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# plugin-hw01 - Assignment 1 Math 417 Winter 2009 Due January...

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Assignment 1 Math 417 — Winter 2009 Due January 16 § 1.1#18. The augmented matrix of the given system is 1 2 3 a 1 3 8 b 1 2 2 c . We reduce it to echelon form. Subtracting the first row from each of the others, we get 1 2 3 a 0 1 5 b - a 0 0 - 1 c - a . Subtracting twice the second row from the first, and multiplying the last row by - 1, we get 1 0 - 7 3 a - 2 b 0 1 5 b - a 0 0 1 a - c . Finally, adding the last row 7 times to the first and - 5 times to the second, we get 1 0 0 10 a - 2 b - 7 c 0 1 0 - 6 a + b + 5 c 0 0 1 a - c . So the solution of the given system is x = 10 a - 2 b - 7 c, y = - 6 a + b + 5 c, z = a - c. § 1.1#24. Let b be the boat’s speed relative to the water (in kilometers per minute, since the problem gives distances in kilometers and times in minutes) and w the speed of the water’s flow. So on the downstream trip, the speed of the boat relative to the land is b + w and on the return trip it is b - w . On each part of the trip, the distance traveled is the speed times the time. So we have 8 = 20( b + w ) = 40( b - w ) . Solving this pair of equations, we get b = 3 / 10 and w = 1 / 10. So the boat’s speed relative to the water is 0 . 3 km / min or 18 km / h, and the water flows at 0 . 1 km / min or 6 km / h. 1

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§ 1.1#26. I’ll start to simplify the equations, by row reduction, but I’ll stop as soon as the answers to the questions become visible, rather than going all the way to rref. The augmented matrix of the given system is 1 1 - 1 2 1 2 1 3 1 1 k 2 - 5 k and subtracting the top row from each of the other two rows produces 1 1 - 1 2 0 1 2 1 0 0 k 2 - 4 k - 2 . Here, we already have two leading ones. If k 2 - 4 isn’t 0, we can divide by it and get a third leading 1 in row 3, column 3. (To get rref, we’d have to subtract multiples of the second and third rows from the higher rows to get 0’s above the leading 1’s.) Then all three variables are leading variables, so there is a unique solution. If, on the other hand, k 2 - 4 = 0, i.e., if k = ± 2, then the following possibilities arise. If k = 2 then the bottom row consists of all 0’s, so the system is consistent, x and y are leading variables, and z is a free variable. Thus the system has infinitely many solutions.
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plugin-hw01 - Assignment 1 Math 417 Winter 2009 Due January...

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