136A2S - row of A with the j th column of B that is to µ a...

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ASSIGNMENT 2 for SECTION 001 Solutions Questions from the textbook Section 3-1: B3 (a) [4 marks] , B4 (b) [3 marks] , B6 [4 marks] , C2 [2 marks] , D4 [3 marks] and D5 [4 marks] Other questions 1. [4 marks] Let A = - 5 2 - 12 5 , P = 1 2 3 4 , D = 1 0 0 - 1 and Q = - 4 2 3 - 1 . Verify that A = 1 2 PDQ, and determine A 100 . We may verify A = 1 2 PDQ directly. Note that we have A = PD ( 1 2 Q ) , and that ( 1 2 Q ) P = I . Hence A 100 = PD 1 2 Q 100 = PD 1 2 Q PD 1 2 Q · · · PD 1 2 Q 100 times = PD 100 1 2 Q = 1 2 PD 100 Q. Now D 100 = 1 0 0 - 1 100 = 1 100 0 0 ( - 1) 100 = I. Thus A 100 = 1 2 PIQ = 1 2 PQ = I. 2. [6 marks] An n × n matrix is a right stochastic matrix if the following properties hold: (i) all the entries are nonnegative; and (ii) the sum of the entries in each row is equal to 1. Prove that the product of two n × n right stochastic matrices is a right stochastic matrix. Let A = a 11 · · · a n 1 . . . . . . . . . a n 1 · · · a nn and B = b 11 · · · b n 1 . . . . . . . . . b n 1 · · · b nn be two right stochastic matrices. We consider the sum of the entries in the i th row of AB . The j th entry in this row is equal to the dot product of the i th row of
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Unformatted text preview: row of A with the j th column of B ; that is, to µ a i 1 ··· a n 1 ¶ b 1 j . . . b nj = a i 1 b 1 j + ··· + a n 1 b nj . 1 Note that, since all the numbers on the right-hand side are nonnegative, the term on the right-hand side is nonnegative. The sum of the entries in the i th row of AB is equal to n X j =1 ( a i 1 b 1 j + ··· + a n 1 b nj ) = a i 1 n X j =1 b 1 j + ··· + a n 1 n X j =1 b nj . Since B is a right stochastic matrix, we have n X j =1 b 1 j = ··· = n X j =1 b nj = 1 . Thus the sum of the entries in the i th row of AB is equal to a i 1 + ··· + a n 1 ; and since A is a right stochastic matrix, this is equal to 1. This the sum of the entries in each row of AB is equal to 1. 2...
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