IntrodLineaAlgebr_solutions

IntrodLineaAlgebr_solutions - INTRODUCTION TO LINEAR...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 3 )and w w 1 ,w 2 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 ±nd the set of all possible v and w satisfying these requirements. On the other hand, the problem just asks you to ±nd one speci±c choice of v and w , not all possible choices. So I think it is easiest to proceed ±rst 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 ±nd a w w 1 2 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 , 0 , 1) · w = w 1 w 3 . There are in±nitely many solutions to this system of equations; one particular solution is w , 2 , 1). There are many others. ± 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 = ± 10 01 . 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 = ± ± 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 = ± . The reasoning is similar: P ± x y = ± 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 ). 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
Background image of page 2
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 } .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/25/2011 for the course PHYS 164 taught by Professor Johnjames during the Fall '09 term at MO St. Louis.

Page1 / 53

IntrodLineaAlgebr_solutions - INTRODUCTION TO LINEAR...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online