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1. Project Report
Linear Algebra
Matlab Final Project
Spring 2006
National University of Computer and
Emerging Science
Islamabad
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2. Project Report
Submitted to:
Sir Sher Afzal Khan
Submitted BY:
Muhammad Rizwan----(060388)
Kamran Ali-----------------(060377)
Mansoor Ahmed-------- (060392)
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3. Project Report
Matrix Algebra ...................................................................................................................................... 5
Addition of two Matrices .................................................................................................................. 5
Algorithm: ..................................................................................................................................... 6
Input: ............................................................................................................................................. 6
Two matrices A and B. ................................................................................................................. 6
Code .............................................................................................................................................. 7
Output ........................................................................................................................................... 8
Subtraction of two Matrices .............................................................................................................. 9
Algorithm: ................................................................................................................................... 10
Input: ........................................................................................................................................... 10
Two matrices A and B. ............................................................................................................... 10
Code ............................................................................................................................................ 10
Output ......................................................................................................................................... 11
Multiplication of two Matrices ....................................................................................................... 12
Algorithm .................................................................................................................................... 13
Input ............................................................................................................................................ 13
Code ............................................................................................................................................ 13
Output ......................................................................................................................................... 15
Transpose of a Matrix ..................................................................................................................... 16
Algorithm .................................................................................................................................... 17
Input ............................................................................................................................................ 17
Matrix A. ..................................................................................................................................... 17
Code ............................................................................................................................................ 17
Output ......................................................................................................................................... 18
Transformation .................................................................................................................................... 18
Reflection through the x-axis .......................................................................................................... 18
Algorithm .................................................................................................................................... 19
Input ............................................................................................................................................ 19
Code ............................................................................................................................................ 19
Output ......................................................................................................................................... 20
Reflection through the y-axis .......................................................................................................... 20
Algorithm .................................................................................................................................... 20
Input ............................................................................................................................................ 20
Code ............................................................................................................................................ 21
Output ......................................................................................................................................... 21
Reflection through the line y = x .................................................................................................... 22
Algorithm .................................................................................................................................... 22
Input ............................................................................................................................................ 22
Code ............................................................................................................................................ 23
Output ......................................................................................................................................... 23
Reflection through the line y = -x ................................................................................................... 23
Algorithm .................................................................................................................................... 24
Input ............................................................................................................................................ 24
Code ............................................................................................................................................ 24
Output ......................................................................................................................................... 25
Reflection through the origin .......................................................................................................... 25
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4. Project Report
Algorithm .................................................................................................................................... 25
Input ............................................................................................................................................ 25
Code ............................................................................................................................................ 26
Output ......................................................................................................................................... 26
Rotation ........................................................................................................................................... 27
Algorithm .................................................................................................................................... 27
Input ............................................................................................................................................ 27
Code ............................................................................................................................................ 28
Output ......................................................................................................................................... 28
Horizontal Contraction and Expansion ........................................................................................... 28
Algorithm: ................................................................................................................................... 29
Input ............................................................................................................................................ 29
Code ............................................................................................................................................ 29
Output ......................................................................................................................................... 30
Vertical Contraction and Expansion ............................................................................................... 30
Algorithm .................................................................................................................................... 31
Input ............................................................................................................................................ 31
Code ............................................................................................................................................ 31
Output ......................................................................................................................................... 32
Horizontal Shear ............................................................................................................................. 32
Algorithm .................................................................................................................................... 32
Input ............................................................................................................................................ 32
Code ............................................................................................................................................ 33
Output ......................................................................................................................................... 33
Vertical Shear.................................................................................................................................. 33
Algorithm .................................................................................................................................... 34
Input ............................................................................................................................................ 34
Code ............................................................................................................................................ 34
Output ......................................................................................................................................... 34
Cryptology .......................................................................................................................................... 35
Encryption and Decryption ............................................................................................................. 35
Algorithm .................................................................................................................................... 35
Input ............................................................................................................................................ 35
Code ............................................................................................................................................ 35
Output ......................................................................................................................................... 36
Clock ................................................................................................................................................... 36
Working .......................................................................................................................................... 36
Input ............................................................................................................................................ 37
Code ............................................................................................................................................ 37
Output ......................................................................................................................................... 38
System of Linear equations ................................................................................................................. 38
Jacobi’s Method .............................................................................................................................. 38
Algorithm .................................................................................................................................... 39
Input ............................................................................................................................................ 39
Code ............................................................................................................................................ 39
Output ......................................................................................................................................... 41
Gauss Seidel’s Method ................................................................................................................... 45
Algorithm .................................................................................................................................... 45
Input ............................................................................................................................................ 45
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5. Project Report
Code ............................................................................................................................................ 45
Output ......................................................................................................................................... 47
Matrix Algebra
Algebra is the branch of mathematical in which, the mathematical operators +,-,*, etc. are involved.
Matrix Algebra is the branch of mathematics in which algebraic expressions involved.
Addition of two Matrices
Let A is an mxn matrix and B is pxq matrix. B can be added to A only when order of A and order of
B are same, in other words m=p and n=q. (A & B are equivalent)
Let aij is the (i,j)th entry of A and bij is (i,j)th entry of B. Let Matrix C is the addition of Matrices A &
B, then (i,j)th entry of C is cij and cij=aij + bij. where i=1,2…m j=1,2,…n.
Example
A=
1 2
3 4
B=
45
67
C=A+B
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6. Project Report
C=
6 8
10 12
Algorithm:
Input:
Two matrices A and B.
