaussian elimination to carry   [ A | I ]   to   [ I | A-1 ].
    or one may quote   [inverse of a upper triangular 2x2 matrix]

  1. Find, where possible, conditions on k, such that the linear system
              x + 2z = 0 ; kx + y + 2z = 3 ; (k+1)y + 2z = 6k
    has

    1.   no solution;
    2.   one solution;
    3.   infinitely many solutions.

    Represent the linear system by an augmented matrix:
    [A|B] = [ 1 0 2 | 0 ] [ k 1 2 | 3 ] [ 0 k+1 2 | 6k ]
    Use row operations to carry the matrix to row-echelon form:
    [row operation]
    [row operation]
    [ 1 0 2 | 0 ] [ 0 1 2(1-k) | 3 ] [ 0 0 2k^2 | 3(k-1) ]
    If   k = 0   then the system becomes
    [ 1 0 2 | 0 ] [ 0 1 2 | 3 ] [ 0 0 0 | 1 ]
    rank A = 2   but   rank (A|b) = 3     implies the system is inconsistent (no solution).
    Otherwise the row-echelon form becomes
    [ 1 0 2 | 0 ] [ 0 1 2(1-k) | 3 ] [ 0 0 1 | 3(k-1)/(2k^2) ]
    rank A = rank (A|b) = n = 3     implies the system has a unique solution.

    There are no other possibilities.   Therefore the system has

      no solution when   k = 0   and  
      a unique solution otherwise.  

  1. Why is the linear system
              x + 2y + z = 0 ; 2x - y + z = 0
    guaranteed to have infinitely many solutions?

      It is an under-determined homogeneous linear system.  

    Solve this linear system.

    Use row operations to carry the homogeneous system to reduced row-echelon form:
    [ 1 2 -1 ] [ 2 -1 1 ]
    [ 1 0 1/5 ] [ 0 1 -3/5 ]
    z   is a free parameter; call it the real number   t.
    Reading the reduced row-echelon form:
    x + z/5 = 0 ; y - 3z/5 = 0 ; z = t
    If one re-defines the free parameter as   z = 5s, then the solution may also be expressed in the simpler form
                (x,y,z) = s(-1,3,5)

    Verify that   (x,y,z) = (1,2,1)   is a solution to the related linear system
              x + 2y + z = 4 ; 2x - y + z = 1
    and hence write down its complete solution.

    1 + 2(2) - 1 = 4 tick mark
    2(1) - 2 + 1 = 1 tick mark
    The complete solution to the inhomogeneous system is the sum of any one particular solution to the inhomogeneous system with the most general solution to the homogeneous system:
    X = Xh + Xp

  1. Show that   ATA   is symmetric for all matrices   A .

    First note that   ATA   is a square matrix for all matrices   A :
    Let the dimensions of   A   be (m×n), then the dimensions of   AT   are (n×m).
    The number of columns in   AT   matches the number of rows in   A   so that
    the product   ATA   is defined and has dimensions   (n×n), which is square.

    For any pair of matrices   A, B   for which the product   BA   is defined,   (BA)T = AT BT.
    Setting   B = AT:     (AT A)T = AT A
    Therefore   ATA   is symmetric for all matrices   A .

      BONUS QUESTION:

  1. Find conditions on   a, b, c, d, such that the matrix A = [ a b ; c d ] commutes with the matrix B = [ 1 2 ; 3 0 ]

    AB = [ a+3b 2a ; c+3d 2c ]
    BA = [ a+2c b+2d ; 3a 3b ]
    Setting BA = AB
    This generates a system of four simultaneous equations:
    b = 2(a-d) , c = 3(a-d)
    (which also satisfies the other two equations).
    Therefore the most general matrix that commutes with   B = [ 1 2 ; 3 0 ] is

    [ a 2(a-d) ; 3(a-d) d ]

    It is easy to verify that
    AB = BA = [7a-6d 2a ; 3a 6a-6d ]

    There are four other valid equivalent alternative answers, depending on which two of   a, b, c, d   are taken as the free parameters.   [The only invalid pair is (b, c).]
            [ a b ; 3b/2 a-b/2 ] ; [ a 2c/3 ; c a-c/3 ] ; [ d+b/2 b ; 3b/2 d ] ; [ d+c/3 2c/3 ; c d ]

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    Created 2009 01 31 and most recently modified 2009 02 14 by Dr. G.H. George