P3-algebralinear-resposta-prof Chileno -pucgo

PUC-GO\nDisciplina: Algebra Linear\nProfessor: Cristão Patrício Nova\nAluno: Ana Carolina Utencio B. Silva\nP3: Algebra Linear\n1. V: Mmn(K)\nW = {A/AT = TA} T ∈ Mmn(K)\nA,B ∈ W\n(A+B)T = AT + BT\n= TA + TB ⇒ (A+B) ∈ W\n= T(A+B)\nC ∈ R ⇒ ∀A ∈ W\n(CA)T = C(AT)\n= C(TA) ⇒ C ∈ A ∈ W\n= T(CA)\nEntão W é subespaço de V 2. R3 ⇒\nB = {(1,0,0),(2,1,1)}\nU = (0,0,0)\nB = {(1,0,0),(2,1,1),(1,0,0)}\n(1) B é LI\nA,B,C ∈ R ⇒ a(1,0,0) + b(2,1,1) + c(1,0,0) = (0,0,0)\n{ a + 2b + c = 0\nb > 0 ⇒ a + b = 0 ⇒ a = 0 ⇒ c = 0\na + b = 0\n(x,y,z) ∈ R3\n(x,y,z) = a(1,0,0) + b(2,1,1) + c(1,0,0)\nI: a + 2b + c = x\nII: b = y → a + b = a + y - z → a = z - y\nI: z - y + 2y + c.x ↔ c + y + z = x ↔ c = x - y - z\n(z - y + y - x - y - z),\n12\" relativa a B {2x - y + 2z = 0\n3x - 2y + z = 0\n-y + 2y + 4z = 0\n3x - 2y + z = 0\n3(-3x) - 2y + z = 0\n-9z + y - 2y = 0\n-8z = 2y ⇒ y = -4z\nx = -3z\ny = -4z\nS = { z ∈ R: z(-3,4,1), z ∈ R }\nDimensão (S) = 1 {u,v,w} 2. I \\exists a,b,c \\in \\mathbb{R} \\newline a u + b v + c w = 0 \\iff a - b = c = 0 \\newline \\exists x,y \\in \\mathbb{R} \\newline \\alpha (u + v) + \\beta (u - w) + \\gamma (v + 2w) = 0 \\newline u(\\alpha + \\beta) + v(\\alpha - \\beta) + w(\\gamma - \\beta) = 0 \\newline \\Rightarrow \\{u + v, u - w, v + 2w\\} \\text{ 1. I } \\newline \\begin{cases} \\alpha + \\beta = a \\newline \\alpha + \\gamma = b \\newline 2\\gamma - \\beta = c \\end{cases} \\implies a = 0 \\Rightarrow \\alpha = -\\beta \\newline a = -b - \\gamma \\newline \\alpha = \\gamma \\newline -\\beta = -\\gamma \\newline \\Rightarrow 2\\beta - \\gamma = 0 \\newline \\gamma = 0 \\Rightarrow \\alpha - \\beta = 0 \\newline \\text{esto es } 1. I