# Ism2 - Chapter 2 Applications of Linear Equations and Matrices(Optional Section 2.1 p 123 2(a No 1 0(b A = 0 0 0 1 0 0 0 0 1 0 4(a Yes 6(a 2 8(a

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Chapter 2 Applications of Linear Equations and Matrices (Optional) Section 2.1, p. 123 2. (a) No. (b) A = 100 010 001 000 . 4. (a) Yes. (b) No such matrix A can be constructed. 6. (a) 2. (b) 3. (c) 2. (d) 1. 8. (a) Odd. (b) Even. (c) Odd. (d) Odd. 10. (a) Yes. (b) No. (c) No. (d) No. T.1. 1 word with weight zero; 2 words with weight one; 1 word with weight two. T.2. 1 word with weight zero; 3 words with weight one; 3 words with weight two; 1 word with weight three. T.3. n words with weight one; n ( n 1) 2 words with weight two. T.4. (a) There are only two code words 000 and 111. Since 000 + 000 = 000, 000 + 111 = 111, and 111 + 111 = 000, it is linear. (b) The code words are 000, 011, 101, and 110. 000+any code word = same code word, 011+011 = 000, 011 + 101 = 110, 011 + 110 = 101, 101 + 110 = 011, so it is linear. T.5. (a) The code words are 0000, 0101, 1010, 1111. (b) Yes. (c) No. If the received word was 11000, no error would be detected, but 1100 is not a code word. ML.1. (a) M = bingen ( 0 , 15 , 4 ) A = 0000000011111111 0000111100001111 0011001100110011 0101010101010101

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48 Chapter 2 (b) s = sum ( M ) s = 0112122312232334 (c) w =[ 0110100110010110 ] w = (d) C M ; w ] A = 0000000011111111 0000111100001111 0011001100110011 0101010101010101 Section 2.2, p. 134 2. (a) P 1 P 2 P 3 P 4 P 5 P 1 01000 P 2 10101 P 3 10010 P 4 P 5 00010 . (b) P 1 P 2 P 3 P 4 P 5 P 6 P 1 011000 P 2 100100 P 3 010000 P 4 001011 P 5 001101 P 6 100000 . 4. (b). 6. (a) One way: P 2 P 5 P 1 . (b) Two ways: P 2 P 5 P 1 P 2 . P 2 P 5 P 3 P 2 . 8. P 2 , P 3 , and P 4 . 10. There is no clique. 12. (a) Strongly connected. (b) Not strongly connected. 14. P 1 , P 2 ,or P 3 . T.1. In a dominance digraph, for each i and j , it is not the case that both P i dominates P j and P j dominates P i . T.2. Let r = 2. For each i and j , b (2) ij , the number of ways P i has two-stage access to P j , is the number of indices k ,1 k n , such that P i has direct access to P k and P k has direct access to P j . This in turn is the number of k such that a ik = 1 and a kj = 1 where A ( G )= ± a ij ² , which is n ³ k =1 a ik a kj = i , j entry of [ A ( G )] 2 .
Section 2.2 49 For r> 2, assume that the theorem has been proved for values up to r 1. Then b ( r ) ij = the number of k such that P i has r 1 stage access to P k and P k has direct access to P j = n ± k =1 b ( r 1) ik · a kj = n ± k =1 ( i , k entry of [ A ( G )] r 1 ) · ( k , j entry of A ( G )) = i , j entry of [ A ( G )] r . T.3. The implication in one direction is proved in the discussion following the theorem. Next suppose P i belongs to the clique { P i ,P j k ,...,P m } . According to the deﬁnition of clique, it contains at least three vertices so we may assume P i , P j and P k all exist in the clique. Then s ij = s ji = s jk = s kj = s ik = s ki = 1 and s (3) ii is a sum of nonnegative integer terms including the positive term which represents three stage access from P i to P j to P k to P i .Thu s s (3) ii is positive.

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## This note was uploaded on 06/25/2008 for the course MATH 22a taught by Professor Chuchel during the Spring '08 term at UC Davis.

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Ism2 - Chapter 2 Applications of Linear Equations and Matrices(Optional Section 2.1 p 123 2(a No 1 0(b A = 0 0 0 1 0 0 0 0 1 0 4(a Yes 6(a 2 8(a

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