Linear Algebra/Comparing Set Descriptions/Solutions

Solutions Problem 1 Decide if the vector is a member of the set. <li> , <li> , <li> , <li> , <li> , <li> , Answer <li> No. <li> Yes. <li> No. <li> Yes. <li> Yes; use Gauss' method to get and . <li> No; use Gauss' method to conclude that there is no solution. Problem 2 Produce two descriptions of this set that are different than this one. Answer One easy thing to do is to double and triple the vector: Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. Answer Instead of showing all three equalities, we can show that the first equals the second, and that the second equals the third. Both equalities are easy, using the methods of this subsection. Problem 4 Show that these sets are equal and that both describe the solution set of this system. Answer That system reduces like this: showing that , and . Problem 5 Decide if the sets are equal. <li> and <li> and <li> and <li> and <li> and Answer For each item, we call the first set and the other . <li> They are equal. To see that , we must show that any element of the first set is in the second, that is, for any vector of the form there is an appropriate such that Restated, given we must find so that this holds. That system reduces to That is, and so any vector in the form for can be stated in the form needed for inclusion in . For , we look for so that these equations hold. Rewrite that as and so <li> These two are equal. To show that , we check that for any we can find an appropriate so that these hold. Use Gauss' method to conclude that and so . For , solve with Gaussian reduction to get and so any member of can be expressed in the form needed for . <li> These sets are equal. To prove that , we must be able to solve for and in terms of . Apply Gaussian reduction to conclude that any pair where will do. For instance, or Thus . For , we solve with Gauss' method to deduce that any vector in is also in . <li> Neither set is a subset of the other. For to hold we must be able to solve for and in terms of and . Gauss' method shows that we can only find an appropriate pair when . That is, has no expression of the form Having shown that is not a subset of , we know so, strictly speaking, we need not go further. But we shall also show that is not a subset of . For to hold, we must be able to solve for and . Apply row reduction to deduce that the only vectors from that are also in are of the form For instance, is in but not in . <li> These sets are equal. First we change the parameters: Now, to show that , we solve with Gauss' method to get that and so . The proof that involves solving with Gaussian reduction to conclude and so any vector in is also in .