Chap 3

SOLUTIONS MANUAL COURTESY

Amir Tahir Khan (2005–CE-152) Saba Khalid*

(2005–CE-145)

Shahzaib Zahoor (2005–CE-133) Sarwat Hasnat

(2005–CE-129)

Haider Ali

(2005–CE-164) SECTION ‘C’

COMPUTER ENGINEERING

This manual contains solutions to all of the problems of Chapter 3 ‘Solution of Linear Systems AX=B’ in Numerical Methods Using MATLAB, Fourth Edition. If you spot an error in a solution or in the wording of a problem, we would greatly appreciate it if you would forward the information via email to us at [email protected]

Submitted To Sir Muhammad Jamil Usmani Lecturer,Mathematics Department, SSUET

INDEX S.NO. 1

EXERCISE

Page

OBJECTIVE

#

Introduction to

3-24

vectors And Matrices 2

Properties of Vectors Upper Triangular

25-33

Gaussian Elimination

34-44

Iterative Methods for Linear Systems

Sarwat Hasnat 2005-CE-129

45-60

And Pivoting 5

Haider Ali 2005-CE-164

linear Systems 4

Amir Tahir Khan 2005-CE-152

And Matrices 3

SOLVED BY

Shahzaib Zahoor Khan 2005-CE-133

61-96

Saba Khalid* 2005-CE-145

2

CHAPTER 3 EXERCISES FOR INTRODUCTION To VECTOR & MATRICES

Q1.

Given the vectors x and Y, find (a) x+y, (b) x- Y, (c) 3x , ( d) II X II (e) 7Y – 4 X, (f) x. y, (g) 117Y – 4 X II.

(i) X = ( 3, - 4 ) and Y = ( - 2, 8)

SOLUTION :(a) X + Y The sum of the vectors X and Y is computed component by component. X + Y = ( X, + Y, X2 + Y2 , Xn + Yn) We have, X = 3, X2 = - 4,Y1 = -2 , Y2 = 8 Putting the values We get, (3,- 4) + (-2,8) = (3+ (-2) , - 4+8) = ( 3-2,4) = (1,4) Ans.

3

(b) X- Y The difference X – Y is formed by taking the difference in each coordinate: X _ Y = (X1 = Y1 , Y2 ------ Xn – Yn )

Putting the values We got,

= (3,-4) – (-2,8) = (3, -(-2) , - 4-8) = (3+2) , - 12) = (5, - 12)

(c) 3 X If C is a real number we define scalar multiplication as follows: CX = ( CX1, CX2 _ _ _ _ (Xn)

Putting the values we get, We get, 3 (3 , - 4) = ( 3 X 3, 3 X- 4) = ( 9 - 12)

(d) || X ||

4

The norm (or length) of the vector X is defined by , ||X ||= (X21 + X22 + ------- X2n )1/2 Putting the values , we get ||3 -4|| = (3)2 + (-4)2 )1/2 = ( 9+16)1/2 =(25)1/2 = (e)

5 Ans.

7Y – 4X

If c & d are scalars then the weighted difference dY – CX becomes, dY – cX =(dY-1 – CX1, dy2 – CX2-----dy2 - Cxn ) Let C = 4 & = 7 Putting the values, we get, 7(-2,8) – 4(3, -4) = (7 ( -2) -4(3) = (8)-4(-4)) = (14-12, 56 + 16)

(f) X. Y The dot product of two vectors , X and Y can be written as, X. Y = x1 y1 + x2 y2 + ------------ xn yn Putting the values , we get,

5

(3,-4) . (-2 ,8) = 3x-2+4 x 8 = - 6- 32 = - 38 Ans.

(g) ||7Y- 4X|| The distance between two points X to Y, ||Y – X|| = ((Y1-X1 )2 +( y2 – X2)2 + ------ (Yn-Xn)2 )1/2 According to condition, || dy – CX|| = (( dy1 – CX1) + ( Y2 – CX2 )2 + ------ (dYn-CXn)2 )1/2 Let C = 4 , d =7 Putting the values We get, || 7 ( -2,8) – ( 3\3-4) || = (( 7 x-2) 4x3)2 + ( 7 X 8 – 4 X -4)2 )1/2 = (( -14-12)2 + (56 +16)2 )1/2 = ((-26)2 + (72)2 )1/2 = (676 + 5184)1/2 = (5860)1/2 =2 or

Q1.

1465

76.55063684 Ans.

ii) X = ( -6, 3,2) & Y = (- 8,5,1) (a) X + Y

6

The sum of the vectors X & Y is computed component by component, X+ Y = (X1+Y1 X2 +Y2, --------- (Xn-Yn) We have X1 = -6, X2 = 3,X3 = 2, Y1 =- 8,y2 = 5 Y3 = 1 Putting the values We get, X + Y = ( - 6-8, 3+5 2+1) X + Y = ( - 14,8,3) Ans. (b)

X-Y

The difference of X – Y is formed by taking the difference in each coordinate, X + Y = (X1 – Y, X2 – y2, ------------- Xn – Yn) Putting the values we get, X – Y ( -6+ 8, 3-5, 2-1) X-Y = (2, - 2 ,1) (c)

3X

If C is a real number we define scalar multiplication as follows: C X = (3X-6, 3 X 3 , 3 X 2) 3 X = ( -18, 9,6) (d) || X || The norm (or length) of the vector X is defined by,

7

|| X || = (X12 + X12 + --------- Xn2)1/2 Putting the values We get, || X || = ((-6)2 + (3)2 + (2)2 )1/2 = ( 36 + 9 + 4)1/2 = (49)1/2 || x|| =7 Ans. (e) 7Y – 4X Let C = 4 & d =7 It C & D are scalars then weighted difference dY – cX becomes, dY – cX = ( dY1 – cX1, dY2 – cX2 ------- dYn – (Xn ) Putting the values We get, 7Y - 4X =(( 7X – 8) –(4X – 6),(7 X 5) – ( 4 X 3 )( 7 X 1) – (4 X 2)) = (- 56 + 24 , 35 – 12, 7 – 8 ) 7Y – 4X = (-32 , 23,-1) Ans. (f) X.Y The dot product of two vectors, X and Y Can be written as, X.Y = X1 Y1 + X2 Y2 + ---------- Xn Yn Putting the values we get, X .Y = ( C-6 X – 8) + ( 3X 5 ) + (2 X 1) = 48+ 15+ 2 X.Y 65 Ans.

8

(g) || 7Y – 4X|| The distance between two points X to Y, ||Y – X|| = ((y1 – X1)2 + ( Y2 –X2)2+ ----- (yn –Xn)2 )1/2 we find, || 7Y - 4X|| = ((7y1 - 4X1 )2 + (7y2 – 4x2)2 + (7Y3 – 4x3)2)1/2 Putting the values We get,

|| 7Y-4X|| = (( 7X -8)2 - (4X-6)2 + (7X5)2 – (4x3)2 + (7X1)2 – (4 X 2)2 )1/2

= ( ( -56) – (-24)-2 +(35)2 – (12)2 + (7)2 - (8)2 )1/2 = (3136 – 576 + 1225 – 144+49-64)1/2 = (3626)1/2 || 7Y -4X|| Q.1

(iii)

= 60.21627687 Ans.

X = ( 4-8,1) & Y = ( 1,-12,-11)

(a) X +Y The sum of the vectors X & Y is computed component by component. X+Y = (X1+ Y

1,

X2 , --------- (Xn + Yn)2 )

We have, X = 4, X X2 – 8, X3 = 1, Y, = 1, Y2 =- 12,Y3 = - 11 We get,

9

X+Y = (4+1, - 8- 12 ,1-11) = (5,-20, -10) Ans. (b) X – Y The difference X – Y is formed by taking the difference in each coordinate: X-Y = (X1+ Y

1,

X2 , --------- (Xn + Yn)2 )

X - Y = (4 - 1, - 8 + 12 ,1+11) = (3,4,12) Ans. (C) 3X It c is a real number , we define scalar multiplication as follows:CX = (CX,CX2------- CXn) Putting the values we get, 3X = ( 3X4, 3X-8, 3X1) 3X = ( 12- 24, 3) Ans. (d) || X || The norm ( or length) of the vector X is defined by, || X || = (X12 + X12 + ------ Xn2 )1/2 Putting the values we get, || X || = ((4)2 + (-8)2 + ------ (1)2 )1/2 = (16+ 64 + 1)1/2 = (81)1/2 || X || = 9 Ans.

10

(e) 7Y – 4X If C & D are scalars then the weighted difference dY – cX becomes, dy- Cx = ( dy1 – CX1 dy2 - (x2, ---------- dyn – CXn ) Putting the values we get, 7Y – 4X = ( 7( 1) – 4(4), 7 (-12) -4 ( -8) , 7 ( -11) – 4 (1)) = (7-16 , - 84 + 32, - 77 – 4) = ( -9 , -52, - 81) Ans. (f) X. Y The dot product of two vectors, X and Y can be written as, X.Y = X1 + X1 + X2 + X2 + ------ Xn Yn Putting the values we get, X. Y = ((x1) + ( -8X -12) + ( 1X -11) ) = ( 4+ 96 – 11) X . Y = 89 Ans.

(g) || 7Y – 4X|| The distance between two points X to Y, ||Y – X || = ( ( Y1 + Y1 )2 + ( Y2 + X2 )2 + ------ ( Xn Yn )2 )1/2 ||7Y – 4 X|| = ( ( 7Y1 + 4X1 )2 + ( 7Y2 + 4X2 )2 )1/2 Putting the values we get, || 7Y -4 X || = (( 7 X 1 – 4 X 4)2 + ( 7x – 12 – 4 x -8) + ( 7X-11 -4 X1)2 )1/2 = ((7-16)2 + (-84 + 32)2 +(-77 – 4))1/2

11

= (( - 9 )2 + ( -52)2 + (-81)2 )1/2 = (9346)1/2 || 7Y -4 X || = 96.67471231 Ans. Q1.