Body of algorithm:
a= Input Enter the rows of A
b= Input Enter the columns of A
for i=1 to number of rows of A
for j=1 to number of columns of A
A(i,j)=Input Enter the array entries of A
End for j
End for i
Display Matrix A
c= Input Enter the rows of B
d= Input Enter the columns of B
for i=1 to number of rows of B
for j=1 to number of columns of B
B(i,j)=Input Enter the array entries of B
End for j
End for i
Display Matrix B
if Size of A ==Size of B
Matrix Addition is possible
for i=1 to number of rows of A
for j=1 to number of columns of B
C(i,j)=A(i,j)+B(i,j);
End for j
End for i
Display Matrix C
else
Matrix Addition is not possible
End for if
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7. Project Report
Code
clc
clear
disp('matrix A')
a= input ('enter the rows of A=');
b= input ('enter the columns of A=');
for i=1:a;
for j=1:b;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j)]);
A(i,j)=input ('enter the entries of A =');
end
end
disp('press any key to see the matrix'),pause
clc
(A)
disp('Matrix B')
c=input('enter the rows of B=');
d=input('enter the columns of B=');
for i=1:c;
for j=1:d;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j)]);
B(i,j)=input('enter the entries of B=');
end
end
disp('press any key to see the matrix'),pause
clc
disp('Matrix A')
(A)
disp('Matrix B')
(B)
if a==c & b==d
disp('Matrix Addition is pissible');
for i=1:a;
for j=1:b;
C(i,j)=A(i,j)+B(i,j);
end
end
disp('press any key to see the Matrix C'),pause
clc
disp('Matrix C=A+B')
disp('Matrix C')
(C)
else
disp('Matrix Addition is not possiblle')
disp('Because size of both matrices are not same')
end
disp('press any key to End'),pause
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8. Project Report
clc
Output
matrix A
enter the rows of A=2
enter the columns of A=2
enter the entry of Row 1 Column 1
enter the entries of A =1
enter the entry of Row 1 Column 2
enter the entries of A =2
enter the entry of Row 2 Column 1
enter the entries of A =3
enter the entry of Row 2 Column 2
enter the entries of A =4
press any key to see the matrix
-----------------------------------------------------------
A=
1 2
3 4
Matrix B
enter the rows of B=2
enter the columns of B=2
enter the entry of Row 1 Column 1
enter the entries of B=5
enter the entry of Row 1 Column 2
enter the entries of B=6
enter the entry of Row 2 Column 1
enter the entries of B=7
enter the entry of Row 2 Column 2
enter the entries of B=8
press any key to see the matrix
----------------------------------------
Matrix A
A=
1 2
3 4
Matrix B
B=
5 6
7 8
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9. Project Report
Matrix Addition is pissible
press any key to see the Matrix C
-------------------------------------------
Matrix C=A+B
Matrix C
C=
6 8
10 12
press any key to End
--------------------------------------------
Subtraction of two Matrices
Subtraction of matrices is same as the addition of matrices. If we want to subtract B from A then we
multiply B with -1 and add it to A
Let A is an mxn matrix and B is mxn matrix. We can find the subtraction A and B by Adding A and
–B. A+ (-B) =A-B.
We can find –B by multiplying B with a constant -1 as (-1) B= -B
Example
A=
1 2
3 4
B=
45
67
C=A-B
C=
-3 -3
-3 -3
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10. Project Report
Algorithm:
Input:
Two matrices A and B.
Body of algorithm:
a= Input Enter the rows of A
b= Input Enter the columns of A
for i=1 to number of rows of A
for j=1 to number of columns of A
A(i,j)=Input Enter the array entries of A
End for j
End for i
Display Matrix A
c= Input Enter the rows of B
d= Input Enter the columns of B
for i=1 to number of rows of B
for j=1 to number of columns of B
B(i,j)=Input Enter the array entries of B
End for j
End for i
Display Matrix B
B=-1*B;
if Size of A ==Size of B
Matrix Addition is possible
for i=1 to number of rows of A or B
for j=1 to number of columns of A or B
C(i,j)=A(i,j)+B(i,j);
End for j
End for i
Display Matrix C
else
Matrix Addition is not possible
End for if
Code
clc
clear
disp('matrix A')
a= input ('enter the rows of A=');
b= input ('enter the columns of A=');
for i=1:a;
for j=1:b;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j)]);
A(i,j)=input ('enter the entries of A =');
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11. Project Report
end
end
disp('press any key to see the matrix'),pause
clc
(A)
disp('Matrix B')
c=input('enter the rows of B=');
d=input('enter the columns of B=');
for i=1:c;
for j=1:d;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j)]);
B(i,j)=input('enter the entries of B=');
end
end
disp('press any key to see the matrix'),pause
clc
disp('Matrix A')
(A)
disp('Matrix B')
(B)
if a==c & b==d
disp('Matrix subtraction is pissible');
B=-1*B;
for i=1:a;
for j=1:b;
C(i,j)=A(i,j)+B(i,j);
end
end
disp('press any key to see the Matrix C'),pause
clc
disp('Matrix C=A+(-B)=A-B')
disp('Matrix C')
(C)
else
disp('Matrix subtraction is not possiblle')
disp('Because size of both matrices are not same')
end
disp('press any key to End'),pause
clc
Output
matrix A
enter the rows of A=2
enter the columns of A=2
enter the entry of Row 1 Column 1
enter the entries of A =1
enter the entry of Row 1 Column 2
enter the entries of A =2
enter the entry of Row 2 Column 1
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12. Project Report
enter the entries of A =3
enter the entry of Row 2 Column 2
enter the entries of A =4
press any key to see the matrix
A=
1 2
3 4
Matrix B
enter the rows of B=2
enter the columns of B=2
enter the entry of Row 1 Column 1
enter the entries of B=4
enter the entry of Row 1 Column 2
enter the entries of B=5
enter the entry of Row 2 Column 1
enter the entries of B=6
enter the entry of Row 2 Column 2
enter the entries of B=7
press any key to see the matrix
Matrix C=A+(-B)=A-B
Matrix C
C=
-3 -3
-3 -3
press any key to End
Multiplication of two Matrices
Let A is an mxn matrix and B is pxq matrix. B can be multiply with A only when number of
columns of A are same as the number of rows of B, in other words n=q. The order of resulting
matrix is m x q.
