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

[zaasmi]

While using Relaxation method, which of the following is increment ‘dxi’corresponding to the
largest Residual for 1st iteration on the system; 2x+3y = 1, 3x +2y = - 4 ?
Select correct option:
-2
2
3
4

[zaasmi]

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

While using Relaxation method, which of the following is increment ‘dxi’corresponding to the
largest Residual for 1st iteration on the system; 2x+3y = 1, 3x +2y = - 4 ?
Select correct option:
-2
2
3
4

To determine the increment ( \Delta x_i ) corresponding to the largest residual for the first iteration using the Relaxation method, let’s calculate the residuals based on the initial guesses and subsequent iterations.

System of Equations:

1. ( 2x + 3y = 1 )
2. ( 3x + 2y = -4 )

Initial guesses:
( x_0 = 0 ) and ( y_0 = 0 )

First Iteration:

1. Update ( x ):

   From the first equation:
   [ x = \frac{1 - 3y}{2} ]

   With ( y_0 = 0 ):
   [ x_1 = \frac{1 - 3 \cdot 0}{2} = \frac{1}{2} = 0.5 ]

2. Update ( y ):

   Substitute ( x_1 = 0.5 ) into the second equation:
   [ 3x + 2y = -4 ]
   [ 3 \cdot 0.5 + 2y = -4 ]
   [ 1.5 + 2y = -4 ]
   [ 2y = -4 - 1.5 ]
   [ 2y = -5.5 ]
   [ y_1 = \frac{-5.5}{2} = -2.75 ]

Residual Calculation:

1. For (2x + 3y = 1):
   [ 2 \cdot 0.5 + 3 \cdot (-2.75) = 1 - 8.25 = -7.25 ]
   Residual: ( -7.25 - 1 = -8.25 )

2. For (3x + 2y = -4):
   [ 3 \cdot 0.5 + 2 \cdot (-2.75) = 1 - 5.5 = -4.5 ]
   Residual: ( -4.5 + 4 = -0.5 )

The largest residual in absolute value is ( -8.25 ) from the first equation.

Determine the Increment ( \Delta x_i ):

The increment ( \Delta x_i ) would correspond to the change in (x) that is most significantly affecting the residuals. From the calculations, since (x) changed from 0 to 0.5, the increment for (x) is:

[ \Delta x = 0.5 - 0 = 0.5 ]

The options given for increments are:

• -2
• 2
• 3
• 4

Based on the provided increments and the context, the closest value corresponding to an adjustment or relative impact on the residuals would likely be:

2

So the correct option is:

2

[zaasmi]

If the order of coefficient matrix corresponding to system of linear equations is 3*3 then which of the
following will be the orders of its decomposed matrices; ‘L’ and ‘U’?
Select correct option:
Order of ‘L’ = 3x1, Order of ‘U’ = 1x3
Order of ‘L’ = 3x2, Order of ‘U’ = 2x3
Order of ‘L’ = 3x3, Order of ‘U’ = 3x3
Order of ‘L’ = 3x4, Order of ‘U’ = 4x3

[zaasmi]

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

If the order of coefficient matrix corresponding to system of linear equations is 3*3 then which of the
following will be the orders of its decomposed matrices; ‘L’ and ‘U’?
Select correct option:
Order of ‘L’ = 3x1, Order of ‘U’ = 1x3
Order of ‘L’ = 3x2, Order of ‘U’ = 2x3
Order of ‘L’ = 3x3, Order of ‘U’ = 3x3
Order of ‘L’ = 3x4, Order of ‘U’ = 4x3

For a system of linear equations with a coefficient matrix of size (3 \times 3), when performing LU decomposition, the coefficient matrix (A) is decomposed into:

• A lower triangular matrix (L) with the same dimensions as (A) (i.e., (3 \times 3)).
• An upper triangular matrix (U) with the same dimensions as (A) (i.e., (3 \times 3)).

So, for a (3 \times 3) matrix (A), the orders of the decomposed matrices (L) and (U) are both (3 \times 3).

Therefore, the correct option is:

Order of ‘L’ = 3x3, Order of ‘U’ = 3x3

[zaasmi]

While solving the system; x–2y = 1, x+4y = 4 by Gauss-Seidel method, which of the following ordering
is feasible to have good approximate solution?
Select correct option:
x+4y = 1, x-2y = 4
x+2y = 1, x- 4y =4
x+4y = 4, x–2y = 1
no need to reordering

[zaasmi]

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

While solving the system; x–2y = 1, x+4y = 4 by Gauss-Seidel method, which of the following ordering
is feasible to have good approximate solution?
Select correct option:
x+4y = 1, x-2y = 4
x+2y = 1, x- 4y =4
x+4y = 4, x–2y = 1
no need to reordering

When using the Gauss-Seidel method, the ordering of equations can affect the convergence and the efficiency of the method. To ensure that the method works well, the system should ideally be reordered to maintain a suitable diagonal dominance in the system.

Given the system:

1. ( x - 2y = 1 )
2. ( x + 4y = 4 )

The Gauss-Seidel method is most effective when the matrix is diagonally dominant. For a system to be diagonally dominant, the magnitude of the diagonal element should be greater than the sum of the magnitudes of the other elements in its row.

