Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# C true only n columns to hold pivots d true only m

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Unformatted text preview: A. So all columns of Œ AB  of A   01 00 2 are already in C .A/. But A D has a larger column space than A D . 00 00 n For square matrices, the column space is R when A is invertible. Solutions to Exercises 29 Problem Set 3.2, page 140 &quot; 122 1 (a) U D 0 0 1 000 46 23 00 # &quot; # 242 Free variables x2 ; x4 ; x5 Free x3 (b) U D 0 4 4 Pivot variables x1 ; x3 Pivot x1 ; x2 000 2 (a) Free variables x2 ; x4 ; x5 and solutions . 2 ; 1; 0; 0; 0/, .0; 0; 2 ; 1; 0/, .0; 0; 3; 0; 1/ (b) Free variable x3 : solution .1; 1; 1/. Special solution for each free variable. 3x5 ; x4 ; x5 ) with x2 ; x4 ; x5 free. The complete solution to B x D 0 is (2x3 ; x3 ; x3 ). The nullspace contains only x D 0 when there are no free variables. &quot; # &quot; # 12000 1 0 1 1 1 , R has the same nullspace as U and A. 4 RD 0 0 1 2 3 , RD 0 00000 0 0 0          13 5 10 135 135 10 5AD D IBD D 2 6 10 21 000 267 21   13 5 D LU . 00 3 3 The complete solution to Ax D 0 is ( 2x2 ; x2 ; 2x4 6 (a) Special solutions .3; 1; 0/ and .5; 0; 1/ (b) .3; 1; 0/. Total of pivot and free is n. x C 3y C 5z D 0; it contains all the vectors .3y C 5z; y; z/ D y.3; 1; 0/ C z.5; 0; 1/ D combination of special solutions. (b) The line through .3; 1; 0/ has equations x C 3y C 5z D 0 and 2x C 6y C 7z D 0. The special solution for the free variable x2 is .3; 1; 0/.       1 3 0 10 1 3 5 with I D Œ 1 ; R D with I D . 8 RD 0 0 0 0 0 1 01 7 (a) The nullspace of A in Problem 5 is the plane 9 (a) False: Any singular square matrix would have free variables 10 11 12 13 14 15 (b) True: An invertible square matrix has no free variables. (c) True (only n columns to hold pivots) (d) True (only m rows to hold pivots) (a) Impossible row 1 (b) A D invertible (c) A D all ones (d) A D 2I; R D I . 3 32 32 2 0001111 1111111 0111111 60 0 0 1 1 1 17 60 0 1 1 1 1 17 60 0 0 0 0 1 17 40 0 0 0 1 1 15 40 0 0 0 0 1 15 40 0 0 0 0 0 05 0000000 0000001 0000000 32 3 2 11011100 01100111 60 0 1 1 1 1 0 07 60 0 0 1 0 1 1 17 4 0 0 0 0 0 0 1 0 5, 4 0 0 0 0 1 1 1 1 5. Notice the identity 00000001 00000000 matrix in the pivot columns of these reduced row echelon forms R. If column 4 of a 3 by 5 matrix is all zero then x4 is a free variable. Its special solution is x D .0; 0; 0; 1; 0/, because 1 will multiply that zero column to give Ax D 0. If column 1 D column 5 then x5 is a free variable. Its special solution is . 1; 0; 0; 0; 1/. If a matrix has n columns and r pivots, there are n r special solutions. The nullspace contains only x D 0 when r D n. The column space is all of Rm when r D m. All important! Solutions to Exercises 30 16 The nullspace contains only x D 0 when A has 5 pivots. Also the column space is R5 , because we can solve Ax D b and every b is in the column space. 17 A D Œ 1 3 1  gives the plane x 3y z D 0; y and z are free variables. The special solutions are .3; 1; 0/ and .1; 0; 1/. &quot;# x 18 Fill in 12 then 4 then 1 to get the complete solution to x 3y z D 12: y D z &quot;# &quot;# &quot;# 12 4 1 0 C y 1 C z 0 D x particular C x nullspace . 0 0 1 19 If LU x D 0, multiply by L 20 21 22 23 24 25 26 30 1 to ﬁnd U x D 0. Then U and LU have the same nullspace. Column 5...
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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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