MATH 2050 Linear Algebra

(Section 4)
2009 Winter

Assignment 5   -   Solutions

[Section 3.1   Determinants]
  1. Find the determinant and the inverse of the matrix
    A = 
[  8 -11 ]
[ -2   3 ]


    det A = 2

    inverse of A = 
[  3/2 11/2 ]
[   1    4  ]


  1. Find the determinant of the matrix
    R = 
[ cos theta  -sin theta  0 ]
[ sin theta   cos theta  0 ]
[     0          0       1 ]


    A row operation could be used to reduce this determinant to triangular form, but a cofactor expansion along the last row or column is much easier:
    [expansion along last row or column]
    det R = 1  for all theta

    det R = 1

    [All matrices representing pure rotations have a determinant of 1.]


  1. Find the determinant of the matrix
    C = 
[ 0  1  2  3 ]
[ 0  0  2  0 ]
[ 3 10 12 20 ]
[ 0  0  0  4 ]


    Use row operations to transform the determinant into upper triangular form:
    [swap rows]
    [swap rows]
    OR one may use a cofactor expansion, starting with any of column 1 or row 2 or row 4.
    Expanding along row 4:
    [expand along row 4]
    [expand along row 2]

    det C = 24


  1. Find the determinant of the matrix
    D = 
[  1  3  5  1  1 ]
[  2  1  4  2  2 ]
[  0  4  3  0  3 ]
[ -1 -6  2 -1  2 ]
[  2  8 -1  2  1 ]


    Columns 1 and 4 of this determinant are identical.
    Immediately we have

    det D = 0


  1. [Textbook, page 114, exercises 3.1, question 8(b)]
    Show that   | 2a+p  2b+q  2c+r |         | a b c |
| 2p+x  2q+y  2r+z |   =   9 | p q r |
| 2x+a  2y+b  2z+c |         | x y z |


    Try to eliminate the " 2a " term from element (1,1) by the row operation R1 – 2 R3.
    However, this introduces an unwanted new term " –4x ", which can be eliminated in turn by the row operation R1 + 4 R2.
    The two row operations together leave a multiple of " p " alone in the (1,1) element.
    [adjusting top row]
    [adjusting top row]
    [adjusting middle row]
    [adjusting bottom row]
    [swapping rows to obtain the desired form]
    Therefore
    | 2a+p  2b+q  2c+r |         | a b c |
| 2p+x  2q+y  2r+z |   =   9 | p q r |
| 2x+a  2y+b  2z+c |         | x y z |


  1. F = 
[ 1  1  0 ]
[ 2  1  c ]
[ c  0  1 ]
    Find the value(s) of   c   for which the matrix   F   is singular.


    Use a cofactor expansion along row 3 (column 3 also works well):
    det F = ...
    det F = (c-1)(c+1)
    OR one may reduce the determinant to triangular form:
    det F = ...
    det F = (c-1)(c+1)
    F   is singular if and only if   det F = 0 ; that is

    c = ±1


  1. If   det X = 2   and   det Y = 3 , then calculate the value of
    det (X^2 Y^(-1) X^T Y^3)   where possible.
    Under what circumstances does   (X^2 Y^(-1) X^T Y^3)   not exist?.


    Provided the two matrices   X   and   Y   are of the same dimensions as each other, all of the products in   (X^2 Y^(-1) X^T Y^3)   exist.
    The fact that   det Y   is not zero guarantees that   Y -1   exists.
    Using the results
    det (AB) = det A det B ,  det AT = det A  
and   det(A^k) = (det A)^k
    det (X^2 Y^(-1) X^T Y^3) = (det X)^3 (det Y)^2

    det (X^2 Y^(-1) X^T Y^3) = 72

    The product   (X^2 Y^(-1) X^T Y^3)   does not exist if and only if the square matrices   X   and   Y   have different dimensions.


  1. The general 3×3 skew-symmetric matrix is
    K = 
[  0  a  b ]
[ -a  0  c ]
[ -b -c  0 ]   (where   a, b, c   are any real numbers).
    Find   det K.


    The determinant may be calculated by a cofactor expansion along any row or down any column.
    Arbitrarily choosing row 1:
    det K = +0 - a(0 + bc) + b(ac - 0) = 0
    OR one may reduce the determinant to triangular form:
    [row operations]
    [row operations]
    There is a row of all zero entries.   Therefore, for all values of   a, b, c ,

    det K = 0

    and all 3×3 skew-symmetric matrices are singular.
    Note that the same is not true for 2×2 skew-symmetric matrices.


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      Created 2009 02 03 and most recently modified 2009 02 03 by Dr. G.H. George