MTH603 - Numerical Analysis

@zainab-ayub said in MTH603 Quiz 2 Solution and Discussion:

Mth603 ka koi student hai tu plz yeh question bta dy kis trha solve ho ga Given the following data x:1 2 5 y:1 4 10 Value of 1st order divided difference f[2 , 5] is

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@zareen said in MTH603 Mid Term Past and Current Solved Paper Discussion:

If n x n matrices A and B are similar, then they have the same eigenvalues (with the same multiplicities).

True
False

Since similar matrices A and B have the same characteristic polynomial, they also have the same eigenvalues. If B = PAP−1 and v = 0 is an eigenvector of A (say Av = λv) then B(Pv) = PAP−1(Pv) = PA(P−1P)v = PAv = λPv. Thus Pv (which is non-zero since P is invertible) is an eigenvector for B with eigenvalue λ.

@zareen said in MTH603 Assignment 1 Solution and Discussion:

Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
2x + y + z = 9;3x −2y + 4z = 9;x +y-2z = 3

System of Linear Equations entered :

[1] 2x + y + z = 9
[2] 3x - 2y + 4z = 9
[3] x + y - 2z = 3

Solve by Substitution :

// Solve equation [3] for the variable y

[3] y = -x + 2z + 3

// Plug this in for variable y in equation [1]

[1] 2x + (-x +2z+3) + z = 9
[1] x + 3z = 6
// Plug this in for variable y in equation [2]

[2] 3x - 2•(-x +2z+3) + 4z = 9
[2] 5x = 15
// Solve equation [2] for the variable x

[2] 5x = 15

[2] x = 3
// Plug this in for variable x in equation [1]

[1] (3) + 3z = 6
[1] 3z = 3
// Solve equation [1] for the variable z

[1] 3z = 3

[1] z = 1
// By now we know this much :

x = 3
y = -x+2z+3
z = 1
// Use the x and z values to solve for y
y = -(3)+2(1)+3 = 2

Solution :
{x,y,z} = {3,2,1}

MTH603 MCQ’s for Final Term.pdf

Solution idea

Spring 2020_MTH603_1_SOL.docx

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/(x-1) + 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 is

Xn+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 = 4

ANS: In Gaussian elimination method we convert the augmented matrix into reduce

Echelon form therefore,

Augmented matrix is

R2- 2R1 , R3 – R1

-1R2 , -1R3
R23
R3-3R2

X + Y+ Z = 6 ;………………….(1)
Y = 2,
Z = 3
Put into eq (1),
we get X = 1 ,