problemset2solutions_1 - MAS 213: Linear Algebra II....

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MAS 213: Linear Algebra II. Problem list for Week #2. Solutions. . Tutorial problems: Problem 1: (Problems 1.4.3(e), 1.4.4(c) in [FIS].) In the following parts you are given a vector space V , a vector v in it, and a subset S of the vector space. Determine if the given vector v lies in the span of the given subset S . If it does, find an explicit representation of v as a linear combination of the vectors in S . (1) V = R 3 , v = (5 , 1 , 5), and S = { (1 , 2 , 3) , ( 2 , 3 , 4) } . (2) V = P 3 ( R ), v = 2 x 3 11 x 2 + 3 x + 2, and S = { x 3 2 x 2 + 3 x 1 , 2 x 3 + x 2 + 3 x 2 } . (3) V = M 2 × 2 ( R ), v = [ 2 1 1 3 ] , and S = {[ 1 1 1 1 ] , [ 0 1 1 1 ] , [ 0 0 1 1 ]} . Solution to problem 1, part 1. The vector v lies in the span of S if and only if there are scalars a, b R such that: (5 , 1 , 5) = a (1 , 2 , 3) + b ( 2 , 3 , 4) . This is the precise meaning of the statement “ v lies in the span of S ”. Scalars a and b solve this equation if and only if they solve the system of three linear equations that we get from the components of these vectors: a 2 b = 5 , 2 a + 3 b = 1 , 3 a 4 b = 5 . These equations can be expressed as an augmented matrix as follows: 1 2 5 2 3 1 3 4 5 . Now we do the usual row manipulations to put this system in row echelon form, in order to decide if there are any solutions. 1 2 5 2 3 1 3 4 5 −→ 1 2 5 0 1 11 0 10 10 −→ 1 0 3 0 0 10 0 1 1 This has no solutions because it contains the equation 0 b = 10, which cannot be satisfied. 1
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2 So v does not lie in the span of S . Solution to problem 1, part 2. In this case the equation we are solving is: 2 x 3 11 x 2 + 3 x + 2 = a ( x 3 2 x 2 + 3 x 1) + b (2 x 3 + x 2 + 3 x 2) . This is equivalent to the system of 4 equations that are obtained by equating coefficients of like powers of x : a + 2 b = 2 , 2 a + b = 11 , 3 a + 3 b = 3 , a 2 b = 2 . Solving this is the usual way: 1 2 2 2 1 11 3 3 3 1 2 2 −→ 1 2 2 0 5 15 0 3 9 0 0 0 −→ 1 0 4 0 0 0 0 1 3 0 0 0 . Thus the system is solved by a = 4 and b = 3. We can check that by writing out the apparent solution: 2 x 3 11 x 2 + 3 x + 2 = 4( x 3 2 x 2 + 3 x 1) 3(2 x 3 + x 2 + 3 x 2) . This checks out. Thus v does lie in the span of S . Solution to problem 1, part 3. We want to understand the solutions to the equation: [ 2 1 1 3 ] = a [ 1 1 1 1 ] + b [ 0 1 1 1 ] + c [ 0 0 1 1 ] . Equating entries in the same positions gives us a system of 4 equations in 3
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This note was uploaded on 12/08/2010 for the course SPMS MAS213 taught by Professor Andrewkricker during the Fall '10 term at Nanyang Technological University.

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problemset2solutions_1 - MAS 213: Linear Algebra II....

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