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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 nontrivial 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 Backsubstitution 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 nontrivial 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 nontrivial 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 nontrivial 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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 Fall '10
 ANDREWKRICKER
 Linear Algebra, Vector Space, Berlin UBahn

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