{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mat1302-final-2003-april-solutions

# mat1302-final-2003-april-solutions - MAT 1302 A...

This preview shows pages 1–4. Sign up to view the full content.

MAT 1302 A Mathematical Methods II Final Exam: Solutions and Marking Scheme April 23, 2003 Q.1. [8 points] (a) [1 pt] Give an example of a matrix in RREF (reduced row echelon form). Marking: 1 point for any correct answer (1’s as pivots, echelon, zeros above, below and to the left of any pivot.) (b) [1 pt] Compute 1 1 + 3 - 4 . Answer: 4 - 3 Marking: 1 point correct answer (c) [3 pts] Give 3 elements in span 1 1 0 , 0 1 1 Marking: 1 point each correct vector; if you gave more than 3, only the first three are checked. (d) [1 pt] Given z = 3 + 4 i compute | z | . Answer: 3 2 + 4 2 = 25 = 5. Marking: 1 point correct answer, deduct .5 for answer ± 5. (e) [2 pts] Give the roots of the polynomial x 2 - 4 x + 8. Answer: The quadratic formula gives x = 4 ± 16 - 4(8) 2 = 4 ± 4 i 2 = 2 ± 2 i as the two roots. Marking: 1 point for correct quadratic formula, 1 point for correct simplification; or 2 point for correct answer by any means. Q.2. [6 points] This question has no partial credit. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 The following is the augmented matrix of a system of linear equations, where k a parameter: 1 2 4 | 2 0 k + 1 4 | 2 0 0 k - 1 | - 1 (a) [2 pts] For which of the following values of k is the system inconsistent: - 1 , 0 , 1 , no values Answer: k = 1 (because last row is 0=1) (b) [2 pts] For which of the following values of k does the system have infinitely many solutions: - 1 , 0 , 1 , no values Answer: k = - 1 (because then 2nd and 3rd rows are multiples, the system is consis- tent, and there is a free variable) (c) [2 pts] For which of the following values of k does the system have a unique solution: - 1 , 0 , 1 , no values Answer: k = 0 (lots of other values are ok, too; it gives 3 pivots in the coefficient matrix) Marking: No partial credit; 2 points each for correct answer. (Circling more than one answer is incorrect = 0 points.) Q.3. [6 points] (a) [1 pt] Define span { v 1 , v 2 } . Answer: span { v 1 , v 2 } = { c 1 v 1 + c 2 v 2 | c 1 , c 2 are scalars } Marking: 1 point for correct definition; .5 for something mostly correct. (b) [5 pts] For which value(s) of the parameter k is the vector 1 k k in span 1 2 3 , 3 2 1 ? Answer: We have to row reduce the following augmented matrix and determine the values of k for which the system is consistent: 1 3 | 1 2 2 | k 3 1 | k - 2 R 1 + R 2 - 3 R 1 + R 3 1 3 | 1 0 - 4 | k - 2 0 - 8 | k - 3 - 2 R 2 + R 3 1 3 | 1 0 - 4 | k - 2 0 0 | - 2( k - 2) + k - 3 This is in echelon form. The system is consistent if and only if 0 = - 2( k - 2)+ k - 3 = - k + 1, in other words, if and only if k = 1.
3 Marking:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

mat1302-final-2003-april-solutions - MAT 1302 A...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online