problemset2solutions_1 - MAS 213 Linear Algebra II Problem...

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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 unknowns: 2 = a, 1 = a + b, 1 = a + b + c, 3 = a + b + c. This clearly has no solutions. In fact, we should have been able to see this right away. If you look at the matrices in S , you will notice that in each of them the bottom two entries are the same. So any matrix we can ‘build’ from these matrices is going to share that property too. But the matrix we were given does not have this property. So it cannot be in the span of the set S .
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3 Problem 2: (Problem 1.4.6 in [FIS].) Consider the following subset of R 3 : S = { (1 , 1 , 1) , (1 , 1 , 0) , (1 , 0 , 0) } . Prove that span( S ) = R
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