aiou solved assighnment matric math 248 spring 2014

Solved Assignments

AIOU

Math (Course code = 248) Assignment No 2

                       Semester spring 2014
Q:1

3A-3B=3(A-B)

L.H.S

3A-3B

3[2    3]-3[1 7]
[1    5] [4    6]

[6    9]]-[3    21]
[3    15] [12   18]

[6-3     9-21]
[3-12 15-18]

[3     -12]
[-9    -3]    Ans

R.H.S

3(A-B)

3[2    3]- [1 7]
[1    5] [4    6]

3[2-1 3-7]
[1-4    5-6]

3[1    -4]
[-3 -1]

[3 -12]
[-9    -3]    Ans

                  so   L.H.S =R.H.S
(B)

[w     x]   =   [2     1]   -   [3     0]
[y     z]         [6   -3]        [-1    5]

[w     x]   =   [2-3     1-0]
[y     z]         [6-(-1) -3-5]

[w     x]   =   [-1     1]
[y     z]         [7 -8]

w= -1
x= 1
y= 7
z= -8

(c)
[1     -3] [1     -1 ]
[1     2] [o     -2 ]

[(1x1)+(-3x0)            (1x-1)+(-3x-2)]
[(1x1)+(2x0)             (1x-1)+(2x-2)]

[1-0     -1+6]
[1+0     -1-4]

[1     5]
[1     -5]   Ans

Q:2
(A)
Sol:

MM^-1

M=[1 2]
[3 4]

M^-1 = adjM

1M1
adj M = adj [1 2]
[3 4]
= [4 -2]
[-3 1]

1M1 = 4x1 -(-2_(-3)_
= 4-(6)
= -2

Putting values in

M^-1= adjM

1M1

M^-1 = [4 -2]

[-3     1]
                 -2
M = [4/-2         -2/-2 ]
          [-3/-2        1/-2 ]

    =   -2           1
   3/2     -1/2

MM^-1= [1 2] [ -2 1]

         [3 4] [ 3/2     -1/2]
= 1x-2 + 2x3/2    1x1+2x-1/2
3x-2 + 4x3/2    3x1+4x-1/2
   = -2+3    1-1
-6+6    3-2
   = 1    0
   0 1 Ans

R.H.S

M^-1M = ?

M^-1 = -2 1

    3/2    -1/2
M = [1 2]
    [3 4]
   M^-1M =   [-2     1   ] [1 2]
    [3/2     -1/2 ] [3 4]
               =   -2x1 + 1x3                  -2x2 + 1x-4
                    3/2 x 2 +-1/2x-3          3/2x2 + -1/2x4
             = -2+3        1-1
                  -6+6             3-2
              =    1           0
                     0           1             Ans

                   so          L.H.S = R.H.S
(B)
          L.H.S

AB^-1= ?

AB =[ 5     2][4    2]
     [2     1][3    -1]
     = 5 x 4 + 2 x 3           5x2 + 2 x - 1
         2 x 4 + 1 x 3           2 x 2 + 1 x -1
   = 20+6               10-2
          8+3                  4-1
    =   26      8
         11      3
AB^-1 =   adj AB
                    |AB|
           = 3        -8
              -11      26
|AB| =   26 x 3 - (-11 x-8)
          = 78 -88
          =   -10
so

AB^-1 =   adj AB
                    |AB|
   =  3        -8
              -11      26
-10
          = 1/-10 [3       -8 ]
              [-11      26]
R.H.S

A^-1B^-1 = ?

A^-1 = adj A

|A|

adj A = 1 -2

- 2 5

|A| = 5 x 1 - 2 x 2

= 5 - 4

= 1

A^-1 = adj A

|A|

= 1 -2

- 2 5

B^-1 = adjB

|B|

adj B = -1 2

3 4

|B| = 4 x -1 - 3 x 2

= -4 - 6

= -10

B^-1 = adjB

|B|

= -1 2

3 4

-10

= 1/-10 [-1 -2]

[ -3 4]

⁼1/-10 [-1 -2] [1 -2]

[-3 4] [- 2 5 ]

      =    [ -1 x 1+ -2 x-2    -1 x -2 +(-2x5)]
[-3 x 1 +4 x -2 -3x-2 + 4x5    ]
= 1/-10 [-1+4    2-10 ]
   [ -3-2    6+20]
   =    [3    -8 ]
   [-11 26]
so
LHS=RHS
(c)
Find the solution set
2x+y=1
5x+3y=2
we will find it by cramer method
Write it in matrix form
|2 1| |x| = |1|
|5 3| |y| = |2|
AX=B
A= |2    1|
   |5    3|
|A| = |2 1|
      |5 3|
= 6-5=1# 0
|D1| = |1    1| = 3-2=1
= |2    3|
|D2| =|2    1| = 4-5=-1
= |5    2|
x = |D1|
|A|
   = 1
y = |D2|
   |A|
    = -1
x = 1
y = -1
   = {1 , -1}