(iv) X = ( 1,-2,4,2) & Y= ( 3-5,- 4,0) X1 = 1, X2 = -2, X3 = 4, X4 = -2, & Y1 = 3, Y2 = -5, Y3 = -4, Y4 = 0,

(a) X + Y The sum of the vectors X & Y is computed component by component, X + Y = (X1 + Y1, X2 + Y2 + ------- Xn + Yn ) Putting the values we get, X + Y = (1+3, - 2- 5, 4-4, 2+0) X + Y = (4,-7,0,2) Ans.

(b) X – Y The difference X – Y is formed by taking the difference in each coordinate: X - Y = (X1 - Y1, X2 - Y2 - ------- Xn - Yn ) Putting the values we get, X + Y = (1- 3, - 2 + 5, 4+4, 2-0) X + Y = (-2,3,8,2) Ans. (c) 3X If C is a real number we define the scalar multiplication as follows:

12

CX = ( CX1, CX2, -------------CXn) Putting the values we get 3X = ( 3 X 1 , 3X -2 , 3 X 4, 3 X 2) 3 X = ( 3, - 6, 12, 6 ) Ans. (d) || X || The norm ( or length) of the vector X is defined by, || X || = (( X1 2 + X2 2 + ----------- Xn 2 ))1/2 Putting the values we get, || X || = ( (1)2 + (-2)2 + (4)2 +(2)2 )1/2 = ( 1+ 4 + 16 + 4 )1/2 = (25)1/2 || X|| = 5 Ans. (e) 7Y – 4X If C & d are scalars then weighted difference dY – Cx becomes, dY –cX = ( dY1 – Cx1 dy2 – CX2 ------ dYn – Cxn ) Putting the values we get, 7Y – 4X = ( (7 X 3 ) – (4 X 1) , ( 7 X -5 ) – ( 4X – 2), 7Y – 4X– (4 X 4) + ( 7 X 0 ) – ( 4 X 2), = ( 21-4,-35+8,-28,-16, 0-8) 7Y – 4X = ( 17, - 27, - 44, - 8 ) Ans.

(f) X.Y The dot product of two vectors X & Y can be written as,

13

X. Y = X1 Y1 + X2 Y2 + ------- Xn Yn Putting the values We get, X . Y = (( 1x 3) + (-2 X – 5) + ( 4 X 4) + ( 2 x 10) ) X . Y = (3+10-16+0) X . Y = - 3 Ans.

(g) || 7Y – 4 X || The distance between two points X to Y, || Y – X || = (( Y1 - X1 )2 + ( Y2 - X2 )2 + ----- ( Yn - Xn )2 Let C = 4 & d = 7 If C & d are scalar then, || 7Y – 4 X || = ((7Y1 - 4X1 )2 + (7Y2 - 4X2 )2 +----- (7Y3 - 4X3 )2 Putting the values We get, ||7Y- 4X|| = ( 7X3- 4X1)2 + ( 7X-5 – 4X-2)2 + ( 7X – 4 – 4 X 4)1/2 = ((17)2 + (- 27) 2 + (- 44) 2 + (- 8) 2 )1/2 = (289 + 7729+1936+64)

1/2

= (3018)1/2 || 7y -4X|| = 54.93632678 Ans. Q.2

Using the law of cosines, it can be shown that the angle θ

between two vectors X and Y is given bythe relation.

14

Cos ( θ) =

X.Y, || X || || Y ||

Find the angle, in radians, between the following vectors: (a) X = (-6,3,2) and Y = ( 2,-2,1) X1 =-6, X2 =3, X3 =2, & Y1 = 2, Y2 = -2, Y3 = 1,

SOLUTION: We know that, X.Y = X1 Y1 + X2 Y2 + X3 Y3 ------------- (1) Putting the values in equation

(1)

We get, = (-6) (2) + (3) ( -2) + ( 2) (1) = - 12 – 6 + 2 X.Y

= - 16

Also, || X || = (X12 X22 + X12 )

1/2

------------ (2)

Putting the values in equation ------------------ (2) We get, = (( - 6)2 + (3)2 + (2)2 )1/2 = (36 + 9 + 4 )1/2 = (49)1/2 || X || = 7 Now,

15

|| Y || = (Y12 Y22 + Y12 )

1/2

------------ (3)

Putting the values in eq. ………… (3) We get, = (2)2 + (-2)2 + (1)2 )1/2 = (4+4 +1)1/2 = (9)1/2 || Y || = 3 Since, Cos θ =

X.Y, || X || || Y || ------------(4)

Putting the values in equation …………. (4) (4)

=>

Cos θ =

− 16 7 x3

− 16 21

θ = Cos-1 ( -0.76190) θ = 2.437045 radians Ans. Q.2 (B) X = (4-8,1) and Y = ( 3,4,12) X1 = 4, X2 = -8, X3 = 1, & Y1 = 3, Y2 = 4, Y3 = 12,

SOLUTION We know that X.Y = X1 , Y1 + X2 , Y2 + X3 , Y3 ------------ (A) Putting the values in equation …………… (A) WE get,

16

= (( 4 X3) (-8) (4) + (1) (12)) = 12- 32 + 12 X. Y =-8 Also , || X || = ( X1 2 + X2 2 + X3 2 )1/2 ------------ (B) Putting the values in equation………………. (B) We get, B => = (42 + ( -8)2 + (1)2 )1/2 = (16+64+1)1/2 = (81)1/2 || X || = 9 Now, || Y || = ( Y1 2 + Y2 2 + Y3 2 )1/3 ------------ (C) Putting the values in equation…………. (C) (C) =>

= (32 + 42 + 122 )1/2

= (9+16+144)1/2 = (169)1/2 || Y || = 13 since Cos θ =

X.Y || X || || Y || ------------(D)

Putting the values in equation (D)

17

We get,

- 8, Cos θ = 9 x 13 ------------(4) - 8, Cos θ = 117 ------------(4) Cos θ = -0.06837 θ = Cos-1 (-0.06837) θ = 1.639225 radians Ans. Q3.

Two vectors X and Y are said to be orthogonal (perpendicular) if the angle between them is

∧

12.

(a)prove that X and Y are both orthogonal if and only if X.Y Proof:Assume that : X , Y ≠ 0 X.Y = 0 if and only if Cos θ =0 If and only if θ = (2n + 1)

∧ 2

If and only if X and y are or trogon Proved Use Part (a) to determine it the following vectors are orthogonal.

(b) X = (-6,4,2) and Y = (6,5,8) X1 =- 6, X2 = 4, X3 =- 2, & Y1 = 6, Y2 = 5, Y3 = 8,

SOLUTION :

18

WE KNOW THAT X.Y = X1 Y1 + X2 Y2 + X3 Y3 Putting the values we get, X.Y = ( -6) (6) + (4) (5) + (2) (8) = - 36+20+16 = - 36+36 X.Y = 0 Hence it is proved that given vectors are orthogonal Q.3 (c) X = (-4,8,3) and Y = (2,5,16) X1 =- 4, X2 = 8, X3 =- 3, & Y1 = 2, Y2 = 5, Y3 = 16,

SOLUTION:We know that X.Y = X1 Y1 + X2 Y2 + X3 Y3 Putting the values we get, X.Y = ( -4) (2) + (8) (5) + (3) (16) = - 8 + 40 + 48 = 90 ( ≠ 0 ) Hence the given vectors are not orthogonal.

Q3.

(d) X = (-5,7,2) and Y = ( 4,1,6)

SOLUTION: We know that

19

X.Y = X1 Y1 + X2 Y2 + X3 Y3 Putting the values we get, X.Y = ( -5) (4) + (7) (1 + (2 (6) = - 20+7+12 X.Y = -1 (≠ 0 ) Hence the given vectors are not orthogonal because X.Y ≠ 0.

Q4.

Find (a) A + B, (b) A-B and (c) 3A-2B for the matrices

A=

−1 9 4 2 −3 −6 0 5 7

,B=

2 −4 9 3 −5 7 8 1 −6

SOLUTION: (a) A + B

A+B=

−1 9 4 2 −3 −6 0 5 7

A+B=

4+2 −1− 4 9 + 9 2+3 −3+3 −6+ 7 0+8 5 +1 7−6

A+B=

− 5 18 6 5 0 1 8 6 1

2 −4 9 3 −5 7 8 1 −6

,+

Ans.

(b) A – B 20

A+B=

−1 9 4 2 −3 −6 0 5 7

A+B=

4−2 −1+ 4 9 − 9 2−3 −3−3 −6−7 0 −8 5 −1 7+6

A+B=

−3 0 2 − 1 2 − 13 − 8 4 13

,-

2 −4 9 3 −5 7 8 1 −6

Ans.

(c) 3A – 2B

3A

=

− 3 27 12 6 − 9 − 18 0 15 21

2B =

4 − 8 18 6 − 10 14 16 2 − 12

3B =

− 3 27 12 6 − 9 − 18 0 15 − 21

3B =

− 3 + 8 27 − 18 12 − 4 6 − 6 − 9 + 10 − 18 − 14 0 − 16 15 − 2 12 + 12

-

14 − 8 18 6 − 10 14 16 2 − 12

21

3A – 2B

Q.5

=

5 9 8 0 1 − 32 − 16 13 133

The Transpose of an MXN matrix A, denoted A" , is the N x matrix obtained from A by converting the rows of A to columns of A' that is , if A = [ bij]NXM , then the elements satisfy the relation. bji = aij for 1 ≤ ι ≤ M,1 ≤ j ≤ N Find the transpose of the following matrices.

