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

[zaasmi]

If we wanted to find the value of a definite integral with an infinite limit, we can instead replace the infinite limit with a variable, and then take the limit as this variable goes to _________.

Chose the correct option :
Constant
Finite
Infinity Click here for detail
Zero

[zaasmi]

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

If we wanted to find the value of a definite integral with an infinite limit, we can instead replace the infinite limit with a variable, and then take the limit as this variable goes to _________.

Chose the correct option :
Constant
Finite
Infinity Click here for detail
Zero

Infinity

Explanation:

• Definite Integral with Infinite Limit: When evaluating a definite integral with an infinite limit, such as (\int_{a}^{\infty} f(x) , dx), we replace the infinite limit with a variable (often denoted as (t)) and then evaluate the integral (\int_{a}^{t} f(x) , dx). After integrating, we take the limit of this result as (t) approaches infinity.

So, the correct option is:

Infinity

[zaasmi]

The Jacobi iteration ______, if A is strictly diagonally dominant.

Converges
Diverges

[zaasmi]

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

The Jacobi iteration ______, if A is strictly diagonally dominant.

Converges
Diverges

Converges

Explanation:

• Jacobi Iteration: This is an iterative method for solving systems of linear equations. The convergence of the Jacobi method depends on the properties of the matrix (A).

• Strict Diagonal Dominance: If matrix (A) is strictly diagonally dominant (i.e., for each row of the matrix, the magnitude of the diagonal entry is greater than the sum of the magnitudes of the other entries in that row), then the Jacobi iteration is guaranteed to converge.

Thus, if (A) is strictly diagonally dominant, the Jacobi iteration converges.

So, the correct option is:

Converges

[zaasmi]

By using determinants, we can easily check that the solution of the given system of linear equation exits and it is unique.

TRUE
FALSE

[zaasmi]

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

By using determinants, we can easily check that the solution of the given system of linear equation exits and it is unique.

TRUE
FALSE

TRUE

Explanation:

• Determinants and Uniqueness: For a system of linear equations (A\mathbf{x} = \mathbf{b}) where (A) is a square matrix, the determinant of (A) helps in determining the existence and uniqueness of the solution.

  • Existence: If (\det(A) \neq 0), the matrix (A) is non-singular (invertible), which implies that the system has a unique solution.

  • Uniqueness: If (\det(A) = 0), the matrix (A) is singular, and the system either has no solution or has infinitely many solutions, depending on the consistency of the system.

Thus, by checking the determinant, we can determine if the solution exists and if it is unique.

So, the statement is:

TRUE

[zaasmi]

The absolute value of a determinant (|detA|) is the product of the absolute values of the eigenvalues of matrix A
TRUE
FALSE

[zaasmi]

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

The absolute value of a determinant (|detA|) is the product of the absolute values of the eigenvalues of matrix A
TRUE
FALSE

TRUE

Explanation:

• Determinant and Eigenvalues: For a square matrix (A), the determinant of (A) is equal to the product of its eigenvalues, considering their algebraic multiplicities.

• Absolute Values: The absolute value of the determinant (|\det(A)|) is indeed equal to the product of the absolute values of the eigenvalues of (A).

So, the statement is:

TRUE

[zaasmi]

Eigenvectors of a symmetric matrix are orthogonal, but only for distinct eigenvalues.

TRUE
FALSE

[zaasmi]

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

Eigenvectors of a symmetric matrix are orthogonal, but only for distinct eigenvalues.

TRUE
FALSE

FALSE

Explanation:

• Eigenvectors of Symmetric Matrices: For a symmetric matrix, eigenvectors corresponding to distinct eigenvalues are orthogonal. However, the orthogonality property also extends to eigenvectors corresponding to the same eigenvalue (i.e., they can be made orthogonal if they are not already).

In summary, for symmetric matrices, eigenvectors corresponding to distinct eigenvalues are orthogonal, and eigenvectors corresponding to the same eigenvalue can be chosen to be orthogonal.

So, the statement is:

FALSE

[zaasmi]

Let A be an n ×n matrix. The number x is an eigenvalue of A if there exists a non-zero vector v such that _______.

Av = xv
Ax=xv
Av + xv=0
Av = Ax1
Av = λv

[zaasmi]

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

Let A be an n ×n matrix. The number x is an eigenvalue of A if there exists a non-zero vector v such that _______.

Av = xv
Ax=xv
Av + xv=0
Av = Ax1
Av = λv

Av = λv

Explanation:

• Eigenvalue and Eigenvector Definition: A number ( \lambda ) is an eigenvalue of an ( n \times n ) matrix ( A ) if there exists a non-zero vector ( v ) such that the equation ( Av = \lambda v ) holds true. Here, ( \lambda ) is the eigenvalue and ( v ) is the corresponding eigenvector.

So, the correct option is:

Av = λv