Q:3
(A)
suppose first angle is = X⁰

              second angle= 100⁰ - X⁰
suppliment of   X⁰      =180⁰ -X⁰
^{suppliment of}^{100⁰ -}^{X⁰ =}^{180⁰- (}^{100⁰ -}^X⁰)
⁼^{180⁰ -}^{100⁰ -x}
⁽^{180⁰ -}^{100⁰ +}^{X⁰ ) -(}^{180⁰ -}^{X⁰) =}100⁰
^{180⁰ -}^{100⁰ +}^{X⁰ -}^{180⁰ -}^{X⁰ =}100⁰
²^{X⁰ -}^{100⁰ =}100⁰
²^{X⁰ =}^{100⁰ +}100⁰
^{X⁰ = 200/2}
^{X⁰ =}100⁰
^{First angle =}100⁰
^{second angle =}^{100⁰ -}^{100⁰ = 0}

^(B)
^{Sum of angles of triangle =}180⁰

Sum of angles of triangle (ABC)=x+90+30
   = x+120
x+120 = 180⁰
x = 180⁰ -120
x = 60   Ans
(c)
largest side of rectangle = 7cm
ratio in large sides = 7 : 1
                               21   3
so
          1/3 = 6/a = 5/b = 4/c =2/d
          6/a=1/3
          a=6x3 =18cm
          5/b=1/3
          b= 5x3 15 cm
    4/c = 1/3
c= 4x3 =12
2/d=1/3
d=2x3 = 6cm
a,b,c,d are sides of rectangle
Q:4
Area of field = 2.5 hectar
   25000 square meter
Suppose lenght of field =3x
width of field =2x
Area = (3x) (2x)
= 6X²
6X² = 25000
X² = 25000/6
= 4166.67
By taking sqaur root
x = 64.55
3x = 3x64.55=193.65
2x=2x64.55= 129.10
2(3x+2x) =2(193.65 +129.10)
   =2(322.75)
   =645.50 meter
(B)
lenght = 4m
width = 3m
hight = 2m
volume = lenght x width x hight
   = 4x3x2
   = 24
(C)
Radius of square= 14cm
Area of square= 14x14/2
=98 Cm²
According to fisa gourse
X²+X² = 14x14
2X² = 14x14
= 14x14/2

area of square = X²
   =98 Cm²
Q:5

(b)

Magnitude of first side = 2ab

Magnitude of second side = a - b

Magnitude of chord = ?

Magnitude of chord = (2ab)² + (a² + b² )

= √ (a² + b² )+ (2ab)²

= √ a⁴ + b⁴ - 2a²b² +4a²b²

= √ a⁴ + b⁴ +(2a²b² )²

= √ (a² + b² )²

= (a² + b²) Ans

Math(code = 248) Assighnment No 1

semester spring 2014

Q:1
(A)
P(1)

P(x) = x² -5x² + 6

X + 1

= (1)² – 5(1)² + 6

1+1

= 1 – 5 + 6

2
= 1 Ans

P (5)= = x -5x² + 6

X + 1

= (5)² – 5(5)² + 6

5 + 1

= 25 – 25 + 6

=     1 Ans
= ( 1 , 1 ) Ans
(B)
Sol:
2X² - 5x -12    x    6X² + 5x - 4
4x + 4x - 3             2x -7x -4
2X² - 8x + 3x - 12 x    6X² + 8x -3x -4
4x + 6x - 2x - 3           2x - 8x + x - 4
2x(x-4) +3 (x-4)      x    2x(3x+4) -1 (3x+4)
2x(2x+3)-1(2x+3)            2x(x-4) + 1(x-4)
(2x+3) (x-4)           x    (2x-1) (3x+4)
(2x-1) (2x+3) x    (2x+1) (x-4)
= 3x +4             Ans
   2x+1
(ii)

= (x+1)² - (x²+1)

(x²+1) (x²+1)

= (x+1)² - (x²+1)

x²+1

⁼^x² + 2x+ 1 - ^x²   -1
^x²   + 1
^{=    2x}
^x² +1 Ans

Attention! Ad blocker Detected!

aiou solved assighnment matric math 248 spring 2014

Solved Assignments

Math (Course code = 248) Assignment No 2

Math(code = 248) Assighnment No 1

Post a Comment

Popular Posts

Tags

Blog Posts

Attention! Ad blocker Detected!

aiou solved assighnment matric math 248 spring 2014

Solved Assignments

Math (Course code = 248) Assignment No 2

Math(code = 248) Assighnment No 1

Share to:

Next

Newer Post

Previous

Older Post

Post a Comment