fylladio3-2005 - Γραµµική Άλγεβρα 3ο...

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Unformatted text preview: Γραµµική Άλγεβρα 3ο Φυλλάδιο Ασκήσεων (ΣΥΣΤΗΜΑΤΑ) ΣΗΜΜΥ 2005-06 1. Να λυθούν τα συστήµατα x1 − 2 x 2 + 3 x3 − 4 x 4 + 2 x5 = −2 x1 + 2 x 2 + 3x3 − x 4 = 1 x1 + 2 x 2 − x3 3x1 + 2 x 2 + x3 − x 4 = 1 2 x1 + 3x 2 + x3 + x 4 = 1 x1 − x 2 + 2 x3 − 3 x 4 και 2 x1 + 2 x 2 + 2 x3 − x 4 = 1 5 x1 + 5 x 2 + 2 x3 − x5 = −3 = 10 . x 2 − x3 + x 4 − 2 x5 = −5 =2 2 x1 + 3 x 2 − x3 + x 4 + 4 x5 = 1 2. Για τις διάφορες τιµές του λ ∈ IR , να λυθούν τα συστήµατα (λ + 3) x + λx + 3 y + 4 z = 3 λx + 5 y + 7 z = 3 (3λ + 3) x + 3. Για τις διάφορες τιµές των x + by + b 2 z = b 3 . και x + cy + c 2 z = c 3 bx + a 2 y + a 2 bz = a 2 b 4. Να υπολογιστούν οι τιµές των a, b, c ∈ IR 5 x1 + 2 x 2 − x3 − 3 x 4 = a ώστε τα συστήµατα (a + 3) x − και 2y + 3z = 4 3 x + (a − 3) y + 9z = b 4x − + 3 x3 + 3 x 4 = c να έχουν λύση. Τα διανύσµατα λy + (λ + 3) z = 3 x + ay + a 2 z = a 3 x + ay + abz = a − x1 z = 2λ . a, b, c ∈ IR , να λυθούν τα συστήµατα x + ay + a 2 z = 1 =b 2z = λ λx + (λ − 1) y + και x + λy + λz = 2 2 x1 + x 2 + x3 y+ 8 y + (a + 14) z = c (a, b, c) ∈ IR 3 σχηµατίζουν διανυσµατικό χώρο; Αν ναι, να υπολογιστεί η διάστασή του. Παράδοση µέχρι 25-11-2005 Σ. Καρανάσιος Endeiktikèc LÔseic FulladÐou 03 1.(i) EÐnai [A|b] = opìte x3 = 5 x4 + 1 , 6 6 (ii) OmoÐwc 1 −2 1 2 1 −1 [A|b] = 0 1 2 3 1 3 2 2 5 2 2 3 2 5 3 −1 1 1 −1 1 1 11 2 −1 1 2 02 ∼ ··· ∼ 1 0 0 0 0 2 1 0 0 0 3 −1 1 5 −3 1 6 −5 1 0 00 0 00 7 x2 = − 6 x4 + 1 , x1 = 5 x4 + 1 , x4 ∈ R 6 6 6 3 −4 2 −2 −1 0 −1 −3 ∼ ··· ∼ 2 −3 0 10 −1 1 −2 −5 −1 1 4 1 1 0 0 0 0 0 100 1 −1 0 0 0 010 0 001 0 000 0 0 0 0 1 ⇒ adÔnato λ34 2. (i) EÐnai D = λ 5 7 = λ2 − 1. DiakrÐnoume tic peript¸seic: 1λλ • D = 0 ⇔ λ = 1, −1. SÔsthma Cramer, ˆra monadik lÔsh h: x = 3(2λ−3) , λ2 −1 −2+3λ , λ2 − 1 2(2λ−3) λ2 −1 y= z= • D = 0 ⇔ λ = 1 λ = −1 'Estw λ = 1. Tìte 1343 1343 [A|b] = 1 5 7 3 ∼ · · · ∼ 0 2 3 0 ⇒ adÔnato 1112 0001 'Estw λ = −1. Tìte −1 3 43 −1 3 4 3 5 7 3 ∼ · · · ∼ 0 2 3 0 ⇒ adÔnato [A|b] = −1 1 −1 −1 2 0005 (ii) 'Omoia, ìpwc sto prohgoÔmeno sÔsthma, èqoume: D = λ2 (λ − 1) • D = 0 ⇔ λ = 0, 1. SÔsthma Cramer, ˆra monadik lÔsh h: x = 2 2 3 9−15λ+3λ2 +λ3 , λ2 (λ−1) λ λ y = −9+12−+λ , z = −9+122 (+3λ −4λ . λ2 (λ 1) λ λ−1) • D = 0 ⇔ λ = 0 λ = 1. Kai stic dÔo peript¸seic ta antÐstoiqa sust mata eÐnai adÔnata. 