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h3 - 2.0[lg 10 Possible answer are 1 —2 3 0(a 0 —1 4 3...

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Unformatted text preview: 2. (.0 [lg]. 10. Possible answer are: 1 —2 3 0 (a) 0 —1 4 3 . 0 2 —5 —2 12. (c) 180. 14. If A321“, then det(AB)= det(A) det(B) = detUn) = 1, so det(A) 75 U and det(B) % U 15. (a) By Corollary 3.3, det(A_1) = 1/det(A). Since A = Ad, we have 1 that“) : det-(A) 2:» (det(A))2 '= 1. Hence det(A) = :|:1. (b) If AT 2 A-l, then det(AT) : det(A‘1). But 1 det(A) det(A) = det(AT) and det(A*1) = hence we have 1 det (A) det(A): :- (det(A))2:1 => det(A)= 17. If A” = A. then det(A2)' = [ri'et(A)]2 =det(A ) so det(A) = 1 Alternate solution: If A2: A and A is nonsing-nlar', then A‘lA2 = A“1_~A= —..Im so A21}. and det(A}— — det(Ifl)— — 1. 30. (a) I3 (h) Only the trivial solution. 32. If A2 = A, then det(A2) = detl(A), so [det(A)]2 = det(A). Thus. det(A)'(det(A) — 1) '= 0. This implies that det(A_) : D or det(A_) : 1. 34. If det(A) % 0. then A is nonsingnlar. Hence. AAAB : A“'1AC.. so. B = C. 16. 6. 18. Let P1__($_1, 3,351), P2(r2, 3,12), P3_(s.3, :93] ”he the vertices of a triangle .T. Then from Equation (2.), we have 1 area- Of T =- — $1 "311 1 - '- 2 List 332' gm 1 .- = 1 we 93 .1 ‘2’ @1312 + 913533 twat-:3 ._ $3112 — yam — metal - .Let A be the matrix representing a-CDuntEI‘ClOCle-Se rotation L through an angle .qb. Thus A = [mags —-sin 933] Sin (3.5 cos 425 and Pi, P2, 1?” are the vertices of L( (311).,the image of T. We have L (E: _ [531903fii5ny151flfii’ m1 sin¢+ y1 cuss ’ L (E? j) = [13-2 (303‘? “yefiifléfi :rg sin (15 + 3,12" cos {i} ’ L([: :D_ [mgcosgii-I—ygsinq‘)" T m3 sin-gin +. 'yg-cos qt ’ Then 1_ __ mlcosefi—ylsinqé $;1sine5-+ylcose5 1"- E det _ wacos e3 — yg sinqb 11223111 {35+ yg-oosgb 1 - era-cose —- yg sin gt s3 sin¢+ ya-rcos <35 1 -: % |(:r1 oos qb- — yl sin g5.)_[:rg. sings + y; cos qt: — 2:3 sin-e1 —- yg cos-e] + (3:2 costs — yg sin Alena-sin e —|— yg cos gt »-- {1318111133 —- ryl cos 915] +(I30osa’) ~-- 313 sin'¢)'[:c1 sin gt +. yl cos qt} — igsinQb — yg cos (fir = is [331112 +9133 +'.$-2313."$3ys ‘- 53134'3 — Efiyl-l :2 area of T. area of L.(T)_ = ...
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