Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 0 b all 8 matrices are rs 34 one reason that r is the

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Unformatted text preview: is sure to have no pivot since it is a combination of earlier columns. With 4 pivots in the other columns, the special solution is s D .1; 0; 1; 0; 1/. The nullspace contains all multiples of this vector s (a line in R5 ). For special solutions .2; 2; 1; 0/ and .3; 1; 0; 1/ with free variables x3 ; x4 : R D   10 2 3 and A can be any invertible 2 by 2 matrix times this R. 01 2 1 &quot; # 100 4 3 is the line through .4; 3; 2; 1/. The nullspace of A D 0 1 0 001 2 # &quot; 10 1=2 2 has .1; 1; 5/ and .0; 3; 1/ in C .A/ and .1; 1; 2/ in N .A/. Which AD 1 3 51 3 other A’s? This construction is impossible: 2 pivot columns and 2 free variables, only 3 columns. # &quot; 1 1 0 0 0 1 0 has .1; 1; 1/ in C .A/ and only the line .c; c; c; c/ in N .A/. AD 1 1 0 0 1   10 01 T . AD has N .A/ D C .A/ and also (a)(b)(c) are all false. Notice rref.A / D 00 00 27 If nullspace D column space (with r pivots) then n r D r . If n D 3 then 3 D 2r is impossible. 28 If A times every column of B is zero, the column space of B is contained in the nullspace     1 1 11 . Here C .B/ equals N .A/. and B D of A. An example is A D 11 1 1 (For B D 0; C .B/ is smaller.) 29 For A D random 3 by 3 matrix, R is almost sure to be I . For 4 by 3, R is most likely to be I with fourth row of zeros. What about a random 3 by 4 matrix? 31 If N .A/ D line through x D .2; 1; 0; 1/; A has three pivots (4 columns and 1 special # &quot; 100 2 1 (add any zero rows). solution). Its reduced echelon form can be R D 0 1 0 001 0 Solutions to Exercises 31  10 3 , R D 01 32 Any zero rows come after these rows: R D Œ 1 33 (a)  1 0 2      0 10 11 01 00 ; , , , 1 00 00 00 00  0 , R D I. 0 (b) All 8 matrices are R’s ! 34 One reason that R is the same for A and A: They have the same nullspace. They also have the same column space, but that is not required for two matrices to share the same R. (R tells us the nullspace and row space.)   y 35 The nullspace of B D Œ A A  contains all vectors x D for y in R4 . y 36 If C x D 0 then Ax D 0 and B x D 0. So N .C / D N .A/ \ N .B/ D intersection. 37 Currents: y1 y3 C y4 D y1 C y2 C Cy5 D y2 C y4 C y6 D y4 y5 y6 D 0. These equations add to 0 D 0. Free variables y3 ; y5 ; y6 : watch for ﬂows around loops. Problem Set 3.3, page 151 1 (a) and (c) are correct; (b) is completely false; (d) is false because R might have 1’s in nonpivot columns. &quot; 4 4 4 4 4 4 2 AD 4 4 4 &quot; 1 2 3 2 3 4 AD 3 4 5 &quot; 1 1 1 1 1 1 AD 1 1 1 &quot; # 120 3 RA D 0 0 1 RB 000 # &quot; 4 4 has R D 4 # &quot; 4 5 has R D 6 # &quot; 1 1 has R D 1  1 0 0 1 0 0 1 0 0  1 0 0 0 1 0 1 0 0 1 0 0 1 2 0 1 0 0  RA 0 # 1 0. 0 # 2 3. 0 # 1 0. 0  0 RA The rank is r D 1; The rank is r D 2; The rank is r D 1 Zero rows go to the bottom    0I I 4 If all pivot variables come last then R D . The nullspace matrix is N D . 00 0 D RA RA RC ! 5 I think R1 D A1 ; R2 D A2 is true. But R1 T ! R2 may have 1’s in some pivots. 6 A and A have the same rank r D number of pivots. But pivcol (t...
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