Math 308 Discussion Problems #2 (Sections ) SOLUTIONS

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1 Math 8 Discussion Problems # (Sections.-.) SOLUTIONS () When Jake works from home, he typically spends 4 minutes of each hour on research, and on teaching, and drinks half a cup of coffee. (The remaining time is spent on the internet.) For each hour he works in the math department, he spends around minutes on research and on teaching, and doesn t drink any coffee. Lastly, if he works at a coffeeshop for an hour, he spends 5 minutes each on research and teaching, and drinks a cup of coffee. (Note: be careful about units of minutes versus hours.) (a) Last week, Jake spent hours working from home, 5 hours working in his office in Padelford Hall, and hours working at Cafe Allegro. Compute what was accomplished, and express the result as a vector equation. Solution. h + 5d + c = hours research = hours teaching. 7 cups of coffee (b) This week, Jake has 5 hours of research to work on and hours of work related to teaching. He also wants cups of coffee, because... of... very important reasons. How much time should he spend working from home, from his office, and from the coffeeshop? 5 Solution. We re trying to solve the equation x h + xd + x c =. The augmented matrix, and its reduced echelon form, are: So I should spend 4 hours working from home, hours working in the office, and 4 hours at the coffeeshop. (c) Describe the situation in part (b) as a vector equation and a matrix equation A t = w. What do the vectors t and w mean in this context? For which other vectors w does the equation A t = w have a solution? Solution. As a vector and matrix equation: 5 t h + td + t c = t t t = The entries of t are the amounts of time spent in each location. The entries of w are the amounts of work done, in hours (or coffee consumed, in cups). Since the echelon form has no zero rows, the system is consistent for any choice of w. This means that it s possible mathematically at least to find a combination w w w.

2 of times at home, at the office, and at the coffeeshop, to accomplish any combination of work and coffee. But in practice there s an important caveat: negative amounts of time don t make sense! So we would also need impose the inequalities t, t, t. We haven t learned how to do this, but the techniques exist (called linear programming) and are important for optimization. (d) Jake tries working in the math department lounge for an hour, and gets minutes of research and minutes of teaching work done, while having time to drink of a cup of coffee. Not bad. But Jake s colleague Vasu claims that there s no need to work in the lounge the other options already give enough flexibility. Is he right? Explain mathematically. Solution. Vasu is right: based on the previous question, we know that l span( h, d, c). Explicitly (if you solve the equations): l = h + d + c. 7 Thus, there is no extra flexibility gained by including l as an option. () (after.) Find a 4 matrix A, in reduced echelon form, with free variable x, such that the general solution of the equation Ax = is 6 x = + s, 6 where s is any real number. Solution. Let s figure out the entries of the matrix. Since it has to be in reduced echelon form, with free variable x, that forces many entries to be or : a A = b Now let s set up the equation A x = b in augmented matrix form: a b 6 After setting x = s, we can solve for the remaining variables to get x 4 = 6, x = bs and x = as: as a bs b x = = + s. s 6 6 Comparing with the initial question, we see that a = and b =.

3 () (after.) Find all values z and z such that (,, ), (,, ), and ( 4, z, z ) do not span R. Solution. By a theorem from class, the vectors will not span R if and only if there s a row of zeroes in the echelon form of the corresponding matrix. So, row-reduce: z.5 z.5 z.5 z. z z.5 z + 6 z + 6 (z 5 ) So, the vectors will fail to span if and only if z + 6 z 5 + =. In other 5 words, if and only if z 5z =. t (4) (after.) (a) Let a =, a =, and a =. Find all values of t for 7 which there will be a unique solution to a x + a x + a x = b for every vector b in R. Explain your answer. Solution. By the Big Theorem, since we have vectors in R, the unique solution for every b property is equivalent to the three vectors spanning R or that they are. Basically, we want the echelon form of the matrix to have three distinct pivots (no free variables or rows of zeroes). So, row-reduce: t t t t t t 7 + t ( t) So, we want the bottom-right entry, which simplifies to.5t.5, to be nonzero. In other words, t. (b) Are the vectors a and a from part (a)? Explain your answer. Solution. Yes, they are. This is because it s easy to tell when a set of two vectors is linearly (in)dependent: they re dependent if and only if they are proportional. In this case, they clearly are not proportional. (c) Let a, a and a be as in (a). Let a 4 = 4. Without doing any further 5 calculations, find all values of t for which there will be a unique solutions to a y + a y + a y + a 4 y 4 = c for every vector c in R. Explain your answer. Solution. The property we re considering: for every c R, there s a unique solution to a y + a y + a y + a 4 y 4 = c this would mean that the vectors {a, a, a, a 4 } are and span R. The spanning part is possible, but the linear independence part is impossible

