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Unformatted text preview: FULLADIO 4 - Gr. 'Algebra Genikì. HMMU 2005-06 1. (a) Na apodeiqjoÔn oi idiìthtec: (i) (AB ) = B A , (ii) (AB ) = B A (1) (b) Na apodeiqjeÐ ìti an oi pÐnakec B kai Γ eÐnai summetrikoÐ tìte gia opoiond pote pÐnaka A, oi pÐnakec A B A kai AΓA eÐnai summetrikoÐ. (g) Na deiqjeÐ ìti gia kje tetragwnikì pÐnaka A isqÔei (Aν ) = (A )ν , ν ∈ N. (d) Na apodeiqjeÐ ìti an o pÐnakac A eÐnai antisummetrikìc tìte o pÐnakac Aν eÐnai summetrikìc an o ν eÐnai rtioc kai antisummetrikìc an o ν eÐnai perittìc. 2. Na upologisjoÔn oi pÐnakec AB kai BA, ìtan A= 8 − 4i 5i −2i . B= 6 3 + 2i 4 + 5i 2 + i 4 + 3i i , 7 6 + 9i 1 − i 3. Na upologistoÔn me qr sh idiot twn oi orÐzousec αβ −α β D1 = −α −β αβ γδ γδ γδ γ −δ kai D2 = λ+α α α β λ+β β γ γ λ+γ . 4. Na lujoÔn oi exis¸seic (me metablht x) λ λ2 x λ2 x λ x λ λ2 =0 kai 1+x 2 3 4 1 2+x 3 4 1 2 3+x 4 1 2 3 4+x = 0. 5. Na upologisjoÔn o antÐstrofoi twn pinkwn: sunϕ −hmϕ 0 B = hmϕ sunϕ 0 , 0 01 1111 1 1 −1 −1 Γ= 1 −1 0 0 . 0 0 1 −1 6. Na lujoÔn ta sust mata: (i) x2 − 2x3 + 3x4 =1 x1 − 3x2 + 2x3 + x4 =0 x1 − x2 − 2x3 + 7x4 =2 x1 − 4x3 + 10x4 =3 (ii) kx1 + x2 + x3 = 1 x1 + kx2 + x3 = 1 x1 + x2 + kx3 =1, k∈R 7. Na prosdioristoÔn ìlec oi timèc twn α, β, γ, δ gia tic opoÐec to sÔsthma x1 + x2 + x3 + x4 2x1 − x2 + 3x3 − x4 4x1 − 5x2 + 7x3 − 5x4 x1 − 2x2 + 2x3 − 2x4 = = = = α β γ δ eÐnai sumbibastì. S. Karansioc Endeiktikèc LÔseic FulladÐou 04 1. (a) (i) 'Estw A = (aij ), B = (bij ), AB = (γij ), (AB ) B = (ζij ), B A = (ηij ). Tìte εij = aji , ζij = bji , γij = δji kai = (δij ), = (εij ), A ηij = ζi1 ε1j + ζi2 ε2j + · · · ζin εnj = b1i aj 1 + b2i aj 2 + · · · bni ajn = aj 1 b1i + · · · ajn bni = γji = δij 'Ara (AB ) = B A . (ii) 'Omoia apodeiknÔetai kai h sqèsh (AB )∗ = B ∗ A∗ . (b) EÐnai (A B A) = A B (A ) = AB A ⇒ A B A summetrikìc pÐnakac. 'Omoia kai o AΓA . (g) Apìdeixh me epagwg . (d) Apì A = −A prokÔptei Aν = (−A )ν = (−1)ν (A )ν = (−1)ν (Aν ) ⇒ Aν = (Aν ) , an ν rtioc kai Aν = −(Aν ) , an ν perittìc. 