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LinAlg Test 1 [Jan 07]

# LinAlg Test 1 [Jan 07] - 1 THE UNIVERSITY OF WESTERN...

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1 THE UNIVERSITY OF WESTERN ONTARIO Department of Applied Mathematics London Ontario Applied Mathematics 025b First Tutorial Test – January 31, 2008 60 min Name: Solution Section 2 (1) Closed book. Only simple calculators are allowed. Print your name above and student number on the page 2. Justify answers by showing sufficient work to get the full marks. Solve each problem in space provided for that specific problem. Use margins or back of each sheet to do your rough work. Instructor – Dr. N. Kiriushcheva PART A. Mark the correct answer. [4] 1.1. The system of equations ax + by + cz = 0 ax + cy = ln c cx + ( a + b ) xy = yz (a) is linear in { x, y, z } and linear in { a, b } ; (b) is linear in { x, y, z } and linear in { b, c } ; (c) is nonlinear in { x, y, z } and linear in { a, b } ; (d) is nonlinear in { x, y, z } and linear in { b, c } ; (e) none of above. Answer: c. 1.2. The solution of the following system of equations x + y = 0 2 x - 3 y = 4 5 x + 6 y = - 2 (a) is x = 4 and y = - 4; (b) is x = 6 and y = - 6; (c) is x = 0 and y = 0; (d) is x = 2 and y = - 2; (e) is none of the above. Answer: e.

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AM 025b Test 1 – January 31, 2008 2 Student #: FOR INSTRUCTOR’S USE ONLY 1 2 3 4 5 6 7 Total 4 8 8 10 10 10 10 60 [8] 2.1. The following equation ( a is a constant) a 2 x - 5 = 25 x + a (a) has exactly one solution, a R ; (b) has infinitely many solutions regardless what a is; (c) has no solution if a = - 5; (d) has a unique solution if a 6 = ± 5. Answer: d. 2.2. If the augmented matrix of the system of linear equations is 1 0 3 0 6 - 3 0 1 2 - 3 0 8 0 0 1 0 0 3 0 0 0 0 1 0 then there (a) is a unique solution to this system; (b) are infinitely many solutions with one free parameter; (c) are infinitely many solutions with two free parameters; (d) are infinitely many solutions with infinity many parameters; (e) is no solution. Answer: b. 2.3. If y ( x ) = ax 3 + bx 2 + cx + d is a curve passing through the point x = - 4 , y = 3, then (a) 4 = - 3 x 3 + 3 x 2 - 3 x + d (b) - 4 = 3 3 a + 3 2 b + 3 c + d (c) 3 = - 64 a + 16 b - 4 c + d (d) 4 = - 3 3 a + 3 2 b - 3 c + d (e) none of the above.
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