ENGR 1405 Engineering Mathematics 1

Faculty of Engineering and Applied Science
2000 Fall

Problem Set 4
Solutions

    [Note:   some browsers do not display symbols correctly.   Read
            “Þ” as “Þ”   (implies),
            “«” as “«”   (two-way arrow),
            “¬” as “¬”   (left arrow),
            “Ö” as “Ö”   (check mark),
            “-” as “-”   (minus)   and
            “ú” as “ú”   (set of real numbers).]

  1. Find the value of the determinant for each of the following matrices.
    [Note:   if your browser shows double bars around each matrix, then read this as single bars = determinant.]

    1.  
      10-1
      -357
      234

      Method 1:         by cofactor expansion along row 1:
      det A =

          |  1   0  -1 |
    | -3   5   7 |  =  +1 * |  5  7 | - 0 + (-1) * | -3  5 |
    |  2   3   4 |          |  3  4 |              |  2  3 |

                    =  (20 - 21) - (-9 - 10)
                    =  -1 + 19
                    =  18

      Method 2:         by row reduction to triangular form:

          |  1   0  -1 |
    | -3   5   7 | 
    |  2   3   4 |

      R2 ¬ R2 + 3 R1,
      R3 ¬ R3 - 2 R1:

          |  1   0  -1 |
 =  |  0   5   4 | 
    |  0   3   6 |

      R2 ¬ R2 / 5:

              |  1   0  -1 |
 =  5 * |  0   1  4/5| 
        |  0   3   6 |

      R3 ¬ R3 - 3 R2:

              |  1   0  -1 |
 =  5 * |  0   1  4/5|  =  5 * 1 * 1 * 18/5  =  18
        |  0   0 18/5|

      Therefore   det A = 18

    2.  
      10-1
      357
      234

      Method 1:         by cofactor expansion along row 1:
      det A =

          |  1   0  -1 |
    |  3   5   7 |  =  +1 * |  5  7 | - 0 + (-1) * |  3  5 |
    |  2   3   4 |          |  3  4 |              |  2  3 |

                    =  (20 - 21) - (9 - 10)
                    =  -1 + 1
                    =  0

      Method 2:         by row reduction to triangular form:

          |  1   0  -1 |
    |  3   5   7 | 
    |  2   3   4 |

      R2 ¬ R2 - 3 R1,
      R3 ¬ R3 - 2 R1:

          |  1   0  -1 |
    |  0   5  10 | 
    |  0   3   6 |
      At this point we notice that   R2 = (5/3) × R3
      Þ     det A = 0

    3.  
      12-14
      3523
      -1214
      5-123

      Cofactor expansion along any row or column will involve the evaluation of four 3×3 determinants, each of which will require the evaluation of three 2×2 determinants.
      Therefore we shall attempt a reduction to triangular form:
      det A =

          |  1   2  -1   4 |
    |  3   5   2   3 |   
    | -1   2   1   4 |   
    |  5  -1   2   3 |

      R2 ¬ R2 - 3 R1,
      R3 ¬ R3 + R1,
      R4 ¬ R4 - 5 R1
      (and extract the common factor found in row 3):

          |  1   2  -1   4 |         |  1   2  -1   4 |
 =  |  0  -1   5  -9 |  =  4 * |  0  -1   5  -9 |
    |  0   4   0   8 |         |  0   1   0   2 |
    |  0 -11   7 -17 |         |  0 -11   7 -17 |

      R2 « R3:

                |  1   2  -1   4 |
 =   -4 * |  0   1   0   2 |
          |  0  -1   5  -9 |
          |  0 -11   7 -17 |

      R3 ¬ R3 + R2,
      R4 ¬ R4 + 11 R2:

                |  1   2  -1   4 |          |  1   2  -1   4 |
 =   -4 * |  0   1   0   2 |  = -20 * |  0   1   0   2 |
          |  0   0   5  -7 |          |  0   0   1 -7/5|
          |  0   0   7   5 |          |  0   0   7   5 |

      R4 ¬ R4 - 7 R3:

                |  1   2  -1   4 |
 =  -20 * |  0   1   0   2 |  
          |  0   0   1 -7/5|
          |  0   0   0 74/5|

 =  -20 * 1 * 1 * 1 * 74/5  =  -296

      Therefore   det A = -296

        Application Problems

  1. A box has co-terminal sides represented by the vectors

    v1 = i + 2 j - 3 k,       v2 = i - 4 j + 2 k,       v3 = 3 i + 2 j + 5 k

    Find the volume of the box by evaluating the determinant of the matrix (V1V2V3),
    where Vi is the column matrix representation of the vector vi.