(a)

− 2 5 12 1 4 −1 7 0 6 11 − 3 8

SOLUTION: Let the given matrix is A then, Transpose of a matrix

Or A1 =

Q.5

(b)

− 2 1 7 11 5 4 0 −3 12 − 1 6 8

Ans.

4 9 2 3 5 7 8 1 6

SOLUTION :

22

Let the given matrix is A then, Transpose of A

Or

Q.6

4 3 8 9 5 1 2 7 6

A1 =

Ans.

The square matrix A of dimension NX N is said to be symmetric if A= A´ (see Exercise 5 for the definition of A´) Determine whether the following square matrices are symmetric.

Or

1 −7 4 −7 2 0 4 0 3

A´ =

Ans.

SOLUTION

1 −7 4 −7 2 0 4 0 3

Let A =

A´ =

1 −7 4 −7 2 0 4 0 3

According to given condition, (A= A´) the given matrix is symmetric.

(b)

4 −7 1 0 2 −7 3 0 4

23

SOLUTION:

Let A =

A´ =

4 −7 1 0 2 −7 3 0 4

4 0 3 −7 2 0 1 −7 4

A ≠ A´ Hence it is not symmetric Q.6

(C) A = [ a ι j ]NXN where  ji aιj =   j − ji + i

j = i  j ≠ i

aij=aji According to given condition, the given matrix is symmetric. Q.6

(d) A = [ a i j ]NXM, where

Cos( ji) aιj =  i − ij − j

i= i≠

j  j

Solution: Cos( ji) aιj =  i − ij − j

i= i≠

j  j

aij=aij According to given condition the given matrix is symmetric.

24

Properties of Vectors And Matrices

25

Q1.

Find AB & BA

−3 2 1 4

A=

5 0 2 −6

B=

SOLUTION

−3 2 1 4

AB =

− 15 + 4 0 − 12 5+8 0−6

=

Ans.

−3 2 1 4

5 0 2 −6

BA =

− 15 10 Ans. − 12 − 20

= Q2

5 0 2 −6

Find AB & BA SOLUTION

1 −2 3 2 0 5

B

3 0 −1 5 3 −2

3 + 2 + 9 − 10 − 6 = 6 + 15 − 10

14 − 16 21 − 10

A=

AB =

3 −6

BA =

−1+ 1

2 − 3 + 25

−1 − 6

3 −6

39 −1

=

9

2

Ans. 9 22

Ans.

−1 − 6 −1

26

Q3 A =

3 1 0 4

1 2 −2 −6

B =

(a) Find (AB) C

&

c=

2 −5 3 4

A (BC)

(b) Find A (B+C) & AB + AC (c) Find (A+B) C & AC + BC (d) Find (AB) ´ and B´ A´ Required matrices for the above

1 0 − 8 24

A=

B+C =

(a)

3 −3 1 −2

(AB) C =

=

=

BC =

A+ B =

1 0 − 8 − 24

8 3 − 22 − 14

AC =

9 − 11 12 16

4 3 −2 −2

2 −5 3 4

1(2) + (0)( 4) 1(−5) + 0(4) − 8(2) + (−24)3 − 8(−5) + (−24)4

2 −5 − 88 − 56

SOLUTION A (BC)

27

8 −3 − 22 − 14

3 1 0 4

3(8) + (1)(−22) 3(3) + 1(−14) 0(8) + 4(−22) 0(3) + 4(−4)

= (b)

=

2 −5 − 88 − 56

A (B + C) and AB + AC SOLUTION A (B + C)

3(3) + 1(1) 3(−3) + (1)(−2) 0(3) + 4(1) 0(−3) + (4)(−2)

=

=

10 − 11 4 −8

* AB + AC =

=

3 −3 1 −2

3 1 0 4

=

1 0 − 8 24

+

9 − 11 12 16

10 − 11 4 −8

(c) Find (A+B) C and AC + BC SOLUTION (A + B) C =

4 3 −2 −2

2 −5 3 4

28

(d)

=

4(2) + 3(3) 4(−5) + 3(4) − 2(2) − 2(3) − 2(−5) − 2(4)

=

15 − 8 − 10 2

(AB)´ & B´ A´

(AB) =

1 0 − 8 − 24

(AB)´ =

1 −8 0 − 24

and B´ A´ =

= Q4.

Ans.

3 0 1 −2 1 4 2 −6

3+0 −6+ 0 1 + 8 − 2 − 24

=

3 −6 9 − 26

Find A2 and B2

A=

−1 − 7 5 −2

2 0 6 5 −4 B = −1 3 −5 2

29

* A2 = AA =

−1 − 7 5 −2

=

− 1(−1) + (−7)5 − 1(−7) + (−7)( 2) 5(−1) + 2(5) 5(−7) + (2)( 2)

=

− 1 − 35 7 − 14 − 5 + 10 − 35 + 4

=

− 36 − 7 5 31

2 0 6 −1 5 −4 3 −5 2

2

* B = BB =

Q.5

−1 − 7 5 −2

2 0 6 −1 5 −4 3 −5 2

=

4 + 0 + 18 0 + 0 − 30 12 + 0 + 12 − 2 − 5 − 12 0 + 25 + 20 − 6 − 20 − 8 6 + 5 + 6 0 + 10 − 10 18 + 20 + 2

=

22 − 30 24 − 19 45 34 17 − 20 40

Find the determinant of following matrices of exist. (a) A =

det A =

−1 − 7 5 2 −1 − 7 5 2

= - 2 + 30 = 28

=

2 0 6 −1 5 −4 3 −5 2

det A =

2 0 6 −1 5 −4 3 −5 2

(b) A

= 2 (10 – 20 ) - 0 + 6 (5 – 15)

30

= 2 (-10) + 6 (-10) = - 20 – 60 = - 80

1 2 3 4 0 6

(c)

it is not square matrix , 50 det. Does not exist.

(d)

A=

1 0 0 0

2 2 0 0

Det A = 1

3 4 5 0

4 6 4 7

2 4 6 0 5 4 0 0 7

-2

0 4 6 0 5 4 0 0 7

= 1 (2 (35) -4(0) + 6(0) - 2 (0-4 (0) +6 (0) + 3 (0 – 0 + 0) - 4 (0) = (( 70 ) -4 (0) -2 (0) – 4 (0) = 70 Ans. Q6

Show that Rx (∝) Rx (- ∝ = I) We know that

31

1 0 0 0 Cos (∝) − Sin(∝) 0 Sin(∝) Cos(∝)

Rx (∝) Rx (- ∝ = i)

1 0 0 0 Cos (∝) − Sin(∝) 0 Sin(∝) Cos(∝)

1+ 0 + 0 0+0+0 0+0+0 2 2 0 + Cos Sin ∝ 0 + Sin ∝ Cos ∝ − Sin ∝ Cos ∝ = 0+0+0 0 + 0 + 0 0 + Sin ∝ Cos ∝ −Sin ∝ Cos ∝ 0 + Cos 2 Sin 2 ∝ Q7 (a)

Show that

Ry (B) Rx (∝)

Q7 (a)

0 Cos( β ) Sin(β ) sin( B) Sin(∝) Cos(∝) − CosBSin ∝ − Cos(d )Sin( β ) Sin(∝) Cosβ Cos (∝)

Rx (∝) Ry (β)

Cos (β ) 0 Sin( β ) 0 1 0 − Sin(β ) 0 Cos ( β )

=

1 0 0 0 Cos (∝) − Sin(∝) 0 Sin(∝) Cos(∝)

=

0 Cosβ Sinβ Sinβ Sin ∝ Cos ∝ − Cosβ Sin ∝ Cos ∝ Sinβ Sin ∝ Cosβ Cos ∝

7(b) Ry (β) Rx (∝)

=

Cosβ 0 Sinβ 0 1 0 − Sinβ 0 Cosβ

1 0 0 0 Cos ∝ − Sin ∝ 0 Sin ∝ Cos ∝

32

=

Cosβ + 0 + 0 0 + 0 + Sin ∝ Sinβ 0 + 0 + Cos ∝ Sinβ 0+0+0 0 + Cos ∝ +0 0 − Sin ∝ +0 − Sinβ + 0 + 0 0 + 0 + Cosβ Sin ∝ 0 + 0 + Cos ∝ Cosβ

=

Cosβ Sinβ Sin ∝ Cos ∝ Sinβ 0 − Sin ∝ Cos ∝ − Sinβ Cosβ Sin ∝ Cos ∝ Cosβ

33

EXERCISES FOR UPPER – TRIANGULAR LINEAR SYSTEMS. Solve upper triangular & find value of determinant of coefficient matrices

Q1.

3X1 - 2X2 + X3 - X4 ---------- (1) 4X2 – X3 + 2X4= - 3 ----------- (2) 2X3 + 3X4 = 11

--------------- (3)

5X9 = 15 ------------------- (4)

SOLUTION From Equation (1) 3X4 = 15 X4 =

15 5

3

Substituting value of 3X1 = 3 is equation ………………. (3) Equation ………….. (3) 2X3 + 3X4 = 11 2X2 + 2(3) =11 2X3 + 9 = 11 2X3 = 11-9 2X3 = 2 2X3 =

2 1 2

X3 = 1 Ans.

34

Substituting value of X3 = 1 , X4 = 3 is equation ………………. (2) Equation …………. (2) 4X2 - X3 + 2X4 = -3 4X2 – 1+2(3) =-3 4X2 – 1+6 = -3 4X2 = -3-6+1 4X2 = -8 4X2 =

8 2 4

X2 = 2 Ans. Substituting value of X2 = -2, X3 = 1, X4 = 3, is equation ……. (1) Equation …………. (1) 3X1 – 2X2 + X3 – X4

=

8

3X1 – 2(-2) +1-3 =8 3X1 + 4+1-3 = 8 3X1 = 8+3-1-4 3X1 = 8+3-1-4 3X1 = 6 X1 =

6 3

2

X1 = 2 Ans For efficient fo determinant matrix . DET = 3 X 4 X 2 X 5

35

= 12 X 10 DET = 120 X= 2, X2 =-2, X3 =1, X4 = 3 det = 120

Q2.