Let aij is the (i,j)th entry of A and bij is (i,j)th entry of B. Let Matrix C is the product of Matrices A &
B, then (i,j)th entry of C is cij and cij=ai1b1j + ai2b2j + a13b3j…… an1 bnj. Where i =1,2…m j=1,2,…n.
Example
A=
1 2
3 4
B=
5 6
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13. Project Report
78
Matrix C=A*B
C=
14 16
28 32
Algorithm
Input
Two Matrices A and B
Body of algorithm:
a= Input Enter the rows of A
b= Input Enter the columns of A
for i=1 to number of rows of A
for j=1 to number of columns of A
A(i,j)=Input Enter the array entries of A
End for j
End for i
Display Matrix A
c= Input Enter the rows of B
d= Input Enter the columns of B
for i=1 to number of rows of B
for j=1 to number of columns of B
B(i,j)=Input Enter the array entries of B
End for j
End for i
Display Matrix B
if number of coloumns of A == number of rows of B
Matrix multiplication is possible
for i=1 to number of rows of A
for j=1to number of coloumns of B
for t=1 to number of coloumns of A or number of rows of B
C(i,j)=0;
C(i,j)=C(i,j)+A(i,t)*B(t,j);
End for t
End for j
End for i
Display matrix C
else
Matrix multiplication is not possible
End for if
Code
clc
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14. Project Report
clear
disp('matrix A')
a= input ('enter the rows of A=');
b= input ('enter the columns of A=');
for i=1:a;
for j=1:b;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j)]);
A(i,j)=input ('enter the entry ');
end
end
disp('press any key to see the matrix'),pause
clc
disp('Matrix A')
(A)
disp('matrix B')
c= input ('enter the rows of B=');
d= input ('enter the columns of B=');
for i=1:c;
for j=1:d;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j)]);
B(i,j)=input ('enter the entry ');
end
end
disp('press any key to see the matrix'),pause
clc
disp('Matrix A')
(A)
disp('Matrix B')
disp(B)
if b==c;
disp('Matrix multiplication is possible')
for i=1:a;
for j=1:d
for t=1:b
C(i,j)=0;
C(i,j)=C(i,j)+A(i,t)*B(t,j);
end
end
end
disp('press any key to see the Matrix C'),pause
clc
disp('Matrix C=A*B')
disp(C)
else
disp('Matrix multiplication is not possible')
disp('Because it does not satisfy the order conditions')
end
disp('press any key to End'),pause
clc
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15. Project Report
Output
matrix A
enter the rows of A=2
enter the columns of A=2
enter the entry of Row 1 Column 1
enter the entry 1
enter the entry of Row 1 Column 2
enter the entry 2
enter the entry of Row 2 Column 1
enter the entry 3
enter the entry of Row 2 Column 2
enter the entry 4
press any key to see the matrix
-------------------------------
Matrix A
A=
1 2
3 4
matrix B
enter the rows of B=2
enter the columns of B=2
enter the entry of Row 1 Column 1
enter the entry 5
enter the entry of Row 1 Column 2
enter the entry 6
enter the entry of Row 2 Column 1
enter the entry 7
enter the entry of Row 2 Column 2
enter the entry 8
press any key to see the matrix
---------------------------------------
Matrix A
A=
1 2
3 4
Matrix B
56
78
Matrix multiplication is possible
press any key to see the Matrix C
------------------------------------------
Matrix A
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16. Project Report
A=
1 2
3 4
Matrix B
56
78
Matrix multiplication is possible
press any key to see the Matrix C
---------------------------------------
Matrix C=A*B
14 16
28 32
press any key to End
---------------------------------
Transpose of a Matrix
Let A is an mxn matrix and B is an nxm matrix. Let aij is the (i,j)th entry of A and Let bij= aji is the
(i,j)th entry of B. Then we can say that B is the transpose matrix of A.and it is written as B=A’.
Example
Matrix A
A=
1 2
3 4
At is transpose of A
Transpose of A=At
At =
1 3
2 4
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17. Project Report
Algorithm
Input
Matrix A.
Body of algorithm:
a= Input Enter the rows of A
b= Input Enter the columns of A
for i=1 to number of rows of A
for j=1 to number of columns of A
A(i,j)=Input Enter the array entries of A
End for j
End for i
Display Matrix A
Let At is the transpose of matrix A
for i=1 to number of coloumns of A
for j=1 to number of rows of A
At(i,j)=A(j,i);
End for j
End end i
Display matrix At
Code
clc
clear
disp('matrix A')
a= input ('enter the rows of A=');
b= input ('enter the columns of A=');
for i=1:a;
for j=1:b;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j)]);
A(i,j)=input ('enter the entry=');
end
end
disp('press any key to see the matrix'),pause
clc
disp('Matrix A')
(A)
disp('press any key to see the transpose of A'),pause
for i=1:b;
for j=1:a;
At(i,j)=A(j,i);
end
end
disp('Transpose of A=At')
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18. Project Report
(At)
disp('press any key to End'),pause
clc
Output
matrix A
enter the rows of A=2
enter the columns of A=2
enter the entry of Row 1 Column 1
enter the entry=1
enter the entry of Row 1 Column 2
enter the entry=2
enter the entry of Row 2 Column 1
enter the entry=3
enter the entry of Row 2 Column 2
enter the entry=4
press any key to see the matrix
---------------------------------------
Matrix A
A=
1 2
3 4
press any key to see the transpose of A
-----------------------------------------
Transpose of A=At
At =
1 3
2 4
press any key to End
-------------------------------------------
Transformation
In mathematics transformation mean when a matrix A as an object that “acts” on a vector v
transform it into Av.
