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IntrodLineaAlgebr_solutions

IntrodLineaAlgebr_solutions - INTRODUCTION TO LINEAR...

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INTRODUCTION TO LINEAR ALGEBRA, Second Edition by Gilbert Strang SOLUTIONS TO SELECTED EXERCISES Christopher Heil Spring 2000 CHAPTER 1 Introduction to Vectors 1.2 #13. Find two vectors v and w that are perpendicular to (1 , 1 , 1) and to each other. Solution There are many ways to go about this. One way would be to write v = ( v 1 , v 2 , v 3 ) and w = ( w 1 , w 2 , w 3 ) and then to write down the equations that v and w must satisfy. These are: v · (1 , 1 , 1) = 0, w · (1 , 1 , 1) = 0, and v · w = 0. This gives a system of equations with the v i and w i as unknowns, that you could then try to solve to find the set of all possible v and w satisfying these requirements. On the other hand, the problem just asks you to find one specific choice of v and w , not all possible choices. So I think it is easiest to proceed first by inspection: v = (1 , 0 , 1) is clearly perpendicular to (1 , 1 , 1) since their dot product is v · (1 , 1 , 1) = 1 · 1 + 0 · 1 1 · 1 = 0. So, we just have to find a w = ( w 1 , w 2 , w 3 ) that is perpendicular to both of these vectors. This w must satisfy w · (1 , 1 , 1) = w 1 + w 2 + w 3 = 0 and v · w = (1 , 0 , 1) · w = w 1 w 3 = 0 . There are infinitely many solutions to this system of equations; one particular solution is w = (1 , 2 , 1). There are many others. 1

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2 STRANG EXERCISE SOLUTIONS, CHAPTER 2 CHAPTER 2 Solving Linear Equations 2.1 #15. (a) What is the 2 × 2 identity matrix? I times x y equals x y . Solution I = 1 0 0 1 . Of course you know this, but why does it work? Because we compute Ix by forming linear combinations of the columns of I , and those columns can be used to form any vector using exactly the components of that vector as the scalars in the linear combination: I x y = 1 0 0 1 x y = x 1 0 + y 0 1 = x 0 + 0 y = x y . (b) What is the 2 × 2 exchange matrix? P times x y equals y x . Solution P = 0 1 1 0 . The reasoning is similar: P x y = 0 1 1 0 x y = x 0 1 + y 1 0 = 0 x + y 0 = y x . Of course you can think about this from an “entrywise” view of matrix/vector multiplication, but often it is very advantageous to think about general problems from a linear combination point of view. 2.2 #19. (Recommended) It is impossible for a system of linear equations to have exactly two solutions. Explain why . NOTE: The wording of this problem is a little ambiguous: If you just write out the answer to parts (a) and (b) of the problem, then you have done the explanation that the problem asks for! (a) If ( x, y, z ) and ( X, Y, Z ) are two solutions of Ax = b , what is another one? Solution Let u = ( x, y, z ) and v = ( X, Y, Z ). The fact that these are SOLUTIONS means that they solve the system of equations; in other words, Au = b and Av = b , with A , b coming from the
STRANG EXERCISE SOLUTIONS, CHAPTER 2 3 system of equations. Intuitively, we would think that any vector w which lies on the line passing through u and v would also be a solution. The equation for this line is = { cu + dv : c, d R and c + d = 1 } .

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IntrodLineaAlgebr_solutions - INTRODUCTION TO LINEAR...

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