    Volume of box = abs(det(A)),   where   det(A) = det(V1V2V3) = 

        |  1  1  3 |
    |  2 -4  2 |
    | -3  2  5 |

    Method 1:         by cofactor expansion along row 1:
    det A =

        1 * | -4  2 | -  1 * |  2  2 |  + 3 * |  2 -4 |
        |  2  5 |        | -3  5 |        | -3  2 |

    =  (-20 - 4) - (10 + 6) + 3(4 - 12)
    =  -24 - 16 - 24
    =  -64

    Method 2:         by row reduction to triangular form:

        |  1  1  3 |
    |  2 -4  2 |
    | -3  2  5 |

    R2 ¬ R2 - 2 R1,
    R3 ¬ R3 + 3 R1:

        |  1  1  3 |         |  1  1  3 |
 =  |  0 -6 -4 |  = -6 * |  0  1 2/3|
    |  0  5 14 |         |  0  5 14 |
    R3 ¬ R3 - 5 R1:
             |  1  1   3 |
 =  -6 * |  0  1  2/3|
         |  0  0 32/3|

 =  -6 * 1 * 1 * 32/3  =  -64

    Therefore   volume of box = | -64 | = 64 (in the appropriate units)

  1. Use matrix determinants to find the polynomial that passes through a given set of points.   To find the polynomial of order "n" that passes through "n+1" given points we must evaluate the determinant of an (n+1)×(n+1) matrix.   For example, for the set of points given by
    S   =   { (x, y) | (1, -1), (3, 5) } ,
    we evaluate
    x0x1 y
    11-1
    135

    to find, in simplified form, the equation   y = 3x - 4.
    Find the appropriate polynomial equation for each of the following sets.

    1. S   =   { (x, y) | (-1, 5), (0, 2), (2, -8) }

      The required determinant is
      x0x1 x2y
      1-1 (-1)25
      10 022
      12 22-8

      The presence of the zeroes in row 3 suggests a cofactor expansion.
      In order to make the collection of coefficients easier, expand initially along row 1:

          | x^0 x^1 x^2  y |     | -1  1  5 |     | 1  1  5 |
    |  1  -1   1   5 |  =  |  0  0  2 |  -  | 1  0  2 | x
    |  1   0   0   2 |     |  2  4 -8 |     | 1  4 -8 |
    |  1   2   4  -8 |

       | 1 -1  5 |  2     | 1 -1  1 |
    +  | 1  0  2 | x   -  | 1  0  0 | y  =  0
       | 1  2 -8 |        | 1  2  4 |

      The coefficient of x0, by cofactor expansion along row 2, is

          | -1  1  5 |                   
  + |  0  0  2 |  = - 0  +  0  -  2 | -1  1 |
    |  2  4 -8 |                    |  2  4 |

    =  -2 * (-4 - 2)  =  +12

      The coefficient of x1, by cofactor expansion along row 2, is

          |  1  1  5 |                   
  - |  1  0  2 |  = -( -1 |  1  5 |  +  0  -  2 |  1  1 | )
    |  1  4 -8 |     (    |  4 -8 |             |  1  4 | )

    =  +1 * (-8 - 20) + 2 * (4 - 1)  =  -28 + 6  =  -22

      The coefficient of x2, by cofactor expansion along row 2, is

          |  1 -1  5 |                   
  + |  1  0  2 |  =  -1 | -1  5 |  +  0  -  2 |  1 -1 |
    |  1  2 -8 |        |  2 -8 |             |  1  2 |

