dy/dx - = 1 - y,y(0) = 0 is an example of

[cyberian]

Which of the following rearrangements make the system of linear equations strictly diagonally dominant: 39 + 2 = -2 62 + 4y + 11z = 1 5r - 23 - 22 = 9 Select one: 2v - 22 = 9 3y = 2 6z + 4y + M = 1 6r + 4y + M = 1 -8v + 2 = -2 51 - %0 - 22 = 0 61 - %v - 28 - 0 Av + M2 = 30 + 2 = -2 No need to rearrange as the system is already diagonally dominant.
Which of the following rearrangements make the system of linear equations strictly diagonally dominant:
39 + 2 = -2
62 + 4y + 11z = 1
5r - 23 - 22 = 9

Select one:
2v - 22 = 9
3y = 2
6z + 4y + M = 1
6r + 4y + M = 1
-8v + 2 = -2
51 - %0 - 22 = 0
61 - %v - 28 - 0 Av + M2 = 30 + 2 = -2

No need to rearrange as the system is already diagonally dominant.

[cyberian]

Which of the following rearrangement make strictly diagonal dominant, the system of linear

[cyberian]

if the relaxation method is applied on the system

[anas khan]

Please Madam Answer This question.
Total Marks 5
Starting Date Friday, May 03, 2024
Closing Date Wednesday, May 08, 2024
Status Open
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Give your opinion on this statement in 200 to 250 words.

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[cyberian]

@anas-khan please post into relevant category

[cyberian]

while using jacobi method for the matrix a=[2 0 0 02 1 0 1 2] o=pi/4

[cyberian]

I’m slightly confused about how to use the power method and the steps to calculate an eigenvalue. - I understand that the power method is defined as U(x+1) = AU(x)/a(x) where “a” is the first component of U(x). I do not understand at all what “U” is. Are we picking any vector we want that minimizes the error? What would I do given the practice problem below?

Apply the power method to
132−4
1 2 3 −4

to obtain three approximations of the largest eigenvalue of A. What is the limiting vector u∞?

[cyberian]

Here is a table of iterations. The actual vector U is horizontal written as entry 1 and 2, and the current approximation to the first eigenvalue is in the third column. The first(largest in absolute value) eigenvalue is negative, so the system needs a certain time to reduce oscillations:

U[1]	U[2]	eigenvalue
1	0	0
1	3	1
1	-9/7	7
1	-57/11	-11/7
1	-261/103	-103/11
1	-1353/419	-419/103
1	-6669/2287	-2287/419

[cyberian]

@cyberian said in MTH603 Quiz 1 Solution and Discussion:

Here is a table of iterations. The actual vector U is horizontal written as entry 1 and 2, and the current approximation to the first eigenvalue is in the third column. The first(largest in absolute value) eigenvalue is negative, so the system needs a certain time to reduce oscillations:

U[1]	U[2]	eigenvalue
1	0	0
1	3	1
1	-9/7	7
1	-57/11	-11/7
1	-261/103	-103/11
1	-1353/419	-419/103
1	-6669/2287	-2287/419

in float reprecentation

U[1]	U[2]	eigenvalue
1.0000000	0	0
1.0000000	3.0000000	1.0000000
1.0000000	-1.2857143	7.0000000
1.0000000	-5.1818182	-1.5714286
1.0000000	-2.5339806	-9.3636364
1.0000000	-3.2291169	-4.0679612
1.0000000	-2.9160472	-5.4582339
1.0000000	-3.0347480	-4.8320944
1.0000000	-2.9862913	-5.0694960
1.0000000	-3.0055137	-4.9725827
1.0000000	-2.9977994	-5.0110274
1.0000000	-3.0008810	-4.9955987

[cyberian]

Using the Jacobi method find all the eigenvalues and the corresponding 2 eigenvectors of the matrix A = 2 1 2 Iterate till the oR- ~1 2 diagonal elements, in magnitude; are less than 0.0005
Using the Jacobi method find all the eigenvalues and the corresponding 2 eigenvectors of the matrix A = 2 1 2 Iterate till the oR- ~1 2 diagonal elements, in magnitude; are less than 0.0005

[cyberian]

while using jacobi method fot the matrix A=[1 1/4 1/4 1/4 1/3 1/2 1/3 1/2 1/5]

[cyberian]

Let A be an n × n matrix. Then λ = 0 is an eigenvalue of A if and only if there exists a non-zero vector v ∈ Rn such that Av = λv = 0. In other words, 0 is an eigenvalue of A if and only if the vector equation Ax = 0 has a non-zero solution x ∈ Rn.

[cyberian]

an eigenvector v is said to be normalized if the coordinate of largest magnitude is equal to aero?

[cyberian]

Two similar matrices have the same eigenvalues, even though they will usually have different eigenvectors. Said more precisely, if B = Ai’AJ. I and x is an eigenvector of A, then M’x is an eigenvector of B = M’AM. So, A1’x is an eigenvector for B, with eigenvalue ).

[cyberian]

b) The eigenvalues of a real symmetric matrix need not be positive. They can be positive, negative, or even zero, depending on the elements and the specific structure of the matrix.

[cyberian]

Jacobian Method in Matrix Form
Let the n system of linear equations be Ax = b. Let us decompose matrix A into a diagonal component D and remainder R such that A = D + R. Iteratively the solution will be obtained using the below equation.