Ans.

5X1 -3X2 - 7X3 + X4 = -14

……………………. (1)

11X2 + 9X3 + 5X4 =22

……………………. (2)

3X3 - 13X4 = - 11

…………………….(3)

7X4 = 14

………………….. (4)

SOLUTION From equation ………….. (4) 7X4 = 14 X4 =

14 7

X4 = 2

2 Ans.

Substituting value of X4 = 2 in equation ………… (3) Equation …………………. (3) 3X3 - 13X4 = -11 3X3 – 13(2) =- 11 3X3 – 26 =- 11 3X3 = -11+26 3X3 = 15 X3 =

5 3 3

36

X3 = 5 Ans.

Substituting value of X3 = 5, X4 = 2 in equation ………… (2) Equation …………………. (2) 11X2 +9X3 + 5X4 = 22 11X2 + 9(5) + 5(2) = 22 11X2 + 45+10 = 22 11X2 = 22-10-45 11X2 = - 33 X2 =

3 3 11

X2 =-3

Substituting value of X2 = 3, X3 = 5, X4 = 2, in equation ………… (1) Equation …………………. (1) 5X1 - 3X2 - 7X3 + X4 = -14 5X1 -3 (-3) -7 (5) + 2 =- 14 5X1 + 9+35+2=-14 5X1 = - 14 -2 – 35 –9 5X1 = -60 X1 =

6 5

-12

X1 = -12 Ans.

37

For determinant of efficient matrix det = 5 x 11 x 3 x 7 = 55 x 21 det =1155 X1 = -12 , X2 = -3, X3 = 5, X4 = 2 Det = 115

4X1 – X2 + 2X3 + 4X4 – X5 = 4 …..…..….. (1)

Q3.

X2 + 6X3 + 2X4 + 7X5 = 0 …..…..….. (2)

−2X 4 − −2X 5 = 10..........(4) 3X 5 = 6.........................(5) SOLUTION 3X 5 = 6 6 2 3 X5 =2 X5 =

Substituting value of X 5 = 2 is equation ………………… (4) Equation…………. => 2X 4 − X 5 = 10 2X 4 − 2 = 10 2X 4 = 10 + 2 2X 4 = 12 −12 6 6 X 4 = −6 X4=

Substituting value of X 4 = −6, X 5 = X 5 = 2 is equation ……………(3)

38

Equation ……………… (3)

X 3 − X 4 − 2X 5 = 3 X 3 − (−6) − 2(2) = 3 X 3 +6−4 = 3 X 3 = 3+4−6 X 3 =1

Substituting value of X 3 = 1, X 4 = −6, X 5 = 0 is equation ……………(2) Equation ……………….. (2)

−2X 2 + 6X 3 + 2X 4 + 7X 5 = 0 −2X 2 + 6(1) + 2(−6) + 7(2) = 0 −2X 2 + 6 − 12 + 14 = 0 −2X 2 = −14 + 12 − 6 2X 2 = −8 2X 2 =

8 4 2

X2 =4 Substituting value of X 2 = 4, X 3 = 1, X 4 = −, X 5 = 2 is equation ……………(1) Equation ……………….. (1)

4X 1 − X 2 + 2X 3 + 2X 4 − X 5 = 4 4X 1 − 4 + 2(1) + 2(−6) − 2 = 4 4X 1 − 4 + 2 − 12 − 2 = 4 4X 1 = 4 + 4π m 2 + 2 + 12 4X 1 = 20 20 5 4 X1 =5 X1 =

for coefficient of Determinant matrix det=

39

4X (−2)X 1X (−2)X 3 = −8X (−2)x 3 = 16X 3 det = 48 X 1 = 52 = 4, X 3 = 1, X 4 = −6, X 5 = 2 det = 48

Q4.

Given

A

a1 a12 0 a 12   0 0

a13  a23  a33 

b1 b12 b13  and B =  0 b12 b23   0 0 b33 

Show that their product C AB is also upper triangular.  a11 (b11 ) + a12 (0) + a13 (0) a11 (b12 ) + a12 (b 22 ) + a13 (0) a11 (b13 ) + a12 (b 23 ) + a13 (b33 )  a (b ) + a (0) + a (0) a (b ) + a (b ) + a (0) 0(b ) + a (b ) + a (b )  22 23 11 12 22 22 23 13 23 23 23 33   11 22  0(b11 ) + 0(0) + a33 (0) 0(b12 ) + 0(b 22 ) + a33 (0) 0(b13 ) + 0(b 23 ) + a33 (b33 ) 

a11b11 a12b12 + a12b 22 a11b13 + a12b 23 + a13b 23   0  a22b 22 a23b 23 + a23b33    0  a33b33 0

Q5.

Solve the lower triangular system AX= B and find det (A).

2X 1 = 6.............................(1) −X 1 + 4X 2 = 5.....................(2) 3X 1 − 2X 2 − X 3 = 4.............(3) X 1 − 2X 2 + 6X 3 + 3X 41 = 2............(4)

40

Solution From equation …………….. (1) 2X 1 = 6.............................(1) 6 2X 1 3 2 X1=3

Substituting value of X 1 = 3 is equation ……………….(2) Equation = > X 1 + 4X 2 = 5 −3 + 4X 2 = 5 4X 2 = 5 + 3 X 2 =8 8 2 4 X2 =2 X2

Substituting value of X 1 = 3, X 2 = 2 is equation ………… (3) 3X 1 − 2X 2 − X 4 = 4 3(3) − 2(2) − X 3 = 4 9−4−X 3 = 4 −X 3 = 4 + 4 − 9 + X 3 = +1 X 3 =1

substituting value of X 1 = 3, X 2 = 2, X 3 = 1 is equation ………….. (4) Equation ………… (4) = >

41

X 1 − 2 X 2 + 6 X 3 + 3X 4 = 2 3 − 2(2) + 6(1) + 3X 4 = 2 3− 4 + 6 + 4−3 3X 4 = −3 −3 1 3 X 4 =1 X4=

for coefficient of determination matrix dt = 2 X 4X (-1) X 3 =-8X3 det = - 24

X 1 = −1, X 2 = 2, X 3 = 1, X 4 = 1 det =- 24 Ans.

Q6.

5X 1 = −10...............(1) X 1 = +3X 2 = ............(2) 3X 1 + 4X 2 + 2X 3 = 2.............(3) −X 1 + 3X 2 − 6X 3 − X 4 = 5............(4) Solution From equation ……………(1) 5X 1 = −10 −10 2 5 X 1 = −2 X1=

Substituting value of X 1 = −2 is equation………………. (2) Equation ………………..(2)

42

X 1 + 3X 2 = 4 −2 + 3X 2 = 4 +2 + 4 = 3X 2 3X 2 = 6 6 2 3 X2 =2 X2=

Substituting value of X 1 = 2, X 2 = 2 is equation………..…(3) Equation ………………..(3)

3X 1 + 4X 2 + 2X 3 = 2 3(−2) + 4(2) + 2X 3 = 2 −6 + 8 + 2X 3 = 2 2X 3 = 2 + 6 − 8 2X 3 = 0 0 2 X3 =0 X3=

Substituting Value of X 1 = 2, X 2 = 2, X 1 = X 3 = 0 is equation…… (4) Equation.. =>

− X 1 + 3X 2 − 6X 3 − X 4 = 5 −(−2) + 3(2) − 6(0) − X 4 = 5 2 + 6 − 0X 4 − 5 +X 4 = 5 − 2 − 6 + X 4 = −3 X 4 =3

for coefficient of determinant matrix dt = 5X3X2X (-1) det = -30

43

X 1 + X 2 + 6X 3 = 7 X 1 = −2, X 2 = 2, X 3 = 0, X 4 = 3 3X 2 + 15X 3 = 9 12X 3 = 12 det =- 30. Ans.

44

Gaussian Elimination and Pivoting Show that A X = B is equivalent to the upper – triangular system U x = y find solution.

1

2X 1 + 4X 2 − 6X 3 = −4

2X 1 + 4X 2 − 6X 3 = −4

X 1 + 5X 2 3X 3 = 10

3X 2 + 6X 3 = 12

X 1 + 3X 2 2X 3 = 5

3X 3 = 3

SOLUTION First we find A x = B is equivalent to 4x= Y UXY  2 4 −6  0 3 6     0 0 3 