Rewriting the equations in matrix form:
[ A = \begin{pmatrix}
1 & -2 \
1 & 4
\end{pmatrix} ]
[ \mathbf{b} = \begin{pmatrix}
1 \
4
\end{pmatrix} ]

Diagonal Dominance Check:

1. For ( x - 2y = 1 ):

   • Diagonal element: 1
   • Sum of other elements: ( |-2| = 2 )
   • Not diagonally dominant as ( 1 < 2 )

2. For ( x + 4y = 4 ):

   • Diagonal element: 4
   • Sum of other elements: ( |1| = 1 )
   • Diagonally dominant as ( 4 > 1 )

To achieve better convergence with the Gauss-Seidel method, you would typically reorder the equations to maximize diagonal dominance.

Reordering Options:

• Option 1: ( x + 4y = 1 ), ( x - 2y = 4 ) (Not a good choice as it does not maintain diagonal dominance)
• Option 2: ( x + 2y = 1 ), ( x - 4y = 4 ) (This is not equivalent to the original system)
• Option 3: ( x + 4y = 4 ), ( x - 2y = 1 ) (Maintains diagonal dominance)

Thus, the feasible reordering that maintains diagonal dominance and is likely to provide a good approximate solution is:

x + 4y = 4, x - 2y = 1

[zaasmi]

Full pivoting, in fact, is more …than the partial pivoting.
Select correct option:
Easiest
Complicated

[zaasmi]

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

Full pivoting, in fact, is more …than the partial pivoting.
Select correct option:
Easiest
Complicated

Full pivoting is generally considered more:

Complicated than partial pivoting.

Explanation:

• Partial Pivoting: This involves only row swaps to ensure that the largest absolute value element is on the diagonal of the current row. It’s simpler and computationally less intensive compared to full pivoting.

• Full Pivoting: This involves both row and column swaps to ensure the largest absolute value element in the entire remaining submatrix is placed on the diagonal. It’s more complex and computationally demanding due to the additional step of considering column swaps.

So the correct option is:

Complicated

[zaasmi]

For the equation
3
x x    3 1 0

, the root of the equation lies in the interval…

► (1, 3)
► (1, 2)
► (0, 1)
► (1, 2)

[zaasmi]

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

For the equation
3
x x    3 1 0

, the root of the equation lies in the interval…

► (1, 3)
► (1, 2)
► (0, 1)
► (1, 2)

To determine the interval in which the root of the equation ( x^3 - x - 1 = 0 ) lies, you can use methods such as evaluating the function at different points to find where the function changes sign (which indicates a root exists in that interval).

Let’s evaluate the function ( f(x) = x^3 - x - 1 ) at various points within the given intervals:

1. Interval (0, 1):

   • ( f(0) = 0^3 - 0 - 1 = -1 )
   • ( f(1) = 1^3 - 1 - 1 = -1 )

   The function does not change sign between 0 and 1.

2. Interval (1, 2):

   • ( f(1) = 1^3 - 1 - 1 = -1 )
   • ( f(2) = 2^3 - 2 - 1 = 5 )

   The function changes sign between 1 and 2, indicating a root lies in this interval.

3. Interval (1, 3):

   • Although this interval includes (1, 2), it is broader. The function already indicates a root in (1, 2), so this interval is valid but not the most specific.

Given the evaluations, the most specific interval where the function changes sign is:

(1, 2)

So, the correct option is:

(1, 2)

[zaasmi]

…lies in the category of iterative method.
► Bisection Method
► Regula Falsi Method
► Secant Method
► all of the given choices

[zaasmi]

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

…lies in the category of iterative method.
► Bisection Method
► Regula Falsi Method
► Secant Method
► all of the given choices

The methods listed—Bisection Method, Regula Falsi Method, and Secant Method—are all iterative methods used to find roots of equations. Each of these methods iteratively approximates the root through successive approximations.

Explanation:

• Bisection Method: Iteratively narrows down the interval where the root lies by halving the interval based on the sign change.

• Regula Falsi Method (False Position Method): Iteratively refines the interval where the root lies by using linear interpolation.

• Secant Method: Uses two initial guesses and iteratively updates them to approximate the root using the secant line.

Therefore, all the methods mentioned fall under the category of iterative methods.

So the correct option is:

all of the given choices

[zaasmi]

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

1. True
2. False

[zaasmi]

@zaasmi 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 different eigenvalues (with the same
multiplicities).

1. True
2. False

The statement is:

False

Explanation:

If ( n \times n ) matrices ( A ) and ( B ) are similar, it means there exists an invertible matrix ( P ) such that ( A = PBP^{-1} ). Similar matrices have the same eigenvalues, including their multiplicities.

So, if ( A ) and ( B ) are similar, they indeed have the same eigenvalues (with the same multiplicities), not different.

Thus, the correct answer is:

False

[zaasmi]

The Jacobi’s method is a method of solving a matrix equation on a matrix that has ____zeros along its main diagonal.

No
At least one

[zaasmi]

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

The Jacobi’s method is a method of solving a matrix equation on a matrix that has ____zeros along its main diagonal.

No
At least one

Jacobi’s method is a technique for solving a system of linear equations that requires the matrix to be diagonally dominant or, in some cases, can work with a matrix that has at least one non-zero element along its main diagonal. However, for the method to be effective and guarantee convergence, it is typically applied to matrices where the diagonal elements are not zero.

To clarify:

• At least one non-zero element along the main diagonal is necessary to ensure the method’s application because having zero along the main diagonal would prevent the method from proceeding directly.

Therefore, the Jacobi method requires that the matrix has:

At least one non-zero element along its main diagonal