problemset3solutions - MAS 213: Linear Algebra II. Problem...

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Unformatted text preview: MAS 213: Linear Algebra II. Problem list for Week #3. Solutions . Problem 1: (Problems 1.6.3 (d)-(e) in [FIS].) Which of the following sets are bases for R 3 ? 1. { (- 1 , 3 , 1) , (2 ,- 4 ,- 3) , (- 3 , 8 , 2) } 2. { (1 ,- 3 ,- 2) , (- 3 , 1 , 3) , (- 2 ,- 10 ,- 2) } Solution to problem 1, part 1. Strictly speaking, we have to show that this is both a spanning set and also that it is linearly independent. But because we already know that R 3 has dimension 3, and because we notice that the set has 3 elements, it is enough for us to prove one of these two conditions. (For example: if we know that it is linearly independent, then a theorem tells us that the set can be enlarged to a basis. But it already has 3 elements! So it would already be a basis.) So we’ll just test whether it is linearly independent. To test that, we have to determine whether there are any non-trivial solutions to the equation: (0 , , 0) = a (- 1 , 3 , 1) + b (2 ,- 4 ,- 3) + c (- 3 , 8 , 2) . We solve this: - 1 2- 3 3- 4 8 1- 3 2 → - 1 2- 3 2- 1- 1- 1 → - 1 2- 3- 1- 1- 3 Back-substitution tells us that there is only the trivial solution a = b = c = 0. Thus the given set is linearly independent, and hence a basis. Solution to problem 1, part 2. Repeating the exercize, we are looking for non-trivial solutions to: (0 , , 0) = a (1 ,- 3 ,- 2) + b (- 3 , 1 , 3) + c (- 2 ,- 10 ,- 2) . 1 We solve: 1- 3- 2- 3 1- 10- 2 3- 2 → 1- 3- 2- 8- 16- 3- 6 → 1- 3- 2 1 2 1 2 → 1- 3- 2 1 2 The general solution to this has an independent variable, so there will be infinitely many solutions. In particular it will have a non-trivial solution, such as: (0 , , 0) = (- 4)(1 ,- 3 ,- 2) + (- 2)(- 3 , 1 , 3) + (1)(- 2 ,- 10 ,- 2) . So the set is linearly dependent, and is not a basis. Problem 2: (Problems 1.6.3 (d)-(e) in [FIS].) Which of the following sets are bases for P 2 ( R )? 1. {- 1 + 2 x + 4 x 2 , 3- 4 x- 10 x 2 ,- 2- 5 x- 6 x 2 } 2. { 1 + 2 x- x 2 , 4- 2 x + x 2 ,- 1 + 18 x- 9 x 2 } Solution to problem 2, part 1. The same discussion in the above problem applies here. That is, we know in advance that P 2 ( R ) has dimension 3, so it is sufficient for us to check whether the given sets are linearly independent or not. So we ask: are there any non-trivial solutions to the equation: 0 = a (- 1 + 2 x + 4 x 2 ) + b (3- 4 x- 10 x 2 ) + c (- 2- 5 x- 6 x 2 ) . Turning this into a system of three equations in three variables and solving: - 1 3- 2 2- 4- 5 4- 10- 6 → - 1 3- 2 0 2- 9 0 2- 14 → - 1 3- 2 0 2- 9 0 0- 5 This only has the unique solution, so the given set is linearly independent, so it is a basis....
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problemset3solutions - MAS 213: Linear Algebra II. Problem...

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