Please discuss idea solution
SOLVED MTH603 Assignment 1 Solution and Discussion

Assignment NO. 1 MTH603 (Fall 2019)
Maximum Marks: 20 Due Date: 24 112019
DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
• To solve this assignment, you should have good command over 01  12 lectures.
• Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in the 01 to 12 lectures.
• Upload assignments properly through LMS, No Assignment will be accepted through email.
• Write your ID on the top of your solution file.
• Don’t use colourful back grounds in your solution files.
• Use Math Type or Equation Editor Etc. for mathematical symbols.
• You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
• Try to make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero mark.
• Also remember that you are supposed to submit your assignment in Word format any other like scan images etc. will not be accepted and we will give zero mark corresponding to these assignments.Question #1: Find the root of the equation, Perform three iteration of the equation,
ln (x −1) + sinx = 0 by using Newton Raphson method.
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
x + y + z = 6; 2x − y + z = 3; x + z = 4 
Assignment No: 01
Question #1: Find the root of the equation, Perform three iteration of the equation,
ln (x −1) + sinx =0 by using Newton Raphson method.Ans: Let f(x) = ln(x+1) + sinx = 0 and f(x) = 1/(x1) + cosx
F (1.5) = ln(0.5) + (1.5) =  0.0667
F(2) = ln(1) + sin(2) = 0.035
Since f (1.5) f (2) < 0 so roots lies in interval [1.5, 2]
Let x0 = 1.75 . x0 can be taken in the interval any real number [ 1.5 , 2 ], we let mid point
of this interval .
As we know Newton Raphson method isXn+1 = xn – f ( xn ) / f(xn)
First iteration
X1 = x0 –f(x0) / f(x0) = 1.75  f(1.75) / f(1.75)
= 1.75 – (0.2571 / 2.3329) = 1.8602
Second iteration:
X2 = x1  f(x) / f(x) = 1.8602 –[ f(1.8602) / f(1.8602)]
= 1.8602  ( 0.1181 / 2.1620 ) = 1.9148
Third iteration:
X3 = x2 f(x2) / f(x2) = 1.9148 –f(1.9148) / f(1.9148)
= 1.9148 – [0.0556/2.0926]
= 1.9414
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
x + y + z = 6;2x − y + z = 3;x + z = 4ANS: In Gaussian elimination method we convert the augmented matrix into reduce
Echelon form therefore,
Augmented matrix is
R2 2R1 , R3 – R1
1R2 , 1R3
R23
R33R2X + Y+ Z = 6 ;………………….(1)
Y = 2,
Z = 3
Put into eq (1),
we get X = 1 , 
This post is deleted! 
This post is deleted!