4 regardless of the value of t we choose, because four vectors in R are always linearly dependent. (5) (after.) (Geometry Question) Consider the infinite system of linear equations in two variables given by ax + by = where (a, b) moves along the unit circle in the plane. (a) How many solutions does this system have? Solution. The unique common solution to all those equations is x = y =. (b) How many equations in the above system give you the same set of solutions? Write down two separate such linear systems, in vector form. Solution. It takes exactly two equations to force the solution to be (x, y) = (, ). We could do, for example, { x + { y = x + y = x + or y = x + y = In the first case, we used the values (a, b) = (, ) and (, ). In the second case, we used (a, b) = (, ) and (, ). (Note that those (a, b) values all correspond to points on the unit circle.) (c) What happens to the infinite linear system if you add to it the equation x + y =? Solution. Nothing changes: there is still the unique solution (x, y) = (, ). (d) What happens to the infinite linear system if one of the equations slightly perturbs to ax + by = c where c is a small positive number? Solution. The system becomes inconsistent, because the other equations still force x = y =, so there s no way to have ax + by = c. Comment: This situation comes up frequently in data analysis, where it s impossible to avoid statistical noise, rounding errors, and so on. We would never expect a quantity ax + by to be exactly zero, so we will need to know when to round it off and accept an approximate solution. (6) For each of the situations described below, give an example (if it s possible) or explain why it s not possible. (a) A set of vectors that does not span R. After adding one more vector, the set does span R. Solution. The simplest example is to start with,, which doesn t span R. Then we add in the vector, and now the vectors do span R.

5 (b) A set of vectors that are. After adding one more vector, the set becomes. Solution. This is impossible. Suppose our original set of vectors has a nontrivial linear combination equalling zero, say x v + + x m v m =. Well, this still counts as a nontrivial linear combination when we add in the a new vector w. In other words, we can write x v + + x m v m + w =. This is nontrivial because one of the x,..., x m coefficients is nonzero (therefore it s fine for the last coefficient to be zero). (c) A set of vectors in R with the following properties (four possibilities): spans R, spans R, doesn t span R, doesn t span R, For each case that is possible, how many vectors could be in the set? (State any constraints, as in there must be at least... or at most... ) Solution. All four cases are possible. Here are some relatively-simple examples: spans R, spans R,,,,,, doesn t span R, doesn t span R,,,,,... Also, by our theorems from class (especially the Unifying Theorem), there are some constraints on the number of vectors: spans R, spans R, exactly vectors at least 4 vectors doesn t span R, doesn t span R, at most vectors any number of vectors

6 (e) A system of equations with a unique solution. After adding another equation to the system, the new system has infinitely-many solutions. Solution. This is impossible. Adding another equation to the system can only reduce the size of the solution set (or, if we re lucky, leave it unchanged). (f) * A system of equations without any solutions. After deleting an equation, the system has infinitely-many solutions. Solution. This is possible. Let s take a system that would have had infinitely-many solutions, then add in an inconsistent equation. For instance: x + y + z = y + z = 5 y + z =. There are no solutions, but if we delete y + z =, we then have infinitely-many.