Let v be in Rn and A is an mxn matrix then the transformation of v is Av and is in Rn. Generally we
denoted it as T:Rm→Rn and T(v) = Av.
Reflection through the x-axis
Let v be vector in R2 and it is in first quadrant when it is reflected through the x-axis it is
transformed in 3rd quadrant. To reflect a vector v through the x-axis we multiply it with
A=
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19. Project Report
1 0
0 -1
Example
V=
1
2
Transformation about x axis T(V)=A*V
T(V) =
1
-2
Algorithm
Input
A vector with x and y co ordinates
Body of algorithm
x=Input Enter the x co ordinate
y=Input Enter the y co ordinate
V=(x , y)
A=
10
0 -1
Display Transformation about x axis
Draw the graph of original vector
Draw the graph of transformed vector
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
A=[1 0;0 -1];
disp('Transformationn about x axis')
T=A*V
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
disp('Red vector is the transformed vector')
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20. Project Report
Output
enter the x=1
enter the y=2
V=
1
2
Transformation about x axis
T=
1
-2
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Reflection through the y-axis
Let v be vector in R2 and it is in 1st quadrant when it is reflected through the y-axis it is transformed
in 2nd quadrant. To reflect a vector v through the x-axis we multiply it with
A=
-1 0
0 1
Example
V=
1
2
Transformation about y axis T(V)=A*V
T (V)=
-1
2
Algorithm
Input
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21. Project Report
A vector with x and y co ordinates
Body of algorithm
x=Input Enter the x co ordinate
y=Input Enter the y co ordinate
V=(x , y)
A=
-1 0
01
Display Transformation about x axis
Draw the graph of original vector
Draw the graph of transformed vector
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
A=[-1 0;0 1];
disp('Transformationn about y axis')
%disp('press any key to view the transformation'),pause
T=A*V
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
disp('Red vector is the transformed vector')
Output
enter the x=1
enter the y=2
V=
1
2
Transformation about y axis
T=
-1
2
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22. Project Report
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Reflection through the line y = x
Let v be vector in R2 and it is in first quadrant when it is reflected through the line y=x. Then y
becomes x and x becomes y in v. To reflect a vector v through the line x=y we multiply it with
A=
0 1
1 0
Example
V=
1
2
Transformation about line y=x axis T(V)=A*V
T (V)=
2
1
Algorithm
Input
A vector with x and y co ordinates
Body of algorithm
x=Input Enter the x co ordinate
y=Input Enter the y co ordinate
V=(x , y)
A=
01
10
Display Transformation about x axis
Draw the graph of original vector
Draw the graph of transformed vector
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23. Project Report
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
A=[0 1;1 0];
disp('Transformationn through line y=x')
T=A*V
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
disp('Red vector is the transformed vector')
Output
enter the x=1
enter the y=2
V=
1
2
Transformationn through line y=x
T=
2
1
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Reflection through the line y = -x
Let v be vector in R2 and it is in first quadrant when it is reflected through the line y=-x. Then it is
transformed into 3rd quadrant in other words y becomes -x and x becomes -y in v and To reflect a
vector v through the line y=-x we multiply it with
A=
0 -1
-1 0
Example
V=
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24. Project Report
1
2
Transformation about y=-x axis T(V)=A*V
T (V)=
-2
-1
Algorithm
Input
A vector with x and y co ordinates
Body of algorithm
x=Input Enter the x co ordinate
y=Input Enter the y co ordinate
V=(x , y)
A=
0 -1
-1 0
Display Transformation about x axis
Draw the graph of original vector
Draw the graph of transformed vector
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
A=[0 -1;-1 0];
disp('Transformationn about y=-x')
%disp('press any key to view the transformation'),pause
T=A*V
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
disp('Red vector is the transformed vector')
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25. Project Report
Output
enter the x=3
enter the y=4
V=
3
4
Transformationn about y=-x
T=
-4
-3
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Reflection through the origin
Let v be vector in R2 and it is in first quadrant when it is reflected through the origin. Then it is
transformed into 3rd quadrant. To reflect a vector v through the origin we multiply it with
A=
-1 0
0 -1
Example
V=
1
2
Transformation about origin axis T(V)=A*V
T (V)=
-2
-1
Algorithm
Input
A vector with x and y co ordinates
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26. Project Report
Body of algorithm
x=Input Enter the x co ordinate
y=Input Enter the y co ordinate
V=(x , y)
A=
-1 0
0 -1
Display Transformation about x axis
Draw the graph of original vector
Draw the graph of transformed vector
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
disp('Transformationn about origin')
A=[-1 0;0 -1];
T=A*V
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
disp('Red vector is the transformed vector')
Output
enter the x=5
enter the y=6
V=
5
6
Transformationn about origin
T=
-5
-6
Blue vector is the original vector
Red vector is the transformed vector
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27. Project Report
Press any key to exit
Rotation
Let V be a vector in R2 we can rotate it through a an angle (a) and find a transformed vector T(V)
To find a transformed vector T(V) we multiply V with matrix A,Matrix A is given below for
Rotation.
A=
Cos a - Sin a
Sin a Cos a
Example
V=
3
4
a = 450
Rotation T(V)=A*V
T(V) =
-0.7071
4.9497
Algorithm
Input
A vector with x and y co ordinates
Angle of rotation.
