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03_hwc1SolnsODDA

# 03_hwc1SolnsODDA - (b Is f invertible If so ﬁnd a formula...

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Unformatted text preview: (b) Is f invertible? If so ﬁnd a formula for the inverse. If not, explain why not. u SOLUTION (a) If v = f w x y , then by the deﬁnition of f , u = x − y , v = −2x and w = x + y . Hence u + v + w = (x − y ) + (−2x) + (x + y ) = 0. (b) There are many u 1 v in IR3 that do not satisfy u + v + w = 0; the vector 1 is one such example. vectors w 1 Therefore, by part (a), f is not onto, and hence is not invertible. 1.11 Consider the function from IR2 to IR2 deﬁned by f (a) Find all values of a, if any, for which f is one to one. (b) Find all values of a, if any, for which f is onto. x y = x2 + ay 2 . x SOLUTION (a) Since y enters the formula for f only through y 2 , you can see that for any y , x x f =f . −y y Thus, f is never one to one, no matter what the value of a is. Next, for (b), we try to solve the corresponding system of equations, namely x2 + ay 2 = u x=v . Using the second equation to eliminate x from the ﬁrst, we have ay 2 = u − x2 . (∗) If a = 0, this has a solution if and only if u = x2 , and x = v , so there is a solution if and only if u = v 2 . This is a small subset of IR2 – a parabola lying in it – and so f is not onto if a = 0. If a = 0, then (∗) becomes y 2 = (u − x2 )/a = (u − v 2 )/a . (∗∗) If the sign of u − v 2 is oposite to that of a, (u − v 2 )/a will be negative, and (∗∗) will have no solution. Thus, f is not onto for any value of a. 1.13 Deﬁne a transformation f from IR3 to IR5 as follows. Deﬁne r f a b c s =t u v 4 (∗ ) 21/september/2005; 22:09 ...
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