• Cyberian's Gold

    The linear equation: x+y=1 has --------- solution/solutions.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    Gauss–Seidel method is also known as method of …………….

    In numerical linear algebra, the Gauss–Seidel method, also known as the Liebmann method or the method of successive displacement, is an iterative method used to solve a system of linear equations.

  • Cyberian's Gold

    Gauss–Seidel method is also known as method of …………….

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    Differences methods are iterative methods. yes or no

    1. Which of the following is not an iterative method? Explanation: Jacobi’s method, Gauss Seidal method and Relaxation method are the iterative methods and Gauss Jordan method is not as it does not involves repetition of a particular set of steps followed by some sequence which is known as iteration.
  • Cyberian's Gold

    Differences methods are iterative methods. yes or no

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    Power method is applicable if the eigen vectors corresponding to eigen values are linearly _______.

    The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal. Power method is applicable if the eigen vectors corresponding to eigen values are linearly independent.

  • Cyberian's Gold

    Power method is applicable if the eigen vectors corresponding to eigen values are linearly _______.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    Two matrices with the same characteristic polynomial need not be similar.

    Two similar matrices have the same characteristic polynomial. The converse however is not true in general: two matrices with the same characteristic polynomial need not be similar. The matrix A and its transpose have the same characteristic polynomial. … In this case A is similar to a matrix in Jordan normal form.

  • Cyberian's Gold

    Two matrices with the same characteristic polynomial need not be similar.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    If A is a nxn triangular matrix (upper triangular, lower triangular) or diagonal matrix , the eigenvalues of A are the diagonal entries of A.

    The eigenvalues of B are 1,4,6 since B is an upper triangular matrix and eigenvalues of an upper triangular matrix are diagonal entries. We claim that the eigenvalues of A and B are the same.

  • Cyberian's Gold

    If A is a nxn triangular matrix (upper triangular, lower triangular) or diagonal matrix , the eigenvalues of A are the diagonal entries of A.

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

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

    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.

    The matrix B has the same A as an eigenvalue. M−1x is the eigenvector. If two matrices are similar, they have the same eigenvalues and the same number of independent eigenvectors (but probably not the same eigenvectors).

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

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

    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.

  • Cyberian's Gold

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

  • Cyberian's Gold

    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

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

    The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal. Power method is applicable if the eigen vectors corresponding to eigen values are linearly independent.

  • Cyberian's Gold

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

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