Body of algorithm
a=Input Angle of rotation
change degree into radian
A=
Cos a - Sin a
Sin a Cos a
Transormation of V under T is below
T=A*V
Draw graph for original vector
Draw graph for transformed vector
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28. Project Report
Code
Output
enter the x=3
enter the y=4
V=
3
4
About how many degree you want to rotate it 45
Given vector is V
Transformation of V under T is below
T=
-0.7071
4.9497
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Horizontal Contraction and Expansion
Let v be vector in R2 to contract or expand it horizontally we multiply it with
A=
k0 contraction: if 0<k<1
01 expansion: if k>1
Example
V=
3
4
Let k=2
T(V) =A*V
T(V) =
6
4
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29. Project Report
Algorithm:
Input
A vector with x and y co ordinates
A scalar k
Body of algorithm
for k=input. Enter the value of scalar k
A=
k0
01
End for k
T=A*V
if k>0&k<1
display horizontal contraction
else
display horizontal expansion
end for if
Draw graph for original vector
Draw graph for transformed vector
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
for k=input('enter the value of k=')
A=[k 0;0 1];
end
disp('press any key to view the transformation'),pause
T=A*V
if k>0&k<1
disp('horizontal contraction')
else
disp('horizontal expansion')
end
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
disp('Red vector is the transformed vector')
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30. Project Report
Output
enter the x=3
enter the y=4
V=
3
4
enter the value of k=2
press any key to view the transformation
T=
6
4
horizontal expansion
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Vertical Contraction and Expansion
Let v be vector in R2 to contract or expand it vertically we multiply it with
A=
1 0 Contraction: if 0<k<1
0 k Expansion: if k>1
Example
V=
3
4
Let k=2
T(V) =A*V
T(V) =
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31. Project Report
3
8
Algorithm
Input
A vector with x and y co ordinates
A scalar
Body of algorithm
for k=input. Enter the value of scalar k
A=
10
0k
End for k
T=A*V
if k>0&k<1
display horizontal contraction
else
display horizontal expansion
end for if
Draw graph for original vector
Draw graph for transformed vector
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
for k=input('enter the value of k=')
A=[1 0;0 k];
end
T=A*[x;y]
if k>0&k<1
disp('vertical contraction')
else
disp('vertical expansion')
end
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
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32. Project Report
disp('Red vector is the transformed vector')
Output
V=
5
3
enter the value of k=2
T=
5
6
vertical expansion
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Horizontal Shear
Let V be a vector and T(V) be its transformed vector and it is horizontally sheared
We can find T(V) by V→A*V and A is given below.
A=
1 k
0 1
Algorithm
Input
A vector with x and y co ordinates
a scalar k
Body of algorithm;
Horizontal Shear
for k=Input Enter the value of scalar k
A=
1k
01
End for For
T=A*V
Draw graph for original vector
Draw graph for transformed vector
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33. Project Report
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
disp('Horizontal Shear')
for k=input('enter the value of k=')
A=[1 k;0 1];
end
T=A*V
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
disp('Red vector is the transformed vector')
Output
enter the x=2
enter the y=1
V=
2
1
Horizontal Shear
enter the value of k=2
T=
4
1
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Vertical Shear
Let V be a vector and T(V) be its transformed vector and it is horizontally sheared
We can find T(V) by V→A*V and A is given below.
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34. Project Report
Algorithm
Input
A vector with x and y co ordinates
a scalar k
Body of algorithm;
Horizontal Shear
for k=Input Enter the value of scalar k
A=
10
k1
End for For
T=A*V
Draw graph for original vector
Draw graph for transformed vector
Code
x=input('enter the x=') ;
y=input('enter the y=');
V=[x;y]
disp('Vertical shear')
for k=input('enter the value of k=')
A=[1 0;k 1];
end
T=A*V
set(gca, 'Xlim',[-15 15],'Ylim',[-15 15])
grid on
line([0 V(1,1)],[0 V(2,1)],'color','b')
line([0 T(1,1)],[0 T(2,1)],'color','r')
disp('Blue vector is the original vector')
disp('Red vector is the transformed vector')
Output
V=
5
4
Vertical shear
enter the value of k=2
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35. Project Report
T=
5
14
Blue vector is the original vector
Red vector is the transformed vector
Press any key to exit
Cryptology
The study of Secret Message is called cryptology.
Encryption and Decryption
Encryption
In this process we transformed a message into a secret message.
Decryption
In this process we transformed a secret message into an original message.
Algorithm
Input
A messge to encrypt and decrypt
Body of Algorithm
Msg = Input message into a string
N=length of string
Key = Anxn
Encrypted message = Key*(Msg');
Decrypted Message = inv(Key)*Encrypted message.
Code
clc
msg = input('Enter a message: ','s')
clc
X=input('Enter your personal I.D');
n = length(msg);
Key = rand(n,n);
EM = Key*(msg');
a=char(round(EM));
disp('press any key to see the encrypted message'),pause
EncryptedMessage= transpose(a)
disp('press any key to see the decrypted message'),pause
Key1=Key;
DEM = inv(Key1)*EM;
w=char(round(DEM));
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36. Project Report
clc
Y=input('Enter your personal I.D');
if X==Y;
disp('Press any key to see the decrypted message'),pause
DecryptenMessage=transpose(w)
else
disp('Your I.D is incorrect')
end
disp('Press any key to exit'),pause
clc
Output
Enter a message:My Name Is Rizwan
msg =
My Name Is Rizwan
Press any kee to see Encrypted Message
------------------------------------------------
Your ID must be in the form of numeric data
Before seen it Enter your personal ID= 123
Press any kee to see Encrypted Message
----------------------------------------------
EncryptedMessage =
Press any kee to see Decrypted Message
-----------------------------------------------
Before seen it Enter your personal ID= 123
Press any kee to see Dencrypted Message
------------------------------------------------
DecryptenMessage =
My Name Is Rizwan
Press any key to end
----------------------------------------------
Clock
Working
Clock works with the rotation of seconds, minutes and hour hand. When
A second hand completes a loop of 60 then a minute hand moves forword by one and when minute
hand moves 60 times an hour hand moves forword by one but hour hand moves five times larger
than second and minute hands.
