Introduction to Linear algebra-Strang-Solutions-Manual_ver13

All cofactors of row 1 are zero a has rank 2 each of

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Unformatted text preview: s row 1 so det A D 0. When the ij entry is i C j , row 3 row 2 D row2 row 1 so A is singular: det A D 0. det A D abc , det B D abcd , det C D a.b a/.c b / by doing elimination. (a) True: det.AB/ D det.A/ det.B/ D 0 (b) False: A row exchange gives det D product of pivots. (c) False: A D 2I and B D I have A B D I but the determinants have 2n 1 ¤ 1 (d) True: det.AB/ D det.A/ det.B/ D det.BA/. A is rectangular so det.AT A/ ¤ .det AT /.det A/: these determinants are not defined. 2 3 d b   @f =@a @f =@c 6 7 Derivatives of f D ln.ad b c/: D 4 ad c b c ad a b c 5 D @f =@b @f =@d ad b c ad b c   d b 1 1 DA . ad b c c a 19 For triangular matrices, just multiply the diagonal entries: det.U / D 6; det.U 2 20 21 22 23 24 25 26 27 28 29 30 a from 2 1 2 , 4:610 4 , 1:610 7, 3:710 12, 5:410 18 , 4:8  10 , 2:7  10 , 9:7  10 , 2:2  10 53 . Pivots are ratios of determinants so the 10th pivot is near 10 10 . The Hilbert matrix is numerically difficult (illconditioned). 31 The Hilbert determinants are 1, 810 25 33 2 2 43 Solutions to Exercises 53 32 Typical determinants of rand.n/ are 106 ; 1025 ; 1079 ; 10218 for n D 50; 100; 200; 400. randn.n/ with normal distribution gives 1031 ; 1078 ; 10186 , Inf which means  21024 . MATLAB allows 1:999999999999999  21023  1:8  10308 but one more 9 gives Inf! 33 I now know that maximizing the determinant for 1, 1 matrices is Hadamard’s problem (1893): see Brenner in American Math. Monthly volume 79 (1972) 626-630. Neil Sloane’s wonderful On-Line Encyclopedia of Integer Sequences (research.att.com/ njas) includes the solution for small n (and more references) when the problem is changed to 0; 1 matrices. That sequence A003432 starts from n D 0 with 1, 1, 1, 2, 3, 5, 9. Then the 1; 1 maximum for size n is 2n 1 times the 0; 1 maximum for size n 1 (so .32/.5/ D 160 for n D 6 in sequence A003433). To reduce the 1; 1 problem from 6 by 6 to the 0; 1 problem for 5 by 5, multiply the six rows by ˙1 to put C1 in column 1. Then subtract row 1 from rows 2 to 6 to get a 5 by 5 submatrix S of 2 ; 0 and divide S by 2. Here is an advanced MATLAB code and a 1; 1 matrix with largest det A D 48 for n D 5: n D 5I p D .n 1/^2I A0 Dones.n/; maxdetD 0; for k D 0 W 2^p 1 Asub D rem(floor(k:  2:^. p C 1 W 0//; 2/I A D A0I A.2 W n; 2 W n/ D 1 reshape(Asub, n 1; n 1/; if abs(det(A// > maxdet, maxdet D abs(det(A)); maxA D A; 2 end end Output: maxA = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 maxdet = 48. 34 Reduce B by row operations to Œ row 3I row 2I row 1 . Then det B D 6 (odd per- mutation). Problem Set 5.2, page 263 1 det A D 1 C 18 C 12 9 4 6 D 12, rows are independent; det B D 0, row 1 C row 2 D row 3; det C D 1, independent rows (det C has one term, odd permutation) 2, independent; det B D 0, dependent; det C D 1, independent. All cofactors of row 1 are zero. A has rank  2. Each of the 6 terms in det A is zero. Column 2 has no pivot. a11 a23 a32 a44 gives 1, because 2 $ 3, a14 a23 a32 a41 gives C1, det A D 1 1 D 0; det B D 2  4  4  2 1 ...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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