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23. a. True. See the first paragraph of the subsection titled Homogeneous Linear Systems. b. False. The equation Ax = 0 gives an implicit description of its solution set. See the subsection entitled Parametric Vector Form. c. False. The equation Ax = 0 always has the trivial solution. The box before Example 1 uses the word nontrivial instead of trivial. d. False. The line goes through p parallel to v. See the paragraph that precedes Fig. 5. e. False. The solution set could be empty! The statement (from Theorem 6) is true only when there exists a vector p such that Ap = b.

So suitable values for  −3  −9   3 x1 and x2 would be 3 and –1 respectively. ) Thus x =    −1 satisfies Ax = 0. 34. Inspect how the columns a1 and a2 of A are related. The second column is –3/2 times the first. Put  3 another way, 3a1 + 2a2 = 0. Thus   satisfies Ax = 0. 2 Note: Exercises 33 and 34 set the stage for the concept of linear dependence. 35. Look for A = [a1 a2 a3] such that 1·a1 + 1a2 + 1·a3 = 0. That is, construct A so that each row sum (the sum of the entries in a row) is zero.

0 0 In particular, x4 is free (and not zero as some may assume). The solution is x1 = 4x2 – 5x6, x3 = x6, x5 = 4x6, with x2, x4, and x6 free. In parametric vector form,  x1   4 x2 − 5 x6   4 x2   0   −5 x6  0   −5 4 x         0       x2  2    x2   0   0  0     1  x3     0   0   x6  0   1 0 x6 x= = =  = x2   + x4   + x6   + + x4  x4     0   x4   0   1  0 0  x5   4 x6   0   0   4 x6  0   4 0                 x6 0   1  0   x6     0   0   x6  ↑ ↑ ↑ u v w Note: The Study Guide discusses two mistakes that students often make on this type of problem.

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A based federer spectral sequence and the rational homotopy of function spaces by Smith S.B.


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