Module Name: Mathematics for Management

Open University Malaysia

Faculty of Business and Management

BBMP 1103

Mathematics for Management

Name: Adam Khaleel

Tutor’ Name: Ali Shareef

Learning Centre: Villa College

Trimester:  September 2011

1. (a)  2A+B=(1 2 2 -2 1 0 )

      A−3B=(-4 -3 -2 -1 0 -1 )

      1× 2A+B=(1 2 2 -2 1 0 )

      2× 2A-6B=(-8 -6 -4 -2 0 -2 )

           -    +      -

      7B=(1 2 2 -2 1 0 )-(-8 -6 -4 -2 0 -2 )

      7B=(9 8 6 0 1 2 )

       ∴    B=(97  87 67 0 17 27 ).

       A-3(97  87 67 0 17 27 )=(-4 -3 -2 -1 0 -1 )

       A-(277  247 187 0 37 67 )=(-4 -3 -2 -1 0 -1 )

       A=(277  247 187 0 37 67 )+(-4 -3 -2 -1 0 -1 )

        ∴   A=(-17  37 47 -7 37 -17 ).  (Zurni et al. 2010).

          (b)        3(-3 8 14 -5 0 -6 )-4(11 -3 7 12 2 -9 )

           =(-9 24 42 -15 0 -18 )-(44 -12 28 48 8 -36 )

           =(-53 36 14 -63 -8 18 ).

   2(-3 8 14 -5 0 -6 )(6 2 7 -5 10 -7 1 11 0 )

                     2×3                3×3

         =2((-18-40+14) (-6+80+154) (-21-56+0) (-30+0-6) (-10+0-66) (-35+0+0) )

         =2(-44 228 -77 -36 -76 -35 )

         =(-88 456 -154 -72 -152 -70 ).  (Zurni et al. 2010).

1. (c) 1st row is selected

There is no inverse if determinant equals to zero.

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

1(4a-12)-2(a+4)+3(6+8)=0

4a-12-2a-8+42=0

2a+22=0

a=-222

a=-11

Therefore if a=-11, the determinant would be zero hence no inverse. (Zurni et al. 2010).

                  (d) |1 1+ac 1+bc 1 1+ad 1+bd 1 1+ae 1+be |1 1+ac 1 1+ad 1 1+ae

                        1(1+ad)(1+be)+1(1+ac)(1+bd)+1(1+bc)(1+ae)

1+ad+be+abde+1+ac+bd+abcd+1+bc+ae+abce

3+ad+be+ac+bd+bc+ae+abde+abcd+abce    →1

1(1+ad)(1+bc)+1(1+ae)(1+bd)+1(1+be)(1+ac)

1+ad+bc+abcd+1+ae+bd+abde+1+be+ac+abce

3+ad+bc+ae+bd+be+ac+abcd+abde+abce    →2

=1-2

=3+ad+be+ac+bd+bc+ae+abde+abcd+abce -3-ad-bc- ae-bd-be-ac-abcd-abde-abce    

= 0.  (Zurni et al. 2010).

1. (e) -0|-2 0 6 0 5 -7 4 9 0 |-2|3 4 2 0 5 -7 4 9 0 |-3|3 4 2 -2 0 6 4 9 0 |+0|3 4 2 -2 0 6 0 5 -7 |

    =-2(3|5 -7 9 0 |+4|4 4 5 -7 |)-3(2|-2 0 4 9 |-6|3 4 -2 0 |)

    =-2(189-152)-3(-36-66)

    =-2(37)-3(-102)

    =-74+306

    =232.  (Zurni et al. 2010).

2.          (a) (0 1 -3 2 3 -1 4 5 -2 )(x1  x2 x3 )=(-5 7 10 )

(b) 0|3 -1 5 -2 |-1|2 -1 4 -2 |-3|2 3 4 5 |

     =0(-6+5)-1(-4+4)-3(10-12)

     =0-0-3(-2)

     = 6.  (Zurni et al. 2010).

(c) Minor matrix

    (0 1 -3 2 3 -1 4 5 -2 )

   ((-6+5) (-4+4) (10-12) (-2+15) (0+12) (0-4) (-1+9) (0+6) (0-2) )

   =(-1 0 -2 13 12 -4 8 6 -2 ).

   (c)             Cofactor matrix

                 ((+×-1) (-×0) (+×-2) (-×+13) (+×+12) (-×-4) (+×+8) (-×+6) (+×-2) )

                  =(-1 0 -2 -13 12 4 8 -6 -2 )

Ad joint matrix

= (Cofactor)T

=(-1 -13 8 0 12 -6 -2 4 -2 )

Inverse matrix

=16(-1 -13 8 0 12 -6 -2 4 -2 )

=(-16 -136 43 0 2 -1 -13 23 -13 ). Zurni, O., Noraziah, H, M., Hawa, I., Fatinah, Z., & Azizan, S. (2010, p.13-16).