2. Blèpe skhsh 5/selÐda 155 kai apant seic sth selÐda 532 tou biblÐou. 3. EÐnai h skhsh 2/selÐda 174 tou biblÐou: γδ γδ γδ γ −δ = αβγ δ 0 2β 2γ 2δ 0 0 2γ 2δ 00 0 −2δ λ+α α α β λ+β β γ γ λ+γ = λ+α+β+γ λ+α+β+γ λ+α+β+γ β λ+β β γ γ λ+γ D1 = D2 = α β −α β −α −β α β 1 1 1 β (λ + α + β + γ ) β λ + β γ γ λ+γ = α2β 2γ (−2δ ) = −8αβγδ. 100 = (λ + α + β + γ ) β λ 0 γ0λ = = (λ + α + β + γ )λ2 . 4. EÐnai h skhsh 3/selÐda 174 tou biblÐou: 0= λ λ2 x λ2 x λ x λ λ2 = 1 λ2 x 1xλ 1 λ λ2 1 λ2 x = (λ + λ + x) 0 x − λ2 λ − x 0 λ − λ2 λ2 − x 2 = √ 1 = (λ + λ2 + x)(−x2 + (λ + λ2 )x − λ4 + λ3 − λ2 ) ⇒ x = −λ − λ2 x = [λ + λ2 ± 3λ(λ − 1)i]. 2 1 2 3 4 1 2+x 3 4 0 = (x + 10) 1 2 3+x 4 1 2 3 4+x 1 0 = (x + 10) 0 0 2 x 0 0 3 0 x 0 4 0 o x = (x + 10)x3 ⇒ x = 0 x = −10. 5. Blèpe skhsh 5/selÐda 175 kai apant seic sth selÐda 533 tou biblÐou. 6. (i) 0 1 −2 3 1 −3 21 [A|B ] = 1 −1 −2 7 1 0 −4 10 1 0 ∼ ··· ∼ 2 3 1 −3 21 0 1 −2 3 0 0 00 0 0 00 0 1 ⇒ x1 − 3x2 + 2x3 + x4 = 0 0 x2 − 2x3 + 3x4 = 1 0 ⇒ x2 = 1 + 2x3 − 3x4 , x1 = · · · = 3 + 4x3 − 8x4 x3 , x4 ∈ R. (ii) D= k11 1k1 11k 111 = (k + 2) 1 k 1 11k 1 1 1 0 = (k + 2) 0 k − 1 0 0 k−1 = (k + 2)(k − 1)2 D 1 • D = 0 ⇔ k = −2, 1 ⇒ monadik lÔsh h: x = Dx , y = Dy , z = Dz ⇒ x = y = z = k+2 D D • D = 0 ⇔ k = −2, k = 1 Gia: 1 −2 1 11 1 −2 1 1 1 ∼ ··· ∼ 0 1 −1 −1 ⇒ sÔsthma adÔnato. ∗ k = −2 ⇒ 1 −2 1 1 −2 1 0 0 0 3 ∗ k = 1 to sÔsthma eÐnai isodÔnamo me thn exÐswsh x + y + z = 1, opìte z = 1 − x − y kai h genik lÔsh eÐnai h (x, y, z ) = (x, y, 1 − x − y ) = x(1, 0, −1) + y (0, 1, −1) + (0, 0, 1), x, y ∈ R. 7. 1111 2 −1 3 −1 [A|b] = 4 −5 7 −5 1 −2 2 −2 α β ∼ ··· ∼ γ δ 1 0 0 0 11 3 −1 00 00 1 3 0 0 α 2α − β . γ + 3β − 10α δ+α−β 'Ara sÔsthma sumbibastì ⇔ γ + 3β − 10α = 0, δ + α − β = 0, opìte ta zhtoÔmena (α, β, γ, δ ) eÐnai ta stoiqeÐa tou dianusmatikoÔ q¸rou V = {(α, β, γ, δ ) : γ + 3β − 10α = 0, δ + α − β = 0} = {(α, β, 10α − 3β, β − α) : α, β ∈ R} = {α(1, 0, 10, −1) + β (0, 1, −3, 1) : α, β ∈ R} = [(1, 0, 10, −1), (0, 1, −3, 1)]. ...
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