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37. Project Report
Input
CPU Clock time
Code
function clock
close all
clc
prog=figure('name','Terminator Clock');
set(prog,'NumberTitle','off');
set(prog,'MenuBar','none');
set(prog,'color','w');
set(prog,'visible','on');
set(prog,'Resize','off');
A=linspace(0,2*pi,5);
B=linspace(0,2*pi,13);
x1=15*cos(A);
y1=15*sin(A);
x2=10*cos(B);
y2=10*sin(B);
plot(x1,y1,'k','linewidth',6)
hold on
plot(x2,y2,'k','linewidth',8)
axis([-12 12 -12 12])
axis off
axis equal
set(gca,'XLim',[-12 12],'YLim',[-12 12])
grid off
text(0,8,'12','FontSize',18,'HorizontalAlignment','center','color','k');
text(4.2,7,'1','FontSize',18,'HorizontalAlignment','center','color','k');
text(7,4,'2','FontSize',18,'HorizontalAlignment','center','color','k');
text(8.3,0,'3','FontSize',18,'HorizontalAlignment','center','color','k');
text(7,-4,'4','FontSize',18,'HorizontalAlignment','center','color','k');
text(4.2,-7,'5','FontSize',18,'HorizontalAlignment','center','color','k');
text(0,-8,'6','FontSize',18,'HorizontalAlignment','center','color','k');
text(-4.2,-7,'7','FontSize',18,'HorizontalAlignment','center','color','k');
text(-7,-4,'8','FontSize',18,'HorizontalAlignment','center','color','k');
text(-8.3,0,'9','FontSize',18,'HorizontalAlignment','center','color','k');
text(-4.2,7,'11','FontSize',18,'HorizontalAlignment','center','color','k');
text(-7,4,'10','FontSize',18,'HorizontalAlignment','center','color','k');
title(Date)
K = clock();
for hr=K(1,4):24
for min=K(1,5):60
for sec=K(1,6):60
timar = timer('TimerFcn','a=15;','StartDelay',1);
start(timar);
Anglesec=pi/2-sec*pi/30;
line([0 7*cos(Anglesec)],[0 7*sin(Anglesec)],'Color','k','linewidth',1.5);
Anglemin=pi/2-min*pi/30;
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38. Project Report
line([0 6.5*cos(Anglemin)],[0 6.5*sin(Anglemin)],'Color','k','linewidth',2.5,'marker','none');
Anglehr=pi/2-hr*pi/6;
line([0 5*cos(Anglehr)],[0 5*sin(Anglehr)],'Color','k','linewidth',2.5);
pause(1);
line([0 7*cos(Anglesec)],[0 7*sin(Anglesec)],'Color','w','marker','none','linewidth',1.5);
line([0 6.5*cos(Anglemin)],[0 6.5*sin(Anglemin)],'Color','w','linewidth',2.5,'marker','none');
line([0 5*cos(Anglehr)],[0 5*sin(Anglehr)],'Color','w','linewidth',2.5,'marker','none');
end
K(1,:)=1;
end
end
end
Output
System of Linear equations
An equation in which the sum of the product of variables is one is called the linear equation. A
system of linear equation can be solved by many methods. There are most important methods are
here.
Jacobi’s Method
Let a11x1+a12x2+a13x3 = b11 , a21x1+a22x2+a23x3 = b21 , a31x1+a32x2+a33x3 = b31 be a
System of linear equation.
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39. Project Report
Let it is diagonally dominant, then
xi+1 = (b11-a12 yi -a13 zi)/a11.
y i+1 = (b21-a21 xi -a23 zi)/a21.
z i+1 = (b31-a31 xi -a32 yi)/a31.
Initially use the value of x=y=z=0, for first iteration. For second iteration we use the values of first
iteration and in the similar way going on until the values of x, y, z converges to particular values.
If the are not converges then the solution is not exist.