     (d) Solutions

=(-16 -136 43 0 2 -1 -13 23 -13 )(-5 7 10 )

           3×3               3×1

=((56-916+403) (0+14-10) (53+143-103) )

∴(x1 x2 x3 )=(-1 4 3 ).(Zurni et al. 2010).

3. (a) (0 1 1 3 -1 1 1 1 -3 )(x y z )(6 -7 -13 )

Determinant of A

0|-1 1 1 -3 |-1|3 1 1 -3 |+1|3 -1 1 1 |

=0-1(-9-1)+1(3+1)

=14

Determinant of A1

(6 1 1 -7 -1 1 -13 1 -3 )

6|-1 1 1 -3 |-1|-7 1 -13 -3 |+1|-7 -1 -13 1 |

=6(3-1)-1(21+13)+1(-7-13)

=-42

Determinant A2

(0 6 1 3 -7 1 1 -13 -3 )

0|-7 1 -13 -3 |-6|3 1 1 -3 |+1|3 -7 1 -13 |

=0-6(-9-1)+1(-39+7)

=28

Determinant A3

(0 1 6 3 -1 -7 1 1 -13 )

0|-1 -7 1 -13 |-1|3 -7 1 -13 |+6|3 -1 1 1 |

=0-1(-39+7)+6(3+1)

=56

   ∴     x=-4214         y=2814               z=5614

              =-3               =2                     =4              (Zurni et al. 2010).

              (b)          y=x2-4x-5

     dydx=2x-4

    Gradient at point A

    M1=2(3)-4=2

    Coordinate of point A

    y=32-4(3)-5=-8

    (x,y)=(3,-8).  Uses of differentiation. (n.d).

    Equation of tangent line

    -8=2(3)+c        c=-14

    y=2x-14

    Equation of perpendicular line

    m1×m2=-1            2×m2=-1

    m2=-12

    -8=-12(3)+c           c=-132

    y=-12x-132

    Coordinate of point B

    -12x-132=x2-4x-5

   (b)    2x2-7x+3=0          (x-3)(2x-1)=0

    x=3                 x=12

       y=(12)2-4(12)-5       y=-274          ∴B=(12,-274)

    Coordinate at x=0

    y=02-4(0)-5    ∴(0,-5)

    Coordinates at y=0

    x2-4x-5=0       (x-5)(x+1)=0

    x=5       x=-1       ∴ (-1,0)     (5,0)

    Coordinate at minimum point

    y=22-4(2)-5       ∴(2,-9).    (Zurni et al. 2010).

 

 (c)

(Zurni et al. 2010).

4. (a)

1. f(x)=x+22

Dom f={xϵR, x≥-2}

      Range f={yϵR, y≥0}

g(x)=3-x5-x

Dom g={xϵR, x≠5 }

Range g={yϵR, -1≤y≤3}

h(x)=x2-6

Dom h={xϵR, }

Range h={yϵR, y≥-6}

2. ff(34)=34+22   =3

3+22    =52    =1.12

gf(47)= 47+22         72=3.5

3-3.55-3.5   =-13

hf(x)=(x+22)2-6

=(x12+2122)2-6

=x+24-6

=x-224.             Operations on Functions. (n.d).

3. y=3-x5-x

    x=3-y5-y

    5x-xy=3-y

    y(x-x)=-5x+3

    g-1(x)=(-5x+3)(1-x).      Definition of Inverse Function. (n.d).

                

        (b)

=x-23y5z3×x4325x12y-23

=x16y53125z3

      (Zurni et al. 2010).

         

         (c)                  

                                  = 25+29-5×4226

=5+9-(5×2)6

=23

    (Zurni et al. 2010).

References

Definition of Inverse Function. (n.d). Retrieved September 26, 2011, from http://www.regentsprep.org/Regents/math/algtrig/ATP8/inverselesson.htm

Operations on Functions. (n.d). Retrieved September 26, 2011, from http://www.purplemath.com/modules/fcnops.htm

Uses of differentiation. (n.d). Retrieved September 26, 2011, from http://www.mathsrevision.net/alevel/pages.php?page=45

Zurni, O., Noraziah, H, M., Hawa, I., Fatinah, Z., & Azizan, S. (2010). Mathematics for Management. Malaysia: Metero Doc. Sdn. Bhd.