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Unformatted text preview: Math 2940 Solutions, Fall 2011
2) To ﬁnd the function with second derivative zero, we integrate zero twice. After one
integration we get a constant C1 . Integrating again we get 1 x + 0 where C0 is another
C0 C1 constant. The vectors in the null space of B are of the form 0 .
011 1 0 0 so 1 0 0 1 = 1 so T (v1 + v2 + v3 ) = 2w1 + w2 + 2w3 .
5) A =
7) The vector 0 is not in the column space of A as its ﬁrst and third entries are diﬀerent
(they are the same for all columns of A). Thus w1 is not in the range of T .
13) (a) Deﬁne T (av1 + bv2 ) = bv1 + av2 . Then T (T (u)) = u for any u so T = T −1 .
(b) Deﬁne T (av1 + bv2 ) = av1 . Then T (T (av1 + bv2 )) = T (av1 ) = av1 .
(c) For (b), if T is invertible (as it is in (a)), composing the equation T 2 = T with T −1 gives
T = I , which is not allowed. 15) (a) M =
(b) M = rs
ad−bc , the inverse of ab
cd (c) If ad − bc = 0, this will not be possible. When this happens 11
. Note 5(1, 4) − 4(1, 5) = (1, 0). Note M −1 =
nates form the ﬁrst column of M −1 . 19) M = ab
cd is not invertible. 5 −1
. The coordi−4
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