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MTH603 Quiz 2 November 2025 Set 4

Numerical Analysis PDF 38 KBUpdated 23 Oct 2026
Quiz MTH603 Quiz 2 November 2025 Set 4 PDF 38 KB
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MTH603 Numerical Analysis
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PDF 38 KB
Updated
23 Oct 2026
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Text from pages 2, 3, 4 of 6. Download the file for all 6 pages.

Question:

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

A. same B. different

Copied: 11/25/2025, 11:20:07 AM

Question:

�L�e�t�[�A�]�b�e�a�3�×�3�r�e�a�l�s�y�m�m�e�t�r�i�c�m�a�t�r�i�x�w�i�t�h�|�a�1�3�|�b�e�t�h�e�n�u�m�e�r�i�c�a�l�l�y�l�a�r�g�e�s�t�o�f�f"�d�i�a�g�o�n�a� �v�a�l�u�e�o�f¸�c�a�n�b�e�f�o�u�n�d�b�y�L�e�t�[�A�]�b�e�a�3�×�3�r�e�a�l�s�y�m�m�e�t�r�i�c�m�a�t�r�i�x�w�i�t�h�|�a�1�3�|�b�e�t�h�e�n�u�m�e�r�i�c�a �u�s�i�n a�g�J�a�c�o�b�i 2�s�m�e�t�h�o�d�t�h�e�v�a�l�u�e�o�f¸�c�a�n�b�e�f�o�u�n�d�b�y�\�b�e�g�i�n�{�a�r�r�a�y�}�{�l�} Let\,\left[ A \right]\,be\,a\,3 \times 3\,real\,symmetric\,matrix\,with\,\left| {{a_{13}}} \right|\,be\,the\,numerically\,\,l\arg est\,off - diagonal\,element\,then\,\\u\sin g\,Jacobi's\,method\,the\,value\,of\,\theta \,can\,be\,found\,by \end{array}

Copied: 11/25/2025, 11:20:21 AM

Question:

Whileusingtherelaxationmethodforfindingthesolutionofthebelowgivensystem,whichofthefollowingincrementwillbeintro �d�u�c�e�d�?�6�x�1"�2�x�2�+�3�x�3�=�1� "�2�x�1�+�7�x�2�+�2�x�3�=�5� �x�1�+�x�2"�5�x�3�="�1�3� �W�h�i�l�e�u�s�i�n�g�t�h�e�r�e�l�a�x�a�t�i�o�n�m�e�t�h�o�d�f�o�r�f�i�n�d�i� �e�b�e�l�o�w�g�i�v�e�n�s�y�s�t�e�m�,�w�h�i�c�h�o�f�t�h�e�f�o�l�l�o�w�i�n�g�i�n�c�r�e�m�e�n�t�w�i�l�l�b�e�i�n�t�r�o�d�u�c�e�d�?�6�x�1"�2�x�2�+�3�x�3�=�1� "�2�x�1�+�7�x�2�+� "�1�3� �\�b�e�g�i�n�{�g�a�t�h�e�r�e�d�} While\,using\,the\,relaxation\,method\,for\,finding\,the\,solution\,of\,the\,below\,given\,system,\, \hfill \\ which\,of\,the\,following\,increment\,will\,be\,introduced? \hfill \\ \begin{array}{*{20}{l}}

{6{x_1} - 2{x_2} + 3{x_3} = 1} \\\ · { - 2{x_1} + 7{x_2} + 2{x_3} = 5} \\\ · {{x_1} + {x_2} - 5{x_3} = - 13} · \end{array} \hfill \\\ · \end{gathered}

A. dx2=R2a22dx2=R2a22d{x_2} = \frac{{{R_2}}}{{{a_{22}}}} B. dx3=R3a33dx3=R3a33d{x_3} = \frac{{{R_3}}}{{{a_{33}}}} C. dx1=R1a11dx1=R1a11d{x_1} = \frac{{{R_1}}}{{{a_{11}}}} D. dx1=R1a22dx1=R1a22d{x_1} = \frac{{{R_1}}}{{{a_{22}}}}

Copied: 11/25/2025, 11:20:42 AM

Question:

The Gauss-Seidel method is applicable to strictly diagonally dominant or symmetric ________ definite matrices A.

A. negative B. positive

Copied: 11/25/2025, 11:21:10 AM

Question:

2&0&0\\0&2&1\\0&1&2

�a�n�d� �'�t�h�e�t�a� ¸� �=�p�i�/�4�'� �t�h�e� �o�r�t�h�o�g�o�n�a�l� �m�a�t�r�i�x� �S�1� �w�i�l�l� �b�e� �g�i�v�e�n� �a�s

�A�.� �S�1�=#¡#£#¢#¢�c�o�sÀ�4�s�i�nÀ�4�0"�s�i�nÀ�4�c�o�sÀ�4�0�0�0�1#¤#¦#¥#¥�S�1�=�[�c�o�s aÀ�4"�s�i�n aÀ�4�0�s�i�n aÀ�4�c

{\cos \frac{\pi }{4}}&{ - \sin \frac{\pi }{4}}&0\\{\sin \frac{\pi }{4}}&{\cos \frac{\pi }{4}}&0\\0&0&1

�B�.� �S�1�=#¡#£#¢#¢�c�o�sÀ�4�0�s�i�nÀ�4�0�1�0"�s�i�nÀ�4�0�c�o�sÀ�4#¤#¦#¥#¥�S�1�=�[�c�o�s aÀ�4�0"�s�i�n aÀ�4�0�1�0�s�i�n a

{\cos \frac{\pi }{4}}&0&{ - \sin \frac{\pi }{4}}\\0&1&0\\{\sin \frac{\pi }{4}}&0&{\cos \frac{\pi }{4}}

�C�.� �S�1�=#¡#£#¢#¢�1�0�0�0�c�o�sÀ�4�s�i�nÀ�4�0"�s�i�nÀ�4�c�o�sÀ�4#¤#¦#¥#¥�S�1�=�[�1�0�0�0�c�o�s aÀ�4"�s�i�n aÀ�4�0�s�i�n

1&0&0\\0&{\cos \frac{\pi }{4}}&{ - \sin \frac{\pi }{4}}\\0&{\sin \frac{\pi }{4}}&{\cos \frac{\pi }{4}}

�D�.� �S�1�=#¡#£#¢#¢�c�o�sÀ�4�0�0�0"�s�i�nÀ�4�0�0�0�1#¤#¦#¥#¥�S�1�=�[�c�o�s aÀ�4�0�0�0"�s�i�n aÀ�4�0�0�0�1�]�{�S�_�1�}� �=� �\ {\cos \frac{\pi }{4}}&0&0\\0&{ - \sin \frac{\pi }{4}}&0\\0&0&1

Copied: 11/25/2025, 11:26:01 AM

Question:

1&{\sqrt 3 }&{\sqrt 2 }\\{\sqrt 3 }&3&2\\{\sqrt 2 }&2&1

�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

Copied: 11/25/2025, 11:27:16 AM

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