    =  -1 * (8 - 10) - 2 * (2 + 1)  =  2 - 6  =  -4

      The coefficient of y, by cofactor expansion along row 2, is

          |  1 -1  1 |                   
  - |  1  0  0 |  = - ( -1 | -1  1 |  +  0  -  0 )
    |  1  2  4 |      (    |  2  4 |             )

    =  +1 * (-4 - 2)  =  -6

      The required polynomial is therefore
              12 - 22 x - 4 x2 - 6 y = 0
      Þ       6 - 11 x - 2 x2 - 3 y = 0
      Þ

      y = (-2 x2 - 11 x + 6) / 3

      As a check, substitute each point in turn into this equation to show that this parabola does pass through all three points:
      6 - 11 x - 2 x2 - 3 y = 0 :
      6 - 11×(-1) - 2×(-1)2 - 3×(5) = 6 + 11 - 2 - 15 = 0.         Ö
      6 - 11×(0) - 2×(0)2 - 3×(2) = 6 + 0 + 0 - 6 = 0.         Ö
      6 - 11×(2) - 2×(2)2 - 3×(-8) = 6 - 22 - 8 + 24 = 0.         Ö

    2. S   =   { (x, y) | (-2, 8), (-1, 3), (1, -5), (2, 9) }

      The required determinant is
      x0x1 x2x3 y
      1-2 (-2)2 (-2)3 8
      1-1 (-1)2 (-1)3 3
      11 12 13 -5
      12 22 23 9

      There are no zero entries in this 5×5 determinant.
      Therefore carry out a partial row reduction to introduce some zero entries:
      R3 ¬ R3 - R2,
      R4 ¬ R4 - R2,
      R5 ¬ R5 - R2:
      x0x1 x2x3 y
      1-2 4 -8 8
      01 -3 7 -5
      03 -3 9 -13
      04 0 16 1

      R2 ¬ R2 + 2 R3,
      R4 ¬ R4 - 3 R3,
      R5 ¬ R5 - 4 R3:
      x0x1 x2x3 y
      10 -2 6 -2
      01 -3 7 -5
      00 6 -12 2
      00 12 -12 21

      = 2 × 3 ×
      x0x1 x2x3 y
      10 -2 6 -2
      01 -3 7 -5
      00 3 -6 1
      00 4 -4 7

      We could continue with the row reduction, but the cofactor expansion is now easier than it would have been earlier.   Row reduction much beyond this point will result in the introduction of non-integer entries.

      [The reduced echelon form of rows 2 to 5 is
      x0x1 x2x3 y
      10 0 0 -25/6
      01 0 0 -65/12
      00 1 0 19/6
      00 0 1 17/12
      , from which
      ([50 + 65 x - 38 x2 - 17 x3] / 12) + y = 0. ]

      In order to make the collection of coefficients easier, expand initially along row 1:

          | x^0 x^1 x^2 x^3  y |
    |  1   0  -2   6  -2 |
    |  0   1  -3   7  -5 |  =  0  = 
    |  0   0   3  -6   1 |
    |  0   0   4  -4   7 |

    |  0  -2   6  -2 |     |  1  -2   6  -2 |  1
    |  1  -3   7  -5 |  -  |  0  -3   7  -5 | x
    |  0   3  -6   1 |     |  0   3  -6   1 |
    |  0   4  -4   7 |     |  0   4  -4   7 |

    |  1   0   6  -2 |  2    |  1   0  -2  -2 |  3
  + |  0   1   7  -5 | x  -  |  0   1  -3  -5 | x
    |  0   0  -6   1 |       |  0   0   3   1 |
    |  0   0  -4   7 |       |  0   0   4   7 |

    |  1   0  -2   6 |
  - |  0   1  -3   7 | y  =  0
    |  0   0   3  -6 |
    |  0   0   4  -4 |

