MTH603 Quiz 1 May 2026

Question:

The linear equation: 0x+0y =2 has -------- solution/solutions.

A. no solution

B. finite many C. unique D. infinite many Copied: 5/7/2026, 12:40:08 PM

Question:

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’?

1. Order of ‘L’ = 3*1, Order of ‘U’ = 1*3
2. Order of ‘L’ = 3*2, Order of ‘U’ = 2*3
3. Order of ‘L’ = 3*4, Order of ‘U’ = 4*3
4. Order of ‘L’ = 3*3, Order of ‘U’ = 3*3

Correct answer: Order of ‘L’ = 3*3, Order of ‘U’ = 3*3

Copied: 5/7/2026, 12:40:38 PM

Question:

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

A. no solution

B. unique C. finite many

D. infinite many

Copied: 5/7/2026, 12:41:26 PM

Question:

Numerical methods for finding the solution of the system of equations are classified as direct and ………… methods

A. Jacobi B. Indirect

C. Iterative

D. None of the given choices Copied: 5/7/2026, 12:41:57 PM

Question:

Full pivoting, in fact, is more ……………than the partial pivoting.

A. Complicated

B. Easiest Copied: 5/7/2026, 12:42:46 PM

Question:

Which of the following rearrangement make strictly diagonal dominant, the system of linear equations; x-3y+z= –2, –6x+4y+11z=1, 5x–2y–2z=9?

1. 5x–2y–2z=9, –6x+4y+11z=1, x–3y+z= –2
2. –6x+4y+11z=1, x–3y+z= –2, 5x–2y–2z=9
3. No need to rearrange as system is already in diagonal dominant form.
4. 5x–2y–2z=9, x–3y+z= –2, –6x+4y+11z=1

Correct answer: 5x–2y–2z=9, x–3y+z= –2, –6x+4y+11z=1

Copied: 5/7/2026, 12:43:24 PM

Question:

While using the Gauss-Seidel Method for finding the solution of the system of equation, the following system x+2y+2z=3x+3y+3z=2x+y+5z=2x+2y+2z=3x+3y+3z=2x+y+5z=2\begin{array}{l} x + 2y + 2z = 3\\x + 3y + 3z = 2\\x + y + 5z = 2 \end{array} can be rewritten as

�A�.� �x�=�3"�2�y"�2�z�y�=�2"�x"�3�z�z�=�2"�y"�x�x�=�3"�2�y"�2�z�y�=�2"�x"�3�z�z�=�2"�y"�x�\�b�e�g�i�n�{�a�r�r�a�y�}�{�l�} x = 3 - 2y - 2z\\y = 2 - x - 3z\\z = 2 - y - x \end{array} �B�.� �x�=�1"�2�3�y"�2�3�z�y�=�1"�3�2�x"�3�2�z�z�=�1"�y�2"�x�2�x�=�1"�2�3�y"�2�3�z�y�=�1"�3�2�x"�3�2�z�z�=�1"�y�2"�x�2�\�b�e�g� x = 1 - \frac{2}{3}y - \frac{2}{3}z\\y = 1 - \frac{3}{2}x - \frac{3}{2}z\\z = 1 - \frac{y}{2} - \frac{x}{2} \end{array} �C�.� �x�=�3"�2�y"�2�z�y�=�2�3"�x�3"�z�z�=�2�5"�x�5"�y�5�x�=�3"�2�y"�2�z�y�=�2�3"�x�3"�z�z�=�2�5"�x�5"�y�5�\�b�e�g�i�n�{�a�r�r�a�y x = 3 - 2y - 2z\\y = \frac{2}{3} - \frac{x}{3} - z\\z = \frac{2}{5} - \frac{x}{5} - \frac{y}{5} \end{array} �D�.� �x�=�3�2"�y"�z�y�=�2"�x�3"�z�z�=�2�5"�x�5"�y�5�x�=�3�2"�y"�z�y�=�2"�x�3"�z�z�=�2�5"�x�5"�y�5�\�b�e�g�i�n�{�a�r�r�a�y�}�{�l�} x = \frac{3}{2} - y - z\\y = 2 - \frac{x}{3} - z\\z = \frac{2}{5} - \frac{x}{5} - \frac{y}{5} \end{array} Copied: 5/7/2026, 12:44:57 PM