Introduction to Linear algebra-Strang-Solutions-Manual_ver13

An example of ll in to remove that 12 choose e32 d 0

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Unformatted text preview: he second product is 0 1 2. The first product is 6 5 4 0 2 3 321 reversed. 13 (a) E times the third column of B is the third column of EB . A column that starts at zero will stay at zero. nonzero row. (b) E could add row 2 to row 3 to change a zero row to a `21 D 1 , E32 has `32 D 2 , E43 has `43 D 3 . Otherwise the E ’s match I . 2 3 4 " # " # 1 4 7 1 4 7 1 2 5! 0 6 12 . The zero became 12, 15 aij D 2i 3j : A D 3 0 3 0 12 24 " # 1 00 1 0. an example of fill-in. To remove that 12, choose E32 D 0 0 21 14 E21 has Solutions to Exercises 13 16 (a) The ages of X and Y are x and y : x 2y D 0 and x C y D 33; x D 22 and y D 11 (b) The line y D mx C c contains x D 2, y D 5 and x D 3, y D 7 when 2m C c D 5 and 3m C c D 7. Then m D 2 is the slope. aC b C c D 4 17 The parabola y D aCbx Ccx 2 goes through the 3 given points when aC 2b C 4c D 8 . aC 3b C 9c D 14 Then a D 2, b D 1, and c D 1. This matrix with columns .1; 1; 1/, .1; 2; 3/, .1; 4; 9/ is a “Vandermonde matrix.” " # " # " # " # 100 1 00 100 100 2 3 a 1 0 , E D 2a 1 0 , F D 0 1 0 : 18 EF D a 1 0 , FE D bc1 b C ac c 1 2b 0 1 0 3c 1 19 20 21 22 " # " # 010 001 PQ D 0 0 1 . In the opposite order, two row exchanges give QP D 1 0 0 , 100 010 If M exchanges rows 2 and 3 then M 2  I (also . M /2 D I ). There are many square D  a b roots of I : Any matrix M D has M 2 D I if a2 C bc D 1. c a    124 10 D (a) Each column of EB is E times a column of B (b) 124 11     124 . All rows of EB are multiples of 1 2 4 . 248         21 11 11 10 . but FE D give EF D and F D No. E D 11 12 01 11 P P (a) a3j xj (b) a21 a11 (c) a21 2 a11 (d) .EAx /1 D .Ax /1 D a1j xj . 23 E.EA/ subtracts 4 times row 1 from row 2 (EEA does the row operation twice). AE subtracts 2 times column 2 of A from column 1 (multiplication by E on the right side acts on columns instead of rows).       2x1 C 3x2 D 1 23 1 2 31 24 A b D ! . The triangular system is 0 5 15 5x2 D 15 4 1 17 Back substitution gives x1 D 5 and x2 D 3. 25 The last equation becomes 0 D 3. If the original 6 is 3, then row 1 C row 2 D row 3. 26 (a) Add two columns b and b and x  D   4 . 1  1 2   1 410 ! 701 0 4 1   10 7 !xD 2 21 27 (a) No solution if d D 0 and c ¤ 0 (b) Many solutions if d D 0 D c . No effect from a; b . 28 A D AI D A.BC / D .AB/C D IC D C . That middle equation is crucial. Solutions to Exercises 14 2 3 2 3 1 0 00 1000 1 0 07 61 60 1 0 07 29 E D 4 subtracts each row from the next row. The result 4 0 1 1 05 0 1 1 05 0 0 11 0121 still has multipliers D 1 in a 3 by 3 Pascal matrix. The product M of all elimination 2 3 1 0 00 1 0 07 61 matrices is 4 . This “alternating sign Pascal matrix” is on page 88. 1 2 1 05 1 3 31 30 Given positive integers with ad b c D 1. Certainly c < a and b < d would be impossible. Also c > a and b > d would be impossible with integers. This leaves   34 row 1 < row 2 OR row 2 < row 1. An example is M D . Multiply by 23         1 1 11 10 11 to get , then multiply twice by to get . This...
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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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