      The coefficient of x0, by cofactor expansion down column 1, is

          |  0  -2   6  -2 |             | -2   6  -2 |
  + |  1  -3   7  -5 |  =  0  -  1 |  3  -6   1 |  +  0  -  0
    |  0   3  -6   1 |             |  4  -4   7 |
    |  0   4  -4   7 |  

  =  -( -2 | -6  1 | - 3 |  6 -2 | + 4 |  6 -2 | )
      (    | -4  7 |     | -4  7 |     | -6  1 | )

  =  -( -2*(-42 + 4) - 3*(42 - 8) + 4*(6 - 12) )

  =  -( 76 - 102 - 24)  =  +50

      The coefficient of x1, by cofactor expansion down column 1, is

          |  1  -2   6  -2 |       | -3   7  -5 |
  - |  0  -3   7  -5 |  =  1 |  3  -6   1 |  -  0  +  0  -  0
    |  0   3  -6   1 |       |  4  -4   7 |
    |  0   4  -4   7 |

  =  -( -3 | -6  1 | - 3 |  7 -5 | + 4 |  7 -5 | )
      (    | -4  7 |     | -4  7 |     | -6  1 | )

  =  -( -3*(-42 + 4) - 3*(49 - 20) + 4*(7 - 30) )

  =  -( 114 - 87 - 92 ) = +65

      The coefficient of x2, by cofactor expansion down column 1 (then column 1 of the 3×3 determinant), is

          |  1   0   6  -2 |       |  1   7  -5 |
  + |  0   1   7  -5 |  =  1 |  0  -6   1 |  =  1 | -6   1 |
    |  0   0  -6   1 |       |  0  -4   7 |       | -4   7 |
    |  0   0  -4   7 |

  = 1(-42 + 4) = -38

      The coefficient of x3, by cofactor expansion down column 1 (then column 1 of the 3×3 determinant), is

          |  1   0  -2  -2 |       |  1  -3  -5 |
  - |  0   1  -3  -5 |  = -1 |  0   3   1 |  = -1 |  3   1 |
    |  0   0   3   1 |       |  0   4   7 |       |  4   7 |
    |  0   0   4   7 |

  = -1(21 - 4) = -17

      The coefficient of y, by cofactor expansion down column 1 (then column 1 of the 3×3 determinant), is

          |  1   0  -2   6 |       |  1  -3   7 |
  + |  0   1  -3   7 |  =  1 |  0   3  -6 |  =  1 |  3  -6 |
    |  0   0   3  -6 |       |  0   4  -4 |       |  4  -4 |
    |  0   0   4  -4 |

  = 1(-12 + 24) = +12

      The required polynomial is therefore
              50 + 65 x - 38 x2 - 17 x3 + 12 y = 0
      Þ

      y = (17 x3 + 38 x2 - 65 x - 50) / 12

      As a check, substitute each point in turn into this equation to show that this curve does pass through all four points:
      50 + 65 x - 38 x2 - 17 x3 + 12 y = 0:
      50 + 65(-2) - 38(-2)2 - 17(-2)3 + 12(8) = 50 - 130 - 152 + 136 + 96 = 0.         Ö
      50 + 65(-1) - 38(-1)2 - 17(-1)3 + 12(3) = 50 - 65 - 38 +17 +36 = 0.         Ö
      50 + 65(1) - 38(1)2 - 17(1)3 + 12(-5) = 50 + 65 - 38 - 17 - 60 = 0.         Ö
      50 + 65(2) - 38(2)2 - 17(2)3 + 12(9) = 50 + 130 - 152 - 136 + 108 = 0.         Ö

  1. One of the problems associated with online commerce and communication is related to the fact that some unauthorized user may be able to access online information.   Because of this, encryption has become very important.   Below we consider a very simplified method of encryption.   First, each letter of the alphabet is numbered according to its place in the alphabet, and a space between words is assigned the value zero.   Second, introduce an invertible encryption matrix, E.   Because this is a simplified approach, choose dim(E) = 2×2.   Third, convert the message to numeric format using the assignment values in step one.   Fourth, break the numeric string message into groups of two numbers and rewrite each numeric pair as a 2×1 column matrix.   If the last column matrix is incomplete, then add a space.   Fifth, left-multiply each column matrix by the encryption matrix, E.   Sixth, reconvert to a numeric string message and send.   Seventh, begin to decode the message by using the fourth step above.   Eighth, left-multiply each column matrix by the inverse of the matrix E.   Ninth, reconvert to a numeric string and finish decoding by using the first step.

