# Iterative Methods-Numerical Analysis-Quiz Solution, Exercises for Numerical Analysis. Biyani Girls College

## Numerical Analysis

Description: This is solution to quiz in class of Numerical Analysis. It was taken by Prof. Anuha Sharma at Biyani Girls College. It includes: AIterative, Methods, Numerical, Analysis, System, Linear, Equations, Infinity, Norm, Solution
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University: Biyani Girls College
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1 Quiz 2, Solution

Question 1

3x1 � x2 + x3 = 1; x1 = 1

3 x2 �

1

3 x3 +

1

3

3x1 + 6x2 + 2x3 = 0; x2 = � 1

2 x1 �

1

3 x3

3x1 + 3x2 + 7x3 = 4; x3 = 4

7 � 3 7 x2 �

3

7 x1

x(0) =  x (0) 1 ; x

(0) 2 ; x

(0) 3

t = (0; 0; 0)

t

 x1 =

1 3x2 �

1 3x3 +

1 3

 x2=0;x3=0

= x1 = 0:333 33 x2 = � 12x1 �

1 3x3

 x1=0:333 33;x3=0

= x2 = �0:166 67 x3 =

4 7 �

3 7x2 �

3 7x1

 x1=0:333 33;x2=�0:166 67

= x3 = 0:5

x(1) =  x (1) 1 ; x

(1) 2 ; x

(1) 3

t = (0:333 33;�0:166 67; 0:5)t

x1 = 1 3x2 �

1 3x3 +

1 3

 x2=�0:166 67;x3=0:5

= x1 = 0:111 11 x2 = � 12x1 �

1 3x3

 x1=0:111 11;x3=0:5

= x2 = �0:222 22 x3 =

4 7 �

3 7x2 �

3 7x1

 x1=0:111 11;x2=�0:222 22

= x3 = 0:619 05

x(2) =  x (2) 1 ; x

(2) 2 ; x

(2) 3

t = (0:111 11;�0:222 22; 0:61905)t

max (j0:111 11� 0:333 33j ; j�0:222 22 + 0:166 67j ; j0:619 05� 0:5j) = 0:222 22

x(2) � x(1) 1 x(2) 1 = max (j0:11111� 0:33333j ; j�0:22222 + 0:16667j ; j0:61905� 0:5j)max (0:52381; 0:222 22; 0:52381) = 0:424 24

1

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Question 2

10x1 � x2 = 9; x1= 1

10 x2 +

9

10

�x1 + 10x2 � 2x3 = 7; x2 = 1

10 x1 +

1

5 x3 +

7

10

�2x2 + 10x3 = 6; x3 = 1

5 x2 +

3

5

x(0) =  x (0) 1 ; x

(0) 2 ; x

(0) 3

t = (0; 0; 0)

t

 x1 =

1 10x2 +

9 10

 x2=0;x3=0

= x1 = 0: x2 =

1 10x1 +

1 5x3 +

7 10

 x1=0:9;x3=0

= x2 = 0:79 x3 =

1 5x2 +

3 5

 x1=0:9;x2=0:79

= x3 = 0:758

x(1) =  x (1) 1 ; x

(1) 2 ; x

(1) 3

t = (0:9; 0:79; 0:758)

t

 x1 =

1 10x2 +

9 10

 x2=0:79;x3=0:758

= x1 = 0:979 x2 =

1 10x1 +

1 5x3 +

7 10

 x1=0:979;x3=0:758

= x2 = 0:949 5 x3 =

1 5x2 +

3 5

 x1=0:979;x2=0:949 5

= x3 = :789 9

x(2) =  x (2) 1 ; x

(2) 2 ; x

(2) 3

t = (0:979 ; 0:949 5; 0:789 9)

t

x(2) � x(1) 1 x(2) 1 = max (j0:979� 0:9j ; j0:949 5� 0:79j ; j0:789 9� 0:758j)max (0:979; 0:949 5; 0:789 9) = 0:162 92 x(2) � x(1)

1 = max (j0:979� 0:9j ; j0:949 5� 0:79j ; j0:789 9� 0:758j) = 0:159 5

2

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