X 1   −4   X  = 12   1    X 3   3 

2X 1 + 4X 2 − 6X 3 = −4 3X 2 + 6X 3 = 12 3X 3 = 3

Y=

X 1   −3 X  =  2   1    X 3   1 

AX=B  2 4 −6  1 5 3     1 3 2 

 −3  −4   2  = 10       1   15 

 −6 + 8 − 6   −4   −3 + 10 + 3 = 10       −3 + 6 + 2   15 

45

 −4   −4  10  = 10       15   15 

Ax = B is equitant to u x = Y  1  2 4 −6   1 5 3  =  1   2  1 3 2  1  2

 0 0  1 0   1 1 3 

0  2 4 −6   1 1 5 3  =  m 1    21  1 3 2   m 31 m 32  U11 m U  12 11  m 31 U11

 2 4 −6  0 3 6     0 0 3 

0 0  1 

 U11 U12  0 U 22   0 0

U12

m12 U11 + U 22

m 31 U12 + m 32 U 22

U13  U 23   U 33 

 m 21 U13 + U 23   m 31 U13 + m 32 U 33  U13

U11 = 2, U12 = 4, U13 = −6

m 21 U11 = 1 m 21 (2) = 1 m 21 =

1 2

m 31 U11 = 1 m 31 (2) = 1 m 31 =

1 2

m 21 U11 + U 22 = 5

m 21 U13 + U 23 = 3

1 (4) + U 22 = 5 2 U 22 = 5 − 2

1 ( )(−6) + U 23 = 5 2 U 23 = 3 + 3

U 22 = 3

U 23 = 6

m 31 U 13 + m 23 U 23 + U 33 = 2

m 31 U 13 + m 23 U 23 + U 33 = 2

1 1 ( )(−6) + (1 / 3)(6) + U 33 = 2 ( )(−6) + (1 / 3)(6) + U 33 = 2 2 2 −3 + 2 + U 33 = 2 −3 + 2 + U 33 = 2 U 33 = 3

U 33 = 3

46

 2 4 −6  1 5 3     1 3 2 

 1  1 =  2 1  2

 0 0  2 4 −6   1 0  x  0 3 6    0 0 3   1 0 3 

 2 4 −6   2 4 −6  1 5 3  = 1 5 3       1 3 2   1 3 2 

2

X 1 + X 2 + 6X 3 = 7

X 1 + X 2 + 6X 3 = 7

3X 2 + 15X 3 = 9

3X 2 + 15X 3 = 9

12X 3 = 12

12X 3 = 12

SOLUTION First we find A x= B is equivalent to u X = Y

UX =Y  2 4 −6   X 1   −4   0 3 6   X  = 12     2    0 0 3   X 3   3 

2X 1 4X 2 − 6X 3 = −7 3X 2 + 15X 3 = 9 12X 3 = 12

Y=

X 1   −3 X  =  2   2    X 3   1 

Ax =B  1 1 6  3  7  −1 2 9   −2  =  2         1 −2 3  1  10 

47

 1 1 6 X 1  7 7  −1 2 9  =  X  =>  2  =  2   2        X 3   1 −2 3 10  10   1 1 6  1 1 6  −1 2 9  =  −1 2 9       1 −2 3  1 −2 3

3)

2X 1 − 2X 2 + 5X 3 = 6

2X 1 − 2X 2 + 5X 3 = 6

2X 1 + 3X 2 + X 3 = 13

5X 2 + 4X 3 = 7

X 1 + 4X 2 − 4X 3 = 3

0.9X 3 = 1.8

SOLUTION Ux=Y  2 −2 5   0 5 −4     0 0 0.9 

X 1  6 X  =  7   2    X 3  1.8

2X 1 − 2X 2 + 5X 3 = 6 5X 2 + 4X 3 = 7 0.9X 3 = 1.8 X3

1.8 = 2,5X 2 − 4(2) = 7 0.9 5X 2 - 8= 7 X2 =3

2X 1 , 2(3) + 5(2) = 6 2X 1 − 6 + 10= 6 X1 =2 X1 =1

X 1  1    Y =  X 2  =  3   X 3   2 

A X= B

48

 2 −2 5   1  6  2 3 1   3  = 13        −1 4 4   2   3 

 2-6+10  6  2+ 9+ 2  = 13     -1+12-8  3  6 6 13 = 13      3   3   2 −2 5  2 3 1 =    −1 4 −4 

 1   1   −1 2

 2 −2 5  2 3 1 =    −1 4 −4 

0  1 m 1  21  m 31 m 32

0 0  1 

 U11 m U  21 11  m 31 U11

U12

  m 21 U13 + U 23  m 21 U13 + m13 U 25 + U 33 

m 21 U12 + U 22

0  2 −2 5   1 0  =  0 5 −4       3 0 0 9 1  5  0

m 31 U 12 +, m 32 U 22

 U11 U12  0 U 22   0 0

m11 U12 + U 22 = 0 m11 U11 = 1 m 21 (−5) = 1 m 21 =

−1 5

−1 )(2) + U 22 = 0 5 −2 + U 22 = 0 5 U 22 = 0.4

(

U13  U 23   U 33 

U13

m 21 U13 + U 23 = 3 −1 )(−1) + U 23 = 3 5 1 + U 23 = 3 5 1 U 22 = 3 − 5 U 23 = 2.8

(

49

m 31 U12 + m 32 U 22 = 1

m 31 U11 = 3

−3 (2) + m 32 (0.4) = 1 5 −6 + m 32 (0.4) = 1 5 11 m 32 = 2

m 23 (−5) = 3 −3 5 U 23 = 2.8 m 31 =

    1 0 0  −5 2 −1    1 0 3  =  −1 1 0     5   3 1 6   −3 11   5 2 1 

m 31 U11 = 3 m 23 (−5) = 3 −3 5 U 23 = 2.8 m 31 =

−1   −5 2  0 0.4 2.8    0 −10   0

 −5 2 −1  −5 2 −1 1 0 3 = 1 0 3      3 1 6   3 1 6 

5

Find the parabola Y = A+B+ (X2 that passes through (1,4),(2,7),(3,14) SOLUTION For each point me obtain equation electing x and Y. A+B+C = 4

(1,4) ………………… (i)

A+2B+4C = 7

(2,7)…………………..(ii)

A+3B+9C = 14

(3,14)…………………(iii)

A=B=C = 4 Subtracted equation (ii) from equation …………..(i) B+3C = 3 ………………… (ii) Subtracted equation (iii) from equation …………..(i) 2B +8C = 10……………..(v)

50

A+B+C =4 B+3C = 3 Subtracted equation (iii) from equation …………..(ii) two times 2C = 4 using back substitution. C =2 B= -3 A=5 The equation of parable is y= 2X2 – 3X+5. 6) Find parabola y = A+Bx + CX2 that passes through (1,6), (2,5), (3,2) Solution: A+B+C = 6

(1,6) ………………… (i)

A+2B+4C = 5

(2,5)…………………..(ii)

A+3B+9C = 2

(3,2)…………………(iii)

Subtracted equation (ii) and equation (iii) from equation (i) to climate A A+B+C = 6………………(i) B+3C = -1 ………………… (ii) 2B+8C =- 4…………. (5) Subtracted equation (5) from equation (4) two times A+B+C = 6………………(i) B+3C = -1 ………………… (ii) 2C =- 2…………. (5)

51

∴

C = -1 B=2 A=5

The equation of parabola is Y = X2 +2X + 5

Q9.

2X 1 + 4X 2 − 4X 3 + 0X 4 = 12 X 1 + 5X 2 − 5X 3 − 3X 4 = 18 2X 1 + 3X 2 + X 3 + 3X 4 = 8 X 1 + 4X 2 − 2X 3 − 2X 4 = 18

SOLUTION 2 1  2  1

4 −4 0  12  5 −5 −3 18 3 1 3 8    4 −2 2   8 

R1 /2

1 1  2  1

2 −2 0   6  5 −5 −3 18 3 1 3 8    4 −2 2   8 

R 2 – R1

1 1  2  1

2 −2 0   6  3 −3 −3 12  3 1 3`   8     4 −2 2   8 

52

2 −2 0   6  3 −3 −3 12  3 1 3 8    4 0 2 2

R 4 – R1

1 1  2  1

R 4 – R1

 1 2 −2 0   6   0 3 −3 −3 12       0 −1 5 3   −4      0 2 0 2   2 

R4 + 2R3

 1 2 −2 0   6   0 3 −3 −3 12       0 −1 5 3   −4       0 0 10 8   −6 

1 0  0  0

2 −2 0   6  3 −3 −3 12  0 4 2 0    0 10 8   −6 

R 4 - 2 R3

1 0  0  0

2 −2 0   6  3 −3 −3 12  0 4 2 0    0 2 4   −6 

R4 – R5 /2

1 0  0  0

2 −2 0   6  3 −3 −3 12  0 4 2 0    0 0 3   −6 

R1 x 2

2 0  0  0

4 −4 0  12  3 −3 −3 12  0 4 2 0    0 0 3   −6 

R4 +

R2 3

53

2X 1 + 4X 3 − 4X 3 + 0X 4 = 12 3X 2 − 3X 3 − 3X 4 = 12 4X 3 + 2X 4 = 0 3X 4 = −6 X 4 = −2 3X 2 − 3(1) − 3(−2) = 12 X 2 =3 2X 1 + 4X 2 − 4X 3 = 12 X1 =2 4X 3 + 2(−2) = 0 4X 3 = 4 X 3 =1

Q.10

X 1 + 2X 2 + 0X 3 − X 4 = 9 2X 1 + 3X 2 − X 3 + 0X 4 = 9 0X 1 + 4X 2 − 2X 3 + 5X 4 = 26 5X 1 + 5X 2 + 2X 3 − 4X 4 = 32 SOLUTION 1 2  0  5

R2 - 2R1

2 0 −1  9  3 −1 0   9  4 2 −5  26     5 2 −4   32   1 2 0 −1  9   2 −1 −1 2   −9       0 4 2 −5  26       5 5 2 −4  32 

54

R4 - 5R1

9  1 2 0 −1  0 −1 −1 2 −9     0 4 2 −5 26    1 −13  0 −5 2

R3 + 4R2

 1 2 0 −1  9   0 −1 −1 2   −9       0 0 −2 3  −10       0 −5 2 1  −13

R4 + 5R2

 1 2 0 −1  9   0 −1 −1 2  −9       0 0 −2 3  −10       0 0 7 −9  −13

R4 + 3R3

 1 2 0 −1  9   0 −1 −1 2   −9       0 0 −2 3  −10      1 0   −13 0 0

R4 +1/2 R3

 1 2 0 −1  0 −1 −1 2     0 0 −2 3    0 0 0 1.5

 9  −9     −10     −3

Hence equation becomes

X 1 + 2X 2 + 0X 3 − 1X 4 = 9............(i ) −X 2 − X 3 + 2X 4 = −9............(ii ) −2X 3 + 3X 4 = −10............(iii ) 1.5X 4 = −3............(iv ) By Equation (iv) X4 = 3/1.5 X4 = -2

55

By equation (iii)

−2X 3 = −10 − 3(−2) −2X 3 = −4 X3 =2 By Equation (ii)

− X 2 = −9 + (2) − 2(−2)

X 2 =3

By equation (i)

X 1 = 9 − 2(3) − 2 X 1 =1 Q8.