Note:-If systemis not diagonally dominant then it may not converge to a particular number
Algorithm
Input
Matrix A and Vector B
Body of Algorithm
Input Matix A and vector B
If A is square Matrix
for i=1:a;
xp(i)=0;
display xp(i) =0
end
t=input number of iterations
for k=1:t;
for i=1:a;
x(i)=B(i,1);
for j=1:b;
if i~=j;
x(i)=x(i)-(A(i,j)*xp(j));
end
end
x(i)=x(i)/A(i,i);
end
for p=1:a;
xp(p)=x(p);
end
else
display message System does not satisfied the conditions of *------ JACOBI METHOD -----*
end
Code
clc
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40. Project Report
clear
a=input('Enter the number of rows of A ');
b=input('Enterthe number of coloumns of A ');
disp('*********************************************');
disp('*********************************************');
disp('Enter the matrix of coefficient of diagonally dominant matrix');
disp('*********************************************');
for i=1:a;
for j=1:b;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j) ' of matrix A']);
A(i,j)=input('Enter the entry ');
end
end
(A)
for i=1:a;
disp(['enter the entry of Row ' int2str(i) ' Column 1 of cloloumn vector B']);
B(i,1)=input('Enter the entry ');
end
(B)
disp('press any key to continue'),pause
clc
if a==b;
for i=1:a;
xp(i)=0;
disp(['Let initial values of x' int2str(i) '=0'])
end
t=input('How many number of iterations you want to proceed== ');
for k=1:t;
disp(['Iteration number is ' int2str(k)]);
for i=1:a;
x(i)=B(i,1);
for j=1:b;
if i~=j;
x(i)=x(i)-(A(i,j)*xp(j));
end
end
x(i)=x(i)/A(i,i);
disp(['Value of x' int2str(i) ' = ' num2str(x(i))]);
end
for p=1:a;
xp(p)=x(p);
end
disp('press any key to see the next iteration'),pause
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41. Project Report
clc
end
disp('*************************************************');
disp('*********************************************');
disp('*********************************************');
disp('The exact solution is below');
disp('*********************************************');
for i=1:a;
disp(['x' int2str(i) ' = ' num2str(x(i))]);
end
disp('press any key to Continue'),pause
clc
disp('NOTE:--If your matrix is diagonally dominent and the solution is not exact Then you should
to proceed it to further iteratoins');
disp('****************************************************************************
********************')
disp('NOTE:--If your matrix is not diagonally dominent, Then the is system is not convergent and
solution may be wrong');
else
disp('*********************************************');
disp('*********************************************');
disp('System does not satisfied the conditions of *------ JACOBI METHOD -----* ');
end
disp('Press any key to End'),pause
clc
Output
Enter the number of rows of A 3
Enterthe number of coloumns of A 3
*********************************************
*********************************************
Enter the matrix of coefficient of diagonally dominant matrix
*********************************************
enter the entry of Row 1 Column 1 of matrix A
Enter the entry 5
enter the entry of Row 1 Column 2 of matrix A
Enter the entry 1
enter the entry of Row 1 Column 3 of matrix A
Enter the entry 1
enter the entry of Row 2 Column 1 of matrix A
Enter the entry 1
enter the entry of Row 2 Column 2 of matrix A
Enter the entry 7
enter the entry of Row 2 Column 3 of matrix A
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42. Project Report
Enter the entry 1
enter the entry of Row 3 Column 1 of matrix A
Enter the entry 1
enter the entry of Row 3 Column 2 of matrix A
Enter the entry 1
enter the entry of Row 3 Column 3 of matrix A
Enter the entry 3
A=
5 1 1
1 7 1
1 1 3
enter the entry of Row 1 Column 1 of cloloumn vector B
Enter the entry 5
enter the entry of Row 2 Column 1 of cloloumn vector B
Enter the entry 7
enter the entry of Row 3 Column 1 of cloloumn vector B
Enter the entry 3
B=
5
7
3
press any key to continue
---------------------------------
Let initial values of x1=0
Let initial values of x2=0
Let initial values of x3=0
How many number of iterations you want to proceed== 16
Iteration number is 1
Value of x1 = 1
Value of x2 = 1
Value of x3 = 1
press any key to see the next iteration
-----------------------------
Iteration number is 2
Value of x1 = 0.6
Value of x2 = 0.71429
Value of x3 = 0.33333
press any key to see the next iteration
-------------------------------------
Iteration number is 3
Value of x1 = 0.79048
Value of x2 = 0.86667
Value of x3 = 0.5619
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43. Project Report
press any key to see the next iteration
-------------------------------
Iteration number is 4
Value of x1 = 0.71429
Value of x2 = 0.8068
Value of x3 = 0.44762
press any key to see the next iteration
--------------------------
Iteration number is 5
Value of x1 = 0.74912
Value of x2 = 0.83401
Value of x3 = 0.49297
press any key to see the next iteration
------------------------
Iteration number is 6
Value of x1 = 0.7346
Value of x2 = 0.82256
Value of x3 = 0.47229
press any key to see the next iteration
------------------------------
Iteration number is 7
Value of x1 = 0.74103
Value of x2 = 0.82759
Value of x3 = 0.48095
press any key to see the next iteration
-------------------------
Iteration number is 8
Value of x1 = 0.73829
Value of x2 = 0.82543
Value of x3 = 0.47713
press any key to see the next iteration
--------------------------------
Iteration number is 9
Value of x1 = 0.73949
Value of x2 = 0.82637
Value of x3 = 0.47876
press any key to see the next iteration
-----------------------------------
Iteration number is 10
Value of x1 = 0.73897
Value of x2 = 0.82596
Value of x3 = 0.47805
press any key to see the next iteration
---------------------------------
Iteration number is 11
Value of x1 = 0.7392
Value of x2 = 0.82614
Value of x3 = 0.47835
press any key to see the next iteration
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44. Project Report
--------------------------------
Iteration number is 12
Value of x1 = 0.7391
Value of x2 = 0.82606
Value of x3 = 0.47822
press any key to see the next iteration
---------------------------
Iteration number is 13
Value of x1 = 0.73914
Value of x2 = 0.8261
Value of x3 = 0.47828
press any key to see the next iteration
-----------------------------
Iteration number is 14
Value of x1 = 0.73913
Value of x2 = 0.82608
Value of x3 = 0.47825
press any key to see the next iteration
------------------------------
Iteration number is 15
Value of x1 = 0.73913
Value of x2 = 0.82609
Value of x3 = 0.47826
press any key to see the next iteration
-------------------------------
Iteration number is 16
Value of x1 = 0.73913
Value of x2 = 0.82609
Value of x3 = 0.47826
press any key to see the next iteration
----------------------------------
*************************************************
*********************************************
*********************************************
The exact solution is below
*********************************************
x1 = 0.73913
x2 = 0.82609
x3 = 0.47826
press any key to Continue
-----------------------------------
NOTE:--If your matrix is diagonally dominant and the solution is not exact Then you should to
proceed it to further iterations
*********************************************************************************
***************
NOTE:--If your matrix is not diagonally dominant, Then the is system is not convergent and solution
may be wrong
Press any key to End
------------------------------------
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45. Project Report
Gauss Seidel’s Method
Gauss elimination method is more better form it converge rapidly as compare to the Jacobi’s Method
There is a little difference from Jacobi that to calculate the value of x1, we initially use y1 and z1
equal to zero. Similarly to calculate the value of y1, we initially use x1 and z1 equal to zero and to
calculate the value of z1, we initially use x1 and y1 equal to zero.For next iteration use the values of x,
y, z from previous iteration.