3. (i) D = a2 − ab = a(a − b) • D = 0 ⇔ a = 0 kai a = b. Tìte • D = 0 ⇔ a = 0, b ∈ R a = b. Gia a = 0 sÔsthma adÔnato. Gia a = b, a = 0 èqoume: 1 a a2 [A|B ] = 1 a a2 a a2 a3 • a = 1, b ∈ R sÔsthma adÔnato. x= a2 (1−b) a− b , y= −b+a2 b , a2 −ab z= 1−a . a2 −ab 1 a a2 1 1 a ∼ ··· ∼ 0 0 0 a − 1 ⇒ a3 0 0 0 a3 − a • a = 1 to sÔsthma eÐnai isodÔnamo me thn exÐswsh x + y + z = 1. 'Ara z = 1 − x − y , opìte genik lÔsh (x, y, z ) = (x, y, 1 − x − y ) = (0, 0, 1) + x(1, 0, −1) + y (0, 1, −1), x, y ∈ R. (ii) D = (a − b)(b − c)(c − a) • D = 0 ⇒ x = abc, y = −ab − ac − bc, z = a + b + c •D=0⇔a=b b=c c=a a = b, b = c ⇒ 1 a a2 a3 1 0 −ac −a2 c − ac2 [A|B ] = 1 a a2 a3 ∼ · · · ∼ 0 1 a + c a2 + ac + c2 ⇒ 1 c c2 c3 00 0 0 x = (ac)z − −a2 c − ac2 , y = −(a + c)z + a2 + ac + c2 , z ∈ R. 'Omoia antimetwpÐzontai, lìgw summetrÐac, kai oi peript¸seic b = c, c = a. a = b = c ⇒ to sÔsthma eÐnai isodÔnamo me thn exÐswsh x + ay + a2 z = a3 , opìte x = a3 − ay − a2 z kai (x, y, z ) = (a3 − ay − a2 z, y, z ) = (a3 , 0, 0) + y (−a, 1, 0) + z (−a2 , 0, 1), y, z ∈ R. 4. (i) EÐnai 5 2 −1 −3 a 1 0 −3 −3 −c 1 0 b ∼ ··· ∼ 0 1 7 6 b + 2c . [A|B ] = 2 1 −1 0 3 3c 00 0 0 a − 2b + c 'Ara to sÔsthma eÐnai sumbibastì an kai mìno an a − 2b + c = 0. Ta dianÔsmata (a, b, c) ∈ R gia ta opoÐa to sÔsthma èqei lÔsh, apoteloÔn dianusmatikì ton q¸ro V = {(a, b, c) : a − 2b + c = 0} = {(2b − c, b, c) : b, c ∈ R} = {b(2, 1, 0) + c(−1, 0, 1) : b, c ∈ R} = [(2, 1, 0), (−1, 0, 1)] me dimV = 2. √ √ (ii) EÐnai: D = (a + 2)(a + 6 − 3)(a + 6 + 3), opìte √ monadik lÔsh gia kˆje b, c ∈ R. • D = 0 ⇒ a = −2, −6 ± 3 ⇒ √ √ • D = 0 ⇔ a = −2, a = −6 − 3, a = −6 + 3 a = −2 ⇒ 1 −2 3 4 1 −2 3 4 1 0 b − 12 . [A|B ] = 3 −5 9 b ∼ · · · ∼ 0 4 −8 12 c 0 0 0 c − 16 Epomènwc to sÔsthma èqei lÔsh an kai mìno an c − 16 = 0 ⇔ c = 16. Ta antÐstoiqa dianÔsmata (a, b, c) = (−2, b, 16) = (−2, 0, 16) + b(0, 1, 0), b ∈ R den apoteloÔn dianusmatikì q¸ro. √ a = −6 − 3 ⇒ √ √ 3 −1 − √ c − b 3 1 1+ √ √ . · · · ∼ 0 4(1 + 3) −3 − 4 3 4 + (3 + 3)(c − √ b) √ √ 0 0 0 −(1 + 3 3) b + 3 3 c + 4 3 √ √ √ Epomènwc to sÔsthma èqei lÔsh an kai mìno an −(1+3 3) b +3 3 c +4 3 = 0. Ta antÐstoiqa √ √ 4 dianÔsmata (−6 − 3, b, 9+9 3 b, − 3 ), b ∈ R, den apoteloÔn dianusmatikì q¸ro. √ 'Omoia gia a = −6 + 3. Tèloc to sÔnolo ìlwn twn (a, b, c) gia ta opoÐa to sÔsthma èqei lÔsh (ènwsh ìlwn twn parapˆnw peript¸sewn) epÐshc den apoteleÐ dianusmatikì q¸ro ...
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This note was uploaded on 10/02/2009 for the course G 001 taught by Professor Shmmygr during the Spring '07 term at National Technical University of Athens, Athens.

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