# 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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1 Quiz 2, Solution
Question 1
3x1x2+x3= 1; x1=1
3x21
3x3+1
3
3x1+ 6x2+ 2x3= 0; x2=1
2x11
3x3
3x1+ 3x2+ 7x3= 4; x3=4
73
7x23
7x1
x(0) =x(0)
1; x(0)
2; x(0)
3t
= (0;0;0)t
x1=1
3x21
3x3+1
3x2=0;x3=0 =x1= 0:333 33
x2=1
2x11
3x3x1=0:333 33;x3=0 =x2=0:166 67
x3=4
73
7x23
7x1x1=0:333 33;x2=0:166 67 =x3= 0:5
x(1) =x(1)
1; x(1)
2; x(1)
3t
= (0:333 33;0:166 67;0:5)t
x1=1
3x21
3x3+1
3x2=0:166 67;x3=0:5=x1= 0:111 11
x2=1
2x11
3x3x1=0:111 11;x3=0:5=x2=0:222 22
x3=4
73
7x23
7x1x1=0:111 11;x2=0:222 22 =x3= 0:619 05
x(2) =x(2)
1; x(2)
2; x(2)
3t
= (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
10x1x2= 9; x1=
1
10x2+9
10
x1+ 10x22x3= 7; x2=1
10x1+1
5x3+7
10
2x2+ 10x3= 6; x3=1
5x2+3
5
x(0) =x(0)
1; x(0)
2; x(0)
3t
= (0;0;0)t
x1=1
10 x2+9
10 x2=0;x3=0 =x1= 0:
x2=1
10 x1+1
5x3+7
10 x1=0:9;x3=0 =x2= 0:79
x3=1
5x2+3
5x1=0:9;x2=0:79 =x3= 0:758
x(1) =x(1)
1; x(1)
2; x(1)
3t
= (0:9;0:79;0:758)t
x1=1
10 x2+9
10 x2=0:79;x3=0:758 =x1= 0:979
x2=1
10 x1+1
5x3+7
10 x1=0:979;x3=0:758 =x2= 0:949 5
x3=1
5x2+3
5x1=0:979;x2=0:949 5 =x3=:789 9
x(2) =x(2)
1; x(2)
2; x(2)
3t
= (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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