4X 1 + 8X 2 + 4X 3 + 0X 4 = 8

X 1 + 5X 2 + 4X 3 + 0X 4 = −4 4X 1 + 4X 2 + 7X 3 + 2X 4 = 10

X 1 + 3X 2 + 0X 3 − 2X 4 = −4

SOLUTION

R1/4

4 1  1  1

8 5 4 3

4 0 8 4 −3 −4  7 2 10   0 −2 −4 

1 1  1  1

2 5 4 3

1 0 2 4 −3 −4  7 2 10   0 −2 −4 

56

R 2 – R1

1 0  1  1

2 3 4 3

1 0 2 3 −3 −6  7 2 8  0 −2 −4 

R 4 – R1

1 0  1  1

2 1 0 2 3 3 −3 −6 4 6 2 8  1 −1 −2 −6

R3 – 2R1

1 0  0  0

2 1 0 2 3 3 −3 −6 0 8 6 20  1 −1 −2 −6

R3 – 2R1

1 0  0  0

2 1 0 2 3 3 −3 −6  0 8 6 20   0 −6 −3 −12 

R4 – 2R2

1 0  0  0

2 1 0 2 3 3 −3 −6  0 8 6 20   0 −6 −3 −12  2

1/2R2 +1/4R3

1 0  0   0

1

3 3 0 8 0 0

2 −3 −6   6 20   1 1 2  0

by equation(iv) ∴ 1 X 4 =1 2

X4 =2

57

by equation (iii)

8X 3 + 6(2) = 20 8X 3 = 20 − 12 8X 3 = 8

X 3 =1 by equation (ii)

3X 2 + 3(1) − 3(2) = −6

X 2 = −1 by equation (i)

X 1 + 2(−1) + 1 = 2 X1 =3 Q11

X 1 + 2X 2 = 7 2X 1 + 3X 2 − X 3 = 9 4X 2 + 2X 3 + 3X 4 = 10 2X 3 − 4X 4 = 12

2R1 – R2

1 2  0 −1  0 4  0 0

0 0 7 1 0 −5 2 3 10   2 −4 12 

4R2 – R3

1 0  0  0

2 1 0 0

0 0 7 1 0 5 2 −3 10   2 −4 12 

4R4 – R3

1 0  0  0

2 1 0 0

0 0 7 1 0 5 2 −3 10   0 −1 2  58

−X 4 = 2

X 4 = −2

Ans.

2X 3 − 3(−2) = 10 2X 3 + 6 = 10

X3 =2 X 2 +2=5 X 2 =3 X 1 + 2(3) + 2 = 5 X 1 = −3 Q12

X1+X 2 =5 2X 1 − X 2 + 5X 3 = −9 3X 2 − 4X 3 + 2X 4 = 19 2X 3 + 6X 4 = 2 SOLUTION

R 2 – R1

1 1 0  2 −1 5   0 3 −4  0 0 2

0 5 0 −9  2 19   6 2

1 1 0  0 −3 5   0 3 −4  0 0 2

0 5 0 −9  2 19   6 2

59

R3 + R2

1 1 0  0 −3 5   0 3 −4  0 0 2

R4 - 2R3

1 1  0 −3  0 0  0 0

0 5 1 0

0 5 0 19  2 19   6 2 0 5 0 19  2 0  2 −34 

2X 4 = −34

X 4 = −17 X 3 + 2(−17) = 0 X 3 = 34 −3X 2 + 5(34) + 0(17) = 19 −3X 2 = 19 − 169 −150 −3 X 2 = 50

X2−

X 1 + 50 = 5 X 1 = −45

60

ITERATIVE METHODS FOR LINEAR SYSTEMS

61

JACOBI Q1,

(a)

4X − y = 15

 X 0 0 initial given Po = O  =   Y 0   0 

X + 5y = 9

After rearing: Same equation order will come: 4X − y = 15 X + 5Y = 9

(1)

(2)

=>

4X = 15 − y 1 X = [15 + y ] 4

=>

5y = 9 − X 1 y = [9 − x ] 4

In general

1 [15 + yk ] 4 1 yk + 1 = [ 9 − xk ] 5

X k +1 =

Now pull K = 0

X 1=

1 [15 + y (0)] 4

1 [15] 4 = 3.75 =

y1=

1 [9 − x 0 ] 5

1 [ 9 − 0] 5 9 = 5 = 1.8 =

62

At k = 1

1 [15 + y 1 ] 4 1 = [15 + 1.8] 4 = 4.2

X2=

1 [9 − x 1 ] 5 1 = [9 − 3.75] 5 1 = [5.25] 5 = 1.05

y2 =

At k = 2

1 [15 + y 2 ] 4 1 = [15 + 1.05] 4 = 4.0125

x2 =

1 [15 + X 2 ] 5 1 = [ 9 − 4.2] 5 1 = [ 4.8] 5 = 0.5333

y2 =

At k = 2

1 [15 + y 3 ] 4 1 = [15 + 0.5333] 4 = 3.88333

X4=

63

1 [9 − X 3 ] 5 1 = [9 − 4.0125] 5 1 = [ 4.9875] 5 = 0.9975

y2 =

TABLE: K

XK

YK

0

3.75.

1.8

1

4.2

1.05

2

4.0125

0.5333

3

3.8833

0.9975

GAUSS – SEIDAL Q1(b)

4X − y = 15

 X 0 initial given Po = O  = Y 0 

X + 5y = 9

0 0  

After rearing: Same equation order will be follow:4X − y = 15 X + 5Y = 9

(1) = >

4X = 15 + y X =

1 [15 + y ] 4

64

(2)

=>

5y = 9 − X y =

1 [9 − X 5

]

as general

1 [15 + yk ] 4 1 yk + 1 = [ 9 − xk + 1] 5

X k +1 =

Now put K = 0

1 [15 + y 0 ] 4 1 = [15] 4 = 3.75

X1=

1 [9 − X 3 ] 5 1 = [9 − 3.75] 5 = 1.05

y1 =

Put K = 1

1 [15 + y 1 ] 4 1 = [15 + 1.05] 4 = 4.0125

X2=

65

1 [9 − X 2 ] 5 1 = [9 − 4.0125] 5 1 = [5.0006] 5 = 1.000125

y2 =

Now Put K = 3 ∴

1 [15 + y 3 ] 4 1 = [15 + 1.000125] 4 = 4.0000

X3=

1 [9 − X 4 ] 5 1 = [9 − 4.000 ] 5 1 = [9 − 4 ] 5 1 = 5 =1

y4 =

Q3.

(a)

K

XK

YK

0

3.75

1.05

1

4.0125

0.9975

2

3.9994

1.000125

3

4.0000

1

−X + 3 y = 1 6X − 2 y = 2

 X 0 Initial given Po = O  = Y 0 

0 0  

After rearing:

66

Equation rearranging

6X − 2 y = 2 −X + 3 y = 1 (1)

=>

6X = 2 + 2 y 1 6X 3 = [ 2 + 2 y ] 6 (2)

=> −X + 3 y = 1 3y = 1 + X =

1 [1 + X 3

]

In general

1 [ 2 + 2 yk ] 6 1 yk+1 = [1 + X k ] 3

Xk+1 =

NOW PUT K = 0 1 [2 + 2y ] 6 1 X = [ 2 + 2(0) ] 3 1 = [ 2 + 0] 6 2 = = 0.3333 6 X1 =

67

1 [1 + X ] 3 1 = [1 + 0] 3 1 = = 0.3333 3

Y1 =

AT K = 1

1 [2 + 2y 1] 6 1 X = [ 2 + 2(0.3333)] 6 1 = [ 2.666] 6 = 0.4444 X2 =

1 [1 + X 1 ] 3 1 = [1.0.333] 3 = 0.4444

y1 =

At K= 2

1 [2 + 2 y 2 ] 6 1 X = [ 2 + 2(0.4444)] 6 1 = [ 2.8888] 6 = 0.4814 X3 =

1 [1 + X 3 ] 3 1 = [1 + 0.4444] 3 1 = [1.4444] 3 = [ 0.4814]

y3 =

At K =3

68

1 [2 + 2y 3 ] 6 1 X = [ 2 + 2(0.4814)] 6 1 = [ 2.2928] 6 = 0.4938 X4 =

1 [1 + X 3 ] 3 1 = [1 + 0.4814] 3 1 = [1.4814] 3 = 0.4938

y4 =

Table

Q3.

(B)

K

XK

YK

0

0.3333

0.3333

1

0.4444

0.4444

2

0.4814

0.4814

3

0.4938

0.4938

−X + 3 y = 1 6X − 2 y = 2

 X 0 0 Initial given Po = O  =   Y 0   0 

After rearranging

6X − 2 y = 2 −X + 3 y = 1 (1)

=>

69

6X = 2 + 2 y 1 X 3 = [2 + 2 y ] 6 (3)

=> −X + 3 y = 1 3y = 1 + X =

1 [1 + X 3

]

In general

1 [ 2 + 2 yk ] 6 1 yk+1 = [1 + X k + 1] 3

Xk+1 =

NOW PUT K = 0 1 [2 + 2y 0 ] 6 1 X = [ 2 + 2(0) ] 6 = 0.3333 X1 =

1 [1 + Xk + 1] 3 1 = [1 + X 1 ] 3 1 = [1 + 0.3333] 3 = 0.4444

y1 =

Put K= 1

70

1 [2 + 2y 1] 6 1 X = [ 2 + 2(0.4444)] 6 = 0.4814 X2 =

1 [1 + X 2 ] 3 1 = [1 + 0.4814] 3 = 0.4938

y2 =

Put K= 2

1 [2 + 2 y 2 ] 6 1 X = [ 2 + 2(0.4938)] 6 1 = [ 2.9876] 6 = 0.4979 X3 =

1 [1 + X 3 ] 3 1 = [1 + 0.4979] 3 = 0.4993

y3 =

At k =3

1 [2 + 2y 3 ] 6 1 X = [ 2 + 2(0.4993)] 6 1 = [ 2.9986] 6 = 0.4998 X4 =

71

1 [1 + X 4 ] 3 1 = [1 + 0.4998] 3 = 0.4999

y4 =

Table

Q2.

