Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Ghi 001 g h i 0 330 000 1 columns of a 2 3 3

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Unformatted text preview: ero matrix! :5    1 11 00 D ; 1 11 00 but AB D :5   1 1 2 has A D I ; BC D 0 1    1 01 10 D D ED . You can find more examples. 0 10 01 # 1 has A2 D 0. Note: Any matrix A D column times row D uvT will 0 2 3 2 3 010 001 6 7 6 7 have A2 D uvT uvT D 0 if vT u D 0. A D 4 0 0 1 5 has A2 D 4 0 0 0 5 000 000 0 22 A D 1  0 DE D 1 " 0 23 A D 0 24 25 26 27 28 29 30 31 but A3 D 0; strictly triangular as in Problem 20. n    n  2 2n 1 11 a an 1 b .A1 /n D , .A2 /n D 2n 1 , .A3 /n D . 0 1 11 0 0 2 32 3 2 3  2 3  2 3  abc 100 a d c 100 010 001 4 d e f 54 0 1 0 5D4 d 5 C4 e 5 C4 f 5 . ghi 001 g h i # "# " #" "# 0 330 000 1   Columns of A 2 3 3 0C 4 1 2 1 D 6 6 0 C 4 8 4 D times rows of B 121 2 1 660 # " 3 30 10 14 4 D AB . 7 81 (a) (row 3 of A)  (column 1 of B )# (row # of A)  (column 2 of B ) # both zero. and " 3 are " " "# 00x x 0xx x   (b) x 0 x x D 0 x x and x 0 0 x D 0 0 x : both upper. 00x x 000 0 ˇ ˇ ˇ     ˇ ˇ ˇ   ˇ ˇ   ˇˇˇ ˇˇˇ ˇˇ A times B ˇˇˇ , ˇˇ A ˇˇˇ , B, ˇˇˇ ˇˇˇ ˇˇ with cuts # " # " 100 100 0 1 0 produce zeros in the 2; 1 and 3; 1 entries. E21 D 1 1 0 and E31 D 001 401 # " # " 100 210 1 1 0 . Then EA D 0 1 1 is the Multiply E ’s to get E D E31 E21 D 013 401 result of both E ’s since .E31 E21 /A D E31 .E21 A/.      11 2 01 in the lower corner of EA. , DD , D c b=a D In 29, c D 53 13 8      A B x Ax B y real part Complex matrix times complex vector D B Ay B x C Ay imaginary part. needs 4 real times real multiplications. Solutions to Exercises 17 32 A times X D Œ x 1 x 2 x 3  will be the identity matrix I D Œ Ax 1 Ax 2 Ax 3 . "# "# " # 3 3 1 00 8 I AD 1 1 0 will have 33 b D 5 gives x D 3x 1 C 5x 2 C 8x 3 D 8 16 0 11 those x 1 D .1; 1; 1/; x 2 D .0; 1; 1/; x 3 D .0; 0; 1/ as columns of its “inverse” A 1 .     aCb aCb a C c b C b when b D c 34 A ones D agrees with ones A D cCd cCd aCc bCd and a D d   ab Then A D . ba 2 3 2 3 0101 2020 aba, ada cba, cda These show 61 0 1 07 6 0 2 0 2 7 bab, bcb dab, dcb 16 2-step 35 A D 4 ; A2 D 4 ; 0 1 0 15 2 0 2 0 5 abc, adc cbc, cdc paths in 1010 0202 bad, bcd dad, dcd the graph 36 Multiplying AB D(m by n)(n by p ) needs mnp multiplications. Then .AB/C needs mpq more. Multiply BC D (n by p)(p by q ) needs npq and then A.BC / needs mnq . (a) If m; n; p; q are 2; 4; 7; 10 we compare .2/.4/.7/ C .2/.7/.10/ D 196 with the larger number .2/.4/.10/ C .4/.7/.10/ D 360. So AB first is better, so that we multiply that 7 by 10 matrix by as few rows as possible. (b) If u; v; w are N by 1, then .uT v/wT needs 2N multiplications but uT .vwT / needs N 2 to find vwT and N 2 more to multiply by the row vector uT . Apologies to use the transpose symbol so early. (c) We are comparing mnp C mpq with mnq C npq . Divide all terms by mnpq : Now we are comparing q 1 Cn 1 with p 1 Cm 1 . This yields a simple important rule. If matrices A and B are multiplying v for AB v, don’t multiply the matrices first. 37 The p...
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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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