    Let 
            [ 2  1 ]
    E = [      ]  .
        [ 3  2 ]
    1. Find the encrypted numeric string for the following message:
              "this is just a test message"
      The associated numeric string for the above is as follows:

      20,8,9,19,0,9,19,0,10,21,19,20,0,1,0,20,5,19,20,0,13,5,19,19,1,7,5,0

      The column vectors to be multiplied by the encryption matrix E are

          [ 20 ]   [  9 ]   [  0 ]
    [  8 ] , [ 19 ] , [  9 ] , etc.

      To save time, augment all fourteen 2×1 matrices into a single 2×14 matrix to be pre-multiplied by E.

      [ 2  1 ] [ 20  9  0 19 10 19  0  0  5 20 13 19  1  5 ]
[ 3  2 ] [  8 19  9  0 21 20  1 20 19  0  5 19  7  0 ]

      [ 48 37  9 38 41 58  1 20 29 40 31 57  9 10 ]
    = [ 76 65 18 57 72 97  2 40 53 60 49 95 17 15 ]

      The encrypted set of numbers is therefore

      48, 76, 37, 65, 9, 18, 38, 57, 41, 72, 58, 97, 1, 2, 20, 40, 29, 53, 40, 60, 31, 49, 57, 95, 9, 17, 10, 15

      See also the associated Excel spreadsheet file.

    2. Determine the inverse of the encryption matrix and decode the following message.

      65,105,42,63,51,79,41,64,19,38,45,69,11,19,57,95,33,60,24,36

      det E = 4 - 3 = 1       Þ     E-1 = adj(E) / 1.

            -1    [ 2 -1 ]
==>  E   =  [-3  2 ]

      The column vectors to be multiplied by the decryption matrix E-1 are

          [  65 ]   [  42 ]   [  51 ]
    [ 105 ] , [  63 ] , [  79 ] , etc.

      As before, to save time, augment all ten 2×1 matrices into a single 2×10 matrix to be pre-multiplied by E-1.

      [ 2 -1 ] [  65  42  51  41  19  45  11  57  33  24 ]
[-3  2 ] [ 105  63  79  64  38  69  19  95  60  36 ]

       [ 25 21 23 18  0 21  3 19  6 12 ]
    =  [ 15  0  5  5 19  3  5 19 21  0 ]

      The decrypted set of numbers is therefore
      25, 15, 21, 0, 23, 5, 18, 5, 0, 19, 21, 3, 3, 5, 19, 19, 6, 21, 12, 0
      The corresponding text (ignoring the trailing space) is

      "you were successful"

      See also the associated Excel spreadsheet file.

  1. In the circuit illustrated below, determine the values of all of the currents:

    Circuit diagram

    Using Kirchoff’s laws:
    No net voltage loss around the upper loop:     2 - 4 I1 - 3 I2 = 0
    No net voltage loss around the lower loop:     - 4 + 3 I2 + 4 I3 = 0
    Charge conservation (flow in = flow out at each node):     I1 + I3 = I2

    The linear system is

    Program to handle the Arithmetic in the Row Reduction of a Linear System
VisualBASIC version (c)1999, Glyn George.

Linear System:

Row 1:    1/1       -1/1        1/1        0/1    
Row 2:    4/1        3/1        0/1        2/1    
Row 3:    0/1        3/1        4/1        4/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  4 / 1 times row 1 from row 2 .

Linear System:

Row 1:    1/1       -1/1        1/1        0/1    
Row 2:    0/1        7/1       -4/1        2/1    
Row 3:    0/1        3/1        4/1        4/1    

-----------------------------------------------------------------------------
Selected row operation:    Divide row 2 by  7 / 1 .

