Introduction to Linear algebra-Strang-Solutions-Manual_ver13

0 1 0 b2 2 b1 525 solvable if b3 2 b1 b2 d 0 49 8 b3

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Unformatted text preview: 1/ or N D 0  2 .c D 2/ or N D 2 by 0 empty matrix. 1         I II 24 A D I I has N D IB D has the same N ; C D I I I has I 00 " # I I I 0. ND 0 I # #" "  11 1124 1023 D (pivot columns) times R. 25 A D 1 2 2 5 D 1 2 0101 13 1326 26 The m by n matrix Z has r ones to start its main diagonal. Otherwise Z is all zeros.     I r by r r by n r IF T ; rref.R / D D 27 R D 0 m r by r m r by n r 00   I0 ;I 28 The row-column reduced echelon form is always 00  0 ; rref.RT R/ D same R 0 is r by r . Problem Set 3.4, page 163 " #" #" # 2 4 6 4 b1 2 4 6 4 b1 2 4 6 4 b1 1 2 5 7 6 b2 ! 0 1 1 2 b2 b1 ! 0 1 1 2 b2 b1 2 3 5 2 b3 0 1 1 2 b3 b1 0 0 0 0 b3 C b2 2b1 Ax D b has a solution when b3 C b2 2 b1 D 0; the column space contains all combinations of .2; 2; 2/ and .4; 5; 3/. This is the plane b3 C b2 2 b1 D 0 (!). The nullspace contains all combinations of s1 D . 1; 1; 1; 0/ and s2 D .2; 2 ; 0; 1/I xcomplete D xp C c1 s1 C c2 s2 I  R " 1 dD0 0  01 11 00 2 2 0 4 1 0 # gives the particular solution xp D .4; 1; 0; 0/: Solutions to Exercises 34 " # " # " # 2 1 3 b1 2 1 3 b1 1 1=2 3=2 5 0 0 2 6 3 9 b 2 ! 0 0 0 b 2 3b 1 Then Œ R d  D 0 0 4 2 6 b3 0 0 0 b 3 2b 1 00 0 0 Ax D b has a solution when b2 3b1 D 0 and b3 2 b1 D 0; C .A/ D line through .2; 6; 4/ which is the intersection of the planes b2 3b1 D 0 and b3 2 b1 D 0; the nullspace contains all combinations of s1 D . 1=2; 1; 0/ and s2 D . 3=2; 0; 1/; particular solution x p D d D .5; 0; 0/ and complete solution x p C c1 s1 C c2 s2 . "# "# 2 3 0 C x2 1 . The matrix is singular but the equations are 3x D complete 1 0 still solvable; b is in the column space. Our particular solution has free variable y D 0. 1 1 D x p C x n D . 2 ; 0; 2 ; 0/ C x2 . 3; 1; 0; 0/ C x4 .0; 0; 2 ; 1/. " #" # 12 2 b1 12 2 b1 4 b2 ! 0 1 0 b2 2 b1 525 solvable if b3 2 b1 b2 D 0. 49 8 b3 00 0 b3 2 b1 b2 Back-substitution gives the particular solution to Ax D b and the special solution to " # "# 5b1 2 b2 2 Ax D 0: x D b2 2 b1 C x3 0 . 0 1   5b1 2 b3 6 (a) Solvable if b2 D 2b1 and 3b1 3b3 C b4 D 0. Then x D D xp b3 2 b1 " # "# 5b1 2 b3 1 (b) Solvable if b2 D 2b1 and 3b1 3b3 C b4 D 0. x D b3 2 b1 C x3 1 . 0 1 " #" # 1 3 1 b1 1 3 1 b2 One more step gives Œ 0 0 0 0  D 1 1 b2 3b1 row 3 2 (row 2) C 4(row 1) 7 3 8 2 b2 ! 0 2 4 0 b3 0 2 2 b3 2 b1 provided b3 2 b2 C4b1 D0. 4x complete 8 (a) Every b is in C .A/: independent rows, only the zero combination gives 0. (b) We need b3 D 2b2 , because .row 3/ 2.row 2/ D 0. " #" #" 1 00 1 2 3 5 b1 12   10 0 0 2 2 b2 2 b1 9L U c D 2 D24 3 11 0 0 0 0 b3 C b2 5b1 36   D A b ; particular x p D . 9; 0; 3; 0/ means 9.1; 2; 3/ C 3.3; 8; 7/ This is Ax p D b.    10 1 2 xD has x p D .2; 4; 0/ and x null D .c; c; c/. 10 01 1 4 # 3 5 b1 8 12 b2 7 13 b3 D .0; 6; 6/. 11 A 1 by 3 system has at least two free variables. But x null in Problem 10 only has one. 12 (a) x 1 x 2 and 0 solve Ax D 0 (b) A.2x 1 2x 2 / D 0; A.2x 1 x 2 / D b 13 (a) The particular solution x is   multiplied by 1      p always  (b...
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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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