(a)

K

XK

YK

0

0.3333

0.4444

1

0.4814

0.4938

2

0.4979

0.4993

3

0.4998

0.4999

8X − 3 y = 10 −X + 4 y = 6

 X 0 Initial given Po = O  = Y 0 

0   0

After rearing: Equation rearranging

8X − 3 y = 10 −X + 4 y = 6 (1)

=>

8X = 10 + 3 y 1 X = [10 + 3 y ] 6 (2)

=>

4y = 6 + X 1 y = [6 + X 3

]

72

In general

1 [10 + 3 yk ] 8 1 yk+1 = [ 6 + X k ] 4

Xk+1 =

Now put K= 0

1 [10 + 3 y 0 ] 8 1 X = [10 + (0) ] 8 1 = [10] 8 = 1.25 X1 =

1 [6 + X 0 ] 4 1 = [ 6 + 0] 4 1 = [ 6] = 1.5 4

y1 =

At k = 1 1 [10 + 3y 1 ] 8 1 X = [10 + 3(1.5) ] 8 1 = [10 + 4.5] 8 1 = [14.5] =1.8125 8

X2 =

1 [6 + X 1 ] 4 1 = [ 6 + 1.25] 4 1 = [ 7.25] = 1.8125 4

y2 =

73

Now AT k = 2 1 [10 + 3y 2 ] 8 1 X = [10 + 3(1.8125)] 8 1 = [10 + 5.4375] 8 1 = [15.4375] = 1.9297 8

X3 =

1 [6 + X 2 ] 4 1 = [ 6 + 1.8125] 4 1 = [ 7.8125] = 1.9531 4

y1 =

At k = 3

1 [10 + 3y 3 ] 8 1 X = [10 + 3(1.9531)] 8 1 = [15.8593] 8 = 1.9824 X4 =

1 [6 + X 3 ] 4 1 = [ 6 + 1.9297 ] 4 1 = [ 7.9297 ] = 1.9824 4

y4 =

Table K

XK

YK

0

1.25

1.5

1

1.8125

1.8125

74

Q2.

(b)

2

1.9297

1.9531

3

1.9824

1.9824

8X − 3 y = 10 −X + 4 y = 6

 X 0 0 Initial given Po = O  =   Y 0   0 

After rearing: The same order will be written;

8X − 3 y = 10 −X + 4 y = 6 (1)

=> 8X − 3 y = 10 8X = 10 + 3 y X =

(2)

1 [10 + 3 y ] 8

=>

4y = 6 + X 1 y = [6 + X 4

]

In general

1 [10 + 3yk ] 8 1 yk+1 = [ 6 + X k + 1] 4

Xk+1 =

Now put K= 0

75

1 [10 + 3 y 0 ] 8 1 X = [10 + 3(0)] 8 1 = [10] 8 = 1.25 X1 =

1 [6 + X 1 ] 4 1 = [ 6 + 1.25] 4 1 = [ 7.25] = 1.8125 4

y1 =

Now At k = 1

1 [10 + 3y 1 ] 8 1 X = [10 + 3(1.8125)] 8 1 = [10 + 5.4375] 8 = 1.9297 X2 =

1 [6 + X 2 ] 4 1 = [ 6 + 1.9297 ] 4 1 = [ 7.9297 ] = 1.9824 4

y2 =

Now AT k = 2 1 [10 + 3y 2 ] 8 1 X = [10 + 3(1.9824)] 8 1 = [10 + 5.9472] 8 1 = [15.9472] = 1.9934 8

X3 =

76

1 [6 + X 3 ] 4 1 = [ 6 + 1.9934] 4 1 = [ 7.9934] = 1.9984 4

y3 =

At k = 3

1 [10 + 3y 3 ] 8 1 X = [10 + 3(1.9984)] 8 1 X = [10 + 5.9952] 8 1 = [15.9952] 8 = 1.9994 X4 =

1 [6 + X 4 ] 4 1 = [ 6 + 1.9994] 4 = 1.9998

y4 =

Table

Q4.

(a)

K

XK

YK

0

1.25

1.8125

1

1.9297

1.9824

2

1.9934

1.9984

3

1.9994

1.9998

2X + 3 y = 1 7X − 2 y = 1

 X 0 0 Initial given Po = O  =   Y 0   0 

After rearing: The same order will be written;

77

2X + 3 y = 1 7X − 2 y = 1 (1)

=>

7 = 1+ 2y 1 X [1 + 2 y ] 7 (2)

=>

3 y = 1 − 2X 1 y = [1 − 2X 3

]

In general

1 [1 + 2 yk ] 7 1 yk+1 = [1 − 2X k ] 3

Xk+1 =

Now put K= 0

1 [1 + 2 y 0 ] 7 1 = [1 + 2(0)] 7 1 = [1] 7 = 0.1428

X1 =

1 [1 − 2X 0 ] 3 1 = [1 − 2(0)] 3 1 = [1] = 0.3333 3

y1 =

Now At k = 1

78

1 [1 + 2 y 1 ] 7 1 X = [1 + 2(0.3333)] 7 1 = [1 + 0.6666] 7 = 0.2380 X2 =

1 [1 + 2X 1 ] 3 1 = [1 + 2(0.1428)] 3 1 = [1 + 0.2856] =0.4285 3

y2 =

Now AT k = 2 1 [1 + 2 y 2 ] 7 1 X = [1 + 2(0.4285) ] 7 = 0.2653

X3 =

1 [1 + 2X 2 ] 3 1 = [1 + 2(0.2380)] 3 1 = [1.476] =0.492 3

y3 =

At k = 3

1 [1 + 2 y 3 ] 7 1 X = [1 + 2(0.492)] 7 1 X = [1.984] 8 = 0.2834 X4 =

79

1 [1 + 2X 3 ] 3 1 = [1 + 2(0.2653) ] 3 = 0.5102

y4 =

Table

Q4.

(b)

K

XK

YK

0

0.1428

0.3333

1

0.2380

0.4285

2

0.2653

0.492

3

0.2834

0.5102

2X + 3 y = 1 7X − 2 y = 1

 X 0 Initial given Po = O  = Y 0 

0 0  

Equation rearranging

7X − 2 y = 1 2X + 3 y = 1 (1)

=>

7 = 1+ 2y 1 X [1 + 2 y ] 7

(2)

=>

3 y = 1 − 2X 1 y = [1 − 2X 3

]

80

In general

1 [1 + 2 yk ] 7 1 yk+1 = [1 − 2X k ] 3

Xk+1 =

Now put K= 0

1 [1 + 2 y 0 ] 7 1 = [1 + 2(0)] 7 1 = [1] 7 = 0.1428

X1 =

1 [1 − 2X 1 ] 3 1 = [1 − 2(0.1428)] 3 1 = [1 − 0.2856 ] 3 1 = [ 0.7144] = 0.2381 3

y1 =

Now At k = 1

1 [1 + 2 y 1 ] 7 1 = [1 + 2(0.2381)] 7 1 = [1.4762 ] 7 = 0.2109

X2 =

1 [1 − 2X 2 ] 3 1 = [1 − 2(0.2109)] 3 1 = [ 0.5782] =0.1927 3

y2 =

81

Now AT k = 2

1 [1 + 2 y 2 ] 7 1 X = [1 + 2(0.1927)] 7 1 = [1.3854] = 0.1979 7 X3 =

1 [1 − 2(0.1979)] 3 1 = [ 0.6042] =0.2014 3

y3 =

At k = 3

1 [1 + 2 y 3 ] 7 1 X = [1 + 2(0.2014)] 7 1 X = [1.4028] 8 = 0.2004 X4 =

1 [1 − 2X 4 ] 3 1 = [1 − 2(0.2004)] 3 1 = [ 0.5992] 3 = 0.1997

y4 =

82

Table

Q6.

