# Linear Algebra with Applications (3rd Edition)

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Unformatted text preview: Maw 1w Homwwk 1 soiutw'm \.\ 20. The total demand for the product of lndustry A is l000 ( . the ('onnunier d ' ‘ (the demand from Industry B). The outi 5 Hum“) plm U. l 1) mt (I must meet this denmnd: {I : 1000 + 01/). Stt" he e' " ‘ ' l : e mg up I snin n (qu ition for Industry B we ()l)l.ellll the system 7 [(3:39 +0951” or l a — 0.10 :1000l 1_ _ l T l T J] l I 70.2“ + {I : 780 V\ neh yields the unique solution (I : ll00 and b : 1000. 28. The thermal equilibrium condition requires that T1 ; T) : and T3 : ‘Tﬁ’lollww‘ 1 l T l i v ~~4Tl + T3 : 7-200 l We can rewrite this system 'rLS T1 m 47} + T3 : ~900 T2 ﬂ 4T}; : “100/ The solution is (Tth T3) : (75.100.125). LIL 171 4 1'4 2 1 i171 : 1 v .111 10. The system reduces to .1:2 7 31'; : 2 % J2 ; 2 + 31:4 1; + 2.1:4 : 3 13 i 73 7 21:4 Let 1‘4 1 171 1 i l‘ .17- 2 3t . . 2 2 .+ . . where t Is an arlntrarv real number. 173 7.5 M 2f ‘ 1'4 1 24. Yes: each elementary row operation is l't‘\'(‘l'>&llil(‘. that is. it (‘élll l)(‘ "undone." For example. the operation of row swapping win be undone by swapping the sznne rows again. The operation of dividing a row liy n scalar ('EU] lie reversed by multiplying the same row liy the sznne senlur. it] +7 .l’g rt :13 + .111 : (ll 1‘1 7+7 2:172 W 3.13 + 4.1'4 : 0 , 35. We need to solve the, system t.“ +913 W 93-3 + 7.“ :. 0 «1?; + :7 which reduces to 1:2 7 1.5.13; : 0 . I \$3 + 2.25.111 I 0 .1'1 —(li25T 1 . . . 5172 . . The SOlllthllS are of the form 13 : 7,) 25f . where 1‘ IS an arbitrary real number. :1" ,1 f “Ll. 4U. 1'1 " ' ‘ We 1111111 111 111111 1111.1113.'111;; 511111 1111111 1111 *1 1113 *1 1111; 't 1 11,1111 1 m 1' + 2 1 2 I ,1 2 1112 '3 1111;; l : 211111111 1S.“'1‘1131\‘1‘t11HU1V1‘1111‘S\'511‘1ll 1 111] + 1113 1 1113 V: 1 111,1 1 21113 + 4111»; : 2 1) 1 12111; + 31113 +111», .; 11111 1111111111‘5111111111111is 1111 r, 1,13 :; 111111 ,,,.4 : I U *1 4. This 111e111rix 11115 121111( ‘2 since its 11111115 1) 1 ‘2 U (1 (J 13‘. 1111111111‘11' 1112111V 51111111011. 111111 2111‘ 211 11121511 1W0 111)\'1()11H 5011111111115. \\ 1111‘ 111 11s 21 111111211 1'1111111111211101101 111 111111 17-; 11.111111111111511 111111111 1'11111111111111011011i; 111111 173 11101111. 1‘111‘11‘101'1‘. 1,1115 11111‘211 5115111111 11115 1111111111‘11’ 1112111y 3111111111115. by 171111 1.31. 1 2 :1 .1 1 11. 1 r111 11 ; 1 7 1x 11 : 11 1 1 111 3 21111213111141 1 1 Since 1111 variables 2111‘ 11111111154. 1111‘ sysLeiu AF : (71211111011 11211'1‘ 1111111111111' 1111111_\' 51111111111115. but it (:011111 have LL 1111111119, 50111111111 (for 131111111111. 1117; 11) 01. 1111 5011111i11115 2111 2111 (1'0111p2111‘ with Exampie 31-). 571. The v1’1‘11111's11f11111‘ 10111111111 113111 10111111 11121111'11111’1111g11 11111111111111 (11111111111115; F1 211111 1'13: 111 Fignrv 1.15 \1'11111'11\\'11 1111111111 1111101111 11115 111111111. Figure 1.15: for Probiem 1.3.54. \ _ 3 2 i): 1 ’ on y : :2— and 3 on y : 32:. 57. Pick a vector on each line, say [1 T1"t7"1' b"t‘f2‘d1'2+b1-7 1611 VVI'l e de 11’1an COHl 111d10110 1 an 3 . a, 1 3 —‘ . . 7 2, b 2 3, so that the desired representation is [11] The unique solution is a _ 4 + 3 2 9 i 3 4. .yygg . .Ay#ll [2] is on the line gt 2, [9] IS on lIHL y— 31. (I, i’ . V l) U ( l) l) ()0. We need : A: _ 1r A» Jr A' :, ‘» ' - n (, 1 3 _ _ ,5 5 3M + “LAB ﬁlm My“; ‘ horn this “6 Ht (1 U l) (j UAW, that (L‘ (‘ and (I (“an he any value. while I) nmst cqnal zero. ill 1 41 l 4 6. Note that :13, 2 + 11:2 5 : 2 5 so that, T is indeed linear. with matrix % 6 3 (5 "‘2 wk)»— CDCI‘JA 24. Compare with EXlellL Figure 2.0: for Problem 21.” . Figure ‘27: for Problem 2.1.26. LV: V‘J “ ' ' ‘ : ‘ ? T(62)]. ' ‘ i H - ' . by Fact 2.1.2.14 [TVQ ~~ 33. we find T(el) and T( 3), then ‘ 34. As in EXBIQSQ r -sin0L “52) = [ cos (Xi Figure 2.13: for Problem 2.1.34. A sin 9 cos 6 ‘ A [cos 9 sin 6 Therefore: A “ image of x, axis image ofxl mus Figure 2.16: for Problem 2.1.42. 1 O b. The image of the point % is the origin. [0] . '2— ‘ r 10 " o 17%11 + I.) :0 (a boive the qulzlf‘lull I 13 : 01' I 1. w: o 1 m U 531] + 1; ~01 I 1 The suitltiorrs are Hf 1110 form .1‘2 ; f where f is rm arbitrary real rilliiiiwr. .I‘jg ll considered in part i).Til{‘H(‘ points are For example. for ‘1‘ v . we [iml the point i l l 7 I -) 011 the lim‘ through lire origin and the observer‘s (we. ...
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