MTH603 Quiz 2 November 2025

Question:

Gauss - Jordan Method is similar to ……….

A. Relaxation Method B. Gaussian elimination method C. Iteration’s method D. Gauss–Seidel method Copied: 11/25/2025, 10:44:32 AM

Question:

Let [A] be a 3x3 real symmetric matrix with |a23||a23|\left| {{a_{23}}} \right| be numerically the largest off-diagonal element of A, then we can construct orthogonal matrix S1 by Jacobi’s method as

�A�.� #¡#£#¢�1�0�0�0�c�o�s¸�s�i�n¸�0"�s�i�n¸�c�o�s¸#¤#¦#¥�[�1�0�0�0�c�o�s a¸"�s�i�n a¸�0�s�i�n a¸�c�o�s a¸�]�\�l�e�f�t�[� �{�\�b�e�g�i�n 1&0&0\\0&{\cos \theta }&{ - \sin \theta }\\0&{\sin \theta }&{\cos \theta }

�B�.� #¡#£#¢�1�0�0�0�c�o�s¸�s�i�n¸�0"�c�o�s¸"�s�i�n¸#¤#¦#¥�[�1�0�0�0�c�o�s a¸"�c�o�s a¸�0�s�i�n a¸"�s�i�n a¸�]�\�l�e�f�t�[� �{�\�b�e�g 1&0&0\\0&{\cos \theta }&{ - \cos \theta }\\0&{\sin \theta }&{ - \sin \theta }

�C�.� #¡#£#¢�c�o�s¸�s�i�n¸�0"�s�i�n¸�c�o�s¸�0�0�0�1#¤#¦#¥�[�c�o�s a¸"�s�i�n a¸�0�s�i�n a¸�c�o�s a¸�0�0�0�1�]�\�l�e�f�t�[� �{�\�b�e�g�i�n {\cos \theta }&{ - \sin \theta }&0\\{\sin \theta }&{\cos \theta }&0\\0&0&1

�D�.� #¡#£#¢�c�o�s¸�0�s�i�n¸�0�1�0"�s�i�n¸�0�c�o�s¸#¤#¦#¥�[�c�o�s a¸�0"�s�i�n a¸�0�1�0�s�i�n a¸�0�c�o�s a¸�]�\�l�e�f�t�[� �{�\�b�e�g�i�n {\cos \theta }&0&{ - \sin \theta }\\0&1&0\\{\sin \theta }&0&{\cos \theta }

Copied: 11/25/2025, 10:44:43 AM

Question:

A 3 x 3 identity matrix have three and __________eigen values.

A. different B. same Copied: 11/25/2025, 10:46:18 AM

Question:

Iterative algorithms can be more rapid than direct methods.

A. TRUE B. FALSE Copied: 11/25/2025, 10:45:15 AM

Question:

The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal.

A. FALSE B. TRUE Copied: 11/25/2025, 10:45:24 AM

Question:

For two matrices A and B, such that “A = Inverse of B”, then which of the following is true?

1. AB = Identity Matrix
2. All choices are true
3. A and B are non-singular
4. B = Inverse of A

Copied: 11/25/2025, 10:46:05 AM

Question:

While using Jacobi method for the matrix �A�=#¡#£#¢�1�1�/�4�1�/�2�1�/�4�1�/�3�1�/�4�1�/�2�1�/�4�1�/�5#¤#¦#¥�A�=�[�1�1�/�4�1�/�2�1�/�4�1�/�3�1�/�4�1�/�2�1�/�4�1�/�5�]�A� �=� �\�l�e�f�t�[� � 1&{1/4}&{1/2}\\{1/4}&{1/3}&{1/4}\\{1/2}&{1/4}&{1/5}

�t�h�e� �v�a�l�u�e� �o�f� �‘�t�h�e�t�a� ¸�’� �c�a�n� �b�e� �f�o�u�n�d� �a�s

�A�.� �t�a�n�2¸�=�2�a�1�3�a�1�1"�a�3�3�t�a�n a�2¸�=�2�a�1�3�a�1�1"�a�3�3�\�t�a�n� �2�\�t�h�e�t�a� � �=� �\�f�r�a�c�{�{�2�{�a�_�{�1�3�}�}�}�}�{� �B�.� �t�a�n�2¸�=�2�a�1�3�a�3�3"�a�1�1�t�a�n a�2¸�=�2�a�1�3�a�3�3"�a�1�1�\�t�a�n� �2�\�t�h�e�t�a� � �=� �\�f�r�a�c�{�{�2�{�a�_�{�1�3�}�}�}�}�{� �C�.� �t�a�n�2¸�=�2�a�1�2�a�1�1"�a�2�2�t�a�n a�2¸�=�2�a�1�2�a�1�1"�a�2�2�\�t�a�n� �2�\�t�h�e�t�a� � �=� �\�f�r�a�c�{�{�2�{�a�_�{�1�2�}�}�}�}�{� �D�.� �t�a�n�2¸�=�2�a�2�3�a�2�2"�a�3�3�t�a�n a�2¸�=�2�a�2�3�a�2�2"�a�3�3�\�t�a�n� �2�\�t�h�e�t�a� � �=� �\�f�r�a�c�{�{�2�{�a�_�{�2�3�}�}�}�}�{� Copied: 11/25/2025, 10:46:30 AM