(a)

K

XK

YK

0

0.1428

0.2381

1

0.2109

0.1927

2

0.1979

0.2014

3

0.2004

0.1997

2X + 3 y − 2 = 11 7X − 2 y + 2 = 10 −X + y + z 2 = 3

 X 0 Initial given Po = O  = Y 0 

0   0

After rearranging

2X + 3 y − 2 = 11 7X − 2 y + 2 = 10 −X + y + z 2 = 3 (1)

=>

5X = [10 + y − z ] X (2)

1 [10 + y − z ] 5

=>

4z = 3 + X − y 1 z = [3 + X y ] 4 In general

83

1 [10 + yk − zk ] 5 1 yk+1 = [11 − 2X k − zk ] 8 1 zk+1 = [3 + X k − yk ] 4 Xk+1 =

Now put K= 0

1 [10 + y 0 − z 0] 5 1 = [10 + 0 − 0] 5 1 = [10] 5 =2

X1 =

1 [10 + y 0 − z 0] 5 1 = [10 + 0 − 0] 5 1 = [10] 5 =2

y1 =

Now At k = 1

1 [11 − 2x 0 + 20] 8 1 = [11] = 1.375 8

y1 =

1 [3 + X 0 − y 0 ] 4 1 = [ 3] 4 = 0.75

z1 =

=

1 (3 + 0 − 0) 4

Now AT k = 2

84

X2 =

1 [10 + y 1 − z 1 ] 5

1 [10 + 1.375 − 0.75) ] 5 = 2.125

=

1 [11 − 2X 1 + z 1 ] 8 1 = [11 − 2(2) + 0.75] = 0.96875 8

y2 =

1 [3 + X 2 − y 2 ] 4 1 = [ 3 + 2.125 − 0.96875] = 1.03906 4

z3 =

At k = 3 X4 =

1 [10 + 2 y 3 − z 3 ] 5

1 [10 + 0.995703 − 1.03906] 5 = 1.983594

=

y4 =

1 [11 − 2X 3 + z 3 ] 8

1 [11 − 2(2.0125) + 0.3906] 8 = 1.0017575

=

1 [3 + X 3 − y 3 ] 4 1 = [ 3 + 2.0125 − 0.95703] 4 1 = [ 4.05547 ] = 1.0138675 4

z4 =

85

Table K

Xk

YK

ZK

0

2

1.375

0.75

1

2.125

0.96875

0.90625

2

2.0125

0.95703

1.03906

3

1.983594

1.0017575

1.0138675

86

Q6.

(b)

2X + 3 y − z = 11 7X − 2 y + z = 10 − X + y + 4z = 3

 X 0   Initial given Po = O Y 0  = z 0   

0   0 0  

After rearranging

2X + 3 y − z = 11 7X − 2 y + z = 10 − X + y + 4z = 3 (1)

=>

5X = [10 + y − z ] X (2)

1 [10 + y − z ] 5

=>

8 y = 11 − 2X + z 1 y = [11 − 2X + z ] 8 (3)

=>

4z = 3 + X − y 1 z = [3 + X − y ] 4 In general

1 [10 + yk − zk ] 5 1 yk+1 = [11 − 2X k + 1 − zk ] 8 1 zk+1 = [3 + X k + 1 − yk ] 4 Xk+1 =

Now put K= 0

87

1 [10 + 0 − 0] 5 1 = [10] 5 =2

X1 =

1 [10 + 2X 1 + z 0 ] 5 1 = [11 − 2(2) + 0] 5 1 = [11 − 4] 5 = 0.875

y1 =

1 [10 + X 1 − y 1 ] 4 1 = [ 3 + 2 − 0.875] 4 = 1.03125

z1 =

Now At k = 1

1 [10 + y 1 + z 1 ] 5 1 = [10 + 0.875 − 1.03125] = 1.96875 5

X1=

1 [11 + yX 2 + z 1 ] 8 1 = [11 − 2(1.96875) + 1.03125] 8 = 1.01171875

y2 =

1 [3 + X 2 − y 2 ] 4 1 = [ 3 + 1.96875 − 1.01171875] = 0.98925812 4

z3 =

Now AT k = 2

88

X3 =

1 [10 + y 2 − z 1 ] 5

1 [10 + 1.01171875 − 0.989257812)] 5 = 2.004492

=

1 [11 − 2X 3 + z 2 ] 8 1 = [11 − 2(2.004492) + 0.989257812] 8 1 = [ 7.980273812] 8 = 0.9975342265

y3 =

1 [3 + X 3 − y 3 ] 4 1 = [ 3 + 2.004492 − 0.9975342265] = 4 1 = [ 4.006957774] 4 = 1.001739443

z3 =

At k = 3

X4 =

1 [10 + y 3 − z 3 ] 5

1 [10 + 0.9975342265 − 1.001739443] 5 1 = [ 9.995794784] 4 = 1.999158957

=

y4 =

1 [11 − 2X 4 + z 3 ] 8

1 [11 − 2(1.999158957) + 1.001739443] 8 1 = [8.003421529] 4 = 1.000427691

=

89

1 [3 + X 4 − y 4 ] 4 1 = [ 3 + 1.999158957 − 1.000427691] 4 1 = [ 3.998731266] = 0.9996828165 4

z4 =

Table

Q6.

(b)

K

Xk

YK

ZK

0

0

0.875

1.03125

1

1.96875

1.01171875

0.98925812

2

2.004492

0.9975342265

1.001739443

3

1.999358957

1.000427691

0.9996828165

X − 5y − z = 8 4X − y − z = 13 2X − y − 6z = −2

 X 0   Initial given Po = O Y 0  = z 0   

0   0 0  

After rearranging

X − 5y − z = 8 4X − y − z = 13 2X − y − 6z = −2 (1)

=>

4X = 13 − y + z 1 X [13 − y + z ] 4 (2)

=>

90

− y = −2 + 6z − 2X ≠ y = −(2 − 6z + 2x ) y = 2-6z+2X (3)

=>

− z = −8 + 5 y − X ≠ z = −(8 − 5 y + X ) z = 8-5y + X In general

1 [13 − yk + zk ] 4 1 yk+1 = [ 2 − 6Xk + 2zk ] 8 zk+1 = 8 − 5 yk + xk Xk+1 =

Now put K= 0

1 [13 − y 0 + z 0 ] 4 1 = [13 − 0 + 0] 4 1 = [13] = 3.25 4

X1 =

y 1 = 2 − 6 z 0 + 2X 0 = 2-6(0)+2(0) =2 z 1 = 8 − 5y 0 + X 0 = 8-5(0)+(0) = 8-0 =8

Now At k = 1

91

1 [13 − y 1 + z 1 ] 4 1 = [13 − 2 + 8] 4 1 = [19] = 4.75 4

X2=

y 2 = 2 − 6z 1 + 2X 1 = 2-6(8) +2(3.25) = 2-48+6.5 = -39.5 y 2 = 8 − 5y 1 + X 1 = 8-5(-39.5) +3.25 = 8+197.5+3.25 = 208.75

Now AT k = 2

X3 =

1 [13 − y 2 + z 2 ] 4

1 [13 − (−39.5) + 208.75] 4 1 = [ 261.25] = 65.3125 4 =

y 3 = 2 − 6z 2 + 2X 2 = 2-5(-39.5) +2(4.75) = 2-1252.5+9.5 = -1241 z 3 = 8 − 5y 2 + X 2 = 8-5(-39.5) +4.75 = 8+197.5+4.75 = 210.25

92

At k = 3 X4 =

1 [13 − y 3 + z 3 ] 4

1 [13 − (−1241) + 210.25] 4 1 = [1464.25] 4 = 366.0625

=

y 4 = 2 − 6z 3 + 2X 3 = 2-5(210.25) +2(65.3125) = 2-1261.5+130.625 = -1128.875 z 4 = 8 − 5y 3 + X 3 = 8-5(-1241) +65.3125 = 8+6205+65.3125 = 6278.3125

Table K

Xk

YK

ZK

0

3.25

2

8

1

4.75

-39.5

208.75

2

65.3125

-1241

210.25

3

366.0625

-1128.875

6278.3125

93

Q7.

(b)

X − 5y − z = 8 4X − y − z = 13 2X − y − 6z = −2

 X 0   Initial given Po = O Y 0  = z 0   

0   0 0  

After rearranging

X − 5y − z = 8 4X − y − z = 13 2X − y − 6z = −2 (1)

=>

4X = 13 − y + z 1 X [13 − y + z ] 4 (2)

=>

− y = −2 − 6z + 2X (3)

=> −z = 8 − 5 y − X

In general 1 [13 − yk + zk ] 4 yk+1 = 2 − 6zk + 2X k + 1

Xk+1 =

zk+1 = 8 − 5 yk + 1 + xk + 1

Now put K= 0

1 [13 − y 0 + z 0 ] 4 1 = [13 − 0 + 0] 4 1 = [13 − 0 + 0] = 3.25 4

X1 =

94

y 1 = 2 − 6z 0 + 2X 1 = 2-6(0)+2(3.25) = 8.5 z 1 = 8 − 5y 1 + X 1 = 8-5(8.5) + (3.25) = 8-42.5+3.25 = -31.25

Now At k = 1 1 [13 − y 1 + z 1 ] 4 1 = [13 − 8.5 + (−31.25)] 4 1 = [13 − 8.5 − 31.25] 4

X2=

=

1 [ −26.75] = 6.6875 4

y 2 = 2 − 6z 1 + 2X 2 = 2-6(-31.25) +2(-6.6875) = 2+187.5-13.375 = 176.125 y 2 = 8 − 5y 2 + X 2 = 8-5(176.125) +(-6.6875) = 8-880.625-6.6875 = -879.3125

Now AT k = 2

95

X3 =

1 [13 − y 2 + z 2 ] 4

1 [13 − (176.125) + (−879.3125)] 4 1 = [13 − 176.125 − 879.3125] 4 1 = [ −1042.4375] = −260.6093 4 =

y 3 = 2 − 6z 2 + 2X 3 = 2-6(-879.3125) +2(-260.6093) = 2+5275.875-521.2186 = 4756.6564 z 3 = 8 − 5y 3 + X 3 = 8-5(4756.6564) +(-260.6093) = 8-23783.282-260.6093 = -24035.8913

At k = 3 X4 =

1 [13 − y 3 + z 3 ] 4

1 [13 − (−1241) + 210.25] 4 1 = [1464.25] 4 = 366.0625

=

y 4 = 2 − 6z 3 + 2X 3 = 2-5(210.25) +2(65.3125) = 2-1261.5+130.625 = -1128.875

96

z 4 = 8 − 5y 3 + X 3 = 8-5(-1241) +65.3125 = 8+6205+65.3125 = 6278.3125

Table K

Xk

YK

ZK

0

3.25

8.5

-31.25

1

-6.6875

176.125

-879.3125

2

-260.6093

4756.6564

-24035.8913

97