Linear System:

Row 1:    1/1       -1/1        1/1        0/1    
Row 2:    0/1        1/1       -4/7        2/7    
Row 3:    0/1        3/1        4/1        4/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  3 / 1 times row 2 from row 3 .

Linear System:

Row 1:    1/1       -1/1        1/1        0/1    
Row 2:    0/1        1/1       -4/7        2/7    
Row 3:    0/1        0/1       40/7       22/7    

-----------------------------------------------------------------------------
Selected row operation:    Divide row 3 by  40 / 7 .

Linear System:

Row 1:    1/1       -1/1        1/1        0/1    
Row 2:    0/1        1/1       -4/7        2/7    
Row 3:    0/1        0/1        1/1       11/20   

-----------------------------------------------------------------------------
Selected row operation:    Subtract -1 / 1 times row 2 from row 1 .

Linear System:

Row 1:    1/1        0/1        3/7        2/7    
Row 2:    0/1        1/1       -4/7        2/7    
Row 3:    0/1        0/1        1/1       11/20   

-----------------------------------------------------------------------------
Selected row operation:    Subtract  3 / 7 times row 3 from row 1 .

Linear System:

Row 1:    1/1        0/1        0/1        1/20   
Row 2:    0/1        1/1       -4/7        2/7    
Row 3:    0/1        0/1        1/1       11/20   

-----------------------------------------------------------------------------
Selected row operation:    Subtract -4 / 7 times row 3 from row 2 .

Linear System:

Row 1:    1/1        0/1        0/1        1/20   
Row 2:    0/1        1/1        0/1        3/5    
Row 3:    0/1        0/1        1/1       11/20   

-----------------------------------------------------------------------------

Program execution terminated.

    Thus the currents are (I1, I2, I3) = (1, 12, 11) / 20 = (0.05, 0.60, 0.55) ampère.

  1. Another application for matrices is to determine numerical approximations.   In this problem we are looking for the numerical approximation for the steady-state temperature distribution in a flat plate when the temperature is known at every point on the boundary of the plate.   We will be considering only the simplest geometry, a rectangular plate.   A visual representation of such a plate is given below.

    Values on grid, row by row, from top edge downwards: 0, 18.75, 25, 18.75, 0 0, g, h, k, 0 0, d, e, f, 0 0, a, b, c, 0 0, 20, 10, 20, 0

    Determine the temperature at each of the points (nodes)   a, b, c, d, e, f, g, h, k   by solving the following system of linear equations.

    4g = 0 + d + h + 18.75     4h = g + e + k + 25 4k = h + f + 0 + 18.75    
    4d = 0 + a + e + g 4e = d + b + f + h 4f = e + c + 0 + k
    4a = 0 + 20 + b + d 4b = a + 10 + c + e     4c = b + 20 + 0 + f

    Normally to find the value of the temperature at these nine points we would have to solve a system of nine equations in nine unknowns.   However because of the symmetry of the boundary temperatures around the vertical mid-line in this particular problem we have the additional set of equations

    c = a f = d k = g

    This last set of equations when properly used will reduce the set of equations to be solved to a set of six equations in six unknowns.

    Replace   c, f and k   where they occur.
    The linear system then reduces to

    -d + 4g - h = 18.75     (1)
    -e - 2g + 4h = 25 (2)
    -d + 4g - h = 18.75     (3)
    -a + 4d - e - g = 0 (4)
    - b - 2d + 4e - h = 0 (5)
    -a + 4d - e - g = 0 (6)
    4a - b - d = 20 (7)
    -2a + 4b - e = 10 (8)
    4a - b - d = 20 (9)

    However, equations (3), (6) and (9) are clearly redundant, as they are identical to equations (1), (4) and (7) respectively.   Upon eliminating the redundant equations, we are left with a linear system of six equations in six unknowns.

    The lengthy calculations of the linsys.exe program are in a separate file.
    The solution vector is

    (a, b, c, d, e, f, g, h, k) = (4065, 4178, 4065, 3122, 4102, 3122, 4321, 5986, 4321) / 448