Algorithm
Input
Matrix A and Vector B
Body of Algorithm
Input Matix A and vector B
If A is square Matrix
for i=2:a;
x(i)=0;
display x(i) =0
end
t=input number of iterations
for k=1:t;
for i=1:a;
x(i)=B(i,1);
for j=1:b;
if i~=j;
x(i)=x(i)-(A(i,j)*x(j));
end
end
x(i)=x(i)/A(i,i);
end
for p=1:a;
xp(p)=x(p);
end
else
display message System does not satisfied the conditions of *------ Gauss Method -----*
end
Code
clc
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46. Project Report
clear
a=input('Enter the number of rows of A ');
b=input('Enterthe number of coloumns of A ');
disp('*********************************************');
disp('*********************************************');
disp('Enter the matrix of coefficient of diagonally dominant matrix');
disp('*********************************************');
for i=1:a;
for j=1:b;
disp(['enter the entry of Row ' int2str(i) ' Column ' int2str(j) 'of matrix A']);
A(i,j)=input('Enter the entry ');
end
end
(A)
for i=1:a;
disp(['enter the entry of Row ' int2str(i) ' Column 1 of cloloumn vector B']);
B(i,1)=input('Enter the entry ');
end
(B)
disp('press any key to continue'),pause
clc
if a==b;
for i=1:a;
x(i)=0;
disp(['Let initial values of x' int2str(i) '=0'])
end
t=input('How many number of iterations you want to proceed ');
for k=1:t;
disp(['Iteration number is ' int2str(k)]);
for i=1:a;
x(i)=B(i,1);
for j=1:b;
if i~=j;
x(i)=x(i)-(A(i,j)*x(j));
end
end
x(i)=x(i)/A(i,i);
disp(['Value of x' int2str(i) ' = ' num2str(x(i))]);
end
disp('press any key to see the next iteration'),pause
clc
end
disp('*************************************************');
disp('*********************************************');
disp('*********************************************');
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47. Project Report
disp('The exact solution is below');
disp('*********************************************');
for i=1:a;
disp(['x' int2str(i) ' = ' num2str(x(i))]);
end
disp('press any key to Continue'),pause
clc
disp('NOTE:--If your matrix is diagonally dominent and the solution is not exact Then you should
to proceed it to further iteratoins');
disp('****************************************************************************
********************')
disp('NOTE:--If your matrix is not diagonally dominent, Then the is system is not convergent and
solution may be wrong');
else
disp('*********************************************');
disp('*********************************************');
disp('System does not satisfied the conditions of *------ Gauss Elimination Method -----* ');
end
disp('Press any key to End'),pause
clc
Output
Enter the number of rows of A 3
Enterthe number of coloumns of A 3
*********************************************
*********************************************
Enter the matrix of coefficient of diagonally dominant matrix
*********************************************
enter the entry of Row 1 Column 1of matrix A
Enter the entry 5
enter the entry of Row 1 Column 2of matrix A
Enter the entry 1
enter the entry of Row 1 Column 3of matrix A
Enter the entry 1
enter the entry of Row 2 Column 1of matrix A
Enter the entry 1
enter the entry of Row 2 Column 2of matrix A
Enter the entry 7
enter the entry of Row 2 Column 3of matrix A
Enter the entry 1
enter the entry of Row 3 Column 1of matrix A
Enter the entry 1
enter the entry of Row 3 Column 2of matrix A
Enter the entry 1
enter the entry of Row 3 Column 3of matrix A
Enter the entry 3
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48. Project Report
A=
5 1 1
1 7 1
1 1 3
enter the entry of Row 1 Column 1 of cloloumn vector B
Enter the entry 5
enter the entry of Row 2 Column 1 of cloloumn vector B
Enter the entry 7
enter the entry of Row 3 Column 1 of cloloumn vector B
Enter the entry 3
B=
5
7
3
press any key to continue
------------------------
Let initial values of x1=0
Let initial values of x2=0
Let initial values of x3=0
How many number of iterations you want to proceed 7
Iteration number is 1
Value of x1 = 1
Value of x2 = 0.85714
Value of x3 = 0.38095
press any key to see the next iteration
-----------------------------
Iteration number is 2
Value of x1 = 0.75238
Value of x2 = 0.8381
Value of x3 = 0.46984
press any key to see the next iteration
-------------------------------
Iteration number is 3
Value of x1 = 0.73841
Value of x2 = 0.82739
Value of x3 = 0.47807
press any key to see the next iteration
--------------------------------
Iteration number is 4
Value of x1 = 0.73891
Value of x2 = 0.82615
Value of x3 = 0.47831
press any key to see the next iteration
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49. Project Report
--------------------------------
Iteration number is 5
Value of x1 = 0.73911
Value of x2 = 0.82608
Value of x3 = 0.47827
press any key to see the next iteration
--------------------------------
Iteration number is 6
Value of x1 = 0.73913
Value of x2 = 0.82609
Value of x3 = 0.47826
press any key to see the next iteration
-----------------------------
Iteration number is 7
Value of x1 = 0.73913
Value of x2 = 0.82609
Value of x3 = 0.47826
press any key to see the next iteration
----------------------------
*************************************************
*********************************************
*********************************************
The exact solution is below
*********************************************
x1 = 0.73913
x2 = 0.82609
x3 = 0.47826
press any key to End
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