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Unformatted text preview: Math 167 homework 2 solutions October 13, 2010 1.5.44: Find a 3 by 3 permutation matrix with P 3 = I (but not P = I ). Find a 4 by 4 permutation matrix with hatwide P with hatwide P 4 negationslash = I . P = 0 0 1 1 0 0 0 1 0 hatwide P = 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1 1.6.2 (a) Find the inverses of the permutation matrices P 1 = 0 0 1 0 1 0 1 0 0 and P 2 = 0 0 1 1 0 0 0 1 0 . P 1 1 = 0 0 1 0 1 0 1 0 0 and P 1 2 = 0 1 0 0 0 1 1 0 0 (b) Explain for permutations why P 1 is always the same as P T . Show that the 1s are in the right places to get PP T = I . First note that any row or column vector of a permutation matrix consists of exactly one nonzero entry, and that entry is 1 (i.e. any such vector is simply a standard coordinate vector). Let x i represent the i th row of P and y j represent the j th column of P 1 . By assumption ( PP 1 ) ij is 1 if i = j and 0 otherwise, thus x T i = y i . 1 1.6.28 If the product M = ABC of three square matrices is invertible, then A , B , C are invertible. Find a formula for B 1 that involves M 1 and A and C . B 1 = CM 1 A 1.6.50: Verify that ( AB ) T equals B T A T but those are different from A T B T : A = parenleftbigg 1 0 2 1 parenrightbigg B = parenleftbigg 1 3 0 1 parenrightbigg AB = parenleftbigg 1 3 2 7 parenrightbigg . In case AB = BA (not generally true!), how do you prove that B T A T = A T B T ? A T = parenleftbigg 1 2 0 1 parenrightbigg B T = parenleftbigg 1 0 3 1 parenrightbigg ( AB ) T = parenleftbigg 1 2 3 7 parenrightbigg = B T A T A T B T = parenleftbigg 7 2 3 1 parenrightbigg negationslash = B T A T If AB = BA , then ( AB ) T = ( BA ) T . By applying the rules for transpose....
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 Fall '10
 RolandW.Freund
 Math, Linear Algebra, Algebra

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