7.2 - Math 2940 Solutions, Fall 2011 Section 7.2 2) To...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 7.2 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 C C C0 C1 constant. The vectors in the null space of B are of the form 0 . 0 2 1 011 011 1 0 0 so 1 0 0 1 = 1 so T (v1 + v2 + v3 ) = 2w1 + w2 + 2w3 . 5) A = 2 1 011 011 1 7) The vector 0 is not in the column space of A as its first and third entries are different 0 (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 . tu d ad−bc −c ad−bc −b ad−bc a 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 = 45 nates form the first column of M −1 . 19) M = ab cd is not invertible. 5 −1 . The coordi−4 1 ...
View Full Document

Ask a homework question - tutors are online