Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Sin 5x5 c the sum of the rst n terms has an

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Unformatted text preview: um must be s D 0.   :6 C :4a :6 :6a a1 k k 1 7 .:5/ ! 0 gives A ! A ; any A D with :4 :4a :4 C :6a :4 C :6a  0 8 If P D cyclic permutation and u0 D .1; 0; 0; 0/ then u1 D .0; 0; 1; 0/; u2 D .0; 1; 0; 0/; 9 10 11 12 13 u3 D .1; 0; 0; 0/; u4 D u0 . The eigenvalues 1; i; 1; i are all on the unit circle. This Markov matrix contains zeros; a positive matrix has one largest eigenvalue  D 1. M 2 is still nonnegative; Œ 1    1 M D Œ 1    1  so multiply on the right by M to find Œ 1    1 M 2 D Œ 1    1  ) columns of M 2 add to 1.  D 1 and a C d 1 from the trace; steady state is a multiple of x 1 D .b; 1 a/. Last row :2; :3; :5 makes A D AT ; rows also add to 1 so .1; : : : ; 1/ is also an eigenvector of A. B has  D 0 and :5 with x 1 D .:3; :2/ and x 2 D . 1; 1/; A has  D 1 so A I has  D 0. e :5t approaches zero and the solution approaches c1 e 0t x 1 D c1 x 1 . x D .1; 1; 1/ is an eigenvector when the row sums are equal; Ax D .:9; :9; :9/ Solutions to Exercises 85 14 .I A/.I CACA2 C   / D .I CACA2 C   / .ACA2 CA3 C   / D I . This says that   0 :5 I C A C A2 C    is .I A/ 1 . When A D ; A2 D 1 I; A3 D 1 A; A4 D 1 I 2 2 4 10    1 1 1 1 C 2 C  2 C 4 C  21 D D .I A/ 1 . and the series adds to 22 1 C 1 C  1 C 1 C  2 2      8 130 :5 1 15 The first two A’s have max < 1; p D and ;I has no inverse. 6 32 :5 0 16  D 1 (Markov), 0 (singular), :2 (from trace). Steady state .:3; :3; :4/ and .30; 30; 40/. 17 No, A has an eigenvalue  D 1 and .I A/ 1 does not exist. " # F1  F2 F3 P1  0 D 3 C 18 The Leslie matrix on page 435 has det.A I / D det 0 P2  2 F1  C F2 P1  C F3 P1 P2 . This is negative for large . It is positive at  D 1 provided that F1 C F2 P1 C F3 P1 P2 > 1. Under this key condition, det.A I / must be zero at some  between 1 and 1. That eigenvalue means that the population grows (under this condition connecting F ’s and P ’s reproduction and survival rates). 19 ƒ times S 1 S has the same diagonal as S 1 S times ƒ because ƒ is diagonal. 20 If B > A > 0 and Ax D max .A/x > 0 then B x > max .A/x and max .B/ > max .A/. Problem Set 8.4, page 446 1 Feasible set D line segment .6; 0/ to .0; 3/; minimum cost at .6; 0/, maximum at .0; 3/. 2 Feasible set has corners .0; 0/; .6; 0/; .2; 2/; .0; 6/. Minimum cost 2x 3 4 5 6 y at .6; 0/. Only two corners .4; 0; 0/ and .0; 2; 0/; let xi ! 1, x2 D 0, and x3 D x1 4. From .0; 0; 2/ move to x D .0; 1; 1:5/ with the constraint x1 C x2 C 2x3 D 4. The new cost is 3.1/ C 8.1:5/ D $15 so r D 1 is the reduced cost. The simplex method also checks x D .1; 0; 1:5/ with cost 5.1/ C 8.1:5/ D $17; r D 1 means more expensive. Cost D 20 at start .4; 0; 0/; keeping x1 Cx2 C2x3 D 4 move to .3; 1; 0/ with cost 18 and r D 2; or move to .2; 0; 1/ with cost 17 and r D 3. Choose x3 as entering variable and move to .0; 0; 2/ with cost 14. Another step will reach .0; 4; 0/ with minimum cost 12. If we reduce the Ph.D. cost to $1 or $2 (below the student cost of $3), the job will go to the Ph.D. with cost vector...
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